PROBLEM 3.140

SOLUTIONS HW #5C - CHAPTER 3 140. 141, 142, 143, 144, 145, 146

PROBLEM 3.141

PROBLEM 3.142

PROBLEM 3.142 (CONTINUED)

PROBLEM 3.143

One pound mass of air undergoes a cycle consisting of the following processes:

Process 1-2: constant-pressure expansion with p = 20 lbf/in.2 from T1 = 500

oR to v2 = 1.4 v1

Process 2-3: adiabatic compression to v3 = v1 and T3 = 820oR

Process 3-1: constant-volume process

Sketch the cycle on a carefully-labeled p- v diagram. Assuming ideal gas behavior, determine

the energy transfers by heat and work for each process, in Btu.

KNOWN: Air undergoes a cycle consisting of three processes.

FIND: Sketch the cycle on a p- v diagram and calculate the energy transfers by heat and work

for each process.

SCHEMATIC AND GIVEN DATA: The following data are given for each process:

Process 1-2: constant-pressure expansion with p = 20 lbf/in.2 from T1 = 500

oR to v2 = 1.4 v1

Process 2-3: adiabatic compression to v3 = v1 and T3 = 820oR

Process 3-1: constant-volume process

ENGINEERING MODEL: (1) the air is a closed system. (2) The

air behaves as an ideal gas. (3) Kinetic and potential energy effects

are negligible.

ANALYSIS: First, fix each principal state. For Process 1-2, the pressure is constant, so

T2 = (v2/ v1)T1 = (1.4)(500) = 700oR. The processes are shown on the accompanying p- v

diagram:

Process 1-2 (constant pressure) → W12 =

= p(V2 – V1) = mR(T2 –T1)

W12 = (1 lb)(

(700 – 500)

=13.71 Btu (out)

Air

m = 1 lb

p

v

(1) (2)

(3)

T1 = 500oR

. T2 = 700

oR

T3 = 820oR

.

.

adiabatic process

PROBLEM 1.143 (CONTINUED)

Energy balance: ΔKE + ΔPE + ΔU = Q12 – W12 . With data from Table A-22

Q12 = m(u2 – u1) + W12 = (1 lb)(119.58 – 85.20)Btu/lb + (13.71 Btu) = 48.09 Btu

Process 2-3: (adiabatic) Q23 = 0

W23 = - m(u3 – u2) = - (1 lb)(140.47 – 119.58)Btu/lb = - 20.89 Btu (in)

Process 3-1: (constant volume) W31 = 0

Q31 = m(u1 – u3) = (1 lb)(85.20 – 140.47)Btu/lb = -55.27 Btu (out)

In Summary

Wcycle = W12 + W23 + W31 = 13.71 + (-20.89) + 0 = - 7.18 Btu

Qcycle = Q12 + Q23 + Q31 = 48.09 + 0 + (-55.27) = -7.18 Btu

Wcycle = Qcycle (as expected)

Note: The cycle is a refrigeration/heat pump cycle.

PROBLEM 3.144

Process 1-2: Constant-temperature expansion at 600 K

from p1 = 0.5 MPa to p2 = 0.4 MPa.

Process 2-3: Polytropic expansion with n = k to p3 =

0.3 MPa.

Process 3-4: Constant-pressure compression to V4 = V3.

Process 4-1: Constant-volume heating.

PROBLEM 3.144 (CONTINUED)

.

PROBLEM 3.145

PROBLEM 3.145 (CONTINUED)

PROBLEM 3.146

A system consisting of 2 kg of carbon dioxide (CO2) gas initially at p1 = 1 bar, T1 = 300 K,

undergoes a power cycle with the following processes:

Process 1-2: constant volume to p2 = 4 bar

Process 2-3: expansion with pv1.28 = constant

Process 3-1: constant-pressure compression

Assuming the ideal gas model and neglecting kinetic and potential energy effects,

(a) sketch the cycle on a p-v diagram and calculate the thermal efficiency.

(b) plot thermal efficiency versus p2/p1 ranging from 1.05 to 4.

KNOWN: A system consisting of carbon dioxide gas undergoes a power cycle made up of three

processes.

FIND: (a) Sketch the cycle on a p-v diagram and calculate the thermal efficiency of the cycle.

(b) Plot the thermal efficiency versus p2/p1 ranging from 1.05 to 4.

SCHEMATIC AND GIVEN DATA: The following data are known for each process

Process 1-2: constant volume from p1 = 1 bar, T1 = 300 K, to p2 = 4 bar

Process 2-3: expansion with pv1.28 = constant

Process 3-1: constant-pressure compression

ENGINEERING MODEL: (1) The CO2 is a closed system. (2) The CO2 behaves as an

ideal gas. (3) Process 2-3 is polytropic with pv1.28 = constant. (4) Kinetic and potential energy

effects are negligible.

ANALYSIS: (a) First, for Process 1-2: V2 = V1 → T2 = (p2/p1)T1 = (4/1)(300K) = 1200 K.

Further

V2 = V1 =

=

= 1.134 m

3/kg

Using the given p-v relation for Process 2-3 with v2 = V2/m = 0.567 m3/kg, and noting that p3 =

p1 =1 bar

v3 =

v2 =

(0.567 m

3/kg) = 1.675 m

3/kg

and

T3 =

=

= 887K

The p-v diagram is

CO2

m = 2 kg

PROBLEM 1.143 (CONTINUED) – PAGE 2

The thermal efficiency is η = Wcycle/Qin, where Wcycle is the net work of the cycle and Qin is the

total heat transfer into the system during the cycle. Next, each process is analyzed.

Process 1-2: W12 = 0 (constant volume)

ΔKE +ΔPE + ΔU = Q12 – W12 → Q12 = m(u2 – u1)

With data from Table A-23

Q12 = (2 kg)

= 1678 kJ (in)

Process 2-3: The work is determined using Eq. 2.17 and the given p-v relation

W23 =

=

= m

=

=

= 422.4 kJ (out)

Using the energy balance with ΔKE =ΔPE = 0: m(u3 – u2) = Q23 – W23

Q23 = m

+ W23 = (2 kg)

+ (422.4 kJ) = -237.8 kJ (out)

Process 3-1: Using Eq. 2.17

W31 =

= mp3(v1 – v3) = mR(T1 – T3) = (2 kg)

= -221.8 kJ (in)

p

v

T2 = 1200K

T2 = 887K

T1 = 300K

(2)

(1) (3) .

.

.

pv1.28 = constant

Ideal gas