e-Notes by A.K.Bhatt, GIT, Belgaum

AVAILABILITY AND IRREVERSIBILITY

Available Energy

The sources of energy can be divided into two groups namely, high-grade energy and low-grade energy. The conversion of high-grade energy to shaft work is exempt from the limitations of the second law, while conversion of low grade energy is subjected to them.

Example: High grade energy:1) Mechanical work 2) electrical energy 3) water power 4) wind power 5) kinetic energy of a jet 6) tidal power.Example: Low grade energy: 1) Heat or thermal energy 2) heat derived from nuclear fission or fusion. 3) Heat derived from combustion of fossil fuels. 4) Solar energy.The high-grade energy in the form of mechanical work or electrical energy is obtained from sources of low-grade energy. The complete conversion of low-grade energy, heat in to high-grade energy, shaft work is impossible. That part of low-grade energy which is available for conversion is refereed to as available energy, while the part which according to the second law must be rejected is known as unavailable energy.

Fig 1: Heat transfer from a constant temperature energy source.

In the previous chapter the concept of efficiency of a device such as turbine, nozzle and compressor are introduced and more correctly termed as first law efficiency, since it is given as he ratio of two energy terms. This chapter gives more meaningful definition of efficiency- second law analysis. Our main goal is to use this analysis to manager our thermal resources and environment better.

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Consider the simple situation shown in figure1 in which there is an energy source Q in the form of heat transfer from a very large source and therefore constant temperature reservoir at temperature T. what is the ultimate potential for producing work?

To answer to this question we imagine that a cyclic heat engine is available as shown in figure (b) to convert the maximum fraction of Q requires that the engine be completely reversible, i.e. a Carnot cycle, and that the lower temperature reservoir be at the lowest temperature possible, often but not necessarily at the ambient temperature. From the first and second laws for the Carnot cycle and the usual consideration of all the Q’s as positive quantities we find

W rev HE = Q –Qo Q / T = Qo / To W rev HE = Q {1-( To / T ) }

The fraction of Q given by the right side of the equation is the available portion of the total energy quantity Q.

Consider the situation shown on the T-S Diagram. The total shaded diagram is Q.The portion of Q that is below To, the environment temperature, can not be converted into work by the heat engine and must instead be thrown away. This portion is therefore the unavailable portion of the energy Q, and the portion lying between the two temperatures T and To is the available energy.

Fig 2: T-S Diagram for a constant temperature energy source.

Let us consider the same situation except that the heat transfer Q is available from a constant pressure source, for ex, a simple heat exchanger as shown in the figure. The Carnot cycle must now be replaced by a sequence of such engines, with the result shown in the figure B the only difference between the first

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and the second example is that the second includes an integral, which corresponds to S

S = ( Q rev / T ) = Qo /To

W rev = Q – To * SNote that this S quantity does not include the standard sign convention. It corresponds to the change of entropy. The equation specifies the available portion of the quantity Q. the portion unavailable for producing work in this circumstance lies below To.

Thus the unavailable energy is the product of lowest temperature of heat rejection ands the change of entropy of the system during the process of supplying heat.

Fig 3: Changing temperature energy source.

Decrease in available energy when the heat is transferred through a finite temperature difference:

Whenever heat id transferred through a finite temperature difference there is a decrease in the availability of the energy so transferred. let us consider a reversible heat engine operating between T1 and To as shown in the figure

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Fig 4: Increase in unavailable energy due to heat transfer through a finite temperature difference.

Then we have Q1 = T1 * S Q2 = To * S W = AE = (T1 – To) S

Let us now assume that q1 is transferred through a finite temperature difference from the reservoir or source at T1 to the engine absorbing heat at T’1 lower than T1 as shown in the figure

Fig 5: Constant Temperature energy source.

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The availability of Q1 as received by the engine at T’1 and to receiving Q1 and rejecting Q’2. Q1 = T1 S = T ’1 S’T1 > T ’1, Hence S’ > S Q2 = Tc S and Q‘2 = To S’

Since S’ > S hence Q’2 > Q2

And hence W = Q1- Q’2 = T’1 S’ - To S’And W = Q1 – Q2 = T1 S – To SHence W’ < W since Q’2 > Q2.

Available energy lost due to irreversible heat transfer through finite temperature difference between source and working fluid during heat addition process is given by

W- W’ = Q’2 - Q2 = To (S’ - S) or decrease in AE = To (S’ - S)

Thus the decrease in available energy is the product of the lowest feasible temperature of heat rejection and the additional entropy change in the system while receiving heat irreversibly compared to the case of reversible heat transfer from the same source. The greater is the temperature difference (T1 – T’1) the greater is the heat rejection Q’2 and greater will be the unavailable part of the energy supplied. Energy is said to be degraded each time when it flows through a finite temperature difference. That’s why the second law is some times called the law of degradation of energy and the energy is said to run down hill.

Availability:

The availability of a given system is defined as the maximum useful work (total work – pdV work) that is obtainable in a process in which the system comes to equilibrium with its surroundings. Availability is thus a composite property depending on the state of the system and surroundings.

Let U,S and V be the initial energy, entropy and volume of a system and Uo So, Vo their final values when the system has come to equilibrium with its environment. The system exchanges heat only with the environment. The process may be either reversible or irreversible. The useful work obtained in the process in the form of equation

W <= (U – ToS + po V) – (Uo – ToSo _ poVo)

Let = U – ToS + poV where is called the availability function and a composite property of both the system and its environment, with U,S,V being properties of the system at some equilibrium state and To po are the temperature

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and pressure of the environment. The decrease in availability function in a process in which the system comes to equilibrium with the environment is

- o = (U – ToS + po V) – (Uo – ToSo _ poVo)

Hence W < = - o

Thus the useful work is equal to or less than the decrease in availability function. The availability A of a given system in a given environment is the maximum useful work obtainable in reversible process. A = W max = - o .

This work is obtained in part from a decrease in the internal energy of the system and in part from the heat withdrawn from the environment.

Let a system be taken from an equilibrium state 1, in which its availability is A1 to a second equilibrium state 2 in which its availability is A2. The end state 2 is not in equilibrium with the environment. The maximum useful work that could be obtained in the process W max = A1 – A2 = (1 - o ) – (2 - o )

W max = (1 - 2)

Availability in steady flow:

The steady stet steady flow equation is given by

H1 + ½ (mv12 ) + mgz1 + Q = H2 + ½ (mv2

2 ) + mgz2 + W The subscript 1 and 2 refer to the entrance and exit respectively. At the exit let the system be in equilibrium with the environment at pressure po and temperature To. Let symbols without subscripts refers to the entrance condition of the system and changes in KE and PE are negligible. Then useful work is given by W = (H-Ho) + Q. the greater the value of Q larger will be the useful work. Thus W will be maximum when Q is a maximum.

Let S and So be the entropies of the systems at the entrance and exit of the device then

(S) system = So – S. and (S)surr = - (Q / To)

From the entropy principle ( So –S ) – (Q - To) > = 0. Therefore Q <= To (So- S) .

The useful work W <= (H-Ho) + To(So-S) <= (H-ToS) + (Ho - ToSo)

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Hence W <= B – Bo where B = H-ToS it is called availability function for steady flow and is also called as Keenan function.

For entrance and exit condition the useful work is maximum when the heat absorbed is a maximum. I.e when the internal irreversibility is zero.

The maximum work obtainable from a system at the entrance of a device when the pressure and temperature at the exit are those of the environment is called the available of the system in steady flow and is given by

A = W max = B- Bo.The alternative names for availability and for unavailable quantity To S are Exergy and Anergy respectively.

Reversible work in a non-flow process:In a non-flow process dm1 = dm2 =0

The entropy equation for flow process reduces to (dW)rev = - d (U1 - ToS + ½ (mv1

2 ) + mgZ)

Between two equilibrium end states 1 and 2

(W) rev1-2 = U1-U2 –To(S1-S2) + m( V12-V2

2)/2 + mg(Z1-Z2)

If the system doesn’t possesses any KE and PE

(W) rev1-2 = U1-U2 –To(S1-S2)

Irreversibility:The actual work done by a system is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process.

I = W max – W This is also some time referred to as degradation or dissipation. For a non flow process between the equilibrium states, when the system exchanges heat only with the environment

I = {(U1-U2) – To(S1-S2) } – {(U1-U2) + Q} = To (S2-S1) – Q= To (S) system + To (S)surr = To[ (S)system + (S)surr ] Hence I > 0.

Similarly for the steady flow process I = Wmax – W

= To (S2-S1) – Q= To (S) system + To (S)surr = To[ (S)system + (S)surr ]

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Thus same expression for irreversibility applies to both flow and non-flow process. The quantity To[ (S)system + (S)surr ] represents an increase in unavailable energy or Anergy.

Second Law efficiency:

With the increased use of availability analysis in recent years a term called second law efficiency has come into more common use. This term refers to comparison of the desired output of a process with the cost or input in terms of the thermodynamic availability. Thus the isentropic turbine efficiency defined by the ratio of actual work output to the work for a hypothetical isentropic expansion. From the same inlet state to the same exit pressure which is called first law efficiency, in that it is a comparison of two energy quantities. The second law efficiency as just described would be the actual work output of the turbine divided by the decrease in availability from the same inlet state to the same exit state. Thus the second law efficiency is Second law = W / (1 - 2)

Where (1 - 2) is the decrease in availability for a steady state steady flow process. Which is equal to the reversible work or maximum work obtainable.In this sense this concept provides a rating or measure of the real process in terms of the actual change of state and is simply another convenient way of utilizing the concept of thermodynamic availability. In a similar manner the second law efficiency of a pump or a compressor is the ratio of the increase in availability to the work input to the device.

NUMERICAL EXAMPLES:

1) In a certain process a vapor while condensing at 420OC transfers heat to water evaporating at 250OC. The resulting steam is used in a power cycle, which rejects heat at 35OC. What is the fraction of the available energy in the heat transfer from the process vapor at 420OC that is lost due to irreversible heat transfer at 250OC.

Fig 6: Increase in unavailable energy

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Solution: ABCD would have been the power cycle if there was no temperature difference between the vapor condensing and the water evaporating, and the area under CD would have been the unavailable energy at 35OC. EFGD is the power cycle when the vapor condenses at 420OC and the water evaporates at 250OC. The unavailable energy becomes the area under DG. Therefore the increase in available energy due to irreversible heat transfer is represented by the area under CG.

Q1= T1 S = T ’1 S’ = S’ / S = T1 /T ’1

W = work done in cycle ABCD = ( T1- To) SW’ = work done in cycle EFGD = ( T’1- To) S’

The fraction of energy that becomes unavailable due to irreversible heat transfer

= (W-W’) / W = [ To(S’-S) ] / [( T1- To) S ]

= To [(S’/S) - 1 ] / ( T1- To) = To( T1- T’1) / T1( T1- To)

= [308( 693-523) ] / [ 523 (693-308) ] = 0.26 Ans

2) Air expands through a turbine from 500 kPa, 520OC to 100 kPa,. 300 OC. during expansion 10KJ of heat is lost to the surroundings which is at 98 kPa, 20 OC. neglecting KE and PE changes determine per kg of air a) the decrease in availability b) the maximum work c) the irreversibility.

Solution: For air the change in entropy is given by S2-S1=[ mCp ln (T2/T1) - mR ln(P2/P1)]

For 1kg of air s2-s1 = Cp ln (T2/T1) - R ln(P2/P1)The change in availability is given by, (1 - 2) = b1-b2 = (h1-Tos1) - (h2 –Tos2)

= (h1-h2) –To (s1-s2) = Cp(T1-T2) – To (Cp ln (T2/T1) - R ln(P2/P1))

= 1.005 (520-300) – 293(1.005 ln(573/793) -0.287 ln (1/5)

= 1.005 (220) – 293 (-0.4619 + 0.3267) = 221.1 + 39.6 = 260.7 kJ/kg

W max = Change in availability = (1 - 2) = 260.7 kJ/kg

From SSSF equation we haveQ + h1 = W + h2 or W = (h1-h2) + Q = 1.005(520-300) –10 =211.1 kJ/kg

Thus the irreversibility I = Wmax – W = 260.7 – 211.1 = 49.6 kJ/kg Ans.

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3) Consider a steam turbine that has throttle governor. The steam in the pipeline flowing through the turbine has a pressure of 3 MPa and a temperature of 360 OC. At certain load the steam is throttled in an adiabatic process to 1.5 MPa. Calculate the availability per kg of steam before and after the process and the reversible work and irreversibility per kg of steam for this process. Assume To= 25OC po=1 bar = 0.1MPa

Solution:The availability at the initial condition = (h1-ho) –To(s1-so) = (3140.9 – 104.9) – 298 (6.7844 – 0.3664 ) = 3036-1912.38 = 1123.62 kJ/kg Ans.

Similarly availability at 1.5MPa after adiabatic throttling = (he-ho) – To(se-so) since in adiabatic throttling he=hi=3140 kJ/kg and adiabatic Q = 0.

Therefore se from the steam table corresponding to he = 3140.9 kJ/kg and 1.5MPa = 7.0833 kJ/kg K.

Thus e = (3140.9 – 104.9) – 298 (7.0833-0.3664) = 3036.13 –2001.46 = 1034.67 kJ/kg

Hence W rev = i - e = 1123.62 – 1034.67 = 88.95 kJ/kg

Also irreversibility I= To –(se-si) = 298(7.0833-6.7844) = 89.07 as W actual is zero.

4) Consider an air compressor that receives ambient air at 100 kPa, 25OC. It compresses the air to a pressure of 1 MPa, where it exits at a temperature of 540 OK. Since the sir and the compressor housing are hotter than ambient it looses 50 kJ per kg air flowing through the compressor. Find the reversible work, reversible heat transfer and irreversibility in the process.

Fig 7

Solution:

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-W

-Q

i e

It is a non-adiabatic compression with no change in KE and PE. Using SSSF equation, which has single inlet and exit. From the ideal gas tables h1= 298.62 kJ/kg and he = 544.69 kJ/kg, si = 6.8629 kJ/kgK and se = 7.4664 kJ/kgK.The energy equation for the actual compressor is

W= h1-h2 + Q = 298.62 –544.69 –50 = -296.07 kJ/kg

The reversible work for the given change of state is given by

W rev = To (s2-s1) - (h2-h1) + Q[1-(To/Th)] = 298.2 (7.4664 - 6.8629) – (0.287 ln 10)- (544.69-298.62) +0 = -263.17 kJ/kg

Thus irreversibility I =w rev-w = -263.17 – (-296.07) = 32.9 kJ/kg.

5) In a steam boiler, hot gases from a fire transfer heat to water which vaporizes at constant temperature. In a certain case, the gases are cooled from 1100oC to 550oC. The specific heat of gases is 1.005kJ/kg K and the latent heat of water at 220oC is 1858.5kJ/kg. All the heat transferred from the gases goes to the water. How much does the total entropy of the combined system of gas and water increase a result of the reversible heat transfer? Obtain the result on the basis of 1 kg of water evaporated. If the temperature of the surroundings is 30oC, find the increase in unavailable energy due to irreversible heat transfer.

Solution.

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Fig 8 : h – s diagram

Gas is cooled from state 1 to state 2 (Fig). For reversible heat transfer, the working fluid in the heat engine would have been heated along 2-1, so that at any instant, the temperature difference between gas and the working fluid is zero. Then 1-b would have been the expansion of the working fluid down to the lowest possible temperature To, and the amount of heat rejection would have been given by the area abcd.

When water evaporates at 220.o C as the gas gets cooled from 1100oC to 550 oC, the resulting power cycle has an unavailable energy represented by the area aefd. The increase in unavailable energy due to irreversible heat transfer is thus given by area befc.

(S) water = Latent heat absorbed /T = 1858.5/(273+220) = 3.77 kJ/kg-K

Q1 = Heat transferred from the gas =Heat absorbed by water during evaporation=mgCpg (1100-550)=1 X 1858.5 kJ

Hence, mgCpg = 1858.5/550 =3.38kJ/o C

S gas = ( dQ /T ) = mg Cpg dT/T

mg Cpg ln Tg2 /Tg1 = 3.38 ln 823 / 1373 = -3.38 * 0.51 = -1.725 kJ/Khence S total = (S) water + (S) gas

= 3.77 – 1.725 = 2.045 kJ/KIncrease in unavailable energy = To (S) total = 303 * 2.045 = 620 kJ.

6) Calculate the available energy in 40 kg of water at 75oC with respect to the surroundings at 5oC, the pressure of water being 1 atm.

Solution

If the water is cooled at a constant pressure of 1 atm from 75oCto 5oC the heat given up may be used as a source for a series of Carnot engines each using

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the surroundings as a sink. It is assumed that the amount of energy received by any engine is small relative to that in the source and the temperature of the source does not change while heat is being exchanged with the engine.

Let us consider that the source has fallen to temperature T, at which level there operates a Carnot engine which takes in heat at this temperature and rejects heat at To = 278 K. If s is the entropy change of water, the work obtainable is

W = - m (T-To ) s where s is negative. W – 40 (T-To) Cp T/T = -40Cp (1-To/T) TWith a very great number of engines in the series the total work (Maximum) obtainable when the water is cooled from 348K to 278 K would be

W (max) = AE = - lim 40 Cp (1- To/T) T= - 40 Cp (1- To/T) T = 40 Cp [(348-278) – 278 ln 348/278]

= 1340 kJ.Q1 = - 4 0 * 4.2 (348-278) = 11760 kJ.

Unavailable energy = Q1 – W max = 11760 – 1340 = 10420 kJ.

7) A 5 kg copper block at a temperature of 200 oC is dropped into an insulated tank containing 100 kg oil at a temperature of 30 oC. Find the increase in entropy of the universe due to this process when copper block and the oil reach thermal equilibrium. Assume that the specific heats of copper and oil are respectively 0.4 kJ/kgK and 2.1 kJ /kgK.

SolutionApplying energy balance, Energy lost by copper block = Energy gained by the oil.Mcu X Cp cu X Temperature difference = m oil X Cpw X Temperature difference.5 X 0.4 X (200 – t ) = 100 X 2.1 (t – 30 ) 212t = 6700 , t = 31.6oC

Heat lost by Copper = 5 X 0.4 X (473 – 304.6) = 336 8 kJ, then associated entropy change is dQ/T 336.8/473 0.7120 kJ/K.

Heat gained by oil = 100 X 2.1 X 1.6 = 1.1089 kJ/K.

Entropy of the universe = S cu +Soil = 0.712 +1.1089 = 1.8209 kJ/kg Ans 8) 0.5 kg of ice block at –10oC is brought into contact with 5 kg copper block at 80oC in an insulated container. Determine the change in entropy of

i) Ice block.ii) Copper block.iii) The universe.

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Given specific heat of ice=2kJ / kg K, specific heat of water = 4.2 kJ/kg K. specific heat of copper =0.5 kJ /kg K, enthalpy of fusion of water at 0oC= 334 kJ/kg. Solution Writing the energy balance equation for the systemEnergy lost by the copper block = Energy gained by the ice5 X 0.5 (80 - t) = 0.5 X 2.0 ( 0+10) + 0.5 X 334 + 0.5 X .4.2 (t - 0)

Therefore the value of t is t = 5OC

Entropy change of copper block = dQ/T = 187.5/(273 +80) = 0.53116kJ/KEntropy change of ice block= dQ/T = (10 +167+10.5)/263 = 0.7129kJ/KEntropy change of the universe = ( S )Cu + ( S)w = 1.24406 Ans

9) Calculate the availability and unavailability of a system that absorbs 15000kJ of heat from a heat source at 500K temperatures while the environment is at 290 K temperature.

Solution: Entropy Change, ds = Q/T=15000/500 =30 kJ/K

Unavailable work =T o ds =290 X 30 =8700 kJ

Available work = Q - To ds =15000 –290 X 30 =6300 kJ

10) 0.2 kg of air initially at 575 K temperature receives 300 kJ of heart reversibly at constant pressure. Determine the available and unavailable energies of the heat added.Take Cp for air = 1.005 kJ / kg K and temperature of surroundings 300 K.

Solution: Let T2 be the temperature of air after the addition of heat at constant pressure. Then

300 = mCp (T2-T1) = 0.2 X 1.005 (T2 –557)=0.201 T2 -115.57T2 = (300+115.57)/0.201 =2067.5K

Entropy change, ds = mCp loge T2/T1 = 0.2 X 1.005 oge 2067.5/575 = 0.2572 kJ /K

Unavailable work = T o dS = 300 X 0.2572 = 77.16 kJAvailable work = Q- T o dS = 300-300 X 0.2572 = 222.84 kJ

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11) A closed system contains 2 kg of air during an adiabatic expansion process there occurs a change in its pressure from 500kPa to 100 kPa and in its temperature from 350 K to 320 K. if the volume doubles during the process make calculations for maximum work, the change in availability and irreversibility. Take for air Cv = 0.718 kJ/kg K and R = 0.287 kJ/kgK. The surrounding conditions may be assumed to be 100 kPa and 300 K.

Solution: When the system undergoes a change from state 1 to 2 during a mass flow process the maximum obtainable work is given by

W max = (U1-U2) –To(S1-S2). Now (U1-U2) =

mCv(T1-T2) = * 0.7185(350-220) = 43.08 kJ.

And S1-S2 = m[Cv ln(T2/T1) – R ln V2/V1] = 2*[0.718 l n 320 / 350 – 0.287 ln 2] = 0.2693 kJ/K. then W max = 43.08 – 300 (-0.2693) = 123.87 kJ.

Change in availability is given by A1 – A2 = (U1-U2) –To(S1-S2) + Po (V1-V2)

V1 = mRT1 / P1 = 2 * 0.287 * 250 / 500. = 0.4018 m3.V2 = 2 V1 = 2* 0.4018 + 0.8036 m3.A1-A2 = 123.87 + 100(0.4018 – 0.8036) = 83.67 kJIt is the measure of maximum useful work or newt work obtainable from the system.Irreversibility = max work – actual workW actual = Q – (U2-U1)

Since the system – process is adiabatic Q = 0 and W actual = -(U2-U)1 = U1-U2

= n mCv(T1-T2) = 2* 0.718 (350-320) = 43.08 kJ.

Irreversibility = 123.87 - 43.08 = 80.79 kJ.

Problem from Availability.

12) A system at 500 K receives 7200kJ/min from a source at 1000 K. The temperature of atmosphere is 300 K. assuming that the temperature of the system and source remain constant during heat transfer find out

The entropy produced during heat transfer.The decrease in available energy after heat transfer.

Temperature of the source T1 = 1000 KTemperature of the system, T2 = 500 KTemperature of atmosphere, To = 300 K

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Heat received by the system, Q =7200kJ/min

Change in entropy of the source during heat transfer = -Q/T1 = -7200/1000 = - 7.2 kJ/min K

Change in entropy of the system during heat transfer = Q/T2 = -7200/500 =14.4kJ/min K

The net change of entropy, S -7.2 + 14.4 = 7.2kJ/min K

Decrease in available energy with source = (1000-300) 7.2 = 5040kJ/minDecrease in available energy with the system = (500 –300) 14.4 = 2880 kJ/min.

13) 15 kg of water is heated in an insulated tank by a churning process from 300 K to 340 K. If the surrounding temperature is 300 K, find the loss in availability for the process

SolutionWork added during churning = increase in enthalpy of water = 15* 4.187*(340 300) = 2512.2 kJ = Enthalpy in the waterAvailability of this energy is given by

m[(u1-u0) – T0 s]s= Cp loge (T1/T0)

s = 4.187 loge (340/300 )

s= 0.524 kJ/kgK Available energy = m[Cv (T1 – T0)- T0 s] =15[4.187 (340 – 300) – 300 X0.524] = 2349.3kJ Ans Loss of availability 2512.2 – 158.7 = 2353.5 kJ AnsThis shows that conversion of work into heat is highly irreversible process, since out of 2512.5 kJ of work energy supplied to increase the temperature, only 158.7 kJ will be available again for conversion into work.

14) 5.0 kg of air at 550 K and 4 bar is enclosed in a closed system. Determine, i) The availability of the system if the surrounding pressure and temperature are 1 bar and 290 K respectively.

ii) If the air is cooled at constant pressure to the atmospheric temperature, find availability and effectiveness.

Availability of the system is = m[(u1 –u0)-T0(s1-s2)] = m[Cv(T1 – T0) - T0s]

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s = Cp *loge (T1/T0) – R loge (p1/p0)s = 1.005 * loge(550/290) – 0.287 * loge(4/1) = 0.643 – 0.397 = 0.246 kJ/kg K

Availability of the system = m[Cv(T1 – T0) - T0s] 5[0.718(550 - 290) –290 X0.246] = 576.7 kJ Ans

ii) Heat transferred during cooling

Q = m X Cp X (T1 –T0) 5.0 X1.005 X(550 - 290) = 1306.5 kJ Heat lost by the systemChange of entropy of the system during cooling S = m X Cp X loge(T1/T0)S = 5.0 X1005 X loge(550/290) = 3.216 kJ/KUnavailable portion of this energy = T0 ()S = 290 X 3.216 = 932.64 kJAvailable energy = 1306.5 - 932.64 = 373.86 kJ Ans

Effectiveness, = Available Energy/ Availability of the system =(373.86/576.7) = 0.648 OR 64.8%

= 64.8% Ans

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SESSION 1

Outcome of the Session:

Microscopic and Macroscopic point of view Thermodynamics Definition Thermodynamic Systems

INTRODUCTION:

Thermodynamics is the science of energy transfer and its effect on the physical properties of substances. The alternate definition is: thermodynamics is the science that deals with work and heat and these properties of substances that bear a relation to heat and work. Like all sciences, the basis of thermodynamics is experimental observation.

Thermodynamics (from the Greek therme, meaning "heat" and dynamis, meaning "power") is a branch of physics that studies the effects of changes in temperature, pressure, and volume on physical systems at the macroscopic scale by analyzing the collective motion of their particles using statisticsThis subject was developed mainly by1) Carnot 2) Mayer 3) Clausius 4) Joule5) Kelvin 6) Maxwell 7) Plank 8) Gibbs

A thermodynamic system is a device or combination of devices containing a quantity of matter that is being studied. A typical thermodynamic system - heat moves from hot (boiler) to cold (condenser), and work is extracted, in this case by a series of pistons.

A typical thermodynamic system

The study of thermodynamics is the basis of such fields as steam power plants, IC Engines, Gas dynamics and aerodynamics, fluid mechanics, Refrigeration and Air conditioning and heat transfer.

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A Steam Power Plant

Thermodynamics deals with four laws. Namely Zeroth law, first law, second lawand Third law of thermodynamics. Fortunately, there is no mathematical proof for any of these laws of thermodynamics, like physical laws, but they are deduced from experimental observations.

Refrigeration Cycle

Thermodynamics deals with three E’s, namely Energy, Equilibrium and Entropy. Thermodynamics also talks about study of materials, chemical reactions, plasmas and other biological reactions.

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Macroscopic and Microscopic point of view:This study deals with macroscopic, as opposed to microscopic or statistical

thermodynamics. In microscopic thermodynamics individual molecule is considered and analysis of collective molecular action. In macroscopic thermodynamics, we concern ourselves with the overall effect of the individual molecular interaction.Macroscopic point of view: The macroscopic level is the level on which we live. We measure most of the quantities on this level.Ex: Temperature measurement, pressure measurement, total volume measurement, specific volume measurement. Thus, microscopic point of view will be used only to explain some phenomena that can’t be understood by macroscopic means.

Microscopic point of view: Considers a system containing a cube of 25 mm containing monoatomic gas at atmospheric pressure and temperature. This volume contains approx. 1020 atoms. To describe the position of each atom in a coordinate system we require three equations. To describe the velocity of each atom we have to specify 3 velocity components. Thus to describe completely the behavior of the system from a microscopic point of view we must deal atleast 6*1020 equations. It is a hopeless computational task.

The other approach that reduces number of variables to a few that can be handled is the macroscopic point of view of Classical Thermodynamics. It concerns with the gross or average effect of many molecules and can be measured by instruments. This measurement is the time-averaged influence of many molecules.In the present study, we concentrate on macroscopic point of view.

Statistical thermodynamics, classical thermodynamics deals with the significance of microscopic point of approach. From the macroscopic point of view it is very clear that continuum has to be there in the system because we are not concerned with the behavior of the individual molecule.

Substance: What follows will be illustrations of the thermodynamics, one must be able to solve problems and to do what the part of the problem must be enumerated. The first consideration is that there must be something performing the energy transformations. This something is called a substance. Ex: In case of IC engine gasoline and air mixture constitutes the substance. In steam turbine the substance is steam.

The substance may be further divided into sub categories, namely pure substance i.e. if it is homogeneous in nature- i.e. if it does not undergo chemical reaction and is not a mechanical mixture of different spices. The other substance is a mixture substance which is not a pure substance.

Thermodynamic System:

A substance does not exist alone. It must be contained. This brings us to the concept of a system.

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In thermodynamics a system is defined as any collection of matter or space of fixed identity, the concept is one of the most important thermodynamics.

Concept of a Boundary

System boundary: When a system is defined, let us say, fluid in a cylinder, what separates the fluid from the cylinder wall and the piston and every thing external to the piston-cylinder? it is the system boundary. Everything not in the system is called the surrounding. Note that piston can be raised or lowered, but the system, matter of fixed identity is constant.The system is further divided into closed system, open system and isolated system.

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_____________________________________________________________

SESSION 2

Outcome of the Session

Thermodynamic System Open and Closed and Isolated System

Control Volume Units, Thermodynamic Properties Extensive and Intensive Property

Thermodynamic System

A Thermodynamic system is defined as a quantity of matter or a region in space upon which attention is concentrated in the analysis of a problem. Everything external to the system is called the surrounding or environment. The system is separated from the surrounding by the system boundary. Boundary may be either fixed or moving. A system and its surrounding together comprise a universe.

A System, Surroundings and Boundary

Open System: The open system is one in which matter crosses the boundary of the system. There may be energy transfer also. Most of th4e engineering devices are generally open systems. Ex: An air compressor in which air enters at low pressure and leave at high pressure and there is energy transfer across the system boundary.

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An Open System

Closed System: A closed system is a system of fixed mass. There is no mass transfer across the system boundary. Ex: A certain quantity of fluid in a cylinder bounded by a piston constitutes a closed system.

A Closed System

Isolated System: The isolated system is one in which there is no interaction between the system and surrounding. It is of the fixed mass and energy and there is no mass or energy transfer across the system boundary.

Control Volume and Control Surface:In thermodynamic analysis of an open system such as air compressor, gas turbine

in which there is a flow of mass into and out of the system, attention is focused on a certain volume in space surrounding the compressor known as control volume, bounded by a surface called the control surface. Matter as well as energy can cross the control surface.

Gas turbine

Homogeneous and Heterogeneous system:A quantity of matter homogeneous throughout in chemical composition and

physical structure is called a phase. Every substance can exist in any one of the three phases viz. Solid, Liquid or gas.

A system consisting of a single phase is called a homogeneous system while a system consisting of more than one phase is known as a heterogeneous system.

THERMODYNAMIC PROPERTIES

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Thermodynamic properties are taken from a macroscopic perspective. We are dealing with quantities that can either directly or indirectly be measured and counted. Therefore, the matter of units becomes an important consideration. Mass, length and time are considered as fundamental physical quantities, they are related by Newton's second law of motion, which states that the force acting on a body is proportional to the product of mass and acceleration in the direction of force.

i.e. F= m * a

Mass – kg Length – m Time – sThis is adopted by CGPM- General Conference of Weights and Measures

In thermodynamics temperature is also considered as fundamental unit in Kelvin.Energy: One of the very important concepts in a study of thermodynamics is the concept of energy. It is defined as the capacity to do work. It is also defined as the capability to produce an effect

When considered from molecular point of view, three general forms of energy become important.

1) Intermolecular potential energy. 2) Molecular kinetic energy 3) Intermolecular energy.

The energy is the important concept which depends on the mass, velocity, intermolecular attraction. In all intermolecular internal energy is most difficult to evaluate.Specific Volume: It is a macroscopic property and defined as the volume occupied by unit mass. It is reciprocal of density and its unit is m3/ kg. The specific volume of a system in a gravitational field may vary from point to point. Specific volume increases as the elevation increases. Thus the definition of specific volume involves the specific volume of a substance at a point in a system.

Pressure: The pressure in a fluid at rest at a given point is the same in all directions. We define pressure as the normal component of force per unit area. Its unit is pascal or N/m2. When dealing with liquids and gases we ordinarily speak of pressure. For solids we speak of stresses.

Two other units not part of international system continue to be widely used are Bar = 105 Pa = 0.1MPa and standard atmosphere is 1 atm = 101325 Pa.

In most thermodynamic investigations, we are concerned with absolute pressure. Most pressure vacuum gauges however read the difference between the absolute pressure and the atmospheric pressure at the gauge. This refers to as gauge pressure.

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Illustration of terms used in pressure measurement.Temperature: It is a fundamental property of thermodynamics and defined as the hotness of the body. Temperature first of all as a sense of hotness or coldness when we touch an object. We also learn that when a hot body and a cold body are brought into contact the hot body becomes cooler and cold body becomes warmer. Because of these difficulties in defining temperature we define equality of temperature.

Property: It is defined as any quantity that depends on the state of the system and is independent of the path ( i.e. the prior history) by which the system arrived at the given state. Conversely the state is specified or described by the properties and later we will consider the number of independent properties a substance can have, i.e, the minimum number of properties that must be specified to fix the state of a substance.Thermodynamic properties can be divided into 2 general classes: intensive and extensive properties:

Intensive property: An intensive property is independent of mass; thus intensive property value remains same even if the matter is divided into two equal parts. Ex: pressure, temperature, density etc.

Extensive Property: The value of an extensive property varies directly with the mass, i.e. if a quantity of matter in given state is divided into 2 equal parts, the properties will have the half the original values. Ex: mass, total volume, total enthalpy, total energy etc.

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Pabs1

Patm

Pabs2

Ordinary Pressure gauge P = Pabs1 – P atm

Ordinary vacuum gaugeP = P atm – P abs2

Barometric reads atmospheric pressure

SESSION 3

Thermodynamic State Point Process, Cycle Cyclic and Non-cyclic Process Quasi-static Process.

Thermodynamic State Point:The state may be identified or described by certain observable macroscopic properties; some familiar one are temperature, pressure and density. The state is specified or described by the properties. The state point can be indicated on a thermodynamic coordinate system. Thermodynamic coordinate system includes pressure volume diagram, temperature volume diagram, temperature entropy diagram, enthalpy entropy diagram, pressure enthalpy diagram so on and so forth.

These co-ordinate system represents thermodynamic state of a substance. Consider a system not undergoing any change. At this point all the properties can be measured or calculated throughout the entire system, which gives us a set of properties that completely describes the condition or the state of the system.

Thermodynamics deals with equilibrium state. When a system undergoes any change then change of state will occur.

Weights

Piston

Gas

Cylinder

Piston Cylinder arrangement

Process: Whenever one or more of the properties of a system change we say that a change in state has occurred. For ex: in a piston and cylinder arrangement, if weight is removed from the piston rises and change in state occurs in which pressure decreases and specific volume increases. The path of succession of states through which the system passes is called process.

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Path: Path is the complete series of states through which the system passes during a change from one given state to other state. It is clear that the transformation of a system from one fixed state to another state is called a process.

The thermodynamic processes that are commonly met within engineering practice are 1) Constant pressure process (Isobaric) 2) Constant volume process(Isochoric) 3)Constant temperature process( Isothermal) 4) Reversible adiabatic process (Isentropic process) 5) Polytropic process 6) Throttling process.

If the system passes through a series of equilibrium states during the process its called reversible process. On the other hand the system passes through a series of non-equilibrium states during a process it is called irreversible process. The state of the processes can not be plotted on the co-ordinate systems since the path of the process is not defined.Generally the system is in equilibrium in the beginning and at the end of the process, the reversible process can be plotted on the coordinate diagram by continuous line and an irreversible process by a dotted line.

Representation of a reversible process An irreversible process

Quasi – static process (very slow process) : Quasi – meaning almost, static meaning infinite slowness. Thus quasi-static process is infinitely slow transition of a system. Infinite slowness is the characteristic feature of a quasi-static process. A quasi-static process is a succession of equilibrium states. It is a reversible process.

Cycle: It is a process whose initial and final states are same. Thus at the end of a cycle all the properties of a working fluid have the same values as they had in the initial states.

There are 2 types of cycles. viz. thermodynamic cycle and mechanical cycle.

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1

2

1

2

P

V

V

Thermodynamic cycle: It is one in which the working substance is recirculated. Ex: water that circulates

through steam power plant and refrigerant that passes through refrigeration plant are the examples of thermodynamic cycle. There is change of phase during the process but the end states do not change

Mechanical cycle:In case of a mechanical cycle the working substance is not re circulated. In an IC

engine air and fuel are burnt in the engine, converted into the products of combustion and are then exhausted into the atmosphere. Hence this type of cycle is called mechanical cycle.

Thermodynamic Equilibrium: The word equilibrium implies a state of balance. In an equilibrium state, there are

no unbalanced potentials within the system or driving forces. Thus, a system in equilibrium experiences no changes when it is isolated from its surroundings.

There are many types of equilibrium. A system is not in thermodynamic equilibrium unless the condition of all the relevant types of equilibrium are satisfied, which includes 1) Thermal equilibrium 2) Mechanical equilibrium 3) Phase equilibrium and 4) Chemical equilibrium.

Thermal Equilibrium: If the temperature is the same throughout the entire system .i.e the system

involves no temperature differential which is the driving force or heat flow then we say system is in thermal equilibrium.Mechanical equilibrium:

It is related to pressure, velocity. A system is in mechanical equilibrium if there is no change in pressure, velocity, specific volume at any point of the system with respect to time. However the pressure may vary within the system with elevation as well as resultant of gravitational effects. However there should not be any imbalance of forces. Then we say the system is in mechanical equilibrium

Phase equilibrium: If a system involves two phases it is in phase equilibrium when the mass of each

phase reaches equilibrium level and stays there.

Chemical equilibrium:If the systems chemical composition does not change with time, i.e., no chemical

reaction occur then we say the system is in chemical equilibrium.

Thus if all thermal, mechanical, phase and chemical equilibrium exist for a system then we say the system exist in thermodynamic equilibrium

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Session 4

Outcome of the Session

Diathermic wall Zeroth Law of Thermodynamics Temperature Concepts International Practical Temperature Scale.

Diathermic wall: A wall which is impermeable to the flow of heat is an adiabatic wall, where as a

wall which permits the flow of heat is a diathermic wall. Thus heat flow takes place through this wall.

Zeroth law of Thermodynamics: It states that when two bodies have equality of temperature with the third body,

they in turn have equality of temperature with each other.When a body A is in thermal equilibrium with body B and also separately with body C then B and C will be in thermal equilibrium with each other.

Equality of temperature – Zeroth law of Thermodynamics.

Temperature scales: Two scales are commonly used for measurement of temperature namely,

Fahrenheit after Gabriel Fahrenheit (1686-1736) and Celsius. The Celsius scale was formerly called the centigrade scale but is now designated the Celsius scale after Anders Celsius (1701-1744), the Swedish astronomer who devised this scale.

In SI units of temperature scale we use absolute temperature scale or absolute scale of temperature which comes from second law of thermodynamics and its unit is Kelvin.

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Table: Thermometers and Thermometric Properties

The absolute scale is related the Celsius scale is the Kelvin scale after William Thompson, 1824-1907, who is also known as Lord Kelvin and is designated K. The relation is K=oC + 273.15

In 1967 the CGPM defined the Kelvin as 1/273.16 of the temperature at the triple point of water.

Thermometer Thermometric property SymbolConstant volume gas thermometer Pressure PConstant pressure gas thermometer Volume V

Electrical Resistance thermometerResistance

R

ThermocoupleThermal e.m.f.

e

Mercury in glass thermometerLength L

In order to obtain a quantitative measure of temperature a reference body is used, and a certain physical characteristic of this body which changes with temperature is selected. The changes in the selected characteristic may be taken as an indication of change in temperature. The selected characteristic is called the thermometric property and the reference body which is used in the determination of temperature is called the thermometer.

A Very common thermometer consists of a small amount of Mercury in an evacuated capillary tube. In this case the extension of the mercury in the tube is used as the thermometric property.

A common Mercury thermometer Mercury-in-glass thermometerPresently temperature of triple point of water which is an easily reproducible state is now the standard fixed point of thermometry.

International practical temperature scale: an international temperature scale was adopted at the seventh general conference on weights and measures held in 1927. It was not to replace the Celsius or ideal gas scales, but to provide a scale that could be easily and rapidly used to calibrate scientific and industrial instruments.

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International practical scale agrees with Celsius scale at the defining fixed points listed in following table.

Table: Temperature of Fixed Points

Temperature in 0CNormal boiling point of oxygen -182.97Triple point of water (standard) +0.01Normal boiling point of water 100.0Normal boiling point of sulphur 444.60Normal melting point antimony 630.50Normal melting point of silver 960.80Normal melting point of gold 1063.0

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Session 5

Outcome of the Session

Numerical Examples with Solutions:

1) A tank contains mixture of 20kg of nitrogen and 20 kg of carbon monoxide. The total tank volume is 20m3. Determine the density and specific volume of the mixture.

Solution:Total mass of the mixture: 20 kg N2 + 20 kg CO = 40 kg mixtureSpecific volume = volume / mass = 20 m3 / 40kg = 0.5 m3/kgDensity of mixture = mass / volume = 1/sp. vol = 1/ 0.5 = 2 kg / m3. Ans.

2) An automobile has a 1200 kg mass and is accelerated to 7m/s2. Determine the force required to perform this acceleration.

Solution:Force required F = m * a = 1200 * 7 =8400 kg m /s2 = 8400 N. Ans.

3) The emf in a thermocouple with the test junction at t0C on gas thermometer scale and reference junction at ice point is given by e = 0.20 t – 5 * 10-4 t2 m V. the millivoltmeter is calibrated at ice and steam points. What will this thermometer read in a place where gas thermometer reads 500C?

Solution:At ice point, when t = 0oC, e= 0 mV.At steam point, when t=100oC, e= 0.20 * 100 – 5* 10-4 * (100)2 = 15 mVAt t = 50oC, e = 0.20 * 50 – 5* 10-4 * (50)2 = 15 mV = 8.75 mVWhen the gas thermometer reads 50oC, the thermocouple will read

t = 100/15 * 8.75 or 58.33 oC. Ans.

4) A barometer to measure absolute pressure shows a mercury column height of 725mm. The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient pressure.

Solution:Ambient pressure = ρ * g * h = 13550 * 9.81* 725 / 1000 = 96371Paor P = 0.9637 bar. Ans.

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5) The temperature t on a Celsius thermometric scale is defined in terms of a property p by the relation p = e(t-B)/A, where A and B are constants. Experiment gives values of p of 1.86 and 6.81 at the ice and steam point respectively. Obtain relation for t and also find the temperature t for the reading of p = 2.5

Solution: At ice point, t = 0oC, p= 1.86 Hence 1.86 = e(0-B)/A

1.86 = e –B/A or eB/A = 1/1.86 and ln eB/A = ln 1/1.86

B/A = - 0.62058 --------------- (1)

At steam point, t=100oC, p= 6.81 6.81 = e(100-B)/Aln 6.81 = (100 - B) /A = 1.9184

1.9184 A + B =100--------------------------(2)

Solving equations (1) and (2)A = 77.05 and B = -47.81 hence p = e [t – (-47.81) ] / 77.05

The temperature at p=2.5, 2.5 = e (t+47.81)/77.05ln 2.5 =( t + 47.81) / 77.05 or t + 47.81 = 0.9163 77.05

or t = 22.79 0C Ans.

6) A hiker is carrying a barometer that measures 101.3 kPa at the base of the mountain. The barometer reads 85 kPa at the top of the mountain. The average air density is 1.21kg/m3. Determine the height of the mountain.

Solution:Pressure at the base of the mountain = ρ1 * g * h1.h1= p / (ρ1 * g ) = 101.3 *1000 / (9.81 * 1.21) = 8534 m.

Also at the top of the mountain p = ρ2 * g * h2.h2 = p / (ρ2 * g ) = 85 *1000 / (9.81 * 1.21) = 7161 m.

Hence the height of the mountain = h1- h2 = 8534 – 7161= 1373 m. Ans.

7) A lunar excursion module (LEM) weighs 1500 kgf at sea level; o earth. What will be its weight on the surface of the moon where g=1.7 m/s2? On the surface of the moon, what will be the force in kgf required to accelerate the module at 10m/s2?

Solution: The mass m of the LEM is given by W = mg/g0

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1500 kgf = m * { (9.806 m/s2 ) / ( 9.806 kg/ kgf * m/s2 )} i.e. m=1500 kgThe weight of the LEM on the moon would be W = 1500 kg * {(1.7 m / s2) / (9.806 kg / kgf * m/s2 ) } Ans.

The force required to accelerate the module at 10 m/s2 = [ 1500 kg / (9.806 kg/kgf * m/s2) ] * 10 m/s2 = 1530 kgf Ans.

8) A cannon ball of 5 kg acts on a piston in a cylinder of 0.15m diameter. As the gunpowder is burnt, a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder is pointing horizontally?

Solution:Mass of the cannon ball = 5 kgDiameter of the cylinder = 0.15 m

Force = Area * Pressure = (π / 4) * (0.15)2 *7* 106 = 123716.25 N Ans.

F = m*a Hence a = F /m = 123716.25 / 5 = 24743.25 m / s2 Ans.

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Pure SubstancesP-T and P-V diagramsTriple point and critical pointSub cooled liquidSaturated liquid, mixture of saturated liquid and vaporSaturated vapor and superheated vapor states of a pure substance with water as example.Enthalpy of change of phase, dryness fraction, T-S and H-S diagrams.Representation of various processes on these diagrams.Steam tables and its use.Throttling Calorimeter, Separating Calorimeter.Throttling and Separating Calorimeter.

Introduction:A pure substance is one that has a homogeneous and in variable chemical composition. It may exist in more than one phase, but the chemical composition is the same in all phases. Thus liquid water, a mixture of liquid water and water vapor, and a mixture of ice and liquid water are all pure substance: every phase has the same chemical composition. On the other and mixture of liquid air and gaseous air is not a pure substance. Because of the composition of the liquid phase is different from that of the vapor phase.

Some times a mixture of gases such as air is considered a pure substance as long as there is no change of phase. Strictly speaking this is not true.

Vapor liquid solid phase equilibrium in a pure substance:

Fig 1: Constant pressure change from liquid to vapor phase for a pure substance

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Consider a system one kg of water contained in a piston cylinder arrangement as shown in the figure. Suppose that the piston and weight maintain a pressure of 0.1 MPa in the cylinder and that initial temperature be 200C. As the heat is transferred to the water the temperature increases appreciably, the specific volume increases slightly under constant pressure. When the temperature 99.60C, additional heat transfer results in a change of phase. I.e. some of the liquid becomes vapor. During this process both temperature and pressure remain constant where as sp. Volume increases considerably. When the last drop of liquid has vaporized further transfer of heat results in an increase in both temperature and sp. volume of the vapor.

The term saturation temperature designates the temperature at which vaporization takes place at a given pressure. Or the pressure is called saturation pressure corresponding to the saturation temperature. Thus for water 99.60C the saturation pressure is 0.1MPa, and for water at 0.1MPa the saturation temperature is 99.60C. Thus there is a definite relation between saturation pressure and saturation temperature.

Pressure Vapor-pressure curve

TemperatureFig 2: Vapor – Pressure curve of a pure substance

If the substance exists as liquid at the saturation temperature and pressure it is called saturated liquid. If the temperature of the liquid is lower than the saturation temperature for the existing pressure it is called either a sub cooled liquid or a compressed liquid.

When the substance exists as a part liquid and part vapor at the saturation temperature a dryness fraction comes into picture. It is also called as quality and it is defined as the ratio of mass of vapor to the total mass. It is denoted by the symbol ’x’. Quality has meaning only when the substance is in saturated state. I.e at the saturation pressure and temperature. The quality x is an intensive property.

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If a substance exists as vapor at the saturation temperature it is called saturated vapor or dry saturated vapor with x=1. When the vapor is at a temperature greater than the saturation temperature at the saturation pressure, it is said to exist as superheated vapor. After that the temperature increases as heat is added at constant pressure.

Temperature – Volume (T-V) diagram for water:

Fig 3: T-V diagram for water

It is clear from the figure that constant pressure lines are ABCD, EFGH, IJKL etc. the peak point of the figure indicated by N is the critical point of water. Thus the critical pressure 22.089 MPa and corresponding critical temperature is 374.14 0C.

A constant pressure process at a pressure greater than the critical pressure is represented by curve PQ. Thus water at 40 MPa, 200C is heated in a constant pressure process, there will be never be two phases present at the state shown. Instead there will be continuously in density at all the times and there will be only one phase present. The question arises is: when do we have a liquid and when do we have a vapor? The answer is that this is not a valid question at super critical pressures. We simply term the substance as fluid. However rather arbitrarily at temperatures below the critical temperatures we usually refer to it as a compressed liquid and at temperatures above the critical temperatures as superheated vapor. It should be noted that however at pressures above the critical pressures we never have a liquid and vapor phase of pure substance existing in equilibrium.

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Table1: Some Critical point dataCritical Temperature, 0C

Critical Pressure, MPa

Critical Volume, m3/kg

Water 374.14 22.089 0.003155Carbon dioxide 031.05 07.39 0.002143Oxygen -118.35 05.08 0.003438Hydrogen -239.85 01.30 0.032192

Consider another experiment with piston cylinder arrangement. Suppose that the cylinder contains one kg of ice at –200C and one bar. When heat is transferred to the ice the pressure remains constant the specific volume increases slightly and the temperature increases until it reaches 00C, at which point the ice melts and temperature remains constant. This state is called saturated solid state. For most substances the specific volume increases during this melting process. But for water specific volume of the liquid is less than the specific volume of the solid.

Sublimation: If the initial pressure of the ice at –200C is 0.26 kPa, heat transferred to the ice results in an increase in the temperature to –100C. At this point however the ice passes directly from the solid phase to the vapor phase. This process is known as sublimation. Further heat transfer results in superheating of the vapor

Triple Point: Consider the ice at 0.6113 kPa and temperature of –200C. Through heat transfer let the temperature increase until it reaches 0.01C. At this point however further heat transfer may cause some of the ice to become vapor and some to become liquid. At this point it is possible to have three phases in equilibrium. This point is called the triple point. Triple point is defined as the state in which all three phases may be present in equilibrium. The pressure and temperature at the triple point for a number of substance is given in following table.

Table 2:Triple Point DataTemperature, 0C Pressure, kPa

Hydrogen -259 7.194Oxygen -219 0.15Nitrogen -210 12.53Carbon Dioxide -56.4 520.8Mercury -39 0.00000013Water 0.01 0.6113Zinc 419 5.066Silver 961 0.01Copper 1083 6.000079

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Consider a solid state as shown in the figure, when the temperature increases with constant pressure the substance passes directly from solid to vapor phase. Along the constant pressure line EF the substance passes from solid to liquid phase at one temperature and then from liquid to vapor phase at higher temperature.

Constant pressure line CD passes through the triple point and it is only at the triple point the three phases exists together in equilibrium. At a pressure above critical pressure such as GH line there is no sharp distinction between liquid and vapor phases. The triple point temperature and critical temperature vary greatly from substance to substance. For ex: critical temperature of helium is 5.3K. Therefore absolute temperature of helium at ambient conditions is over 50 times greater than the critical temperature. On the other hand water has a critical temperature 374.14 0C (647.29K) and at ambient conditions the temperature of water is less than ½ the critical temperature.

Allotropic transformationIt should be pointed out that a pure substance can exist in a number of different solid phases. A transition from one solid phase to another is called an allotropic transformation. This can be well understood by the following figure.

Temperature

Fig 4: P-T Diagram for water.

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Fig 5: P-T Diagram for iron

Independent properties of a pure substance:One important reason for introducing the concept of pure substance is that

the state of a simple compressible pure substance is defined by two independent properties. For ex: if the specific volume and temperature of a super heated steam are specified the state of the steam is determined.

To understand the significance of the term independent property, consider the saturate liquid and vapor state of a pure substance. These two states have the same pressure and temperature but they are definitely not the same state. In a saturation state therefore pressure and temperature are not independent properties. Two independent properties such as pressure -specific volume, pressure and quality, temperature-specific volumes are required to specify a saturation state of a pure substance.

Equations of state for the vapor phase of a simple compressible substance:

From the experimental observations it has been established that the equation of state under low-density gases is given by PV=RuT. The ideal gas equation of state and compressibility factor equation are good approximations at low-density conditions. Therefore ideal gas equation of state is very convenient to use in thermodynamic calculations. The question comes what is low density? Or what range of density will the ideal gas equation of state hold with accuracy?

The analysis gives the pressure temperature deviations from ideal gas behavior. To answer5 the question the concept of compressibility factor Z is introduced and is defined by the relation Z= PV/RuT. For ideal gas Z=1 and the deviation of Z from unity is a measure of the deviation of actual relation from the ideal gas equation of state.

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Fig 6: Compressibility of Nitrogen

It shows a skeleton compressibility charge for nitrogen. Three observations can be made from this chart.

All the temperature Z 1 as P 0. i.e. as the pressure approaches zero the P-V-T behavior closely approaches that predicted by the ideal gas equation of state. Note also that at temperatures of 300K and above i.e. above room temperature the compressibility factor is near unity up to a pressure of 10MPa. This means that the ideal gas equation of state can be used for nitrogen over this range with considerable accuracy.

Now suppose we reduce the temperature from 300K but keep the pressure constant at 4MPa, the density will increase and we note a sharp decrease below unity in the value of compressibility factor. Values of Z <1 mean that the actual density is greater than would be predicted by ideal gas behavior. As the temperature reduced from 300K and pressure remains constant at 400MPa the molecules are brought closer together, there is an attractive force between the molecules. The lower the temperature the greater is the intermolecular attractive force. Thus more attractive force more density, which is greater than, would be predicted by the ideal gas. (Ideal gas intermolecular attraction is negligible)

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On comparison of various compressibility charts for other pure substances it is observed that the diagrams are all similar in the characteristics described above for nitrogen at least in qualitative sense. Quantitatively the diagram are all different sine critical temperature, critical pressure of different substances vary over a wide range. There is a way to put all these substances on a common basis called reduced properties w.r.t. the values at critical point like

Reduced pressure = Pr = P / Pc and reduced temperature Tr = T / Tc

These equations state that the reduced property for a given state ids the value of this property on this state divided by the value of the same property at the critical point.

Tables of Thermodynamic properties:

Tables of thermodynamic properties of many substances are available and in general these tables have the same form. The steam table is specially selected because they are a vehicle for presenting thermodynamic tables and steam is used extensively in power plants and industrial purposes. The table consists of four separate tables namely:

1) Saturated steam table 2) Saturated water table 3) Temperature basis table and 4) Pressure basis table.

It contains values of enthalpy, internal energy, specific volume and entropy. If the values of T and P are given other all values can be directly taken from the table. Therefore we should first learn how to use the table. We note that all these are independent variables.

To finding the correct table other nuisance of everyday use of the table in the interpolation. I.e. when one of the stated values is not exactly equal to a value listed in the table. Recently computerized tables are in use, which eliminates these problems. But the students nevertheless must learn to understand the significance, construction and limitations of the tables.

The actual steam table as presented in various books is a summary table based on complicated curve fit to the behavior of water. Here we concentrate three properties namely T,P and v and note that other properties listed namely u, h, s are presented. It is further noted that the separation of phases in terms of values of P and T is actually described by the relation illustrated in the pressure-temperature diagram.

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The specific volume of a substance having given quality can be found out by the definition of quality. The volume is the sum of volume of liquid and vapor.

V= Vliq + V vap.

In terms of masses: mv = mliq * vf + m vap* vg By introducing the quality x we have v = (1-x) vf + x vg

We know that vfg = vg - vf

We can write the specific volume equation for wet steam as v = vf + x vfg

In superheated region pressure and temperature are independent properties and therefore for each pressure a large number of temperature is given and for each temperature four thermodynamics properties are listed namely specific volume, enthalpy, entropy and internal energy.

Temperature – specific volume diagram :

Figure shows a T-v plot for water with an indication of percent error in assuming ideal gas behavior along the saturated vapor curve and also in several area of superheated region. Generally a slight decrease and only a small error is made if one same that the volume of a compressed liquid is equal to the specific volume of the saturated liquid at the same temperature. It is generally accepted procedure particularly when compressed liquid data are not available.

Fig 7: Temperature – specific volume diagram for water

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In the modern scenario computerized table are available. The main program operated with a visual interface in the window environment on a PC which is generally user friendly. The program operates on DOS environment, which covers not only the tables of water, but it covers most of the pure substances used in engineering industries. The generalized chart with compressibility factor is also included so it is possible to get the value of Z a more little accurately than reading the graph. It is useful in the case of two-phase mixture, like saturated liquid and vapor values are needed.

Thermodynamic Surfaces:

The matter discussed in this chapter can be well summarized by a consideration of a pressure, specific volume, temperature surface (PVT surface). Two such surfaces are shown in the figure for a substance such as water in which the specific volume increases during freezing and the other in which a substance, its specific decreases during freezing.

Fig 8 : PvT Surface -- Water

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Fig 9 : PvT Surface – Water

In these diagrams the pressure, specific volume and temperature are plotted on a mutually perpendicular coordinates and each possible equilibrium state is thus represented by a point on the surface. This follows directly from the fact that a pure substance has only two independent intensive properties. All points along a quasi equilibrium surface lie on the PvT surface since such a process always passes through equilibrium states.

The regions of the surfaces that represent a single surface- the solid, liquid and vapor faces are indicated. These surfaces are curved. The two phase regions – the solid-liquid, solid-vapor, liquid-vapor regions are ruled by surfaces. It is understood that they are made up of straight lines parallel to the specific volume axis. This of course follows from the fact that in the two-phase region lines of constant pressure are also lines of constant temperature, although the specific volume may change. The triple point actually appears as the triple line on the PVT surface, since the pressure and temperature of the triple point are fixed but the specific volume may vary depending upon the proportion of each phase.

It is also of interest to note that the pressure temperature and pressure volume projections of these surfaces. We have already considered the P-T diagram for water. It is on this diagram we observe the triple point. Various lines of constant temperature are shown on the P-v diagram and corresponding constant temperature sections are identically seen on the P-v-T surfaces. The critical isotherm as a point of intersection at the critical point.

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Fig 10 : PvT Surface for a Substance which Contracts Upon Freezing

Fig 11 : PvT Surface for a Substance which Expands Upon Freezing

One notice that for a substance such as water which expands on freezing, the freezing temperature decreases with an increase in pressure. For a substance that contracts on freezing, the freezing temperature increases as the pressure increases. Thus as the pressure of the vapor is increased along the constant temperature line a substance that expands on freezing first becomes solid and then liquid. For substance that contract on freezing, the corresponding constant temperature line indicates that as the pressure of the vapor increases, it first becomes liquid and then solid.

T-s Diagram for a pure substance:

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Consider the heating of 1 kg of ice at –50C to steam at 2500C. The pressure being maintained at 1 atm. It is observed that the entropy of steam increases in different regimes of heating namely

1) Entropy increase of ice to saturated freezing temperature2) Entropy increase of ice as it melts into water.3) Entropy increase of water as it is heated from 0Oc to 100OC.4) Entropy increase of water as it is vaporized at 100OC absorbing latent heat

of vaporization.5) Entropy increase of vapor as it is heated from 100OC to 250OC.

Fig 12: T – s Diagram for water Constant pressure lines

These entropy changes are shown in T-S graph. It is a constant pressure process. If during heating process the pressure had been maintained constant at 2 atm, a similar curve would be obtained. If these states for different pressures are joined the phase equilibrium diagram of a pure substance on the T-s coordinate would be obtained as shown below.

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Fig 13: T –s Diagram for water

Most often liquid vapor transformation only are of interest, the following figure the liquid, the vapor and the transition zones only at a particular pressure, sf – specific entropy of saturated water and sg - specific entropy of saturated vapor. The entropy change of the system during the phase change from a liquid to vapor at that constant

pressure is sfg. (= sg – sf). The value of sfg

decreases as pressure increases.

And becomes zero at the critical point.

Fig 14: Phase equilibrium diagram –T – s Diagram

H-S diagram (Mollier diagram) for a pure substance: From the first and second law of thermodynamics following property relations are obtained.Tds = dh – v dp and ( h / s )p = TThese equations form the basis of h-s diagram of a pure substance. The slope of an isobar on the h-s coordinates is equal to the absolute temperature. If the temperature remains constant the slope will remain constant. If the temperature increases the slope of the isobar will increase.

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Consider once again the heating of ice at –5OC to steam at 250OC the pressure being maintained constant at 1 atm. The slope of the isobar of 1 bar on the h-s coordinates first increases as the temperature of the ice increases from –5OC to 0OC. the slope then remains constant as ice melts to water at 0OC. Te slope of isobar again increases as the temperature of water rises from 0OC to 100OC. the slope again remains constant as water vaporizes at constant temperature. Finally the slope of the isobar continues to increase as the temperature of steam increases to 250OC and beyond (as shown in the figure below.

Fig 15: h –s diagram for water - constant pressure lines.

Similarly the isobars of different pressures can be drawn on h-s diagrams as shown in the figure below.

Fig 16: h –s for water.This figure shows the phase equilibrium diagram of a pure substance on the h-s coordinate indicating the saturated solid line, saturated liquid line saturated vapor line, the various phases and the transition zones.

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Fig 17: h-s diagram for water:

This figure is the Mollier diagram indicating only the liquid and vapor phases. As the pressure increases the saturation temperature increases, slope of the isobar increases. Hence the constant pressure lines diverge from one another and the critical isobar is at a tangent to the critical point. In the vapor region the states of equal slopes at various pressures are joined by lines as shown, which are the constant temperature lines. Here at a particular pressure hf is the specific enthalpy of saturated water and hg is specific enthalpy of saturated vapor and hfg

(= hg - hf) is the latent heat of vaporization at that pressure. As the pressure increases hfg decreases and at the critical pressure hfg becomes zero.

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Dryness fraction and various equations:

Dryness fraction is defined as the ratio of mass of dry steam to total mass of steam. It is denoted by x and is also called as quality of steam.

x = mv / (mv +ml ) where mv and ml are the masses of vapor and liquid respectively.Let V be the toal volume of a liquid vapor mixture of quality x in which Vf volume of saturated liquid and Vg volume of saturated vapor, the corresponding being m, mf, mg respectively.

We have m = mf + mg and V= Vf +Vg

Therefore mv = mf * vf + mg*vg v= (1-x) vf + x vg.

Similarly s = (1-x) sf + x sg h= (1-x) hf + x hg u= (1-x) uf + x ug.Same equations are written as v = vf + x vfg, h = hf + x hfg u = uf + x ufg s = sf + x sfg

If the condition of the steam is superheated then we have Degree of superheat, which is difference between the superheated temperature to the saturation temperature.

super = Tsuper –Tsaturation

The other properties are calculated as v super = vsat * Tsuper / Tsat

hsuper = hg + Cp (Tsuper-Tsat)ssuper = sg + Cp log(Tsuper/Tsat)

Measurement of Steam Quality:

The state of a pure substance gets fixed if two independent properties are given. Thus the pure substance is said to have two degrees of freedom.

Fig 18: T – s and h – s diagram

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Figure shows the values of pressure and temperature would fix up the state. But when the substance is in the saturation state or two phase region the measured values of pressure and temperature could apply equally well to saturated liquid point f and saturated vapor point g. or two mixtures of any quality points x1, x2 or x3. of the two properties, P and T only one is independent; the other is a dependent property. If the pressure is given the saturation temperature gets automatically fixed for the substance. In order to fix up the state of the mixture one more property such as specific volume, enthalpy or composition of mixture or internal energy is required to be known. Since it is relatively difficult measure the specific volume of the mixture devices such as calorimeters are used for determining the quality or the enthalpy of the mixture.

In the measurement of quality, the object is always to bring the state of the substance from the two phase region to the single phase region or superheated region where, the pressure and temperature are independent and measured to fix the state, either by adiabatic throttling or e3lectric heating.

There are four types of calorimeter are common in use namely1) Separating Calorimeter2) Throttling Calorimeter3) Combined separating and throttling calorimeter.4) Electrical calorimeter.

Separating Calorimeter: the steam whose dryness fraction is to be determined is very wet then separating calorimeter gives the quality of the steam. A known quantity of steam is passed through a separating calorimeter as shown. The steam is made to change direction suddenly, the water being denser than the dry steam is separated out. The quantity of water, which is separated out, is measured at the separator. The dry steam coming out of the separator is sent through a condenser where it is condensed separately.The dryness fraction of the steam is calculated by weighing the mass of the water and mass of dry steam after condensation separately.

Throttling Calorimeter: sample of wet steam of mass m at pressure p1 is taken from the steam main through a perforated sampling tube as shown in the figure.

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Fig 19: Throttling calorimeter

Then it is throttled by the partially opened valve to a pressure p2 measured by a mercury manometer and temperature T2 so that after throttling the steam is in the superheated region. The process is shown on T-s and H-s diagram.

The steady flow energy equation gives the enthalpy after throttling as equal to enthalpy before throttling. It is a irreversible process hence joined by a dotted line. Thus the initial state of the steam is p1 , t1 and its dryness fraction is x1 and the final state of the superheated steam is p2 x2.

Fig 20: T – s and h – s diagram for throttling calorimeter

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Now h1 =h2

hf1+x1 hfg1 = h2

x1 =( h2 – hf1 ) / hfg1

With P2 and T2 being known, h2 can be found out from the superheated steam table. The values of hf, hfg are taken from saturated steam table., thus quality of the wet steam x1 can be calculated.

Combined separating and throttling calorimeter:When the steam is very wet and the pressure after throttling is not low enough to take the steam to the superheated region then a combined separating and throttling calorimeter is used for the measurement of quality. Steam from the main is first passed through a separator as shown in the figure, where some part of the moisture separates due to sudden change in direction and falls by gravity and partially dry vapor is then throttled and taken to the superheated region.

Fig 21: Schematic diagram for separating and throttling calorimeter.

As shown in the figure process 1-2 represents moisture separation from the wet sample of steam at constant pressure P1 and process 2-3 represents throttling to pressure P2 withP2 and T3 being measured, h3 can be found out from the superheated steam table.

h3= h2 = hf1 + x2 hfg1

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Therefore x2, the quality of the steam after partial moisture separation can be evaluated. If m kg of steam is taken through the sampling tube in y seconds, m1

kg is separated and m2 kg is throttled and then condensed to water and collected, we have

m=m1+m2.

The mass of dry vapor will be at state2 is x2m2.Therefore the quality of the sample of the steam at state1 which ids x1 is given byx1= x2m2 / (m1 + m2)

There is one more method of measurement of quality of wet steam by using electric calorimeter as shown in the figure. The sample of steam is passed in steady flow through an electric heater. Electrical energy input Q should be sufficient to taken the steam to the superheated region where pressure P2 and temperature T2 are measured. If I is the current flowing through the heater in amperes and V the voltage across the coil at steady state Q=VI. If m is the mass of steam taken in t seconds under steady flow condition then the steady flow energy equation for heater is given by m1h1 + Q = m1h2

Fig 22: Electric calorimeter

Hence h1 + Q / m1 = h2. With h2,Q and m1 =being known h1 can be computed. Thus h1= hf1 + x1 hfg1.

Hence x1 can be calculated.

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1) A steam boiler initially contains 5 m3 of steam and 5 m3 of water at 1 MPa. Steam is taken out at constant pressure until 4 m3 of water is left. What is the heat transferred during the process?Solution:

Fig 23: Steam drum.

At 1 MPa vf = 0.001127 and vg= 0.1944 m3 / kg.hg=2778.1 kJ/kg uf = 761.68 , ufg = 1822 , ug = 2583.6 kJ/kg

The initial mass of saturated water and steam in the boiler = (Vf /vf) + (Vg / vg) =

[(5/0.001127) + (5/0.1944) ] = [(4.45 * 103) + (25.7) ] kg

Final mass of saturated water and steam = (4/0.001127) +(6/0.1944) = [ (3.55 * 103) + 30.8] kg

Mass of steam taken out of the boiler, ms = [4.45 * 103 + 24.7] - [ (3.55 * 103) + 30.8]

= 894.9 kg

Making an energy balance we have initial energy stored in saturated water and steam + heat transfer from external source = final energy stored in saturated water and steam + energy leaving the steam or

U1 + Q =U1 + ms*hg Assuming that the steam taken out is dry.Hence 4.45 * 103 * 761.68+27.7*2583.6+ Q = 3.55 * 103 *761.68 +30.8 *2583.6 +894.9 * 2778.1Q = 2425000-685500+13176 Q = 1752676 kJ.

2) Steam flows in a pipeline at 1.5Mpa. After expanding to 0.1MPa in a throttling calorimeter, the temperature is found to be 120 o C. Find the quality of steam in the pipe line. What is the maximum moisture at1.5 MPa

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that can be determined with this set-up if least 5 o C of superheat is required after throttling for accurate readings?

Fig 24 : h –s Diagram

Solution.At state 2 when p = 0.1 MPa and t = 120 o C by interpolation, h2 = 2716.2 kJ/kg and p = 1.5 MPa hf = 844.89 and hfg = 1947.3 kJ/kg and h1 = h2 hf1+ x1 hfg1 = h2

844.89 + x1 *1947.3 = 2716.2x1 = 1871.3 / 1947.3 = 0.963 Ans.When p = 0.1MPa and t = 99.63 + 5 = 104.63 oC, h3=2685.5 kJ/kg

Since h3=h4 2685.5 = 844.89 + x4 *1947.3x4 = 1840.6 / 1947.3 = 0.948The maximum moisture that can be determined with this set up is only 5.2% Ans3) The following data were obtained with a separating and throttling calorimeter:Pressure in pipeline :1.5 MPaCondition after throttling:0.1 MPa.110o C During 5 min moisture collected in the separator:0.150 litre at 70 oC, Steam condensed after throttling during 5 minFind the quality of steam in the pipeline

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Fig 25: h –s DiagramSolution :

AT 0.1 MPa, 110 oC, h3 = 2696.2 kJ/kgNow h3=h2 = hf2 + x2 hfg2

2696.2=844.89 + x2 1947.3x2 = 1851.31/1947.3 = 0.955

If m1 = mass of moisture collected in the separator in 5 min and m2= mass of steam condensed after throttling in 5 min then

x1 = (x2 m2) / (m1 + m2) at 70oC vf = 0.001023 m3/kgm1= 0.1462 kg and m2 = 3.24 kgHence x1= 0.955*3.24 / (0.1462 +3.24) = 0.915 Ans.

4) A vessel having a volume of .4 m3 contains 2.0 kg of liquid water and vapor mixture in equilibrium at a pressure of 600 kPa. Calculate a) The volume and mass of liquid b) The volume and mass of vapor.

Solution: The specific Volume is calculated first. v =(0.4/2 ) = 0.2 m3/kgThe quality of steam can now be calculated 0.2 = 0.001101 + x * 0.3146, x = 0.6322Therefore mass of liquid is 2.0*(1 - 0.6322) = 0.7356 kgMass of vapor is 2.0(0.6322) = 1.2644 kgVolume of liquid is ml vf

= 0.7356(0.001101) =0.0008m3

Volume of vapor is mvvg = 1.2644(0.3157)=0.3992 kg

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5) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vessel until it becomes saturated vapor. Calculate the heat transferred per kg steam.

Solution:

From the steam table at 1 bar pressure, ts = 99.62oC,vf =0.001043m3/kg,vfg=1.69296m3/kg, uf =417.33 kJ/kg, ufg =2088.72kJ/kg,

Volume of one kg of given state of vapor = vf + x vfg = 0.001043 + 0.523 * 1.69296 = 0.8864 m3/kg

Enthalpy correspond to the state of steam, u= uf + x ufg = 417.33 + 0.523 * 2088.72 =1509.73kJ/kgv (0.8864 m3/kg ) correspond to the saturated condition of the steam, from steam table, which is vg, The pressure found to be 2 bar and ug =2529.49kJ/kgHeat added = 2529.49– 1509.73 = 1019.76 kJ/kg Ans

.6) A rigid vessel contains one kg of mixture of saturated water and saturated steam at a pressure of 0.15MPa. When mixture is heated the state passes trough the critical point. Determine,a) Volume of the vesselb) The mass of the liquid and vaporc) The temperature of the mixture when the pressure rises to 3 MPa.d) The heat transfer required to produce the final state.Solution. Vc: Critical volume = 0.003155 m3/kg (From the steam table.)

v = vf + x X vfg 0.003155 = 0.001035 + x1 * 1.15828 , x1 = 0.00183Hence, mass of vapor = x = 0.00183 kg

Mass of liquid = (1-x1) = 0.998 kg.u1 = uf + x1 X ufg ;

u1 = 466.92 + 0.00183 X 2052072 = 470.68 kJ/kgSaturation temperature correspond to 3 MPa, is 233.9OC, the temperature of the mixture

x2 = (0.003155 – 0.001216)/0.06546 = 0.02962 Heat transfer in constant volume process = u2 – u1,

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u2 = 1004.76 + 0.02962 X 1599.34 = 1052.13 kJ/kgHeat transfer = 1052.13 – 470.68 = 581.45 kJ/kg.

7) Steam initially at 0.3MPa, 250OC is cooled at constant volume. Findi)  At what temperature will the steam become saturated vapor?ii) What is the quality at 80OC? What is the heat transferred per kg of steam in cooling from 250 OC to 80OC?

Solution: At 300kPa, 250 OC, from the steam table , it is a superheated condition. v = 0.79636 m3/kg , u = 2728.69 kJ/kg

vg = 0.79636 m3/kg , Ps = 225 kPa, ts = 124 OC at 80 OC

0.79636 = 0.001029 + x 3.45612x = (0.79636 - 0.001029) /3.45612 = 0.23u2 = 520.45+0.23(2013.1) = 483.463 kJ/kg

Heat transfer = change in internal energy as the process is const. volume, Heat transfer = u1 – u2 = 2728.69 – 983.463 = 1745.227 kJ/kg

8) State whether the following samples of steam are wet, dry or superheated: Justify your answer.

I) Pressure = 1 MPa absolute enthalpy = 2880 kJ/kgII) Pressure=500kPa absolute, volume =0.35m3 /kgIII) Temperature = 200oC Pressure = 1.2 MPa.IV) Temperature = 100oC, entropy =6.88kJ/kg K.V) Pressure= 10 kPa, enthalpy = 2584 kJ/kg.

Try as homework.

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