Complete Analysis of a Refrigerator Goal Solve for the performance coefficient o

Question

Complete Analysis of a Refrigerator Goal Solve for the performance coefficient of a refrigerator using a five-step process the includes: 1. Making a state table. 2. Making a process table. 3. Calculating the totals for Work, Heat, and Internal-Energy-Change. 4. Identifying the heat input (cold reservoir) and output (hot reservoir). 5. Calculating the performance coefficient of the refrigerator. Problem Shown in the figure to the right is a cyclic process undergone by a refrigerator. Your refrigerator shall use 3.0 moles of helium gas 5.0. (monatomic). During the process a->b, the volume increases by a factor of T 300 K P 100,000 Pa Refrigerator Cycle Solution (1) Fill in the State Table (all pressures in Pascals, all volumes in cubic meters, all temperatures in K). Volume Temperature Pressure a 10000074826 X 300 b 100000 37413X 1500 c 500000 74826 X 1500 Work Heat du a->b 299304 74826 44895.6 b->c-12047 |X-12047-??? (2) Fill in the Process Table (all entries in Joules). ca 44895.6448956 Work 1.624 Heat = 1.6304 XJ du-0 (3) Find the Totals: (4) Find the heat input (from "cold reservoir) and Q-hot the heat output (to "hot reservoir") Q-cold - 74826 (5) Find the performance coefficient of the refrigerator: coefficient . 485

Explanation / Answer

Pa = 100000 Pa
Ta = 300 K
Using ideal gas equation, PaVa = nRTa,
Va = nRTa/Pa
= [3 x 8.314 x 300] / 100000
= 0.074826 m3.

Given that Vb = 5Va
= 5 x 0.074826 = 0.37413 m3.
Pa = Pb = 100000 Pa
Using idela gas equation, PbVb = nRTb
Tb = PbVb/nR
= [100000 x 0.37413] / [3 x 8.314]
= 1500 K

Since the process b to c is isothermal, Ta = Tb = 1500 K
From the figure, Va = Vc = 0.074826 m3.
Using ideal gas equation, Pc = nRTc/Vc
= [3 x 8.314 x 1500] / 0.074826
= 500000 Pa

a-b process (isobaric)
Work done during a to b is the area under a-b curve
W = PdV
= 100000 x [0.37413 - 0.074826]
= 29930.4 J
dQ = nCpdT
For monoatomic gas, Cp = 5/2R
= 3 x 5/2 R x [1500-300]
= 74826 J
dU = nCvdT
= 3 x 3/2R x [1500-300]
= 44895.6 J

b-c process (isothermal)
Work done = nRT ln(Vc/Vb)
= 3 x 8.314 x 1500 x ln(1/5)
= - 60213.9 J
dU = 0 for isothermal process since dT = 0
dQ = work done = - 60213.9 J

c-a process (isochoric)
Work done = 0 since dV = 0
dQ = nCvdT
= 3 x 3/2 R x [300-1500]
= - 44895.6 J
dQ = dU = - 44895.6 J

Total work = 29930.4 J - 60213.9 J + 0
= -30283.5 J
Total heat, dQ = 74826 - 60213.9 - 44895.6
= - 30283.5 J
dU = 44895.6 - 44895.6 = 0

COP = Q/W
= 1

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.