Question

The following data is for an absorption refrigeration system:
Temperature of evaporator = -15°C
Ambient temperature = 33°C
Actual COP = 35% of maximum COP
Load = 25 tonnes
Heat is supplied to the generator by condensing steam at 0.3 MPa, 80% quality.
What is the flow rate of the steam required for this purpose?

Take
$T_{sat}={130}^{\circ }C,h_{fg}=2250kJ/kg,at0.3MPa.$

• A. 5.88 kg/min
• B. 5.14 kg/min
• C. 7.81 kg/min
• D. 6.44 kg/min

Answer 1

The correct option is D 6.44 kg/min

$T_1\text{= generator temperature}$

= 130 + 273 = 403 K

$T_2\text{= Condenser and absorber temperature = absorber temperature}$

= 33 + 273 = 306 K

$T_R\text{= Evaporator temperature = -15 + 273 = 258 K}$

$(COP{)}_{max}=\left(\frac{T_1-T_2}{T_2-T_R}\right)\frac{T_R}{T_1}$

$=\left(\frac{403-306}{306-258}\right)\times \frac{258}{403}=1.2973$

$\Rightarrow (COP{)}_{actual}=0.35\times 1.2973=0.4528$

$and(COP{)}_{act}=\frac{Q_c}{Q_g}$

$Q_g=\frac{Q_c}{(COP{)}_{actual}}$

$\Rightarrow \frac{25\times 3.5\times 60\times 60}{0.4528\times 3600}=193.24kW$

Heat transferred by 1 kg of steam on condensation

$=(h_f+x{h}_{fg})=h_f=x{h}_{fg}$

= 0.80 × 2250

Steam flow rate

$=\frac{193.24}{0.80\times 2250}=0.1073kg/s=6.44kg/min$