Goal Solve for the performance coefficient of a refrigerator using a five-step p

Question

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.

Solution

(1) Fill in the State Table (all pressures in Pascals, all volumes in cubic meters, all temperatures in K).

Pressure Volume Temp

a ____ ___ ____

b ____ ___ ____

c  ____ ___ ____

(2) Fill in the Process Table (all entries in Joules).

Work Heat dU

a->b  ____ ___ ____

b->c  ____ ___ ____

c->a  ____ ___ ____

(3) Find the Totals:

Work=____J

Heat= ____ J

dU= _____J

(4) Find the heat input (from "cold reservoir") and the heat output (to "hot reservoir"

Q-hot=____J

Q-cold=____J

(5) Find the performance coefficient of the refrigerator:

coefficent=_____

Explanation / Answer

Given: n = 6 ; Ta = 300 K ; Pa = 105 Pa Cv = 3/2*R ; Cp = Cv + R = 5/2*R

1. Using gas equation Pa*Va = n*R*Ta   hence Va = 6*8.314*300/105 = 0.15 m3

Now process a-->b is isobaric so Pb = Pa = 105 Pa    and we are given Vb = 2*Va = 0.3 m3

So Va/Ta = Vb/Tb   i,e Tb = 2*300 = 600 K

Now process b-->c is isothermal so Tc = Tb = 600 K also since process c-->a isochoric hence Vc = Va= 0.15 m3

therefore Pb*Vb = Pc*Vc   i,e Pc = 2*105 Pa

2. For the process a-->b (Isobaric)

Wab = Pa*(Vb - Va) = 105*0.15 = 1.5*104 J ;   Hab = n*Cp*T = 6*5/2*8.314*(600-300) = 37413 J

Uab = n*Cv*T = 6*3/2*8.314*(600-300) = 22447.8 J

For the process b--->c (isothermal)

Wbc  = n*R*Tc ln(Vc/Vb) = 6*8.314*600*ln(1/2) = -20746.2 J ;   Ubc = 0 ; Hbc = Wbc  = -20746.2 J

For the process c-->a (Isochoric)

Wca = 0 ; Hca  =   Uca  = n*Cv*(Ta - Tc) = -6*3/2*8.314*300 = -22447.8 J

3. Total Work Wtot = Wab + Wbc + Wca = 15000 -20746.2 = -5746.2 J

Total heat Htot = Hab + Hbc + Hca  = 37413 - 20746.2 - 22447.8 = -5781 J

Total change in internal energy Utot = Uab + Ubc + Uca  = 22447.8-22447.8 = 0

4. Heat Input Q-cold = 37413 J

Heat output Q-hot = 43194 J

5. Performance coeffiecient of refrigerator (COPR) = Q-cold/(Q-hot - Q-cold) = 37413/5781 = 6.47

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

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