File:Exercise 10_1.

EES 09-04-2021 10:16:09 Page 1

EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

Given

P 1 = 100

T 1 = ConvertTemp K ; C ; 300

Q H = 900

fluid$ = 'Air'

r = 8,5

start from state 1

u 1 = u fluid$ ; T = T 1

s 1 = s fluid$ ; T = T 1 ; P = P 1

v 1 = v fluid$ ; T = T 1 ; P = P 1

Isentropic compression to state 2

s2 = s1

v1
v2 =
r

u 2 = u fluid$ ; s = s 2 ; v = v 2

T 2 = T fluid$ ; u = u 2

P 2 = P fluid$ ; T = T 2 ; v = v 2

Heat addition to state 3

u3 = u2 + QH

v3 = v2

T 3 = T fluid$ ; u = u 3

P 3 = P fluid$ ; T = T 3 ; v = v 3

s 3 = s fluid$ ; P = P 3 ; u = u 3

Isentropic expansion

s4 = s3

v4 = r ꞏ v3

P 4 = P fluid$ ; T = T 4 ; v = v 4

u 4 = u fluid$ ; s = s 4 ; v = v 4

T 4 = T fluid$ ; u = u 4

Cycle outputs

W comp = u2 – u1
File:Exercise 10_1.EES 09-04-2021 10:16:09 Page 2
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W ex = u3 – u4

W net = W ex – W comp

W net
 Otto =
QH

SOLUTION
Unit Settings: SI C kPa kJ mass deg
Otto = 0,5287 fluid$ = 'Air' QH = 900
r = 8,5 Wcomp = 289,4 Wex = 765,2
Wnet = 475,9 [kJ/kg]

No unit problems were detected.

Arrays Table: Main

Ti Pi si vi ui

[mixed]

1 26,85 100 5,706 0,8611 214,3

2 415,4 1951 5,706 0,1013 503,7
3 1439 4851 6,495 0,1013 1404
4 582,9 285,3 6,495 0,8611 638,5
File:Exercise 10_4.EES 09-04-2021 10:16:48 Page 1
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"Given"
P[1] = 100
T[1] = 295
T[3] = 2000
fluid$ = 'Air'
r = 9

"start from state 1"

u[1] = intenergy(fluid$; T=T[1])
s[1] = entropy(fluid$; T=T[1]; P=P[1])
v[1] = volume(fluid$;T=T[1];P=P[1])

"Isentropic compression to state 2"

s[2] = s[1]
v[2] = v[1] / r
u[2] = intenergy(fluid$;s=s[2]; v=v[2])
T[2] = temperature(fluid$;u=u[2])
P[2] = pressure(fluid$;T=T[2]; v=v[2])

"Heat addition to state 3"

u[3] = intenergy(fluid$; T=T[3])

v[3] = v[2]
P[3] = pressure(fluid$;T=T[3]; v=v[3])
s[3] = entropy(fluid$; P=P[3]; u=u[3])

"Isentropic expansion"
s[4] = s[3]
v[4] = r*v[3]
P[4] = pressure(fluid$;T=T[4]; v=v[4])
u[4] = intenergy(fluid$;s=s[4]; v=v[4])
T[4] = temperature(fluid$;u=u[4])

"Cycle outputs"
Q_H = u[3] - u[2]
W_comp = u[2] - u[1]
W_ex = u[3] - u[4]
W_net = W_ex - W_comp
eta_Otto = W_net/Q_H

SOLUTION
Unit Settings: SI K kPa kJ mass deg
Otto = 0,5307 fluid$ = 'Air' QH = 1172
r =9 Wcomp = 295,9 Wex = 917,6
Wnet = 621,7 [kJ/kg]

14 potential unit problems were detected.

Arrays Table: Main

Ti Pi si vi ui

[mixed]

1 295 100 5,689 0,8468 210,7

2 692,4 2112 5,689 0,09409 506,7
3 2000 6102 6,622 0,09409 1678
4 1001 339,5 6,622 0,8468 760,6
File:Exercise 10_11.EES 09-04-2021 10:18:37 Page 1
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Given

P 1 = 95

T 1 = 300

P 3 = 6500

T 3 = 2200

fluid$ = 'Air'

start from state 1

u 1 = u fluid$ ; T = T 1

s 1 = s fluid$ ; T = T 1 ; P = P 1

v 1 = v fluid$ ; T = T 1 ; P = P 1

Isentropic compression to state 2

s2 = s1

P2 = P3

v 2 = v fluid$ ; s = s 2 ; P = P 2

u 2 = u fluid$ ; s = s 2 ; P = P 2

h 2 = h fluid$ ; s = s 2 ; P = P 2

T 2 = T fluid$ ; u = u 2

Heat addition at constant pressure state 3

v3 = rc ꞏ v2

v 3 = v fluid$ ; T = T 3 ; P = P 3

u 3 = u fluid$ ; T = T 3

h 3 = h fluid$ ; T = T 3

s 3 = s fluid$ ; P = P 3 ; u = u 3

W comb = P2 ꞏ v3 – v2

Isentropic expansion

s4 = s3

v4 = v1

u 4 = u fluid$ ; s = s 4 ; v = v 4

T 4 = T fluid$ ; u = u 4

P 4 = P fluid$ ; T = T 4 ; v = v 4

Cycle outputs
File:Exercise 10_11.EES 09-04-2021 10:18:37 Page 2
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

v1
r =
v2

QH = h3 – h2

QC = u4 – u1

W comp = u2 – u1

W ex = W comb + u 3 – u 4

W net = W ex – W comp

W net
 Diesel =
QH

SOLUTION
Unit Settings: SI K kPa kJ mass deg
Diesel = 0,5808 fluid$ = 'Air' QC = 632,4
QH = 1508 r = 21,5 rc = 2,304
Wcomb = 357,4 Wcomp = 506,7 Wex = 1383
Wnet = 876,1 [kJ/kg]

16 potential unit problems were detected.

Arrays Table: Main

Ti Pi si vi ui hi

[mixed]

1 300 95 5,72 0,9065 214,3

2 954,9 6500 5,72 0,04217 721 995,1
3 2200 6500 6,723 0,09715 1872 2504
4 1101 348,7 6,723 0,9065 846,7
File:Exercise 10_16.EES 09-04-2021 10:55:39 Page 1
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"Given"
P[1] = 100
T[1] = converttemp(K; C; 300)
r = 18
r_c = 2
fluid$ = 'Air'

"start from state 1"

u[1] = intenergy(fluid$; T=T[1])
s[1] = entropy(fluid$; T=T[1]; P=p[1])
v[1] = volume(fluid$;T=T[1];P=P[1])

"Isentropic compression to state 2"

s[2] = s[1]
v[2] = v[1] / r
u[2] = intenergy(fluid$;s=s[2]; v=v[2])
h[2] = enthalpy(fluid$;s=s[2]; v=v[2])
T[2] = temperature(fluid$;u=u[2])
P[2] = pressure(fluid$;T=T[2]; v=v[2])

"Heat addition at constant pressure state 3"

P[3] = P[2]
v[3] = r_c* v[2]
T[3] = temperature(fluid$;P= P[3]; v=v[3])
u[3] = intenergy(fluid$; T=T[3])
h[3] = enthalpy(fluid$; T=T[3])
s[3] = entropy(fluid$; P=P[3]; u=u[3])
W_comb = P[2] * (v[3] - v[2])

"Isentropic expansion"
s[4] = s[3]
v[4] = v[1]
u[4] = intenergy(fluid$;s=s[4]; v=v[4])
T[4] = temperature(fluid$;u=u[4])
P[4] = pressure(fluid$;T=T[4]; v=v[4])

"Cycle outputs"
Q_H = h[3]-h[2]
Q_C = u[4] - u[1]
W_comp = u[2] - u[1]
W_ex = W_comb + u[3] - u[4]
W_net = W_ex - W_comp
eta_Diesel = W_net/Q_H
W_check = Q_H - Q_C
eta_2 = W_check / Q_H

SOLUTION
Unit Settings: SI C kPa kJ mass deg
2 = 0,579 Diesel = 0,579 fluid$ = 'Air'
QC = 449,8 QH = 1068 r = 18
rc = 2 Wcheck = 618,6 Wcomb = 257,9
Wcomp = 459,5 Wex = 1078 Wnet = 618,6 [kJ/kg]

16 potential unit problems were detected.

Arrays Table: Main

Ti Pi si vi ui hi

[mixed]

1 26,85 100 5,706 0,8611 214,3

2 625,5 5392 5,706 0,04784 673,8 931,7
3 1524 5392 6,525 0,09568 1484 2000
4 613,9 295,7 6,525 0,8611 664,2
File:Exercise 10_23.EES 09-04-2021 11:51:48 Page 1
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Given

P 1 = 100

T 1 = 300

P 2 = 800

T 3 = 1200

fluid$ = 'Air'

m = 4

start from state 1

s 1 = s fluid$ ; T = T 1 ; P = P 1

h 1 = h fluid$ ; T = T 1

Isentropic compression to state 2

s2 = s1

h 2 = h fluid$ ; s = s 2 ; P = P 2

Heat addition at constant pressure state 3

P3 = P2

h 3 = h fluid$ ; T = T 3

s 3 = s fluid$ ; P = P 3 ; T = T 3

Isentropic expansion to the state 4

s4 = s3

P4 = P1

h 4 = h fluid$ ; s = s 4 ; P = P 4

Find

W comp = h2 – h1

W out = h3 – h4

W net = W out – W comp

Q in = h 3 – h 2

W net
 =
Q in

W net = W net ꞏ m
File:Exercise 10_23.EES 09-04-2021 11:51:48 Page 2
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SOLUTION
Unit Settings: SI K kPa kJ mass deg
 = 0,428 fluid$ = 'Air' m=4
Qin = 733,3 Wcomp = 244,3 Wnet = 1255 [kW]
Wnet = 313,8 Wout = 558,1

7 potential unit problems were detected.

Arrays Table: Main

Ti si Pi hi

1 300 5,706 100 300,4

2 5,706 800 544,8
3 1200 6,585 800 1278
4 6,585 100 719,9
 Solutions for selected exercises
(Lecture 8)

Exercise 8.22

The percentage change in work required to remove heat to when changing the refrigerator
temperature from -5 ℃ to -8 ℃ when the surrounding air is at 25 ℃.

By assuming an reverse Carnot cycle and constant temperature of the surrounding air, the
percentage change in work required to remove an equal amount of heat, can be expressed
with just the COP of each refrigerator

Since initial refrigerator temperatures and the temperature of surroundings are given, COP’s
can be calculated as

1 1
𝐶𝐶𝐶𝐶𝑃𝑃𝑅𝑅,1 = = = 8.9383
𝑇𝑇𝐻𝐻 298.15 𝐾𝐾� − 1
�𝑇𝑇 − 1 268.15 𝐾𝐾
𝐶𝐶,1
1 1
𝐶𝐶𝐶𝐶𝑃𝑃𝑅𝑅,2 = = = 8.0348
𝑇𝑇𝐻𝐻 298.15 𝐾𝐾� − 1
�𝑇𝑇 − 1 263.15 𝐾𝐾
𝐶𝐶,2

Using both COPs in the first equation for the final percentage change in work leads to

𝐶𝐶𝐶𝐶𝑃𝑃𝑅𝑅,1 8.9383
∆𝑊𝑊𝑝𝑝 = � − 1� 𝑥𝑥 100% = � − 1� 𝑥𝑥 100% = 11.24%
𝐶𝐶𝐶𝐶𝑃𝑃𝑅𝑅,2 8.0348

Exercise 8.23

Please show that a heat pump and a refrigerator between same reservoirs, both operate on a
reverse Carnot cycle, transfer the same amount of heat.

The statement is proven by comparing the definition of COP for a heat pump and a refrigerator
with an energy balance
Please note that this is valid for Carnot heat pump and Carnot refrigerator.

Exercise 8.38

The mass flow rate of a refrigerant R134a in a Carnot refrigerator which takes heat a -4 ℃ and
rejects heat at 24 ℃ needs to be determined for an assumed power input of 380 W.

An energy balance around the isothermal condenser gives

In order to find the rate of heat rejected from the refrigerator, the COP is calculated as
follows

Since power input and rate of heat rejected are known now, the rate of heat transfer from the
heat sink is

Using the specific enthalpies h2 and h3 of satured R134a from Appendix A9a, The resulting
mass flow rate of refrigerant is
Exercise 9.36

Specific enthalpy of saturated R-134a at 0.20 MPa for gas h2 = 241.30 kJ/kg (A8b), specific
entropy of saturated R-134a at 0.20 MPa for gas s2 = 0.9253 kJ/kg-K (A8b), specific enthalpy of
saturated R-134a at 1.2 MPa for fluid h4 = 115.76 kJ/kg (A8b).
The properties of the refrigerant entering the compressor are those of saturated R-134a vapour at
P2 = 0.20 MPa, h2 = 241.30 kJ/kg and s2 = 0.9253 kJ/kg-K.

Since we know that the refrigeration cycle is ideal, compression through the compressor is
isentropic so that s3 = s2 = 0.9253 kJ/kg-K, so refrigerant leaving the compressor is superheated
with specific enthalpy interpolated from Appendix 9c at P3 = 1.2 MPa:
𝑘𝐽 𝑘𝐽 𝑘𝐽
ℎ3 − 275.52 0.9253 − 0.9164
𝑘𝑔 𝑘𝑔 𝐾 𝑘𝑔 𝐾
=
𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽
287.44 − 275.52 0.9527 − 0.9164
𝑘𝑔 𝑘𝑔 𝑘𝑔 𝐾 𝑘𝑔 𝐾

And h3 = 278.44 kJ/kg

The specific enthalpy of refrigerant leaving the condenser is that of saturated liquid R- 134a at P4
= 1.2 MPa, h4 = 115.76 kJ/kg.

Expansion through the throttling valve in an ideal reverse Rankine cycle is a constant enthalpy
process, so specific enthalpy at the evaporator inlet is h1 = h4 = 115.76 kJ/kg. The mass flow rate
of refrigerant through the cycle can be found using an energy balance around the compressor,
with power supplied by the car:

𝑊̇𝑐𝑜𝑚𝑝 = 𝑚̇𝑟𝑒𝑓 (ℎ3 − ℎ2 )

𝑊̇𝑐𝑜𝑚𝑝 1.2 𝑘𝑊 𝑘𝑔
𝑚̇𝑟𝑒𝑓 = = = 0.0323
ℎ3 − ℎ2 278.44 𝑘𝐽 𝑘𝐽 𝑠
− 241.30
𝑘𝑔 𝑘𝑔

then the rate of heat input to the refrigerator from the surrounding air is

𝑘𝑔 𝑘𝐽 𝑘𝐽
𝑄̇𝐶 = 𝑚̇𝑟𝑒𝑓 (ℎ2 − ℎ1 ) = 0.0323 (241.30 − 115.76 ) = 4.056 𝑘𝑊
𝑠 𝑘𝑔 𝑘𝑔

For air at Tavg = 25°C, the average specific heat at constant pressure is cp = 1.004 kJ/kgK, so the
mass flow rate of air that can be cooled by the refrigerator is

𝑄̇𝐶 4.056 𝑘𝑊 𝑘𝑔
𝑚̇𝑎𝑖𝑟 = = 0.2002
𝑐𝑝 ∆𝑇 1.004 𝑘𝐽 (35 − 15)℃ 𝑠
𝑘𝑔 𝐾

The refrigerator can cool air at a maximum mass flow rate of 0.202 kg/s.
File:Exercise 9_36.EES 26-03-2021 08:54:26 Page 1
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

Given

Refrigerant side

Fluid$ = 'R134a'

P 2 = 200

P 3 = 1200

x2 = 1

x4 = 0

W comp = 1200

Air side

T i = 35

T o = 15

State 2

h 2 = h Fluid$ ; P = P 2 ; x = x 2

s 2 = s Fluid$ ; P = P 2 ; x = x 2

T 2 = T Fluid$ ; P = P 2 ; s = s 2

State 3

s3 = s2

h 3 = h Fluid$ ; s = s 3 ; P = P 3

T 3 = T Fluid$ ; P = P 3 ; s = s 3

State 4

P4 = P3

h 4 = h Fluid$ ; P = P 4 ; x = x 4

s 4 = s Fluid$ ; P = P 4 ; x = x 4

T 4 = T sat Fluid$ ; P = P 4

State 1

h1 = h4

P1 = P2

s 1 = s Fluid$ ; P = P 1 ; h = h 1

T1 = T2

Cycle performance - refrigerant side

File:Exercise 9_36.EES 26-03-2021 08:54:26 Page 2
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W comp = m ꞏ h3 – h2

QC = m ꞏ h2 – h1

QC
COP =
W comp

Air side analysis

Q C = Q air

h i;air = h Air ; T = T i

h o;air = h Air ; T = T o

Q air = m air ꞏ h i;air – h o;air

SOLUTION
Unit Settings: SI C kPa J mass deg
COP = 3,389 Fluid$ = 'R134a'
hi,air = 308635 ho,air = 288542
m = 0,0321 mair = 0,2024
Qair = 4067 QC = 4067
Ti = 35 To = 15
Wcomp = 1200

11 potential unit problems were detected.

File:Exercise 9_37.EES 26-03-2021 08:55:52 Page 1
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

Given

Fluid$ = 'R134a'

T 2 = – 12

T 4 = 20

Q C = 8000

x2 = 1

x4 = 0

State 2

P 2 = P sat Fluid$ ; T = T 2

h 2 = h Fluid$ ; T = T 2 ; x = x 2

s 2 = s Fluid$ ; T = T 2 ; x = x 2

State 3

s3 = s2

P3 = P4

h 3 = h Fluid$ ; s = s 3 ; P = P 3

T 3 = T Fluid$ ; P = P 3 ; s = s 3

State 4

P 4 = P sat Fluid$ ; T = T 4

h 4 = h Fluid$ ; T = T 4 ; x = x 4

s 4 = s Fluid$ ; T = T 4 ; x = x 4

State 1

h1 = h4

P1 = P2

s 1 = s Fluid$ ; P = P 1 ; h = h 1

T1 = T2

W in = m ꞏ h3 – h2

QC = m ꞏ h2 – h1

QC
COP =
W in

T 2 + 273,15
COP Carnot =
T4 – T2
File:Exercise 9_37.EES 26-03-2021 08:55:52 Page 2
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SOLUTION
Unit Settings: SI C kPa J mass deg
COP = 7,055 COPCarnot = 8,161
Fluid$ = 'R134a' m = 0,04879
QC = 8000 Win = 1134 [W]

11 potential unit problems were detected.

Arrays Table: Main

xi Ti si Pi hi

1 -12 311,2 185,4 79323

2 1 -12 939,1 185,4 243305
3 25 939,1 572,1 266548
4 0 20 300,6 572,1 79323

125 R134a

100

75
T [°C]

50

800 kPa

25
500 kPa

0
217,1 kPa

0,2 0,4 0,6 0,8

-25
0 150 300 450 600 750 900 1050 1200
s [J/kg-K]
File:Lecture 8 extra exercise.EES 27-03-2021 08:40:21 Page 1
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

Given

Fluid$ = 'R134a'

T2 = 0

P 2 = 240

P 3 = 450

P 5 = 900

x2 = 1

x4 = 1

x6 = 0

State 2

h 2 = h Fluid$ ; x = 1 ; P = P 2

s 2 = s Fluid$ ; x = 1 ; P = P 2

State 3

s3 = s2

h 3 = h Fluid$ ; s = s 3 ; P = P 3

T 3 = T Fluid$ ; P = P 3 ; s = s 3

State 4

P4 = P3

h 4 = h Fluid$ ; P = P 4 ; x = x 4

s 4 = s Fluid$ ; P = P 4 ; x = x 4

T 4 = T Fluid$ ; P = P 4 ; s = s 4

State 5

s5 = s4

h 5 = h Fluid$ ; P = P 5 ; s = s 5

T 5 = T Fluid$ ; P = P 5 ; s = s 5

State 6

P6 = P5

h 6 = h Fluid$ ; x = x 6 ; P = P 6

s 6 = s Fluid$ ; x = x 6 ; P = P 6

T 6 = T Fluid$ ; P = P 6 ; s = s 6

State 1
File:Lecture 8 extra exercise.EES 27-03-2021 08:40:21 Page 2
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h1 = h6

P1 = P2

s 1 = s Fluid$ ; P = P 1 ; h = h 1

T 1 = T Fluid$ ; P = P 1 ; s = s 1

W comp;Lo = h3 – h2

W comp;Hi = h5 – h4

W in = W comp;Lo + W comp;Hi

QC = h2 – h1

QC
COP =
W in

T 2 + 273,15
COP Carnot =
T6 – T2

SOLUTION
Unit Settings: SI C kPa J mass deg
COP = 5,347 COPCarnot = 7,693
Fluid$ = 'R134a' QC = 145674
Wcomp,Hi = 14348 Wcomp,Lo = 12898
Win = 27245 [W]

16 potential unit problems were detected.

Arrays Table: Main

xi Ti si Pi hi

1 -5,384 390,6 240 101605

2 1 0 934,6 240 247279
3 15,23 934,6 450 260177
4 1 12,46 925,3 450 257526
5 37,88 925,3 900 271874
6 0 35,51 373,8 900 101605
Lecture 7 Exercises, no answers

7a

8.1

8.2

8.8
8.12

The thermal efficiency of a Carnot cycle is:

Thus, we can get Tc from following formula:

Exercise 9.1
Steam enters the turbine in an ideal Rankine cycle at 8 MPa and leaves at 10 kPa. What is the work
done by the turbine per kilogram of steam?

Solution:
At a Rankine cycle (slide 5, Lecture 7b) the expansion process in the turbine is isentropic,
therefore, s3=s4 (1).

At state 3, it is known from the description of the exercise that the fluid in state 3 is saturated water
at 8 MPa. Therefore, it is possible to calculate the Specific Entropy and Specific Enthalpy of that
fluid using the values from A8b.
 h3 = 2758.0 kJ/kg (2)
 s3 = 5.7432 kJ/kgK (3)
From (1), (3)  s4= 5.7432 kJ/kgK (4)

Also, using the A8b, it is obtained:

Specific enthalpy for saturated water at 10 kPa for fluid hf = 191.83 kJ/kg and gas hg = 2584.7
kJ/kg (5).
Specific entropy for saturated water at 10 kPa for fluid sf = 0.6493 kJ/kgK and gas sg = 8.1502
kJ/kgK (6).

Using (4), (5), (6), it is possible to calculate the quality of the steam in the exit of the turbine, and
more specifically:
kJ kJ
s4 −sf 5.7432kgK−0.6493kgK
x4 = = kJ kJ = 0.679105 (7)
sg −sf 8.1502kgK−0.6493kgK

The enthalpy at the turbine exit is then calculated,:

h4 = hf + x4 * (hg - hf) (8)
h4 = 191.83 kJ/kg + 0.679105 * (2584.7 kJ/kg -191.83 kJ/kg) = 1816.84 kJ/kg (9)

[1]
The work output of the turbine can be finally calculated:
wt = (h3-h4) = 2758 kJ/kg - 1816.84 kJ/kg = 941.160 kJ/kg.

Exercise 9.2
An ideal Rankine cycle has a condenser operating at a pressure of 50 kPa. If the steam quality at the
outlet of the turbine is required to be 80%, what should the boiler pressure be?

Solution:
In an ideal Rankine cycle (slide 5, Lecure 7b), the pressure of the boiler in the inlet and the outlet
is equal, therefore, P2=P3.
It is also known that in that cycle the expansion through the turbine is isentropic, therefore, the
entropy in the inlet (state 3) and the outlet (state 4) of the turbine is equal (s3=s4).
From A8b, it is also known that the specific entropy of saturated water at 50 kPa for fluid is sf = 1.0910
kJ/kgK and gas sg = 7.5939 kJ/kgK.

Combining all the above-mentioned information, it is possible to calculate the value of s4:
s4= sf + x4 *(sg – sf)  s4 = 1.0910 kJ/kgK + 0.80 * (7.5939 kJ/kgK - 1.0910 kJ/kgK)
= 6.2933 kJ/kgK.

Since the fluid in state 3 is saturated vapour, it is possible to calculate the pressure of the boiler,
using the values of entropy of saturated gas and by applying interpolation on them.

Therefore:
kJ kJ
P3 −2.25 MPa 6.2933 kgK−6.2575 kgK
= kJ kJ  P3 = 2.2746 MPa
2.50 MPa −2.25 MPa 6.2972 kgK−6.2575kgK

[2]
9.5

High temperature of heat source is 350 C, and low temperature of heat sink is 20 C. Steam in state 1 is in
liquid phase, quality x[1]=0. Steam in state 3 is saturated gas, x[3] = 1.Thus we have:
"Given"
T[3] = 350 [C]
x[3] = 1
T[4] = 20 [C]
x[1] = 0

The max thermal efficiency of Carnot cycle is:

Code in EES is:

"Calculate the efficiency of a Carnot cycle"
eta_Carnot = 1- (T[4]+273.15)/(T[3] + 273.15)

For Rankine cycle, TS plot is shown as below:

State 1 is saturated liquid, its temperature is the same with state 4. So, the property of state 1 is:
"State 1"
T[1] = T[4]
P[1]=p_sat(Water,T=T[1])
h[1]=enthalpy(Water,T=T[1],x=x[1])
s[1] = entropy(Water,T=T[1], x=x[1])
v[1] = volume(Water,T=T[1], x=x[1])

State 1 water is pumped to a high pressure state 2, but state 2 has same pressure with state 3. So we
can find the enthalpy and entropy of state 2 water:
"State 2"
P[2] = P[3]
h[2] = h[1] + v[1]*(P[2]-P[1])
With the information given, the thermal property of state 3 is:
"State 3"
h[3] = enthalpy(Water,T=T[3],x=x[3])
s[3] = entropy(Water,P=P[3], x=x[3])
p[3] = p_sat(Water,T=T[3])

The process from state 3 to state 4 is an isentropic process, so s[4] = s[3]. Further, the property of state
4 is:
"State 4"
s[4] = s[3]
h[4] = enthalpy(Water,s = s[4],T=T[4])
P[4] = p_sat(Water,T=T[4])

Finally, the thermal efficiency of Rankine cycle can be calculated from net work and heat involved:
"Calculate power and efficiency of the Rankine cycle"
w_dot_turb = (h[3] - h[4])
w_dot_pump = (h[2]-h[1])
q_dot_H = (h[3] - h[2])
q_dot_C = (h[4] - h[1])

w_dot_net = w_dot_turb - w_dot_pump

eta = w_dot_net / q_dot_H
 Solutions for selected exercises
(Lecture 6a and 6b)

Exercise 7.1

Similar to example 7.1 in lecture 6b, the saturation pressure Psat,2 needs to be determined, here
for T2 = 205 ℃. By using the Clausius-Clapeyron equation

we are able to
 first solve for the constant C in state 1 with T1 = 200 ℃ and
 then solve for the saturation pressure in state 2 with T2 = 205 ℃ .

Therefore, the average enthalpy of vaporization is estimated by means of the saturated water
property table:

This allows us to calculate C and Psat,2:

The value of 1723.0 kPa from Appendix 8a is close to our estimated saturation pressure of
1704.3 kPa. Please note that R in kJ/(kg*K) is used instead of R in kJ/(kmol*K).
Exercise 7.4

For this exercise, the saturation pressure of water vapour in equilibrium with ice at -20 ℃ needs
to be determined.

Since temperature, pressure and latent heat of sublimation is given for water at the triple point,
the Clausius-Clapeyron equation can be used to
 first find the constant C at the triple point and
 then solve for the saturation pressure at T2 = -20 ℃ .

The calculation is similar to the one in exercise 7.1 and leads to the following results:

Again, please note that R in kJ/(kg*K) is used instead of R in kJ/(kmol*K).

Exercise 7.8

The goal is to determine the minimum amount of time required to evaporate five kilograms of
water which are boiling in an open pot placed over a 500W heater.

Keep in mind that water boils at different temperatures based on the atmospheric pressure.
When assuming the atmospheric pressure to be 100 kPa, the mass flow rate of evaporated water
can be estimated to:

By considering the initial mass of 5 kilograms, the total duration is calculated as:
Exercise 7.11

In this exercise, you need to identify the phase and specific enthalpy of water at temperature of
300 °C and pressure of 300 kPa.

There are two ways to identify the phase:

1. Determine the saturation pressure of water at 300 °C and compare to 300 kPa.
2. Determine the saturation temperature of water at 300 kPa and compare to 300 °C.

In both ways, superheated steam can be identified since the saturation pressure of water at 300
°C is 8.581 MPa (>> 300 kPa) and the saturation temperature of water at 300 kPa is 133.52 °C
(<< 300°C).

The superheated steam tables in Appendix 8c give the enthalpy h = 3069.3 kJ/kg.

Exercise 7.26

2 L of saturated liquid at 100 kPa in a cyclinder with a frictionless piston is heated until its
quality is 80%. How much heat is added?

The mass can be found by comparing the volume with the specific volume of liquid saturated
water at 100 kPa (Appendix A8b).

By using the final quality and specific enthalpy of saturated water at 100 kPa for liquid and
gas, the specific enthalpy in the final state is computed.

The product of mass and specific enthakpy difference between initial and final state gives the
required heat.
Exercise 7.46

Superheated steam at 0.5 MPa and 600 °C enters an insulated chamber with 2.5 kg/s. Liquid
water at 5 MPa and 20 °C enters with unknown mass flow but condenses the steam in order to
have saturated liquid at 0.5 MPa exiting the chamber.

In conclusion, the insulated chamber is characterized by two incoming streams and one exiting
stream. The following energy balance for an open system shows that the specific heat capacities
of the various substances at the given pressures are needed to calculate the missing mass flow
of liquid water.

Determining the h1,w (Appendix A8d) , h1,s (Appendix A8c) and h2 (Appendix A8b) allows the
calculation of liquid water mass flow.

Exercise 7.78

Use both the ideal gas equation and van der Waals equation to determine the pressure of 2 kg
Oxygen occupying a volume of 0.1 m3 at a temperature of 250 K.

First, the amount of oxygen in moles is calculated.

The specific molar volume is calculated by diving the given volume by the amount of oxygen
in moles.

By using the ideal gas law, the pressure PA can be calculated.

Table 7.5 within chapter 7 helps to calculate the the pressure PB with the van der Waals
equation.

For this state of oxygen, the ideal gas equation approximation is close to the value calculated
with the van der Waals equation. Please note that, different to exercises 7.1 and 7.4, Ru in
kJ/(kmol*K) is used.

Exercise 7.79

Use both the ideal gas equation and the compressibility charts to determine the density of
carbon dioxide at a pressure of 6.5 MPa and temperature of 30 °C.

First, the ideal gas equation is used to compute the density (inverse of the specific volume)
with R from Appendix A1.

The reduced temperature TR and reduced pressure PR are calculated with critical pressure Pc
and critical temperature Tc from Appendix A6.

According to Appendix 10a, the compressibility factor is Z = 0.55.

By using the definition of the compressibility factor, we can solve for the specific volume.

For carbon dioxide in the given state, the density calculated by the ideal gas equation
significantly differs to the more accurate value calculated from the compressibility charts.
 Solutions for Lecture 4 on heat transfer

4a.exercise 1

The relationship between charging rate P and charging voltage V is P = V*I, where I is
current. This relation can be written in EES as:
W_dot_el = V_batt * i_batt
Beside, the time it took to charge the battery is:
t_charge = Q_max / W_dot_el
Thus, by setting up some parameters, the EES solution of this problem is:

Q_max = 80*convert(kW, W)
V_batt = 800
W_dot_el = 50 * convert(kW,W)
t_charge = Q_max / W_dot_el
W_dot_el = V_batt * i_batt
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4a ex2

L = 0,4 [m]

th = 0,005 [m]

k glass = 1,4

T in = 70

T out = 10

2
A glass = L

T in – T out
Q = A glass ꞏ k glass ꞏ
th
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EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

4a ex3

D = 0,05

A = 3,142 ꞏ D

T cyl = 50

T fl = 20

h conv = 200

Q = A ꞏ h conv ꞏ T fl – T cyl
File:Temperature of Earth.EES 21-02-2022 10:27:39 Page 1
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

Given

Q' in = 1000

6
r earth = 6,378 x 10

 = 1

T space = 4

Incoming energy

2
A earth = 3,142 ꞏ r earth

Q in = Q' in ꞏ A earth

Outgoing energy

2
A rad = 4 ꞏ 3,142 ꞏ r earth

4 4
Q rad =  ꞏ A rad ꞏ 5,670E–08 [W/m2-K4] ꞏ T earth – T space

Q rad = Q in

T earth;C = ConvertTemp K ; C ; T earth

File:Lecture 4b ex1.EES 21-02-2022 10:30:27 Page 1
EES Ver. 10.836: #0780: for use by students and faculty, Technical University of Denmark, Lyngby,Denmark

4b ex1

L = 0,4 [m]

th = 0,005 [m]

k glass = 1,4

T in = 70

T out = 10

2
A glass = L

th
R cond =
k glass ꞏ A glass

T in – T out
Q =
R cond

th ins = 0,005

k ins = 0,1

th ins
R ins =
k ins ꞏ A glass

R tot = R cond + R ins

T in – T out
Q ins =
R tot
4b.exercise 2

Above all, the mix box following first law of thermodynamic, thus:
m_dot_1*h_1 + m_dot_2 * h_2 - Q_dot_loss = 4* m_dot_out * h_mix
where m_dot_1 is fresh air mass flow rate, h_1 is fresh air enthalpy, m_dot_2 is recirculated
air mass flow rate, h_2 is recirculated air enthalpy, Q_dot_loss is energy loss, is m_dot_out
outlet air mass flow rate, h_mix is outlet air enthalpy.

The fresh air mass flow rate is:

m_dot_1 = V_dot_ind * N_people * convert(m3/hr, m3/s) * rho_1
where V_dot_ind is fresh air need per person, N_people is number of people, rho_1 is
density of fresh air. The density and enthalpy of air corresponding to both temperature T_1
and pressure P of fresh air:
rho_1 = density(Air_ha,T=T_1,P=P)
h_1 = enthalpy(Air_ha, T = T_1, P = P)

The mass flow rate of recirculated air is:

m_dot_2 = V_dot_recirc * convert(m3/hr, m3/s) * rho_2
where V_dot_recirc is 4000 m3/hr, rho_2 and h_2 are the density and enthalpy of
recirculated air at temperature T_2:
rho_2 = density(Air_ha,T=T_2,P=P)
h_2 = enthalpy(Air_ha, T = T_2, P = P)

For mass balance:

m_dot_out = (m_dot_1 + m_dot_2) / 4
After mixing, the temperature inside the box is T_mix, and the enthalpy is:
h_mix = enthalpy(Air_ha, T = T_mix, P = P)

To close the energy balance of the system, we still need to solve heat transfer process and
find out Q_dot_loss and T_mix. Notice the insulation layer is installed on the outside of the
pipe. Thus, the thermal resistance of insulation layer is:
R_ins = ln(D_out/D_box) / (L_box * 2 * pi * k_ins)
Further, the thermal resistance values for convection on the inner and outer surfaces is:
R_conv_in = 1/(h_in* pi * D_box * L_box)
R_conv_out = 1 / (h_out * pi * D_out * L_box)

The thermal resistance of the pipe is the sum of three terms above:
R_tot = R_conv_in + R_ins + R_conv_out
At last, the heat loss is:
Q_dot_loss = (T_mix - T_out) / R_tot

By substituting the heat loss into the energy balance of the system, we only have one variable
T_mix left, which is easy to be solved. To sum up, the EES solution can be written as follow.

"Given"
N_people = 200
V_dot_ind = 35
V_dot_recirc = 4000
T_1 = 12
T_2 = 22
P = 1*convert(atm, kPa)

"Heat losses"
T_out = 12
L_box = 1
D_box = 2.5
h_in = 100
h_out = 25
k_ins = 0.1
th_ins = 100 * convert(mm,m)
D_out = D_box + 2* th_ins
R_conv_in = 1/(h_in* pi * D_box * L_box)
R_conv_out = 1 / (h_out * pi * D_out * L_box)
R_ins = ln(D_out/D_box) / (L_box * 2 * pi * k_ins)
R_tot = R_conv_in + R_ins + R_conv_out
Q_dot_loss = (T_mix - T_out) / R_tot

"Mass balance"
rho_1 = density(Air_ha,T=T_1,P=P)
m_dot_1 = V_dot_ind * N_people * convert(m3/hr, m3/s) * rho_1

rho_2 = density(Air_ha,T=T_2,P=P)
m_dot_2 = V_dot_recirc * convert(m3/hr, m3/s) * rho_2

m_dot_out = (m_dot_1 + m_dot_2) / 4

"1st Law Energy Balance"

h_1 = enthalpy(Air_ha, T = T_1, P = P)
h_2 = enthalpy(Air_ha, T = T_2, P = P)
h_mix = enthalpy(Air_ha, T = T_mix, P = P)

m_dot_1*h_1 + m_dot_2 * h_2 - Q_dot_loss = 4* m_dot_out * h_mix

 Solutions for selected exercises in Lecture 3
(Chapter 4 in “Energy, Entropy and Engines”)

4.60

Notice, the “34 kW electrical heater” means Q = 34 kW. Based on first law:

Thus, the work done in the heater W is:

Be very careful about the unit. Many students mess up the J/kg and kJ/kg and get the wrong
answer.

4.66

In this question, we should considered internal energy only:

The enthalpy of air can be interpolated from Appendix 7. The enthalpy of water can be
interpolated from Appendix 8. Another equivalent approach is.

where Cp,a is 1.0806 kJ/(kg ℃) (Appendix 4), Cw is 4.18 kJ/(kg ℃) (Appendix 3). As a result:
4.68

Air enters a nozzle at 300 kPa and 350 K with a velocity of 30 m/s and leaves at 100 kPa with a
velocity of 200 m/s. The heat loss from the nozzle is 20 kJ/kg of airflow through it. What is the exit
temperature of the air?

Energy balance:

There is no work done and no potential energy. Consequently, you can rearrange for h2:

Initial air specific enthalpy at T1 = 350 K from Appendix 7: h1 = 350.49 kJ/kg.

Please note that kJ are converted in J by adding 103 to the denominator of the middle term.

4.82

A gas turbine operates witj 0.1 kg/s of helium that enters at 8 MPa and 600 K and leaves at 200
kPa and 350 K. If the turbine power output is 120 kW, find the rate of heat loss from the turbine.
Neglect kinetic energy changes.

Energy balance:

No potential and no kinetic energy. Specific heat of helium at constant pressure cp = 5.193
kJ/kg°C assumed (Appendix 1).

Please note that a consideration of task-specific temperature (600 K, 350K) and pressure (8
MPa, 200 kPa) leads to a more realistic value for the specific heat cp but is not included here.
4.76.

Nitrogen flowing at a rate of 60 kg/min is compressed from 150 kPa and 30°C to 800 kPa and
150°C. The compressor is cooled at a rate of 15 kJ/kg of gas during operation. What is the power
required to drive the compressor?

Energy balance:

𝑄̇ + 𝑊̇ + 𝑚̇ ℎ + + 𝑔𝑧 = 𝑄̇ + 𝑊̇ + 𝑚̇ ℎ + + 𝑔𝑧

Kinetic energy and potential energy are equal to zero. Specific heat of nitrogen at constant
pressure cp = 1.04179 kJ/kg°C at Tavg = 363.15 K interpolated from Appendix 4.

𝑊̇ + 𝑚̇(ℎ) = 𝑄̇ + 𝑚̇(ℎ)

/
𝑊̇ = 1.04179 °C(150°C − 30°C) + 15 = 140 𝑘𝑊
/

4.77.

An air compressor takes air at 315 K and compresses it. For every kilogram of air passing through
the compressor it does 200 kJ of work and loses 10 kJ of heat. Find the temperature of the air
exiting the compressor. Do not assume that the specific heat of air is constant.

Specific enthalpy of air at 315 K, hin = 315.27 kJ/kg (Appendix 7).

𝑊̇ + 𝑚̇(ℎ) = 𝑄̇ + 𝑚̇(ℎ)

(ℎ ) ̇
= 𝑊̇ + 𝑚̇(ℎ) −𝑄 = 200 + 315.27 − 10 = 505 𝑘𝐽/𝑘𝑔

Interpolating from the air tables, Appendix 7, the final temperature is Tout = 502.18 K.

4.80.

Argon enters an adiabatic turbine at 2 MPa and 600°C at a rate of 2.5 kg/s and leaves at 150 kPa.
Determine the exit temperature of the gas when the turbine is generating 300 kW. Neglect kinetic
energy changes.

Specific heat of argon at constant pressure cp = 0.520 kJ/kg°C (Appendix 1).

𝑚̇(ℎ) = 𝑊̇ + (ℎ)

𝑚̇𝐶 𝑇 = 𝑊̇ + 𝑚̇𝐶 𝑇

̇
𝑇 = ̇
+𝑇 = + 600 = 369°C
. ×
4.81.

Air enters an adiabatic turbine at 800 kPa and 870 K with a velocity of 60 m/s, and leaves at 120
kPa and 520 K with a velocity of 100 m/s. The inlet area of the turbine is 90 cm2. What is the
power output?

Gas constant of air R = 0.2870 kJ/kgK (Appendix 1), specific enthalpy of air at 520 K h2 = 523.63
kJ/kg (Appendix 7).

𝑚̇ ℎ + = 𝑊̇ + 𝑚̇ ℎ +

𝑅𝑇 0.2870 × 870
𝑣 = = = 0.312 𝑚 /𝑘𝑔
𝑃 800

𝐴𝑖𝑛 𝑉𝑖𝑛 90 × 10−4 × 60

𝑚̇ = = = 1.73 𝑘𝑔/𝑠
𝑣𝑖𝑛 0.312

Via interpolation h1 = 899.415 kJ/kg (Appendix 7).

𝑊̇ = 𝑚̇ ℎ + − 𝑚̇ ℎ + = 1.73 (899.415 − 523.63 + = 644.6 𝑘𝑊

 Solutions for selected exercises:
Chapter 4 in “Energy, Entropy and Engines”

Exercise 4.4

The torque of the pullley can be found in Eq. 4.36:

By deploying the formula of the shaft power corresponding to rotation speed, we get the result as:

Notice the unit of the rotation speed is [1/min].

Exercise 4.5

The force to the piston is F = P*A. Therefore, the work done to the piston is:

Exercise 4.7

This is a question about potential energy. When the piston goes up, the water will leave the container.
Thus, the potential energy of the water in the container is reducing. The force on the piston is also
reducing in this process.

The force on the piston is:

The work can be calculated via integration over the entire depth:

Exercise 4.8 (lecture)

Exercise 4.10
 A piston compresses air in cylinder.

The work done can be calculated with

Since mass of air m = 2 kg, gas constant of air R = 0.2870 kJ/(kgK) and T = T1 = T2 = 20 °C are known, the
only missing information is V1/V2. The knowledge of P1 = 100 kPa and P2 = 300kPa makes it to possible to
use the ideal gas equation to determine V1/V2 with

By using the known properties, the following work done results

Exercise 4.13

To calculate the work done during a polytropic process for argon, an integration of the boundary work
equation (done in chapter 4) is needed. Therefore, the final volume of the gas needs to be determined
with the polytropic process equation

Since now V2 is known, the work done can be calculated with

Based on standard conventions, the work done is negative since the gas does work on the surrounding.
4.14)
Air at 800 K and 1 MPa is expanded in a polytropic process for which PV1.6 = constant until the
pressure reaches 0.1 MPa. Find the final gas temperature and the work done per unit mass of air.

SOLUTION:
Based on the description of the exercise, it can be assumed that the mass of the system (Air volume)
remains constant during the polytropic process and, in addition, the air behaves as an ideal gas.
Gas constant of air R = 0.2870 kJ/kgK (Found in Appendix, A1).

Using the ideal gas law for the initial and the final state of the system, we write:
𝑚
Initial state: 𝑃1 𝑉1 = 𝑅𝑢 𝑇1 (1)
𝑚
Final state: 𝑃2 𝑉2 = 𝑅𝑢 𝑇2 (2)

m = Constant and combining (1) and (2), we write:

𝑃1 𝑉1 𝑃2 𝑉2 𝑃1 𝑉1 𝑃2 𝑉2 𝑇2 𝑃 𝑉
• =𝑚𝑅, = 𝑚𝑅 ↔ = ↔ = 𝑃2𝑉2 (3)
𝑇1 𝑇2 𝑇1 𝑇2 𝑇1 1 1
Also, PV1.6 is constant, hence we can write:
1
𝑉2 𝑃 1.6
• 𝑃1 𝑉11.6 = 𝑃2 𝑉21.6 ↔ = (𝑃1 ) (𝟒)
𝑉1 2

Replacing (4) to (3), we write:

1 1.6−1 0.6
𝑃 𝑃 1.6 𝑃2 0.1 𝑀𝑃𝑎 1.6
•
1.6
𝑇2 = 𝑇1 2 ( 1 ) = 𝑇1 ( ) = 800 𝐾( ) = 𝟑𝟑𝟕. 𝟑𝟓𝟕 𝑲
𝑃1 𝑃2 𝑃1 1 𝑀𝑃𝑎

The work function of a polytropic process can be obtained by integrating the boundary work
equation (eq. 4.30, page 97 of the book).
𝑃2 𝑉2 −𝑃1 𝑉1 (𝑈𝑠𝑖𝑛𝑔 𝐼𝑑𝑒𝑎𝑙 𝑔𝑎𝑠 𝑙𝑎𝑤) 𝑚𝑅𝑇2 −𝑚𝑅𝑇1 𝑚𝑅(𝑇2 −𝑇1 )
• 𝑊12 = ⇔ 𝑊12 = =
𝑛−1 𝑛−1 𝑛−1

The work per unit mass that the system does on the surroundings is calculated as follows:
 𝑊12 𝑅(𝑇2 −𝑇1 ) 0.2870 𝑘𝐽/𝑘𝑔𝐾(337.357𝐾 − 800 𝐾)
𝑤12 = = = = −𝟐𝟐𝟏. 𝟐𝟗𝟖 𝒌𝑱/𝒌𝒈
𝑚 𝑛−1 1.6 − 1

Final solutions:
1. Argon temperature = 337 K
2. Work per unit mass = 221.3 kJ/kg

4.15)
A cylinder containing 0.05 m3 of carbon dioxide at 200 kPa and 100°C is expanded in a polytropic
process to 100 kPa and 20°C. Determine the work done.

SOLUTION:
Based on the description of the exercise, it can be assumed that the mass of the system (CO2
volume) remains constant during the polytropic process and, in addition, CO2 behaves as an ideal
gas.
Gas constant of carbon dioxide, R = 0.1889 kJ/kgK (Found in Appendix, A1).

The work function of a polytropic process can be obtained by integrating the boundary work
equation (eq. 4.30, page 97 of the book).
𝑃2 𝑉2 −𝑃1 𝑉1
• 𝑊12 = (1)
𝑛−1
The system is experiencing a polytropic process, hence, we can use the main polytropic equation
to describe the initial and the final state of the system.
𝑃1 𝑉 𝑛 𝑃 𝑉 𝑛 𝑃 𝑉
• 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 ↔ = (𝑉1 ) ↔ 𝑙𝑛 (𝑃1 ) = 𝑙𝑛 (𝑉1 ) ↔ 𝑙 𝑛 (𝑃1 ) = 𝑛 𝑙𝑛 𝑉1 ↔
𝑃2 2 2 2 2 2

𝑙𝑛 (𝑃1 /𝑃2)
𝑛= (2)
𝑙𝑛 (𝑉2 /𝑉1 )

Applying the ideal gas law in the initial state of the system, the mass of CO2 can be calculated:
 𝑃1 𝑉1 200 𝑘𝑃𝑎 𝑥 0.05 𝑚3
• 𝑚𝐶𝑂2 = = = 0.14187 kg
𝑅 𝑇1 0.1889 𝑘𝐽/𝑘𝑔𝐾 𝑥 373.15 𝐾

Using the value of mCO2 in the ideal gas law for the final state of the system, the volume of CO2
can also be calculated:

𝑚𝑅𝑇2 0.14187 𝑘𝑔 𝑥 0.1889 𝑘𝐽/𝑘𝑔𝐾 𝑥 293.15 𝐾

• 𝑉2 = = = 0.078562 𝑚3
𝑃2 100 𝑘𝑃𝑎

Using the value of V2 to equation (2), we can write:

𝑙𝑛 (𝑃1 /𝑃2) 𝑙𝑛 (200 𝑘𝑃𝑎/100 𝑘𝑃𝑎

• 𝑛= = 𝑙𝑛 (0.0785𝑚3/0.05𝑚3 ) = 1.5367
𝑙𝑛 (𝑉2 /𝑉1 )

All the values of the parameters of equation (1) are known, hence, we write:
𝑃2 𝑉2 −𝑃1 𝑉1 100 𝑘𝑃𝑎 𝑥 0.0785 𝑚3 −200 𝑘𝑃𝑎 𝑥 0.05 𝑚3
• 𝑊12 = = = −𝟑. 𝟗𝟗𝟒𝟒 𝒌𝑱
𝑛−1 1.5367−1
 Solutions problems chapter 2 and 3

• Chapter 2: 2.5, 2.8, 2.12, 2.27, 2.28, 2.37

2.5: What is the weight of an object with a mass of 150 kg on a planet where g = 4.1 m/s2?

The definition of the weight is given by equation 2.5 on p. 22 of Chapter 2:

𝐹𝐹𝑤𝑤 = 𝑚𝑚𝑚𝑚

𝐹𝐹𝑤𝑤 = 150 × 4.1 = 𝟔𝟔𝟔𝟔𝟔𝟔 𝑵𝑵

2.8: A 5 kg box sliding across the floor with an initial velocity of 8 m/s is decelerated by friction to
3 m/s over 5 s. What is the force of friction acting on it?

The definition of the acceleration is in the equation 2.13, p. 28.

∆𝑉𝑉 3−8
𝑎𝑎 = = = −1 𝑚𝑚/𝑠𝑠 2
∆𝑡𝑡 5

The frictional is the only force acting on the box. We can use equation 2.5:

𝐹𝐹𝑤𝑤 = 𝑚𝑚𝑚𝑚 = 5 × (−1) = −5 𝑁𝑁

The force is in opposite direction of the sliding of the box, that is why we obtain a negative value, but
we can say that the force decelerating the box is 5 N.

2.12: Acceleration due to gravity at Earth’s surface is g = 9.80665 m/s2 and decreases by approximately
3.3 × 10–6 m/s2 for each metre of height above the ground. What is the potential energy of a 100 kg
mass raised to an altitude of 1000 m: (a) assuming constant g, (b) accounting for the decrease in g
with height?

(a) Assuming the g is constant we have only one variable: the height. We can use equation 2.11,
on p. 25.

∆𝑃𝑃𝑃𝑃 = 𝑚𝑚𝑚𝑚∆𝑧𝑧 = 100 × 9.80665 × 1000 = 𝟗𝟗𝟗𝟗𝟗𝟗 𝟔𝟔𝟔𝟔𝟔𝟔 𝐉𝐉

(b) When taking into account that g decreases with height, we have 2 variables, therefore, we
have to use equation 2.10 from p. 25.

𝑑𝑑(𝑃𝑃𝑃𝑃) = 𝑑𝑑(𝑚𝑚𝑚𝑚𝑚𝑚)
ℎ ℎ
∆𝑃𝑃𝑃𝑃 = 𝑚𝑚 � 𝑔𝑔𝑔𝑔𝑔𝑔 = 100 × � (9.80665 − 3.3 × 10−6 × z)dz
0 0

∆PE = 100 × (9.80665 h − 1.65 × 10−6 h2 ) = 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝑱𝑱

2.27: Tank A has a volume of 0.5 m3 and contains air with density 1.2 kg / m3 while tank B has a
volume of 0.8 m3 and contains air with density 0.9 kg / m3. The two tanks are connected to each other
and their contents mixed. What is the final air density in the tanks?

First, we have to calculate the mass of air in each tank:

𝑚𝑚𝐴𝐴 = 𝜌𝜌𝐴𝐴 𝑉𝑉𝐴𝐴 = 1.2 × 0.5 = 0.6 𝑘𝑘𝑘𝑘

𝑚𝑚𝐵𝐵 = 𝜌𝜌𝐵𝐵 𝑉𝑉𝐵𝐵 = 0.9 × 0.8 = 0.72 𝑘𝑘𝑘𝑘

The total mass of air, combining the two tanks, is:

𝑚𝑚𝑡𝑡 = 𝑚𝑚𝐴𝐴 + 𝑚𝑚𝐵𝐵 = 0.6 + 0.72 = 1.32 𝑘𝑘𝑘𝑘

The total volume is:

𝑉𝑉𝑡𝑡 = 𝑉𝑉𝐴𝐴 + 𝑉𝑉𝐵𝐵 = 0.5 + 0.8 = 1.3 𝑚𝑚3

The density of air in the combined tank is then:

𝑚𝑚𝑡𝑡 1.32
𝜌𝜌 = = = 𝟏𝟏. 𝟎𝟎𝟎𝟎 𝒌𝒌𝒌𝒌/𝒎𝒎𝟑𝟑
𝑉𝑉𝑡𝑡 1.3

2.28: A 0.5 m3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water
at 25 °C. Find the weight of the liquid.

The masses of each liquid in the container are:

𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝜌𝜌𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑉𝑉 × 0.9 = 1000 × 0.5 × 0.9 = 450 𝑘𝑘𝑘𝑘

𝑚𝑚𝑒𝑒𝑒𝑒ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝜌𝜌𝑒𝑒𝑒𝑒ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉 × 0.1 = 783 × 0.5 × 0.1 = 39.15 𝑘𝑘𝑘𝑘

The total mass is :

𝑚𝑚𝑡𝑡 = 𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑒𝑒𝑒𝑒ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 450 + 39.15 = 489.15 𝑘𝑘𝑘𝑘

𝐹𝐹𝑤𝑤 = 𝑚𝑚𝑡𝑡 𝑔𝑔 = 489.15 × 9.81 = 𝟒𝟒 𝟕𝟕𝟕𝟕𝟕𝟕. 𝟔𝟔 𝑵𝑵

2.37: The inside of a house is at 20 °C while the exterior air is at –5 °C. The temperature of the walls
of the house does not change with time. Are the walls at equilibrium?

On pages 35 and 36, you can find the definition of steady state and equilibrium.

As the temperatures inside the house and outside are different and the system is not isolated, there
is heat transfer from the inside to the outside of the house at a constant rate. As the walls keep
interacting with the inside and outside of the house, the walls are in steady state but do not reach
equilibrium (it is illustrated on Figure 2.13 (b) on p. 35 and Figure 2.14 p. 36). Heat must be constantly
added to maintain the temperature difference.
 • Chapter 3: 3.3, 3.7 , 3.9, 3.15, 3.20, 3.27, 3.39

3.3: If 5 g of methane is burned in the chemical reaction CH4 + 2O2 -> CO2 + 2H2O, what mass of oxygen
is consumed and what are the masses of the combustion products?

On p. 52, you can find the relation between mass, moles and molar mass:

𝑚𝑚 = 𝑁𝑁 × 𝑀𝑀
As we have the mass of methane we can calculate the mole of methane:
𝑚𝑚𝐶𝐶𝐶𝐶4 5
𝑁𝑁𝐶𝐶𝐶𝐶4 = = = 0.31 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑀𝑀𝐶𝐶𝐶𝐶4 16

From the equation we know that for one mole of methane, two moles of oxygen was used and one
mole of carbon dioxyde was produced and two moles of water.

𝑁𝑁𝑂𝑂2 = 2 × 𝑁𝑁𝐶𝐶𝐶𝐶4 = 0.62 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑁𝑁𝐶𝐶𝐶𝐶2 = 𝑁𝑁𝐶𝐶𝐶𝐶4 = 0.31 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑁𝑁𝐻𝐻2 𝑂𝑂 = 2 × 𝑁𝑁𝐶𝐶𝐶𝐶4 = 0.62 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑚𝑚𝑂𝑂2 = 𝑁𝑁𝑂𝑂2 × 𝑀𝑀𝑂𝑂2 = 𝟏𝟏𝟏𝟏. 𝟗𝟗 𝒈𝒈

𝑚𝑚𝐶𝐶𝐶𝐶2 = 𝑁𝑁𝐶𝐶𝑂𝑂2 × 𝑀𝑀𝐶𝐶𝐶𝐶2 = 𝟏𝟏𝟏𝟏. 𝟕𝟕 𝒈𝒈

𝑚𝑚𝐻𝐻2 𝑂𝑂 = 𝑁𝑁𝐻𝐻2 𝑂𝑂 × 𝑀𝑀𝐻𝐻2 𝑂𝑂 = 𝟏𝟏𝟏𝟏. 𝟐𝟐 𝒈𝒈

3.7: A cubical container, 10 cm long along each edge, is filled with a gas at a pressure of 350 kPa.
Determine the force that the gas exerts on each wall of the container.

First, we calculate the surface area of each wall of the container:

𝐴𝐴 = 0.12 = 0.01 𝑚𝑚2

Then, to find the force that the gas exerts on each wall of the container we use the formula 3.2 on
p. 54.

𝐹𝐹
𝑃𝑃 =
𝐴𝐴

𝐹𝐹 = 𝑃𝑃 × 𝐴𝐴 = 350 × 103 × 0.01 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝑵𝑵

3.9: A cylindrical drum, 1 m high, is filled with water (ρ = 1000 kg/m3) to a depth of 0.2 m. The rest of
the drum is then filled with oil (ρ = 850 kg/m3). What is the gauge pressure on the bottom of the drum?

The gauge pressure is in this case the sum of the pressure exerted by water and oil:

𝑃𝑃 = 𝜌𝜌𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔ℎ𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 + 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝑔𝑔ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = 9.81 × (1000 × 0.2 + 850 × 0.8) = 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖. 𝟖𝟖 𝑷𝑷𝑷𝑷
3.15: What is the volume in liters of 1 gmol of an ideal gas at standard temperature and pressure,
defined as 0 °C and 101.325 kPa?

The ideal gas equation has to be used, it can be found on p. 58, equation 3.10:

𝑁𝑁𝑁𝑁𝑁𝑁 10−3 × 8.314 × 103 × 273.15

𝑉𝑉 = = = 0.0224 𝑚𝑚3 = 𝟐𝟐𝟐𝟐. 𝟒𝟒 𝑳𝑳
𝑃𝑃 101325

3.20: An air bubble, 1 mm in diameter, is released at the bottom of a lake 20 m deep where the
temperature is 10 °C. It rises to the surface of the lake where the temperature is 25 °C. What will the
radius of the bubble be at the surface?

The pressure at the bottom of the lake is calculated as follow:

𝑃𝑃20 = 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔ℎ = 297.53 𝑘𝑘𝑘𝑘𝑘𝑘

Pressure at the top of the lake is the atmospheric pressure: P0 = 101.325 kPa
The air bubble is supposed to be an ideal gas with constant mass:

𝑃𝑃20 𝑉𝑉20 𝑃𝑃0 𝑉𝑉0

=
𝑇𝑇20 𝑇𝑇0

𝑉𝑉0 𝑃𝑃20 𝑇𝑇0

= = 3.09
𝑉𝑉20 𝑃𝑃0 𝑇𝑇20

The volume of a sphere is:

4
𝑉𝑉 = 𝜋𝜋𝑟𝑟 3
3

Therefore:

3
𝑟𝑟0 = √3.09 × 𝑟𝑟20 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒎𝒎𝒎𝒎

3.27: An evacuated 20 L container is filled with gas until the pressure inside reaches 800 kPa at 25 °C.
By weighing the container before and after filling it is determined that its mass increased by 26 g.
What gas was it filled with?

𝑚𝑚𝑚𝑚𝑚𝑚 26 × 10−3 × 8314 × 298.15

𝑀𝑀 = = = 𝟒𝟒. 𝟎𝟎𝟎𝟎 𝒌𝒌𝒌𝒌/𝒌𝒌𝒌𝒌𝒌𝒌𝒌𝒌
𝑃𝑃𝑃𝑃 800 × 103 × 20 × 10−3

3.39: Five kilograms of air, initially at 0 °C, are heated until the temperature reaches 50 °C. The internal
energy increases by 180 kJ during this process. Find the specific heat of air.

The relation between internal energy and specific heat is given by equation 3.30, p. 65.
 𝑈𝑈2 − 𝑈𝑈1 = 𝑚𝑚𝑚𝑚(𝑇𝑇2 − 𝑇𝑇1 )

𝑈𝑈2 − 𝑈𝑈1 180 𝑘𝑘𝑘𝑘 𝒌𝒌𝒌𝒌

𝑐𝑐 = = = 𝟎𝟎. 𝟕𝟕𝟕𝟕
𝑚𝑚(𝑇𝑇2 − 𝑇𝑇1 ) 5 𝑘𝑘𝑘𝑘 (50 𝐶𝐶 − 0 𝐶𝐶) 𝒌𝒌𝒌𝒌 𝑪𝑪