Answer to Question #257808 in Molecular Physics | Thermodynamics for Mwalimu

Question

3. A refrigerator operating on standard vapour compression cycle has a co￾efficiency performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated

liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are

62.55 kJ/kg and 201.45 kJ/kg respectively. The saturated refrigerant vapour leaving evaporator

has an enthalpy of 187.53 kJ/kg. Find the refrigerant temperature at compressor discharge. The

cp of refrigerant vapour may be taken to be 0.6155 kJ/kg°C

Answer 1

Solution;

Given;

$C.O.P=6.5$

$W=50kW$

$h'_3=201.45kJ/kg$

$h_{f_4}=h_1=62.55kJ/kg$

$h_2=187.53kJ/kg$

$C_p=0.6155kJ/kg°c$

$t=35°c$

To find the temperature $t_3$ at compressor discharge;

Refrigerating capacity;

=W×C.O.P

=$50×6.5$ =$325kW$

Heat extracted per kg of refrigerant;

=$h_2-h_1$

=$187.53-62.55$ =$124.98kJ/kg$

Refrigerant flow rate;

=$\frac{325}{124.98}$ =$2.6kg/s$

Compressor power ;

$=50kW$

Therefore;

Heat input per kg;

=$\frac{50}{2.6}=19.228kJ/kg$

Enthalpy of vapour after compression;

=$h_2+19.228$

=$187.53+19.228$ =$206.758kJ/kg$

Superheat;

=$206.758-h'_3$ =$206.758-201.45$

=$5.308kJ/kg$

But;

$5.308=1×C_p(t_3-t)$

$5.308=1×0.6155(t_3-35)$

Hence;

$t_3=\frac{5.308}{0.6155}+35$ =$43.62°c$

Ans;

43.62°c