1. Basic state values of matter

Example 1.1

The pressure inside a boiler is p p = 115.105 Pa and pv = 9.44.104 Pa inside a condenser. Calculate the absolute pressure inside the boiler and condenser if the barometric pressure pb = 0.098 MPa. Find the vacuum in the condenser in % !

Solution:

pp = 11.5 MPa: pv = 94.4 kPa; pb = 0.098 MPa; pa boil = ? ; pa cond = ?

a) The absolute pressure inside the boiler pa boil = pb + pp = 0.098 + 11.5 = 11.598 MPa

The absolute pressure in the condenser pa cond = pb - pv = 0.098 - 0.0944 = 0.0036 MPa

b) The vacuum inside the condenser in % :

1

Example 1.2

A pressure vessel of a volume of 0.1 m3 contains 1.25 kg of oxygen. What is the density and specific volume?

2. THE IDEAL-GAS EQUATION OF STATE f(p, v, T)=0

Example 2.1

Calculate the densitz and specific volume of carbon dioxide CO2 under normal physical conditions.

Example 2.2

A pressure vessel contains nitrogen at a temperature of t = 20 °C and pressure p = 2.2 MPa. The maximum allowed overpressure is 6 MPa.What is the maximum temperature that the nitrogen can be heated to if the barometric pressure is pb = 0,1 MPa?

x [%] = x.100 = 96.3 %

Solution:

V = 0.1 m3; m = 1. 25 kg 02; ρ = ? ; v = ?

air density

specific volume of oxygen

Solution :

ρ = ? ; v = ?

Normal conditions are: pressure p = 0.101325 MPa, temperature t = 0 °C.

The gas constant for CO2 r = 188,97 J/(kg.K); absolute temperature T = t + 273.15 = 273.15 K

Using the equation of state p v = r T ;

Then

Density

Solution:

t = 20 °C ; p = 2.2 MPa ; ppmax = 6 MPa ; pb = 0.1 MPa; t2 = ?

p1 = 2.2 MPa, T1 = t1 + 273 = 20 + 273 = 293 K

p2 = pa max = ppmax + pb = 6 + 0.1 = 6.1 MPa

The isochoric process

2

3. Ideal-gas mixtures

Example 3.1

1 kg of dry air comprises 23.2 mass % oxygen and 76.8 % nitrogen. Find the volume fractions of air (i.e. the individual volumes), the gasconstant, the mean virtual molar mass and the partial pressures of oxygen and nitrogen if the air pressure is p = 0.1013 MPa.

4. I. LAWS OF THERMODYNAMICS

Example 4.1

A steel component weighing 0.2 kg heated inside a furnace is then inserted inside a calorimeter containing 0.5 kg of water at a temperatureof t = 20 °C. After stabilisation, the temperature inside the calometer is 75 °C. Calculate the temperature of the component before it wasinserted into the calorimeter. The specific heat capacitz of steel is c = 0.469 kJ/(kg.K).

t2 = T2 - 273 = 812 - 273 = 539 °C

Solution:

wO2= 0.232; wN2 = 0.768; p = 0.1013 MPa ; xO2 = ?; xN2 = ?; r = ?; M = ?; pO2 = ?; pN2 = ?

Volume composition of air

The partiál pressure of oxygen is po2 = xO2 p = 0.209. 0.1013 = 0.02118 MPa

The partiál pressure of nitrogen is pN2 = xN2 p = 0.791. 0.1013 = 0.08012 MPa

The volume fraction for air

The volume fraction for nitrogen

The gas constant for air

The mean virtual molar mass for air

or

3

5. Reversible thermodynamic processes of ideal gasses

Example 5.1

A closed container contains 0.6 m3 of air with a pressure of p1 = 4.9.105 Pa and temperature t1 = 20 °C. What is the pressure and temperature after losing 104.5 kJ of heat (i.e. cooling)?

Example 5.2

How much heat must be received by 0.5 kg of oxygen at a pressure of p1 = 2.94.10 5 Pa and temperature t1 = 35 °C in order to carry out work A12 = 27 900 J under a constant pressure, and what will the final volume and temperature be?

Solution:

mFe = 0.2 kg; mH2O = 0.5 kg; t1H2O =20 °C ; t2H2O = 75 °C; cFe = 0.469 kJ/(kg.K)

The heat transferred from the steel to the water is

QH2O =mH2O . cH2O (t2H2O - t1H2O) = 0.5 . 4186.8 (75 - 20) = 115.103 J

From the heat balance

QFe = mFe . cFe (tFe - t2H2O) = QH2O

we can calculate

Solution:

V= 0.6 m3 ; p1 = 4.9.105 Pa ; t1 = 20 °C ; Q12 = -104.5 kJ; p2 = ?; t2 = ?

In an isochoric process ; where p2and T2 are the sought values. Here, it is necessary to use the First Law of Thermodynamics for an isochoric process

Q = m.. Δu = m . cv (t2 - t1), kde is the mass of air

Pressure

Solution:

m = 0.5 kg; p1 = 2.94.105 Pa; t1 = 35 °C; A12 = 27 900 J; V2 = ? ; t 2 = ?

The work of the isobaric process (p2 = p1 = p1,2 )

4

Example 5.3

490 kJ of heat is rejected in an isothermic compression of 0.3 m3 of air with a pressure of p1 = 106 Pa and temperature t = 300 °C. Calculate the final volume and pressure!

Example 5.4

1 kg of air with an initial temperature of t1 = 30 °C and pressure p1 = 0.0981.106 Pa is adiabatically compressed to p2 = 0.981.l06Pa. Deterermine the final volume, the final temperature and volume work done on this system (κ = l.4 ).

The heat received

Q12 = m.cp (t2 - t1) = 0.5 . 0.917.103 (238 - 35) = 0.931.103 J = 0.931 kJ

, where

In ab isobaric process

;

Solution :

V1 = 0.3 m3; p1 = 106 Pa; t = 300 °C; Q12 = -490 kJ; V2 = ? ; p2= ?

From i sothermic work

we get

The pressure ratio

The final pressure

The final volume (from the isothermic equation p1V1 = p2V2)

Solution :

t1 = 30 °C ; p1 = 0.0981.106 Pa ; p2 = 0.981.106 Pa ; v2 = ? ; t2= ? ; a12 = ?

t2 = T2 - 273 = 585.5 - 273 = 312.5 C

From the temperature and pressure ratio in an adiabatic process we get the final temeprature

5

Example 5.5

1.5 kg of air is polytropically compressed from p1= 0.088.106 Pa and t1 = 18 °C to p2 = 0.981.106 Pa and t2 = 125 °C. What is the polytropic, final volume the work done on the system the heat lost and the change in internal energy..

Example 5.6

Air at a temperature of 127 °C is compressed isothermically to a quarter of the initial volume and then adiabatically expands to its originalpressure. Calculate the temperature of the air after the adiabatic expansion!

The work done on the system

The final volume (from the equation of state)

Solution :

m= 1.5 kg; p1 = 0.088.106 Pa; t1 = 18 °C ; p2 = 0.981.106 Pa ; t2 = 125 °C ; n = ?; V2 = ?; A12 = ?; Q12 = ?; ΔU = ?

The polytropic exponent n = 1.149

The change in internal energy (from the First Law of Thermodynamics)

ΔU = Q12 - A12 = - 193.103 - (309.103) = 116.103 J

The temperature-pressure ratio in a polytropic process is

and the logarithm is

The final volume

The work in a polytropic process

which is work consumed

The heat lost

Solution:

t 1,2 = 127 °C; v2 = v1/4 ; p3 = p1 ; t3 = ?

In an isothermic process from 1 to 2: p1/ p2 = v2/ v1

In an adiabatic process from 2 -3: tj.

The temperature

6

6. II. The law of thermodynamics, entropy, T-s diagram

Example 6.1

1 kg of oxygen at a pressure p1 = 4.9.105 Pa and temperature t1 = 127 °C expands isobarically to double its initial volume, then is

compressed isothermically to p2 = 39.2.105 Pa. Calculate the change in entropy of the oxygen!

t3 = T3 - 273 = 269 - 273 = - 4 °C

Solution:

p1 = 4.9.105 Pa ; t 1 = 127 °C; v2 = 2 v1 ; p3 = p2 = 39,2.105 Pa; Δs = ?

The change in entropy of an isobaric process

Δs12 = c p1n (T2/T1) = cp1n (v2/v1) = cp 1n (2.v1/v1) = 0,917.1031n 2 = 0,917.103.0,6931 = 0,6353. 103 J/(kg.K)

The change in entropy of an isothermic process

Δs(23) = q2,3/ T2,3 , kde T2,3 = T1.( v2/v1) = 400 (2.v1/v1) = 800 K

The total change in entropy Δs(13) = Δs(12) + Δs(23) = (0.6353 - 0.52 ).103 = 0.1153.103 J/(kg.K)

7

Example 6.2

1.5 kg of air is compressed polytropically from pressure p1 = 0.88.105 Pa and tempreature t1= 18 °C to pressure p2 = 9.81.105 Pa and temperature t2.= 125 °C. Find the entropy change!

Example 6.3

Inside a recuperation exchanger with a gas turbine the air is heated by the exhaust gases from temperature t1 = 140 °C to t2 = 270 °C. The gases are cooled from temperature t3 = 340 °C to temperature t4 = 210 °C. The gas is considered ideal having the properties of air. The temperature of the ambient to = 20 °C. There is no heat loss. Calculate the exergy loss!

solution:

m = 1.5 kg; p1 = 0.88.105 Pa ; t1 = 18 °C ; p2 = 9,81.105 Pa ; t2 = 125 °C Δs = ?

The change in entropy in a polytropic process

ΔS = m Δs = 1.5 (- 0.3765.103) = - 0.5645.103 J/K

(Note: The exponent of this polytropic process was determined in Exercise 5.47)

Solution:

t1 = 140 °C ; t2 = 270 °C ; t3 = 340 °C ; t4 = 210 °C ; t0 = 20 °C; Δe = ?

Energy loss Δe = T0 Δs, where Δs is the change in entropy Δs = Δs1 + Δs2

Δs = ( -0.2395 + 0.275 ).103 =0.0355.103 J/(kg.K)

8

7. Carnot cycle

Example 7.1

1 kg of air is executing the Carnot cycle between the temperatures tH= 327 °C and tC = 27 °C. The highest temperature is 2.106 Pa; the

lowest is 1.2.105 Pa. Calculate the work, thermal efficiancy, the received and rejected heat and the exergic efficiency if the ambienttemperatrure is to= 20 °C !

The exergy loss Δe = 293. 0.0355.103 = 10.4.103 J/kg = 10.4 kJ/kg

Soluition :

tH = tmax = 327 °C; tC = tmin = 27 °C , pmax =2.106 Pa, pmin = 1,2.105 Pa ; ao = ? ; ηt = ?; qH = ? /qC/ = ? ; ηE = ?

The pressure in point 2 of the cycle

The pressure in point 4 of the cycle

The exergetic efficiency is given by the ratio of the actually exerted and maximum work

ηE = a/a max ; a max = qH - To Δs = qH To (qH/T1) ; a0 - qH - │qC│

The heat received

The heat rejected

The thermal efficiency of the cycle

9

8. Cycle of IC-engine

Example 8.1

The working gas in internal combustion engines with a combined input of heat is air. If the pressure p1 = 0.0981.106 Pa, temperature t1 = 30 °C, compression ratio ε = 7, pressure ratio Ψ = 2, cutoff ratio ϕ = 1.2, calculate the state values in the characteristic points of the cycle, the input heat, the work done by the system and the thermal efficiency of the cycle. The specific heat capacity is constant.

Solution:

p1 = 0.0981.106 Pa; t1 = 30 °C; ε = 7; Ψ = 2; ϕ = 1,2 ; q1 = ? ; aC =? : ηt = ?

x quantities given, the rest calculated

Point 1

Point 2

v2= v1 / ε = 0.887 / 7= 0.127 m3/kg

Quantities

Point p [MPa] v [m3/kg] T [K]

1 0.0981x 0.887 303x

2 1.491 0.127 659

3 2.982 0.127 1318

4 2.982 0.1525 1580

5 0.254 0.887 785

10

9. Compressor

Example 9.1

A twostage compressor sucks in air at a temperature of t1 = 20 °C and a pressure of 0.0981 MPa and compresses it to a pressure of p´2 = 6 MPa. Calculate the power of the engine necessary for driving the compressor and the amount of cooling water for both stages of thecompressor and also for the intercooler if the ratio of the output and input pressures is the same for both stages and the mechanical

Pressure p2 = p1. ε κ = 0.0981.106. 71,4 = 1.4 91.106 Pa

Point 3

Pressure p3 = p2.Ψ = 1.491.106. 2 = 2.982.106 Pa

specific volume v3 = v2 = 0.127 m3 /kg

Temperature

Point 4

Pressure p4 = p 3 = 2.982.106 Pa

specific volume v4 = v3 ϕ = 0,127. 1,2 = 0,1525 m3/kg

Temperature T4 = ϕ.T3 = 1.2. 1318 = 1580 K

Point 5

specific volume v5 = v1 = 0.887 m3/kg

Heat received

qH = cv( T3 - T2)+ cp ( T4 - T3) = 0,714.103.( 1318 - 659 )+1005.(1580-1318)= 733.8.103 J/kg

Heat lost

Cycle work

Thermal efficiency of the c ycle

Temperature

Pressure

Temperature

11

efficiency of each stage is 0.7. The temperature of the water increases by 15 K. Compression in both stages is polytropic with an exponentn = 1.3. The power of the compressor is V = 0.14 m3 /s.

Solution :

t1 = 20 °C; p1 = 0.0981 MPa, p2 = 6 MPa, ηk = 0,7, Δt H2O = 15 K; n =1.3; V = 0.14 m3/s ; PNT = ?, PVT = ?, m NT = ?, m VT = ?, mMCH = ?

The pressure ratio in both stages is

12

10. Gas flow

Example 10.1

The temperature t´1 = t1 = 20 °C

The volume V1= V´1 = V = 0.14 m3/s

The power necessary for compression of the air in the low-pressure stage is

and for the high-pressure stage is PVT= PNT= 36.17.103 W

The power of the engine for both stages is

The temperature after compression in both stages is

The heat loss from both stages during compression is

The amount of cooling water for the low-pressure (high-pressure) stage of the compressor is

The heat loss from the intercooler is

The amount of cooling water for the intercooler is

The total amount of cooling water that is necessary is mH2O = m H2O NT + mH2O VT + m H2O MCH = 0.11045 + 0.11045 + 0.4655 = 0.6844 kg/s

13

A pressure vessel contains oxygen at a pressure of p1 = 5 MPa. The gas leaves through a nozzle into the ambient with a pressure of p2 = 4 MPa. The initial temperature of the oxygen is 100 °C. Calculate the theoreatical speed upon exit and the mass flow rate (kg/s), if the areaof the end of the nozzle is S = 20.10-6 m2 . Find the speed and mass flow rate of the oxygen upon exit in to the ambient which has a pressure of p´2 = 0.098 MPa.

11.Cycle of gas turbines

Example 11.1

Calculate the thermal efficiency of a cycle of a gas turbine with a compression ration of 10, which has a heat input at a constant pressure, κ= 1.4.

12. REAL GASES AND VAPOURS

Solution:

p1 = 5 MPa; p2 = 4 MPa , t1 = 100 °C, S = 20.10-6 m,2 p´2 = 0.098 MPa, w2 = ?, m = ? w2´ = ?, m´ = ?

a) The pressure ratio

Mass flow

b) The pressure ratio

Mass flow

Solution:

ε = 10; κ = 1.4 ; ηt = ?

The thermal efficiency of a turbine with heat input with p=const.

ηt = 1-( 1/ εκ-1) = 1 - (1/10 1,4 -1) = 1 - ( 1/ 2,51 ) = 1 - 0.3985 = 0.6015

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Example 12.1

Calculate the state of water vapour with a pressure of p = 0.59 MPa and a temperature t = 190 °C.

Example 12.2

What is the heat received by 1 kg of water vapour with a constant pressure of p = 1.47 MPa if the vapour quality increases from x1 = 0.8 to x 2 = 0.96 ?

Solution :

p = 0. 59 MPa ; t = 190 °C; the state of the water vapour = ?

From the table for saturated water H20 it is possible to determine the temperature of the saturated vapour with a pressure of 0.59 MPa, t23 = 158.14 °C < t = 190 °C.

This is a case of superheated vapour and, using the table for superheated water vapour (i.e. the i-s diagram), it is possible to find the specific volume v, the enthalpy i and entropy s of the vapour for the pressure p and temperature t.

The superheating of the water vapour is Δt = t - t23 = 190 - 158.14 = 31.86 K

Solution:

p = 1.47 MPa ; x1 = 0.8 ; x2 = 0.96 ; q12 = ? ( dp = 0 )

The heat received q12 = i2 - i1 is equal to the enthalpy difference.

a) The calculation using the i-s diagram for water vapour:

15

Example 12.3

Calculate the heat received by 6 kg of water vapour with a volume of 0.6 m3 at a temperature of p1 = 0.6 MPa, so that the pressureincreases to p2 = 1 MPa at a constant volume. Find also the final quality of vapour.

For pressure p and quality x1 from the i-s diagram: i1 = 2390 kJ/kg . For p and x2: i2 = 2702 kJ/kg. The heat received q = i2 -i1 = 2702 -2390 = 312 kJ/kg.

b) The calculation using the tables for saturated water H2O:

Enthalpy i1 = i´1 + x1(i´´1- i´1)

i2 = i´2 + x2 (i´´2 - i´2 )

for p = const. i´1 = i´2 = i', i´´1 = i´´2=i"

The heat received:

q12 = (x2 - x1)( i´´- i´) =(x2-x1).l23

from the table for pressure p l23 = 1951.25 kJ/kg

q12 = ( 0.96 - 0.8 ).1951.25 = 0.16.1951.25 = 312.2 kJ/kg

Solution :

m = 6 kg ; V = 0.6 m3; p1 = 0.6 MPa ; p2 = 1 MPa ; q12 =? ; x2 = ?

From the condition v1 = v2 = v and considering the fact that in the initial and final states (1 and 2) the vapour is wet

v = v´1 + x1 (v ´´ 1 - v´1)

a) Using the tables for water vapour, the specific volume of the vapour is:

16

Example 12.4

1 kg of water vapour at a pressure of p1 = 3 MPa and temperature t1,2 = 300 °C is compressed to one fifth of the original volume while

v = v´2 + x2( v´´2 - v´2) we get the vapour quality

x1 = ( v - v´1) / ( v´´1- v´1) ; x2 = ( v - v´2) / ( v´´2 - v´2)

From the table for saturated water we substitute for the values of specific volumes of saturated liquid and vapour

;

The heat received at a constant volume of 1 kg of vapour is q12 = u2 - u1 = (i2 - p2v2)-(i1 - p1 v1 ) = ( i2 - i1 ) - v.( p2 - p1)

The enthalpy is

i1 = i´1 + x1123,1 = 670.5 + 0.3145.2086 = 1326.5 kJ/kg

i2 = i´2 + x2123,2 = 762.7 + 0.5112.2015 = 1792.7 kJ/kg

q12 = (1792.7 - 1326.5) - 0.1 .106 (1 - 0.6) = 466.2 - 40 = 426.2 103 J/kg

The total heat received is

Q12 = m . q12 = 6.426.2 = 2557.2 kJ

b) The calculation using the i-s diagram for water vapour is:

The enthalpy in point 1 for pressure p1 and the specific volume is v is i1 =1325 kJ/kg.

The enthalpy in point 2 for pressure p2 and the specific volume v is i2 = 1795 kJ/kg

The vapour quality in point 2 is x2 = 0.51

The heat received is q12 = (i2 - i1) - v.( p2 -p1) = (1795 - 1325 ) - 0.1.106 . (1 - 0.4) = 470 - 40 = 430.103 J/kg.

17

maintaining the temperature. Calculate the final state of the vapour, the work done on the szstem (i.e. consumed) and the heat lost.Illustrate the process in a p-v, T-s and i-s diagram for water vapour.

Solution:

p1 = 3 MPa; t1 = t1 = t = 300 °C ; v2 = v1/5; state 2 = ? ; a12 =? ; q12 = ?

18

a) The calculation using tables for p1, t - state 1

v1 = 0.08119 m3/kg

i1 = 2988 kJ/kg

s1 = 6.53 kJ/(kg.K)

The specific volume v2 = v1/5 = 0.08119/5 = 0.016238 m3/kg

Since the volume of saturated vapour t = 300 °C is 0.02164 m3/kg, the final state 2 falls into the area of wet vapour, p2 = 8.592 MPa.

v2 = v´ 2 + x2( v´´2 - v´2) → the quality in point 2

The enthalpy in point 2 i2 = i´2 + x2l23,2 = 1344.9 + 0.7335 . 1404.2 = 2373.9 kJ/kg

The e ntropy in point 2 s2 = s´2 + x2( s´´2 - s´2) = 3.2548 + 0.7335 (5.7049 - 3.2548) = 5.0518 kJ/(kg.K)

The difference in internal energies Δu = u2 - u1 = (i2- i1) -(p2v2 - plv1) = (2373.9 - 2988).103-(8.592.0.016238 - 3.0.08119).106 = -

510.4.103 J/kg

The lost heat q12 = T(s2 - s1) = 573 (5.0518 - 6.53)= -847 kJ/kg

The consumed work a12 = q12 - Δu = -847 - (-510.4 ) = -336.6 kJ/kg

b) The calculation using the i-s diagram for water vapour: for p1 and t we get v1 = 0.08 m3/kg , i1= 2990 kJ/kg, s1 = 6.53 kJ/(kg.K), v2 =

v1/5 = 0,08/5 = 0.016 m3/kg

The difference in internal energies Δu = u2 - u1 = (i2 - i1) - (p2v2 - p1v1) = ( 2375 - 2990).103 - (8.6 . 0.016 - 3 . 0.08).106 = -615.103 +

0.1024.106 = - 512.6.103 J/kg

For temperature t and v2 we obtain x2 = 0.73, p2 = 8.6 MPa from the i-s diagram i2 = 2375 kJ/kg, s2 = 5.05 kJ/(kg.K)

19

Example 12.5

1 kg of water vapour at a pressure of p1= 0.3 MPa and temperature t1 = 300 °C expands adiabatically to pressure p2 = 0.05 MPa. Calculate the final parameters of the vapour, the volume work and the enthalpy change. Plot a T-s and i-s diagram.

The lost heat q12 = T(s2 - s1) = 573 (5.05 - 6.53) = - 573 . 1.48 = - 848 kJ/kg

The consumed work a12 = q12 - Δu = -848 - (-512.6) = -335.4 kJ/kg

Solution:

p1 = 0.3 MPa ; t1 = 300 °C ; p2 = 0.05 MPa ; state 2= ? ; a12 = ? ; Δi = ?

Using the tables for water vapour:

For p1, t1 we find i1 = 2988 kJ/kg ; v1 = 0.08119 m3/kg;, s1 = 6.53 kJ/(kg.K) based on the condition for an adiabatic process s2 = s1 and regarding the fact that s2 < s´´2 at a pressure of p2 (this is actually a case of wet vapour)

20

Example 12.6

Calculate the temperature at which ice melts under a pair of skates with blades 0.25 m long and 1 mm thick (the weight of the skater is m= 70 kg) if 1/10 of the surface area of the blades is touching the ice. The specific heat for melting ice is l12 = 334 kJ/kg, the density of the

ice is ρL= 0.9168.103 kg/m3, the density of water is ρv = 1.0002.103 kg/m3.

13. VAPOUR FLOW

Example 13.1

Calculate the output speed of water vapour and the diameter of the nozzle of a Lavalovy jet, if the output quantity of the water vapour is m= 216 kg/h at a pressure p1 = 0.588 MPa and the temperature t1 = 220 °C. The output preessure is p2 = 0.049 MPa.

The specific volume v2 = v´2+ x2(v´´2 - v´2) = 0.0010299 + 0.836 (3.239 - 0.0010299) = 2.708 m3/kg

Enthalpy i2 = i´2 + x2l23,2 = 340.6 + 0.836. 2304 = 2266.2 kJ/kg

(Note: l23,2 = 2304 kJ/kg)

The difference in enthalpies Δi = i2 - i1 = 2266.2 - 2988 = -721.8 kJ/kg

The difference in internal energies Δu = u2 - u1 = Δi - (p2v2 - plv1 ) = -721.8.103 -( 0.05. 2.708 - 3. 0.08119 ).106 = -721.8.103 +

108.17.103 = - 613.63.103 J/kg

The volume work a12 = -Δu = 613,63 kJ/kg

Solution:

L = 0.25 m; δ = 1 mm ; m = 70 kg ; K = 0.1 ; l 12= 334 kJ/kg; ρL = 0. 9168.103 kg/m3; ρV = 1.0002.103 kg/m3 ; t12 = ?

Let T12 = 273 K (i.e. the temperature of melting ice at barometric pressure p) and dT related to it. The elementary change dT and dpare replaced by

ΔT = T12(p+Δp) - T12,p = > T12(p+Δp) = T12,p + ΔT = 273 - 2.02 = 271.98 K, t2 = -2.02 °C

The temperature is calculated using the Clausius-Clapeyron equation

Solution:

m = 216 kg/h; p1 = 0.588 MPa ; t1 = 220 °C ; p2 = 0.049 MPa ,w2 = ? ; d2 = ?

(using the i-s diagram for water vapour)

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14. Cycle of thermal power station

Example 14.1

The cycle of a thermal power station works with additional heating of water vapour. Pressure p1 = 12 MPa, temperature t1 = 530 ˚C, pressure p´2 = 3500 Pa. The heating is carried out at a pressure of p´1 = 2.3 MPa to a temperature of t´1 = 480 ˚C. Calculate the increase in the quality of the vapour upon exit from the turbine and the increase in thermal efficiency of the cycle with additional heating compared tothat without.

For pressure p1 and temperature t1 we calculate the enthalpz i1= 2890 kJ/kg

Entalpie v bodě 2 i2 = 2420 kJ/kg

The output speed of water vapour

The diameter of the output noyyle of the jet

Solution:

p1 = 12 MPa; t1 = 530 ˚C; p´2 = 3500 Pa, p´ 1 = 2.3 MPa; t´1 = 480 ˚C; Δx = ?; Δηt = ?

22

From the i-s diagram for water vapour for states 1, 2, l´, 2´ and 2" we calculate i1 = 3426 kJ/kg ; i2 = 2956 kJ/kg i1' = 3420 k J/kg;

i2' = 2180 kJ/kg; i2" = 1972 kJ/kg and the quality of the vapourx2 = 0,85 ; x2 = 0,763.

Using the tables for water vapour we obtain i3 = 111.86 kJ/kg

The increase in the thermal efficiency Δηt = η;t2 - ηt1 = 0.453 - 0.439 = 0.014

The increase in the quality of the vapourΔx = x2´ - x2´´ = 0,85 - 0,763 = 0,087

The thermal efficiency of the cycle without additional heating

The thermal efficiency of the cycle with additional heating

23

15. Refrigerating

Example 15.1

An ammonia refrigerating station with a power output of Q = 110 kW works at an evaporation temperature of t1 = -15 °C. Saturated vapour (see figure below) leaves the evaporator. The condensation temperature t3 = 30 °C, then follows undercooling of the condensate to temperature t´3 = 25 °C. Calculate the theoretical level of refrigeration of the cycle, the amount of refrigerant NH3 in circulation and the theoretical power output of the motor necessary for driving the compressor of the station.

16. MOIST (ATMOSPHERIC) AIR

Example 16.1

1000 kg moist air with a temperature of t1 = 40 °C and relative humidity ϕ1= 0.2 with a pressure of p = 0.098 MPa is moistened with 5 kg of water with a temperature tw= 20 °C. Calculate the state of the air after moistening.

Solution:

Q = 110 kW; t1 = -15 °C; t3 = 30 °C, t3' = 25 °C ; qvth = ? ; mNH3 = ?, P = ?

From the i-log p NH3 diagram - for the given temperatures - we obtain the enthalpies

i1= 1662 kJ/kg,

i2 = 1894 kJ/kg ;

i3´ = 536 kJ/kg

The theoretical level of cooling

qvth = i1 - i4 = i1 - i3' = 1662 - 536 = 1126 kJ/kg

The amount of refrigerant in circulation

m NH3 = Q / qvth = 110/1126 = 0.0977 kg/s

The theoretical power output for driving the compressore is P = m( i2 - i1) = 0.0977 ( 1894 - 1662 ) = 22.66 kW

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17. Heat transfer

Example 17.1

Calculate the thickness of a concrete wall with a heat conductivity of λ = 1.28 W/(m.K), through which heat flows with a magnitude q =349 W/m2. The temperatures on the two sides of the wall are t1 = 20 °C, t2 = -20 °C.

Solution :

m = 1000 kg ; t1 = 40 °C ; ϕ1 = 0.2 ; p =0.098 MPa, mw = 5 kg, tw = 20 °C ; the state after moistening = ?

The specific humidity can be determined using the i-x diagram for state 1: x1 = 0.00948 kg/kg s.v.

The amount of dry air

The increase in specific humidity

Δx = x2 - x1 = mw/ mv = 5 / 991 = 0.005048 kg/kg s.v.

x2= x1 + Δx = 0.00948 + 0.005048 = 0.014528 kg/kg s.v.

During moistening, the slope of the straight line is equal to the enthalpy of the moistening medium (i.e. water)

iw = 83.7 kJ/kg. For iw and x2 we get for state 2: i2 = 65 kJ/kg s.v. ; t2 =27.9 °C

Solution :

λ = 1.28 W/(m.K); q = 349 W/m2; t1 = 20 °C ; t2 = -20 °C ; δ = ?

From the heat flow:

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Example 17.2

The lining inside a boiler comprises two layers (δ1 = 0.35 m, λ1 = 1.4 W/(m.K), δ2 = 0.25 m, λ2 = 0.55 W/(m.K)). The temperature of the inside surface is t1 = 900 °C, the temperature of the outside surface is t3 = 90 °C. Calculate the heat lost and the temperature between thelayers.

Example 17.3

Calculate the heat load q [W/m] and the heat diffusion coefficient k [W/(m.K)], temperatures tS1 , tS2 of the wall of the pipe of the heat exchanger. The dimensions of the pipe are: d1 = 0.1m, d2 = 0.108 m. The mean temperatures of the media are ti = 350 °C, te = 1500 °C.

The heat transfer coefficients on the inside surface of the pipe are α1 = 300, α2 = 110 W/(m2.K). The heat conductivity coefficient of the material of the pipe is λ = 45 W/(m.K).

Example 17.4

Horizontal piping of a diameter of d = 0.057 m is heated by a natural flow of air. The mean temperature of the surface of the piping is ts = 4 °C, the temperature of the surrounding air is to = 36 °C. Calculate the heat transfer coefficient from the air to the wall of the piping.

Solution:

δ1 = 0.35 m, λ1 = 1.4W/(m.K), δ2 = 0.25 m, λ2 = 0.55 W/(m.K), t1 = 900 °C, t3 = 90 °C ; q = ? , t2 = ?

The heat flow through the double wall is

From the heat flow through the first layer

we get

Solution:

d1 = 0.1 m ; d2 = 0.108 m ; ti = 350 °C ; t e = 1500 °C , α1= 300 W/(m2 .K), α2 = 110 W/(m2.K) ; λ = 45 W/(m.K) ; q = ? ; k = ? ; tS1= ? tS2= ?

The heat flow through the double cylindrical wall is

From the heat flow from the ambient 1 to the wall of the exchanger q = α1 2 π r1 (t1- ti) we get

The temperature is

Solution :

d = 0.057 m ; ts = 4 °C, to = 36 °C , α = ?

For the heat transfer by natural convection the Nu number is calculated as Nu = C ( Gr Pr )n

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Příklad 17.5

Insulated piping covered with a layer of paint passes through a room whose walls have a temperature of 27 °C. The diamater of the pipingis d = 0.1 m, the length of the piping is l = 8 m. Calculate the heat transferred by radiation if the surface temperature is 7 °C. How does thisheat change if the piping passes through a 0.2 x 0.2 m casing whose internal surface is covered by aluminium paint.

Example 17.6

Inside a heat exchanger, black oil is cooled from t1 = 300 °C to t2 = 200 °C and crude oil is simultaneously heated from t´1 = 25 °C to t´2 = 175 °C. Calculate the mean temperature gradient inside this exchanger in the case of a) parrallel flow, b)counterflow. What is thedifference between the surface areas of the exchanger in the cases of a) and b) if, in both cases, the heat transfers and the heat transfercoefficients are the same.

The "best-guess" (i.e. mean) temperature is t = ( ts + to)/ 2 = ( 36 + 4 ) / 2 = 20 °C

Δt = ts - to = 36 - 4 = 32 °C

The physical parameters of air at this temperature are λ = 2.596.10-2 W/(m.K)

υ = 15.1.10-6 m2/s Pr = 0.703 β = 1/293 1/K

Gr Pr = 8.75.105..0..703 = 6.15.105

This value determines the transition mode, for which C = 0.54 , n = 1/4

Grashof's number is

Then

The heat transfe coefficient is

Solution:

t 1= 27 °C ; d = 0,1 m ; l = 8 m, t 2 = 7 °C ; Q = ? ; Q´= ? - for the 0.2 x 0.2 m casing

The heat radiated is (for ε1= 0.39 ) Q = ε 1 co S1 [(T1 /100)4 - (T2 /100)4] = 0.39. 5.7.π.0.1 . [ ( 300 / 100 )4-( 280/100 )4 ] = 112 W

Q ´= εn co S l [ (T1,/100)4 - (T'2/100)4] = 0.59. 5.7. π . 0.1 . 8.[(300 / 100)4- (280/100)4] = 430 W

for ε1 =0.92 ; ε2= 0.39

where

Solution:

t1 = 300 °C; t2 = 200 °C; t1 = 25 °C; t'2 = 175 °C ; Δts = ?; Δtp = ?; Ss/Sp = ?

a ) The parallel flow exchanger

The mean logarithmic temperature gradient is

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b) The counterflow exchanger

The ratio of the surface areas in the case of parallel flow and counterflow with the same heat transfer and heat transfer coefficients is

Ss / Sp = Δtp /Δts = 150.75 / 104.2 = 1.447

The mean logarithmic temperature gradient is

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