thermal engineering (mg university) part1

8/12/2019 Thermal Engineering (MG UNIVERSITY) PART1

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THERMAL ENGINEERING (MG UNIVERSITY) PART 1

Module 1

Steam Engineering: Properties of steam - wet, dry and superheated steam - dryness

fraction - enthalpy and internal energy - entropy of steam - temperature entropy diagram -

process - Mollier chart - Rankine cycle for wet, dry and superheated steam. Steam

Generators - classification - modern steam generators - boiler mountings and accessories.

Steam Engineering

Formation of steam -

Consider a cylinder fitted with a piston which can move freely upwards and downwards

in it.

(a) Let 1 kg of water at 0oC under the piston

Let the piston is loaded with load w to ensure heating at constant pressure.

Now if heat is imparted to water, a rise in temperature will be noticed and this rise will

continue till boiling point is reached.

B.P of water, at normal atmospheric pressure of 1.01325 bar is 100oC. But it increases

with the increase in pressure.

(b) The volume of water will increase slightly with the increase in temperature, but increase

in volume of water (or work) is generally neglected for all types of calculations.

The boiling temperature is known as the temperature of formation of steam or saturationtemperature.


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(c) Now, if supply of heat to water is continued, it will be notices that rise of temperature

after the boiling point is reached nil but piston starts moving upwards which indicates that there

is increase in volume which is only possible if steam formation occurs.

The heat being supplied does not show any rise of temperature but changes water into

vapour state (steam) and is known asLatent heator hidden heat.

So long as the steam is in contact with water, it is called wet steam.

(d) If heating of steam is further progressed such that all the water particles associated with

steam are evaporated, the steam so obtained is called dry and saturated steam.

If vg m3is the volume of 1 kg of dry and saturated steam then work done on the piston

will be

P (Vg- Vf), where P is the constant pressure (due to weight W on the piston).

(e) If the supply of heat to the dry and saturated steam is continued at constant pressure,

there will be increase in temperature and volume of steam.

The steam so obtained is called super heated steam and it behaves like a perfect gas.


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Temperature Vs Total Heat Graph during steam formation

A represents the initial condition of water at 0oC and pressure p (in bar)

During the formation of the super heated steam, from water at freezing point, the heat is

absorbed in the following 3 stages.

The heating of water upto boiling temperature or saturation temperature (ts) is shown by

AB.

APknown as sensible heat, liquid heat or total heat of water.

The change of state from liquid to steam is sown by BC PQ, latent heat of vaporisation.

The super heating process is CD.

QR known as the heat of superheat.

LINE, AR represents the total heat of the super heated steam.

If the pressure is increased, the boiling temperature also increases.

The line passing through the points A, B, E, K Saturated liquid line.

The line passing through the points L, F, C Dry saturated steam line.

[Some times, these terms are briefly written as liquid line and dry steam line. but the

word saturated is always understood].


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Note:

When the pressure and saturation temperature increases, the latent heat of vaporisation

decreases, it becomes ZERO at a point (N), where liquid and dry steam lines meet.

The point N is known as critical point and at this point, the liquid and vapour phases

merge, and become identical in every respect.

The temperature corresponding to critical point N is known as critical temperatureand

the pressure is known as critical pressure.

For steam, the critical temperature is 374.15oC and critical pressure is 220.9 bar

Pc = 220.9 bar

Tc = 374.15oC

At critical point and above, there is no definite transition from liquid to vapour and two

phases cannot be distinguished visually. The latent heat of vaporisation is zero at critical point

and has no meaning at pressure higher than critical.

At T = 273.16 k and P = 0.006113 bar ice, water and steam co-exist in the

thermodynamic equilibrium in a closed vessel and bcf (Belleni - 200) is called triple point line.

At lower pressures than this, ice sublimates to steam.

IMPORTANT TERMS RELATING STEAM FORMATION

1. Sensible Heat of water (hf)

It is defined as the quantity of heat absorbed by 1 kg of water when it is heated from

0oC (freezing point) to boiling point.

If i kg of water is heated from 0oC to 100

oC the sensible heat added to it will be 4.18

100 = 418 kJ

But if water is at say 20oC initially then sensible heat added will be 4.18 (100-20) =

334.7 kJ

This type of heat is denoted by letter h fand its value can be directly read from the steam

tables.


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The value of specific heat of water may be taken as 4.18 kJ/kg K at low pressures but at

high pressures it is different from this value.

2. Latent Heat or Hidden Heat (hfg)

It is the amount of heat required to convert water at a given temperature and pressure

into steam at the same temperature and pressure.

The value of L.H is not constant and varies according to pressure variation.

3. Dryness Fraction (x)

It is related with wet steam

Mass of dry saturated vapour to the total mass of the mixture.

x =g g

g f

m m

m m m

mg= Mass of actual Dry steam

mf= Mass of water in suspension

m = Mass of mixture = mg + mf

eg:- If in 1 kg of wet steam 0.9 kg is the dry steam and 0.1 kg water particles then x =

0.9.

No steam can be completely dry and saturated, so long as it is in contact with the water

from which it is being formed.

The steam is called saturated when the molecules escaping from the liquid become

equal to the molecules returning to it.

Saturated steam may be dry or wet. When the saturated vapour contains particles ofliquid evenly distributed over the entire mass of vapour, it is called wet saturated steam.

Wet steam is characterised by its dryness fraction.

Dryness fraction, x =mass of day saturated vapour

mass of mixture


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=mg

m

x =mass of dry vapour in the mixture

mass of the mixture

Q. Calculate the dryness fraction of steam which has 1.25 kg of water in suspension with 40

kg of steam

=g

g f

m

m m=

400.97

40 1.25

4. Total heat or enthalpy of wet steam (h)

It is defined as the quantity of heat required to convert 1 kg of water at 0oC into steam at

constant pressure.

5. Total heat of dry saturated steam

If steam is dry saturated, x = 1 and hg= hf+ hfg

6. Superheated steam

Total heat of super heating is always carried out at constant pressure.

It represents the quantity of heat required to convert 1 kg of water at 0oC into super

heated steam at constant pressure.

sup f fg ps sup sh h h c T T

The value of specific heat of steam at constant pressure Cpsdepends upon the degree of

superheat and the pressure of steam generation. Its average value is taken from 2 to 2.1 kJ/kg K.

Water boils at 12oC if pressure on the surface of water is kept at 0.014 bar.

7oC if pressure 0.01 bar.

Advantages obtained by using super heated steam

1. By super heating steam, its heat content and have its capacity to do work is increased

without having increase its pressure.

2. High temperature use of super heated steam results in an increase in thermal


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efficiency.

3. Super heating is done in a super heater which obtains its heat from waste furnace

gases which would have otherwise passed uselessly up the chimney.

Volume of wet and dry steam

If steam has a dryness fraction of x.

1 kg of this steam will contain x kg of dry steam and (1 - x) kg of water.

Let ,

fv volume of 1 kg of water

gv volume of 1 kg of perfect dry steam

fv = specific volume of saturated liquid

fgv = specific volume of evaporation

gv = specific volume of dry steam, then

[specific volume of a fluid is the volume occupied by a unit mass of the fluid]

Volume of 1 kg of wet steam = volume of dry steam + volume of water

[Since vf is very small as compared to gv , therefore the expression (1 - x) vfmay be

neglected.

Volume of 1 kg of wet steam = 3gx v m

g fx v 1 x v

g f fx v v x v

f g fv x v v

f fgv x v

f fg fg fgv x v v v


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f fg fgv v 1 x v

g fgv 1 x v

Super heated Steam

The superheated steam behaves like a perfect gas and therefore, its volume can be

worked out by applying Charles law to steam at the beginning and at the end of super heating

process.

vg= Specific volume of dry steam at pressure P

Ts= Saturation temperature in K

Tsup= Temperature of super heated steam in K

Vsup = Volume of 1 kg of super heated steam at pressure P.

Theng sup

s sup

PV PV

T T

g supsup

s

V TV

T

Internal Energy of steam

The actual Heat energy above the freezing point of water stored in steam is known as

internal energy of steam.

The work of evaporation is not stored in the steam as it is utilised in during external

work.

So the internal energy of steam could be found by subtracting work of evaporation from

the total heat.

u = h - pv

For wet steam

f fg gu h x h px v


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= f fg gh h 100px v kJ/kg

Pressure on the piston in bar

= P 105 N/m2

1 bar = 105 N/m2

For dry saturated steam

f fg gu h h p v

g gh 100pv kJ/kg

For super heated steam

f fg ps sup s supu h h C T T PV

g ps sup s suph C T T 100PV

Entropy of steam

1. The entropy of water at 0oC is taken as zero. The water is heated and evaporated at

constant pressure. The steam is also super heated at constant pressure in super heaters.

2. So the entropy of steam can be calculated from the formula for the change of entropy

at constant pressure.

Entropy of water

ps

C dTdQd

T T

The total increase in entropy of water from freezing point to boiling point, may be

obtained by integrating the above expression within the limits 273 K and TsK.

ss T p dTso 273

s

Cd

T

s sf p e p

T TS C log 2.3C log

273 273


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The value of Sfmay be directly seen from the steam tables

Entropy Increase during Evaporation

When the water is completely evaporated into steam, it absorbs full latent heat (h fg) at

constant temperature T, corresponding to the given pressure.

Entropy =Heat absorbed

Absolute temperature

Increase of entropy during evaporation

fgfg

hS

T

If the steam is wet with dryness fraction x, the evaporation will be partial.

i.e., if evaporation is partial,

Heat absorbed = x hfg

Increase of entropy, fgfgx h

ST

Entropy of wet and dry steam

Entropy of wet and dry steam =

Entropy of water + Entropy during evaporation

=fg

f f fg

x hS S x S

T (wet steam)

=fg

f f fg g

hS S S S

T (dry steam)

Entropy of super heated steam

Heat absorbed; dQ = CpsdT

ps dTs

Cd

T [value taken 1.67 kJ/kg K to 2.5 kJ/kg K]


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sup sup

g s

S T

s pS T

dTd C

T or

sup supsup g ps e p

T TS S C log 2.3C log

T T

where sup gS S is the increase in entropy.

Entropy of 1 kg of superheated steam is

supsup g psT

S S 2.3C logT

TEMPERATURE - ENTROPY (T.S) DIAGRAM

STEAM TABLES

The generation of steam at different pressures has been studied experimentally and

various properties of steam have been obtained at different conditions. The properties have been

listed in tables called steam tables. The steam tables are available for

1. Saturated water and steam - on pressure basis.

2. Saturated water and steam - on temperature basis.

3. Super heated steam - on pressure and temperature basis for enthalpy, entropy and

specific volume.

4. Supercritical steam - on pressure and temperature basis above 221.2 bar and


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374.15oC for enthalpy, entropy and specific volume.

Some important points regarding Steam Tables

(a) The steam table gives values for 1 kg of water and 1 kg of steam.

(b) The steam table gives values of properties from the triple point of water to the

critical point of steam.

(c) For getting values of thermodynamic properties, either saturation pressure or

saturation temperature need to be known. Pressure based steam table (i.e., extreme left pressure

column is placed) is used when pressure value is known, similarly temperature based steam table

is used when temperature value is known.

(d) At low pressure the volume of saturated liquid is very small as compared to the

volume of dry steam and usually the specific volume of the liquid is neglected. but at very high

pressure the volume of liquid is comparable and should not be neglected.

(e) The specific enthalpy and specific entropy at 0oC are both taken as zero and

measurements are made from 0oC onwards.

(f) In computing properties for wet steam it should be noted that only hfg and sfg are

affected by dryness fraction but hfand sfare not affected by dryness fraction. This means that for

steam with dryness fraction x,

g f fgh h x h

g f fgS S xS

Property Table

Property Wet steam Dry steam Super heated steam

Volume f g1 x v x v gv supg

s

Tv .

T

Enthalpyf gfh x h f fg gh h h g ps sup sah C T T


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Entropyf fgS xS f fg gS S S sup

g ps ns

TS C l

T

Enthalpy - Entropy chart (Mollier chart)

Most of the thermodynamic systems deal with flow of steam in steady condition where

change in enthalpy is encountered.

The most convenient method of computing change in enthalpy is the enthalpy-entropy

chart.

Saturated liquid region is not required for solving engineering problems and therefore

only a part of chart near saturated vapour region and super heat region is shown.

This chart is very useful for solving problems on nozzles and steam power plants.

1. Dryness fraction lines

2. Constant volume lines

3. Constant pressure line

4. Isothermal lines

5. Isentropic lines

6. Throttling lines

RANKINE CYCLE

M.Rankine (1820-1872), a Professor at Glasgow University


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It is also a reversible cycle but it differs from the Carnot cycle in the following respects:

(i) The condensation process is allowed to proceed to completion; the exhaust steam

from the engine/turbine is completely condensed. At the end of condensation process the

working fluid is only liquid and not a mixture of liquid and vapour.

(ii) The pressure of liquid water can be easily raised to the boiler pressure (pressure at

which steam is being generated in the boiler) by employing a small sized pump.

In addition, the steam may be super heated in the boiler so as to obtain exhaust steam of

higher quality. That will prevent pitting and erosion of turbine blades.

Steam power plant working on ideal Rankine cycle

The various elements are:

A boiler which generates steam at constant pressure

An engine or turbine in which steam expands isentropically and work is done.

A condenser in which heat is removed from the exhaust steam and it is completely

converted into water at constant pressure

A hot well in which the under state is collected

A pump which raises the pressure of liquid water to the boiler pressure and pumps it

into the boiler for conversion into steam.


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Consider a steady flow conditions at all states and 1 kg of steam is circulating through

the cycle.

The heat supplied by the boiler per kg of steam generated

Heat absorbed = Q1= (h2h1) = (h2h4) - (h1h4)

where,

Wp= (h1h4) is called pump work per kg of steam.

Heat rejected into the condenser = Q2= (h3h4)

Net work done per kg of steam = Q1- Q2

= (h2h4) - Wp- (h3h4)

= (h2h3) - Wp

= WT- WP

Where,

WT= Turbine work = (h2h3) = isentropic enthalpy drop during expansion

Rankine efficiency = R1

Network done W

Heat supplied Q

=

1 2 P

1 3 P

h h W

h h W

The pump work (WP) is very small as compared to turbine work (h2 h3) and heat

added (h2h1), therefore it can be fairly neglected.


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WP= ( P1- P2) V4

P1= Boiler pressure, P2= Condenser pressure

V4= Specific volume of saturated liquid at condenser pressure.

The field pump handles liquid water which is in compressed, which means with the

increase in pressure its density or specific volume undergoes a little change. Using general

property relation for reversible adiabatic compression, we get,

Tds = dh - vdp

ds = 0

dh = v dp

h = v P ... (since change in specific volume is negligible)

hf2- hf3= V1(P1- P2)

When P is in bar and v is in m3/kg, we have

hf2- hf3= V4(P1- P2) 105J/kg

The Rankine efficiency without pump work is

1 2R

1 f 3

h h

h f

............ (1)

State 3 (i.e., at the end of isentropic expansion) must be known then only h 3 can be

determined. State 3 is located from the steam table by equating entropy S2and S3or by drawing

a vertical line on the Mollier chart from State 1 to condenser pressure.

Modified Rankine Cycle (Steam Engine Cycle)

In the steam engine the expansion is not continued up to the point 2 as the stroke will be

too long and as the work obtained is very small at the tail end of the stroke which is not even

sufficient to overcome the frictional resistances near the end of the stroke. Therefore in actual

practice the expansion is terminated at point 5 instead of 2 and the steam is released at constant

volume. This causes a sudden pressure drop from P2 to P2 to Pb (back pressure) at constant

volume due to the steam communicating with outside atmosphere. This is represented by 56 fig.

This reduces the stroke length of the engine without any appreciable change in the work done.


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Specific Steam Consumption (S.S.C)

It is defined as the steam consumption (kg/s) to produce unit power (kW)

S.S.C = 1 2

Mass flow rate per hour kg /s 3600kg/kWhr

Net power output kW h h

(h1- h2) kJ work is obtained from 1 kg of steam.

1 kW hr = 3600 kJ

S.S.C = 1 2

3600kg/kWhr

h h

In case of steam power plant, the specific steam consumption is an indicator of the

relative size of the plant.

Work ratio (Wr): It is the ratio of network done to the turbine work.

1 2 P

r1 2

h h WW

h h


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Relative Efficiency or Efficiency Ratio

Relative Efficiency =Thermal Efficiency

Ranking Efficiency

Q. A simple Rankine cycle steam power plant operates between the temperature of 260 oC

and 95oC. The steam is supplied to the turbine at a dry saturated condition. In the turbine

it expands in an isentropic manner. Determine the efficiency of the Rankine cycle followed

by the turbine and the efficiency of the carnot cycle operating between these two

temperature limits. Draw the T - S and H - S diagrams.

Solution:

T1= 260oC = 260 + 273 = 533 K ; T2= 95

oC = 95 + 273 = 368 K.

From steam table, At 260oC, P2= 46.94 bar 1 95

oC, P2= 0.845 bar.

The initial and final conditions of steam are shown in the H-S diagram.

h1= 2800 kJ/kg;

h2= 2170 kJ/kg;

From steam tables at temperature 95oC,

hf3= 398 kJ/kg

Efficiency of Rankine cycle, 1 2R1 f 3

h h

h f

=2800 2170

2800 398

= 0.262 = 26.2%

Efficiency of Carnot cycle, 1 2c1

T T

T

=533 368

533

= 0.3096 = 30.96%


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Ranking cycle for wet dry and super heated steam

The value of h1and h2may be determined by using steam tables

h1= hg= 2796.4 kJ/kg ; Sg= 6.001 kJ/kg

hf3= hf = 398 kJ/kg = 2270.2 kJ/kg

Sf3= Sf = 1.25 kJ/kg ; Sfg= 6.167 kJ/kg K

Dryness fraction at 2

S1= S2

6.001 = 1.25 + x (6.167) x = 0.77

h2= hf+ x hfg

= 398 + 0.77 2270 - 2 = 2146 kJ/kg

Specific Steam Consumption

It is the mass of steam that must be supplied to a steam engine or turbine in order to

develop a unit amount of work or power out put.

The amount of work or power out put is usually expresses in kilowatt hour (kWh).

W = J/s

S.S.C =1 2

Mass flow rate per hour kg/ s kg 3600

Net power out put kW kWS h h

= 1 2

3600kg/kWhr

h h

Q. A steam power plant uses steam at a pressure of 50 bar and temperature 500oC and

exhausted into a condenser where a pressure of 0.05 bar is maintained. The mass flow rate

of the steam is 150 kg/sec. determine (a) the Rankine engine efficiency (b) Power developed

(c) specific steam consumption (d) Heat rejected into the condenser per hour (e) Carnot

efficiency.

P1= 50 bar, P2= 0.05 bar

From steam tables:


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50 bar 263.99oC (Saturation Temperature)

Page - 44 (Properties of super heated steam)

h1= 3433.8 and S1= 6.9770 kJ

S1= S2

6.977 = f 3 fgS xS

6.977 = 0.4764 + x 7.9187, x 0.82

h2 = hf2= x hfg= 137.8 + 0.82 2423.8 = 2125.316 kJ/kg

Vf3= 1.005 10-3

m3/kg

hf3= 137.82 kJ/kg

(a) Rankine Engine Efficiency = 1 2

1 3

h h 3433.8 2125.316

h h 3433.8 137.82

= 0.3969 = 39.69%

(b) Power developed = ms work done per kg = 150 (h1- h2)

= 150 1308.384

= 196257.6 kW = 196.257 mW

(c) S.S.C =1 2

3600 3600

h h 1308.384

= 2.751 kg/kW hr

(d) Heat Rejected into the condenser = Q2= ms(h2- h3)

= 150 (2125.316 - 137.8)

= 298127.4 kJ/s

(e) Carnot efficiency, c =

2

1

273 32.9T1 1

T 273 263.9

= 0.43 = 13%

P1 Boiler Pressure, P2= Condenser Pr


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V3 Specific volume of saturated liquid at the condenser pressure

WP = (P1- P2) V3

(f) Ranking cycle efficiency,

1 2 PR

1 f 3 P

h h W

h h W

=

3433.7 2125.316 50 0.05 /10

3433.7 137.8 50 0.05 /10

=1308.384 4.995

100 39.6%3295.9 4.995

Q. Dry saturated steam at 10 bar is supplied to a prime mover and the exhaust takes place

at 0.2 bar. Determine the Rankine Efficiency, efficiency ratio and specific steam

consumption of the prime mover, if the indicated thermal efficiency is 20%. Also find the

percentage change in the Rankine efficiency, if steam is initially 90% dry.

From Mollier chart, h1= 2775 kJ/kg, h2= 2150 kJ/kg

From steam tables, we find that enthalpy of water at 0.2 bar,

hf3= 251.5 kJ/kg

Rankine Efficiency,

1 2

R1 f 3

h h

h h

=2775 2150

2775 251.5

= 0.247 or 24.7%

Efficiency ratio =Indicated thermal efficiency

Ranking efficiency

0.20.247

= 0.81 or 81%

Specific Steam Consumption =1 2

3600

h h


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=3600

2775 2150= 5.76 kg/kWh

Percentage change in the Rankine efficiency if the steam is initially 90% dry

h1= 2580 kJ/kg, h2= 2030 kJ/kg

Rankine efficiency,

2 3R

2 f 3

h h

h h

=

25080 2030

2580 251.5

= 0.236 or 23.6%

Percentage change in Rankine Efficiency

= 24.7 23.6 100 4.45%24.7

Q. In a Rankine cycle, the steam at turbine inlet is saturated at a pressure of 30 bar and the

exhaust pressure is 0.25 bar. Determine, (i) Pump-Work (ii) Turbine power. (iii)

Rankine efficiency (iv) condenser heat flow (v) dryness at the end of expansion.

Assume flow rate of 10 Kg/s.

P1= 30 bar

P2 = .05 bar

(i) Pump work per 1 Kg.

p 4 3 fW m P P V

51 (30 .04) .00102 10 3KJ

Power required for the pump10 3KJ

30KWsec

(ii) Turbine Power

From steam table for 30 bar, dry sale steam

h1 = kg, 2803 KJ/Kg

at (1) entropy


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S1 = Sg1= 7.831 KJ/kgK

at .2 steam is wet steam

2 f 2 2 2 2S S x Sfg 0.893 x x ...

Since 1-2 is an isentropic process

S1= S2

7.831= 0.893 + x2 .........

x2= 0.763

Enthalpy at 2, (wet steam of x2 dry)

2 f 2 2 2

h h x hfg

= 272 + 0.763 2346

Turbine power = 10 (2803-2062) KJ/s. = 7410 KW

(iii) Rankine Efficiency

1 2 p

1 3 p

h h W

h (h w )

=

(2803 2062) 3

2803 272 3

=0.292 or 29.2%

(iv) heat flow rate in the condenser

= m(h2h3) = 10 (2062272) = 17900 KW

(v) Dryness at the end of expansion = 0.763 = 76.3%

Thermodynamic Processes of steam

Constant volume process

V1 = x1Vg1, V2= x2Vg2

(i) W 12= 0 dv = 0

(ii) U1= h1100P1V1= h1100P1X1Vg1

U2 = h2P2V2100 = h2P2X2Vg2100.....(Wet)


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= h2P2Vg2100.........(dry saturated)

= h2P2Vsup 100 ...........(super heated)

(iii) heat absorbed, q12= du + w 12= U2U1

Applying first law energy equation

2

1

Q u pdv

2 1 2 1U U P(V V )

if initially steam is wel. V1= X1Vg1

Finally super heated V2= Vsup

Constant Temperature Process

in wet steam region (hynerbolic in super heated steam region) will be a constant pressure process also during Condensation & evaporationQ = h2h1

W = P1(V2V1)

Limited to wet steam region onlyHyperbolic Process

Process PV = C Hyperbolic process is also an isothermal process in the superheated steam

regions.

2 22

11 1

vcW pdv dv c log

v v

21

1

VW P V1log

V


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Q u w

22 1 1 11

VU U P V loge

V

22 2 2 1 1 1 1 11

Vh P V h P V P V log

V

22 1 1 11

VQ h h P V log

V

Isentropic Process

Q u w

Q = O adiabalic

W = U1U2

Steady flow reversible

W = h1h2

1 1 1 2 2 2

1 2

u P V Q W U P V

h h

Polytropic Process

steam follows the low PVn= C

Work done 1 1 2 2P V P V

Wn 1

Applying first law energy equation to the non flow process.

Q u W

= 1 1 2 22 1P V P V

U Un 1

1 1 2 22 2 2 1 1 1P V P V

h P V h P Vn 1


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2 1 1 1 2 21

h h P V P V 1n 1

2 1 1 1 2 2n

Q h h P V P Vn 1

Throttling Process

Const. enthalpy in the absence of heat and work transfer enthalpy remainsconstant.

h1= h2

during throttling pressure always fallsProcess Wo Qn

Isochoric O U2U1

Isobaric 2 1P V V 2 1h h

Isothermal 2 1P(V V ) h2h1

Hyper bolic 21 11

VP V log e

V

22 1 1 11

Vh h P V log e

V

Isentropic U2U1 0

Polytropic 1 1 2 2P V P V

n 1

2 1 1 1 2 2n

h h P V P Vn 1

throttling process h1 = h2

thermal engineering (mg university) part1

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