Question

In a absorption type refrigerator, the heat is supplied to NH${}_3$ generator by condensing steam at 120${}^{\circ }$C. The temperature to be maintained in the refrigerator is -4${}^{\circ }$C. The temperature of the atmosphere is 30${}^{\circ }$C. The refrigerator load is 20 tons and actual COP is 80% of maximum COP. Heat rejected in condenser is 35 kW. What is the heat rejected in absorber? Neglecting pump work.

• A. 35 kW
• B. 83.27 kW
• C. 48.27 kW
• D. 70 kW

Answer 1

The correct option is B 83.27 kW

$\text{Maximum COP}=CO{P}_{max}=\frac{T_e(T_g-T_C)}{T_g(T_c-T_e)}$

$T_e=\text{Evaporator temperature = -4+273=269K}$

$T_g=\text{Generator temperature = 120+273=393K}$

$T_c=\text{Condensation temperature = 30+273=303K}$

$CO{P}_{max}=\frac{269(393-303)}{393(303-269)}=1.812$

$CO{P}_{actual}=0.8\times CO{P}_{max}$

$=0.8\times 1.812=1.45$

$CO{P}_{max}=\frac{Q_e}{Q_g}=\frac{\text{Heat absorbed in evaporator}}{\text{Heat absorbed in generator}}$

$1.45=\frac{20\times 3.5}{Q_g}$

$Q_g=48.27kW$

From energy balance,

${\text{Where, Q}}_C=\text{Heat rejected in condenser}$

${\text{Where, Q}}_A=\text{Heat rejected in absorber}$

$Q_g+Q_e=Q_C+Q_A$

$48.27+20\times 3.5=35+Q_A$

$Q_A=83.27kW$