Question

An absorption refrigeration system having evaporator temperature of -5°C and ambient temperature of 25°C. The heat is supplied to generator from steam at 0.1 MPa, 0.95 quality. The saturated temperature of steam at 0.1 MPa is 100°C and $h_{fg}=2000kJ/kg.$ What is the steam flow rate required if the actual COP is 80% of maximum COP and the refrigerator effect is 60 kW?

• A. 0.041 kg/s
• B. 0.01 kg/s
• C. 0.022 kg/s
• D. 0.06 kg/s

Answer 1

The correct option is C 0.022 kg/s

$T_E=268K$

$T_a=273+25=298K$

$T_C=373K$

${\left(COP\right)}_{max}=\left(\frac{T_E}{T_d-T_E}\right)\times \left(\frac{T_G-T_d}{T_G}\right)=\left(\frac{268}{298-268}\right)\times \left(\frac{373-298}{373}\right)=1.796$

${\left(COP\right)}_a=$ Actual COP = 0.8 (COP) max
= 1.437

$1.437=\frac{RE}{W_{in}}=\frac{60}{Q_G}$

$Q_G=41.75kW$

$m_s\times 0.95\times h_{fg}=Q_G=41.75$

$m_s=0.022kg/s$