A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four processes. The sum of all heat transferred during a cycle is – 340 kJ. The system completes 200 cycles per min. Complete the following table showing the method for each item, and compute the net rate of work output in kW.

Process Q (kJ/min) W (kJ/min) ΔE (kJ/min)

1—2 0 4340 —

2—3 42000 0 —

3—4 -4200 — -73200

4—1 — — —

Question

A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four processes. The sum of all heat transferred during a cycle is – 340 kJ. The system completes 200 cycles per min. Complete the following table showing the method for each item, and compute the net rate of work output in kW.

Process	Q (kJ/min)	W (kJ/min)	ΔE (kJ/min)
1—2	0	4340	—
2—3	42000	0	—
3—4	-4200	—	-73200
4—1	—	—	—

Process	Q (kJ/min)	W (kJ/min)	ΔE (kJ/min)
1—2	0	4340	—
2—3	42000	0	—
3—4	-4200	—	-73200
4—1	—	—	—

Accepted Answer

Sum of all heat transferred during the cycle = – 340 kJ.

Number of cycles completed by the system = 200 cycles/min.

Process 1—2 :

Q = Δ E + W

0 = Δ E + 4340

∴ Δ E = – 4340 kJ/min.

Process 2—3 :

Q = Δ E + W

42000 = Δ E + 0

Δ E = 42000 kJ/min.

Process 3—4 :

Q = Δ E + W

– 4200 = – 73200 + W

∴ W = 69000 kJ/min.

Process 4—1 :

[latex]\underset{ ext { cycle }}{\Sigma Q}=-340[/latex] kJ

The system completes 200 cycles/min

[latex]\because \quad Q_{1-2}=Q_{2-3}+Q_{3-4}+Q_{4-1}=-340 imes 200=-68000[/latex] kJ/min

or [latex]0+42000+(-4200)+Q_{4-1}=-68000[/latex]

[latex]Q_{4-1}=-105800[/latex] kJ/min.

Now, ∫ dE = 0, since cyclic integral of any property is zero.

[latex]\Delta E_{1-2}+\Delta E_{2-3}+\Delta E_{3-4}+\Delta E_{4-1}=0[/latex]

[latex]-4340+42000+(-73200)+\Delta E_{4-1}=0[/latex]

∴ [latex]\Delta E_{4-1}=35540[/latex] kJ/min.

∴ [latex]W_{4-1}=Q_{4-1}-\Delta E_{4-1}[/latex]

= – 105800 – 35540 = – 141340 kJ/min

The completed table is given below :

Process	Q (kJ/min)	W (kJ/min)	ΔE (kJ/min)
1—2	0	4340	-4340
2—3	42000	0	42000
3—4	-4200	69000	-73200
4—1	-105800	-141340	35540

Since [latex]\underset{ ext { cycle }}{\Sigma Q}=\underset{ ext { cycle }}{\Sigma W}[/latex]

Rate of work output = – 68000 kJ/min [latex]=-\frac{68000}{60}[/latex] kJ/s or kW

= 1133.33 kW.

Question 4.12: A fluid system, contained in a piston and cylinder machine, ...

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