The thermal efficiency of the diesel engine.

Question

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Answer 1

Question

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

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Answer 2

Question

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

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Answer 3

Question

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

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Answer 4

Question

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

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Answer 5

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

Answer 6

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

Answer 7

Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

Compression ratio (r) is 20.

Initial temperature of the cycle (T1) is 20°C.

Initial pressure of the cycle (P1) is 95 kPa.

Maximum temperature of the cycle (T3) is 2200 K.

Calculation:

Draw the P−v diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 53P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

R=0.287 kJ/kg⋅K

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of 293 K .

u1=209.06 kJ/kgvr1=659.2

Calculate the relative specific volume at state 2(vr2).

vr2=V2V1vr1

vr2=1rvr1=120(659.2)=32.96

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of 32.96.

T2=912.7 Kh2=947.17 kJ/kg

Calculate the cut-off ratio (rc).

P3v3T3=P2v2T2

T3=v3v2T2=(rc)T2

rc=T3T2=2200 K912.7 K=1.854

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of 2200 K.

u3=1872.4 kJ/kgh3=2503.2 kJ/kg

Calculate the heat addition in the constant pressure heat addition process 2-3(qin).

qin=h3−h2=2503.2 kJ/kg−941.17 kJ/kg=1556 kJ/kg⋅K

Calculate the temperature at state 4(T4).

T4=T3(V3V4)n−1=T3(2.410V2V4)n−1=T3(2.410r)n−1

=(2200 K)(2.41020)1.35−1=1049 K

Calculate the work output for the polytrophic process 3-4(w34,out).

w34,out=R(T4−T3)1−n=(0.287 kJ/kg⋅K)(1049 K−2200 K)1−1.35=943.8 kJ/kg

Calculate the heat rejected for the polytrophic process 3-4(q34,out).

q34,out=−w34,out−(u4−u3)=−(943.8 kJ/kg)−(801.13 kJ/kg−1872.4 kJ/kg)=127.5 kJ/kg

Calculate the heat rejected by the constant volume heat rejection process 4-1(q41,out).

q41,out=u4−u1=801.13 kJ/kg−209.06 kJ/kg =592.1 kJ/kg

Calculate the total heat rejected by the cycle (qout).

qout=q34,out+q41,out=127.5 kJ/kg +592.1 kJ/kg =719.6 kJ/kg

Calculate the thermal efficiency of the diesel engine (ηth).

ηth=1−qoutqin=1−719.6 kJ/kg 1556 kJ/kg⋅K=0.5375=53.75%

Thus, the thermal efficiency of the diesel engine is 63.5%.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The mean effective pressure.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

Answer 13

b)

To determine

The mean effective pressure.

Answer 14

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific volume at state 1(v1).

v1=RT1P1=(0.287 kPa⋅m3/kg⋅K)(293 K)95 kPa=0.885 m3/kg

Calculate the mean effective pressure (MEP).

MEP=wnet,outv1−v2=qin−qoutv1(1−1/r)=1556 kJ/kg⋅K−719.6 kJ/kg⋅K(0.885 m3/kg)(1−1/20)

=836.4 kJ/kg⋅K(0.885 m3/kg)(1−1/20)(kPa⋅m3kJ)=995 kPa

Thus, the mean effective pressure is 933 kPa.

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 9 - Prob. 1P Ch. 9 - What is the difference between air-standard...Ch. 9 - Prob. 3P Ch. 9 - Prob. 4P Ch. 9 - Prob. 5P Ch. 9 - Prob. 6P Ch. 9 - Prob. 7P Ch. 9 - Prob. 8P Ch. 9 - Prob. 9P Ch. 9 - Prob. 10P

The thermal efficiency of the diesel engine.

The thermal efficiency of the diesel engine.

Explanation of Solution

The mean effective pressure.

Explanation of Solution

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