In a vapor-compression refrigeration cycle, ammonia exits the evaporator as satu

Question

In a vapor-compression refrigeration cycle, ammonia exits the evaporator as saturated vapor at -22 degree C. There are irreversibilities in the compressor. The refrigerant enters the condenser at 16 bar and 160 degree C, and saturated liquid exits at 16 bar. There is no significant heat transfer between the compressor and its surroundings, and the refrigerant passes through the evaporator with a negligible change in pressure. Calculate the coefficient of performance, beta, and the isentropic compressor efficiency, defined as: eta_c = (dot W_c/dot m)_ideal/(dot W_c/dot m)_actual

Explanation / Answer

Use Refrigration 134 a cycle

State 1 at

T1= -22 C and x= 1 bar

h1= 214.70 kJ/ kg

s1= 0.9351 kJ/kgK

state 2 compressor exit

P2= 16bar

s2=s1=0.9351 kJ/kgK

h2= 275.25 kJ/kg

T2 = 58 C

State 3 condeser exit

P3= 16 bar x3= 0

h3 = 133.00 kJ/kg ( liquid state)

State 4

T4 = T1 = -22 C

h4 = h3

COP= (h1-h4)/ (h2-h1)

= ( 214.7- 133)/(275.25-214.7)

= 1.349

Here the efficiency of the isentropic compression is 90 percent from the ideal cycle

.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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