1 10/01/2007
ENGRD 221 – Prof. N. Zabaras SOLUTIONS TO HOMEWORK 5
Problem 1 (Carnot Cycle): One kilogram of air as an ideal gas executes a Carnot power cycle having a thermal efficiency of 60%. The heat transfer to the air during the isothermal expansion is 40KJ. At the end of the isothermal expansion, the pressure is 5.6 bar and the volume is 0.3m 0.3m3 .Determine (a) The maximum and minimum temperatures for the cycle, in K. (b) The pressure and volume at the beginning of the isothermal expansion in bar and m3 , respectively. (c) The work and heat transfer for each of the four processes in KJ. (d) Sketch the cycle on p-v coordinates.
1 kg of air undergoes a Carnot cycle for which, η = 0.6 . Find: Determine the minimum and maximum temperatures, the pressure and volume at the beginning of the isothermal expansion, the work and heat transfer for each process, and sketch the cycle on pv coordinates. Known :
Schematic and given data:
Assumption: The system shown in the schematic consists of air modeled as an ideal gas. Analysis: (a) using the ideal gas model equation of state T2 = Then
P2V 2 mR
⇒ T2 = 585.4K
since η = 1−
T C T H
⇒ TC = TH (1 − η ) = T2 (1 − η ) = 234.2 K
T3 = T4 = T C (b) For process 1-2, Q12 = 40 KJ . An energy balance leads m(u2 − u1 ) = Q12 − W12 But since internal energy of and ideal gas depends on temperature and T1 = T 2 and W12 = Q12 . Further W12 =
∫
2
1
pdV =
∫
2
1
mRT H V
dV = mRT H ln
values:
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V2 . Solving and inserting V 1
1 10/01/2007
ENGRD 221 – Prof. N. Zabaras ln
V2 V1
=
W 12 T H
= 0.2381 ⇒ V1 = 0.24m3
Since T1 = T 2 p1V1 = mRT , p2V2 = mRT ⇒ p2V2 = p1V1 ⇒ p1 =
p2V 2 V 1
= 7 bar
(c) For process 2-3: Q23 = 0 . An energy balance reduces to W23 = m( u2 − u3 ) With data from Table A-22, W23 = 1kg (423.7 − 167.0) = 256.7 kJ . For process 3-4 W34 = Q34 . Also Eq. 5.6 is applicable | Q34 | TC
=
| Q12 | T H
⇒| Q34 |= 0.4(40 KJ ) = 16 KJ ⇒ Q34 = −16 KJ , W34 = −16 KJ
For process 4-1, Q41 = 0 . An energy balance reduces to W41 = m( u4 − u1 ) = m( u3 − u2 ) = −256.7 KJ (d)
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ENGRD 221 – Prof. N. Zabaras Problem 2 (Carnot Cycle): : The pressure-volume diagram of a Carnot power cycle executed by an ideal gas with constant specific heat ratio of k is shown here. Demonstrate that (a) V4V2 = V1V3
T 2 ⎛ p 2 ⎞ =⎜ ⎟ T 3 ⎜⎝ p3 ⎠⎟
(b)
⎛ V ⎞ = ⎜⎜ 3 ⎟⎟ T 3 ⎝ V 2 ⎠
T 2
(c)
k −1 k
k −1
Known: A Carnot cycle is executed by an ideal gas with constant specific heat ratio k. Find: Show that: V4V2 = V1V3
⎛ p ⎞ = ⎜⎜ 2 ⎟⎟ T 3 ⎝ p 3 ⎠
T 2
⎛ V ⎞ = ⎜⎜ 3 ⎟⎟ T 3 ⎝ V 2 ⎠
T 2
k −1 k
k −1
Assumptions: 1) The system shown in the figure consists of an ideal gas. 2) The specific heat ratio k is constant (required for part (b) only). 3) The system undergoes a Carnot cycle. Analysis: (a) The thermal efficiency of the cycle can be written as: W cycle η cycle = Qin where Wcycle = QH – QC. For the cycle shown above, QH = Q12 and QC = Q34. Since the internal energy of an ideal gas depends only on temperature, and energy balance for process 1-2 reduces to U2-U1 = Q12 – W12 where U2 = U1. Thus, Q12 = W12. Therefore, Q W η cycle = 1 − 34 = 1 − 34 Q12 W 12 Furthermore, W 12 =
∫
2
1
pdV =
∫
2
mRT H V
1
⎛ V 2 ⎞ ⎟⎟ V ⎝ 1 ⎠
dV = mRT H ln⎜⎜
Similarly for W34. W 34 =
∫
4
3
pdV =
4
mRT C
3
V
∫
⎛ V 3 ⎞ ⎟⎟ V ⎝ 4 ⎠
dV = mRT C ln⎜⎜
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1 10/01/2007
ENGRD 221 – Prof. N. Zabaras Thus, the thermal efficiency is V mRT C ln⎛ ⎜ 3
η cycle
⎞ V 4 ⎠⎟ ⎝ = 1− V ⎞ mRT H ln⎛ ⎜ 2 V ⎟ ⎝ 1 ⎠
However, for a Carnot cycle the thermal efficiency is also defined as η cycle = 1 −
T C T H
Thus, with algebra we find that V ln⎛ ⎜ 3
⎞ ⎟ ⎝ V 4 ⎠ = 1 ⇒ V V = V V 4 2 3 1 V 2 ⎞ ln⎛ ⎜ V ⎟ ⎝ 1 ⎠
(b) and (c) Process 2-3 is adiabatic; therefore a energy balance in differential form reads: dU = δ Q − δ W
0
where δW = pdV and with assumption (1) dU = mc vdT. Collecting these results and using PV = mRT and c v = R we find that: k − 1 1 d ln T = −d ln V k − 1 Integrating, and assuming k is a constant (assumption (2)) gives:
⎛ T ⎞ ⎛ V ⎞ ln⎜⎜ 3 ⎟⎟ = − ln⎜⎜ 3 ⎟⎟ ⎝ T 2 ⎠ ⎝ V 2 ⎠
k −1
⎛ V ⎞ ⇒ = ⎜⎜ 3 ⎟⎟ T 3 ⎝ V 2 ⎠ T 2
k −1
Finally, using V = mRT / p
⎛ p T ⎞ = ⎜⎜ 2 2 ⎟⎟ T 3 ⎝ p 3 T 3 ⎠
T 2
k − 1
⎛ p ⎞ ⇒ = ⎜⎜ 2 ⎟⎟ T 3 ⎝ p3 ⎠ T 2
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k −1 k
1 10/01/2007
ENGRD 221 – Prof. N. Zabaras
Problem 3: ’ Known: A system undergoes a cycle while receiving QH at T H and discharging QC at ’ T C. QH and QC are with hot and cold reservoirs at T H and TC, respectively. ’
’
Find: (a) Determine an expression for Wcycle in terms of QH, T H, T C and σ. ’ ’ (b) State relationships of T H to TH and T C to TC. (c) Obtain an expression for Wcycle when there are (i) no internal irreversibilities (ii) no irreversibilities.
Schematic and Given Data: Analysis: (a) An energy balance gives Wcycle = QH – QC. ….(1) An entropy balance gives Q Q ΔS cycle = H − C + σ cycle ….(2) T ' H T 'C where σ cycle is the amount of entropy produced within the system. ΔS = 0, since the system undergoes a cycle. Solving Eq.(2) for QC and substituting in Eq. (1) for W cycle, ⎛ T ' ⎞ ….(3) Wcycle = QH ⎜⎜1 − C ⎟⎟⎟ − T 'C σ cycle ⎜⎝ T ' H ⎠⎟ ’
(b) For heat transfer to occur from the hot reservoir to the system, TH ≥ T H. ’ For heat transfer to occur from the system to the cold reservoir, T C ≥TC. (c) If there were no irreversibilities within the system during the cycle, the term ⎛ ⎞ vanishes in Eq.(3) leaving, W = Q ⎜⎜1 − T 'C ⎟⎟ ….(4) cycle H ⎜⎝ T ' H ⎠⎟⎟
σ cycle
External irreversibilities are associated with heat transfer between the reservoirs and the ’ ’ system. If these are also absent, T H = T H and T C =TC. Eq.(4) then becomes, ⎛ T ⎞ Wcycle = QH ⎜⎜1 − C ⎟⎟⎟ which is the maximum theoretical work that can be obtained. ⎜⎝ T H ⎠⎟
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1 10/01/2007
ENGRD 221 – Prof. N. Zabaras
Problem 4 (Compute entropy increase): A cylindrical rod of length L insulated on its lateral surface is initially in contact at one end with a wall at temperature T H and at the other end with
a wall at a lower temperature T C . The temperature within the rod initially varies linearly with position z according to T ( z ) = T H − (
T H − T C
) z . The rod is then insulated on its ends and L eventually comes to a final equilibrium state where the temperature is T f . Evaluate T f in terms of T H and T C and show that the amount of entropy produced is
⎛
T C
⎝
T H − TC
σ = mc ⎜1 + ln T f +
ln TC −
T H
⎞
ln TH ⎟ TH − TC ⎠
where c is the specific heat of the rod. Known: The temperature within the rod initially varies linearly with position z according T − T C to T ( z ) = T H − ( H ) z . The rod is then insulated on its ends and eventually comes to a L final equilibrium state where the temperature is T f . Find: Evaluate T in terms of T H and T C and the amount of entropy produced. Schematic and Given Data:
Analysis: The final temperature can be determined using an energy balance which reduces to give ΔU = Q − W = 0 . Each element of rod dz changes temperature from T(z)
to the final temperature T , and thus contributes to the change in internal energy
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1 10/01/2007
ENGRD 221 – Prof. N. Zabaras L
L
L
ΔU = ∫ du = ∫ c(T f − T ( z)) dm = ∫ c(Tf − T ( z)) ρ Adz 0
0
0
L
T H − T C
0
L
= ∫ c(T f − TH + (
) z) ρ Adz L
⎡ T H − T C z 2 ⎤ = ρ Ac ⎢(T f − TH ) z + ⎥ L 2 ⎦0 ⎣ L
T − T ⎤ ⎡ = ρ AcL ⎢ (T f − T H ) + H C ⎥ 2 ⎦0 ⎣ Since ΔU = Q − W = 0 , we have T f =
T H + T C
. 2 To find the entropy production, an entropy balance reduces to give 2 δ Q ΔS = + σ = σ 1 T
∫
0
Eq6.24 ⇒ dS = dm × c ln L
T f T ( z)
= ρ Ac ln T f
L
⇒ σ = ΔS = ∫ dS = ∫ ρ Ac ln 0
0
T ( z)
T f T ( z)
dz
dz
L L = ρ Ac ∫ ⎡⎣ ln T f − ln T ( z) ⎤⎦ dz = ρ Ac ⎡⎢ L ln Tf − ∫ ln T ( z) dz ⎤⎥ 0 0 ⎣ ⎦
Since T ( z ) = T H − (
T H − T C L
) z , dT = −(
T H − T C L
) dz ⇒ dz = −
L T H − T C
dT
So L
∫
0
ln T ( z ) dT =
∫
TC
T H
ln T ( z)
L − L dT = T H − TC TH − TC
∫
T H
T C
ln T ( z) dT =
⎡ T ln T H TC ln T C ⎤ = L ⎢ H − − 1⎥ − − T T T T H C ⎣ H C ⎦ Substitute into σ = ρ Ac ⎡ L ln T f −
⎢⎣
L
∫
0
ln T ( z) dz ⎤
⎥⎦
We have
⎛
T C
⎝
T H − TC
σ = mc ⎜1 + ln T f +
ln TC −
T H
⎞
ln TH ⎟ TH − TC ⎠
Page 7 of 11
L
T H
[T ln T − T ]T
TH − TC
C
1 10/01/2007
ENGRD 221 – Prof. N. Zabaras
Problem 5 (*): At steady state, an insulated mixing chamber receives two liquid streams 1 and m 2 , of the same substance at temperatures T 1 and T 2 and mass flow rates m
3 . Using an incompressible substance respectively. A single stream exits at T 3 and m model with constant specific heat c, obtain an expression for 1 / m 3 . (a) T 3 in terms of T 1, T 2, and the ratio of mass flow rates m (b) the rate of entropy production per unit of mass exiting the chamber in terms of c, 1 / m 3 . T 1,/T 2, and m Known: An insulated mixing chamber of steady state rece ives liquid streams of the same 1 , T1 and, m 2 ,T2. A single stream exits at m 3 , T3. substance at m Find: a) Evaluate T3 b) Evaluate σ/ m 3 Assumptions: 1. The control volume is at steady state. 2. No heat or work transfer for the control volume. 3. The liquid streams are modeled as incompressible with constant specific heat c and negligible effects of pressure. 1 + m 2 = m 3 . Energy balance equation Analysis: a) At steady state, mass balance reads m results in 1 h1 + m 2 h2 - m 3 h3. Combining with mass balance gives, 0= m
1 (h1 – h2) + m 3 (h2 – h3) 0= m 1 c(T1 – T2) + m 3 c(T2 – T3) => T3 = T2 + ( m 1 / m 3 )(T1 – T2) 0= m
=>
b) Entropy balance reduces at steady state (Equation 6.39 with assumptions of no heat transfer) reduces to: s1 + m s2 - m s3 + σ ; Using the mass balance and rearranging, we obtain: 0= m 1
2
3
1 (s1 – s2) + m 3 (s2 – s3) + σ => 0 = m 1 c ln(T1/T2) + m 3 c ln(T2/T3) + σ 3 => (using equation 6.24 from the text) 0 = m
Solving for σ , we obtain the following: m 3
m 3 σ
⎡ m 1 ⎛T2 ⎟⎞ ⎛⎜ T 3 ⎟⎞⎤ ⎜ ⎢ ⎥ =c ⎢ m 3 ln ⎜⎜⎝ T1 ⎟⎟⎟⎠ + ln ⎝⎜⎜T2 ⎟⎟⎟⎠⎥ ⎣ ⎦
Replacing T3 = T2 + m1 (T1 – T2), we obtain the desired result. m 3
⎡ m ⎛ T ⎞ ⎛ m ⎛ T ⎞ m ⎞⎤ = c ⎢ 1 ln⎜⎜ 2 ⎟⎟ + ln⎜⎜1 + 1 ⎜⎜ 1 ⎟⎟ − 1 ⎟⎟⎥ m 3 ⎝ m 3 ⎝ T 2 ⎠ m 3 ⎠⎦ ⎣ m 3 ⎝ T 1 ⎠
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1 10/01/2007
ENGRD 221 – Prof. N. Zabaras
Problem 6(Use tables): Employing the ideal gas model determine the change in specific entropy between the indicated states, in kJ/(kg K). Solve two ways: Use the appropriate ideal gas table, and a constant specific heat value from Table A-20. (a) air, p1 = 100kPa, T1 = 20 oC, p2 = 100 kPa, T2 = 100 o C
(b) air, p1 = 1bar , T1 = 27 oC, p2 = 3bar , T2 = 377 o C (c) carbon dioxide,
1
= 150kPa, T1 = 30 oC, p2 = 300 kPa, T2 = 300 o C
(d) carbon monoxide, T1 = 300 K , v1 = 1.1m3 / kg , T2 = 500 K , v2 = 0.75 m3 / kg (e) nitrogen, p1 = 2 MPa, T1 = 800 K , p2 = 1MPa, T2 = 300 K case
Ideal Gas Table
Δ s = s 0 (T2 ) − s 0 (T1 ) − R ln (a)
Constant Specific Heat T p Δ s = c p ln 2 − R ln 2 T1 p1
p2 p1
With s 0 data from Table A22 Δ s = 1.92119 − 1.678298 − 0 = 0.24289kJ / kgK
With c p at 333K from Table A20
Δ s = 1.007 ln
373
− 0 = 0.2431kJ / kgK 293 With c p at 475K from Table A20
(b)
With s 0 data from Table A22 8.314 2 Δ s = 2.49364 − 1.70203 − ln = 0.47632kJ / kgK Δ s = 1.0245 ln 650 − 8.314 ln 3 = 0.47684kJ / kg 28.97 1 300 28.97 1
(c)
With s 0 data from Table A23 and M from Table A1 300 241.033 − 214.284 − 8.314ln 150 = 0.4769kJ / kg Δ s = 44.01 With pv=RT p2 T2 V 1 500 1.1 = = = 2.444 p1 T1 V 2 300 0.75
(d)
(e)
With c p at 438K from Table A20
Δ s = 0.9686 ln
573 303
−
8.314 44.01
ln
300 150
= 0.4862kJ / k
With c p at 400K from Table A20
Δ s = 1.047 ln
500 300
−
8.314 28.01
ln 2.444 = 0.2696kJ / k
With s 0 data from Table A23 and M from Table A1 212.719 − 197.723 − 8.314ln 2.444 Δ s = = 0.2701kJ / k 28.01 With c p at 550K from Table A20 With s 0 data from Table A23 and M from Table A1 1 300 8.314 1 191.682 − 220.907 − 8.314ln ln = −0.8389kJ / kgK Δ s = 1.065ln − 2 800 28.02 2 Δ s = = −0.8373kJ / kgK 28.02
o
Problem 7 (Reversible process): One kilogram of water initially at 160 C, 1.5 bar undergoes an isothermal, internally reversible compression process to the saturated liquid state. Determine the work and heat transfer, in each in kJ. Sketch the process on p-v and T-s coordinates. Associate the work and heat transfer with areas on these diagrams.
Known: One kg of water undergoes an isothermal process between two specified states.
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1 10/01/2007
ENGRD 221 – Prof. N. Zabaras Find: Determine the heat transfer and the work. Sketch the process on p-v and T-s coordinates. Schematic & Given Data:
Assumptions: 1. As shown in the accompanying figure, the system consists of 1 kg of water. 2. The compression takes place isothermally and without internal irreversibilities. 3. There is no change in kinetic and potential energy between the end states. 2
Analysis: Using assumption 2, Eq.(6.25) becomes Q =
∫ TdS
=
mT ( s2 − s1 ) . Then with
1
data from Tables A-2 and A-4, Q = 1kg (433 K )(1.9427 − 7.4665) kJ / kg. K = −2391.8 kJ The magnitude of the heat transfer is represented by area 1-2-a-b-1 on the T-s diagram. The energy balance reduces to give W = Q – m(u2 – u1). With data from Tables A-2 and A-4, W = -2391.8 – 1 kg(674.86 – 2595.2) kJ/kg = -471.5 kJ. 2
Alternaltely, W
=
∫ PdV . The magnitude of the work is represented by the area 1-2-a-b1
1 on the p-v diagram.
Problem 8 (Entropy change): For each of the following systems, specify whether the entropy change during the indicated process is positive, negative, zero, or indeterminate. (a) One kilogram of water vapor undergoing an adiabatic compression process (b) Two pounds mass of nitrogen heated in an internally reversible process (c) One kilogram of Refrigerant 134a undergoing an adiabatic process during which it is stirred by a paddle wheel. (d) One-pound mass of carbon dioxide cooled isothermally. (e) Two pounds mass of oxygen modeled as an ideal gas undergoing a constant pressure process to a higher temperature (f) Two kilograms of argon modeled as an ideal gas undergoing an isothermal process to a lower pressure.
(a)
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ENGRD 221 – Prof. N. Zabaras
ΔS = ∫
2
1
δ Q T
+ σ = σ ≥ 0 . It can be + or 0 depending on the nature of the process.
0
Indeterminate! (b) Nitrogen heated internally reversibly. 2 δQ 2 δ Q ΔS = + σ = > 0 . Increase! 1 T 1 T
∫
∫
(c) R134a stirred adiabatically 2 δ Q ΔS = + σ = σ > 0 1 T
∫
0
Stir would increase entropy. (d) CO2 cooled isothermally 2 δ Q ΔS = + σ . Indeterminate! 1 T >= 0
∫
<0
(e) Ideal gas undergoing a constant pressure process T2>T1 Using Eq. 6.21(a) p Δ s = s 0 (T2 ) − s 0 (T1 ) − R ln 2 = s 0 (T2 ) − s 0 (T 1 ) > 0 Increase! p1 (f) Constant temperature p2
0 Increase! p1 p1
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