Question #257810

A refrigeration system of 10.5 tonnes capacity at an evaporator tem￾perature of – 12°C and a condenser temperature of 27°C is needed in a food storage locker. The refrigerant ammonia is sub-cooled by 6°C before entering the expansion valve. The vapour is 0.95 dry as it leaves the evaporator coil. The compression in the compressor is of adiabatic type. Using p-h chart find : (i) Condition of volume at outlet of the compressor (ii) Condition of vapour at entrance to evaporator (iii) C.O.P. (iv) Power required, in kW. Neglect valve throttling and clearance effect

Expert's answer

TE=TsatPe=12,TC=TsatPc=28_TE=T_{satPe}=−12℃,T_C=T_{satPc}=28℃


Capacity=10 TR, x1=0.95


h1=hf12+x1×hfg12h_1=h_{f−12℃}+x_1×h_{fg−12℃}


hfg12=hghf=1447.74144.929=1302.811 kJkgh_{fg−12℃}=h_g−h_f=1447.74−144.929 =1302.811~\frac{kJ} {kg}


(from psychrometric chart booklet for N3)


h1=144.929+0.95×1302.811=1382.6 kJkgh_1=144.929+0.95×1302.811 =1382.6~\frac{kJ} {kg}


Condition of refrigerant at outlet of compressor


T2=67°C,h2=1598 kJkg,v2=0.15 m3kg,p2=10.5 barT_2=67°C,h_2=1598~\frac{kJ}{kg},v_2=0.15~\frac{m^3}{kg},p_2=10.5~ bar

Condition of refrigerant at inlet to evaporator


T4=12°C,h4=310 kJkg,p4=2.65 barT_4=−12°C,h_4=310~\frac{kJ}{kg},p_4=2.65 ~bar

COP of given system = R.E.Wc=h1h4h2h1=4.98\frac{R.E.}{W_c}=\frac{h_1−h_4}{h_2−h_1}=4.98

Also, 

Q˙a=m˙R×(h1h4)Q̇_a=ṁ_R×(h_1−h_4)

For 10 TR capacity, 10x3.5=m˙R×(1382.6310)10x3.5=ṁ_R×(1382.6−310)

mass flow rate, m˙R=0.0326 kgsecṁ_R=0.0326~\frac{kg}{sec}

Now, power required = m˙R×(h2h1)=7.0287 kW.ṁ_R×(h_2−h_1)=7.0287 ~kW.


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