A food storage locker has a capacity of 12 TR and operates between the evaporation temperature of -8$^{\circ}$C and condensation temperature of 30$^{\circ}$C.

The refrigerant R-12 is sub-cooled by 5˚C before throttling and the vapour is superheated by 2°C before leaving the evaporator. Assuming a twin cylinder single acting compressor running at 1000 RPM with L:D ratio of 1:5. Calculate:

i. COP of system

ii. Specific power required

iii. Diameter

iv. Stroke of compressor having volumetric efficiency of 0.945.

Draw P-h and T-s diagram.

Subject : Refrigeration and Air Conditioning

Topic : Vapour Compression Refrigeration System

Difficulty : Medium


Q = 12 TR ;$T_1’=-8°C = - 8 + 273 = 265 K ;T_2’= 30° C = 30 + 273 =303 K ;T_3’-T_3=5° C;$
$T_1= -2° C = -2+273 = 271 K ;h_{f1}=28.72kJ/kg ;h_{f3}’=64.59 kJ/kg ; $
$h_1’= 184.07 kJ/kg ;h_2’= 199.62 kJ/kg ;s_{f1}= 0.1149 kJ/kg K ;s_{f3}=0.2400 kJ/kg K ;$
$s_1’= 0.7007 kJ/kg K ;s_2’=0.6853 kJ/kg K ;v_1’= 0.079 m^3/kg;v_2’= 0.0235 m^3/kg ;$
$c_{pl}= 1.235 kJ/kg K ;c_{pv}= 0.733 kJ/kg K$
The T-s and p-h diagrams are shown in (a) and (b) respectively,
1. Coefficient of performance
First of all, let us find the temperature of superheated vapour at point 2 $(T_2)$.
We know that entropy at point 1,


$=0.7007+2.3\times0.733\log\frac{271}{265}=0.7171$ .............(i)

And entropy at point 2,


$=0.6853+2.3\times0.733\log\frac{T_2}{303}=0.7171$ .............(ii)

Since the entropy at point 1 is equal to entropy at point 2,
therefore equating equations (i) and (ii),
$0.7171=0.6853+1.686\log\frac{T_2}{303}$ or $\log\frac{T_2}{303}=\frac{0.7171-0.6853}{0.6853}=0.0188$

$\frac{T_2}{303}=1.0444$ .......(Taking antilog of 0.0188)
Therefore $T_2=1.0444\times303=316.4 K$

We know that enthalpy at point 1,

$=184.07+0.733(271-265)=188.47 kJ/kg$
Enthalpy at point 2,

$=199.62+0.7333(316.4-303)=209.44 kJ/kg$

and enthalpy of liquid refrigerant at point 3,
$h_{f3}=h_{f3}’+c_{pl}(T_3’-T_3)$ $=64.59-1.235\times 5=58.42 kJ/kg$
Therefore C.O.P
$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{188.47-58.42}{209.44-188.47}=\frac{130.05}{20.97}=6.2$ (Ans)

2. Theoretical power per tonne of refrigeration
We know that the heat extracted or refrigerating effect per kg of the refrigerant, $R_E=h_1-h_{f3}=188.47-58.42=130.05 kJ/kg$

and the refrigerating capacity of the system
Q=12 TR=$12\times210=2520 kJ/min$ .......(Given)
Therefore Mass flow of the refrigerant,
$m_R=\frac{Q}{R_E}=\frac{2520}{130.05}=19.4 kg/min$

Work done during compression of the refrigerant
$m_r(h_2-h_1=19.4(209.44-188.47)=406.82 kJ/min$
Therefore Theoretical power per tonne of refrigeration
$=\frac{406.82}{60\times 12}=0.565 kW/TR$ (Ans)

3. Bore and stroke of compressor
Let D = Bore of compressor
L = Stroke of compressor = 1.5 D, and ........(given)
N = Speed of compressor = 1000 r.p.m. .........(given)
First of all, let us find the specific volume at suction to the compressor, i.e. at point 1. Applying Charles' law ,
$\frac{v_1}{T_1}=\frac{v_1’}{T_1’} or v_1=v_1’\times\frac{T_1}{T_1’}=0.0790\times\frac{271}{265}=0.081 m^3/kg$

(a)When there is no clearance
We know that theoretical suction volume or piston displacement per minute
$=m_R\times v_1=19.4\times 0.081=1.57 m^3/min$

and theoretical suction volume or piston displacement per cylinder per minute
=$\frac{1.57}{2}=0.785 m^3/min$ ...............(iii)

Also theoretical suction volume or piston displacement per minute
= Piston area $\times$ Stroke$\times$ R.P.M.
=$\frac{\pi}{4}\times D^2\times L\times N=\frac{\pi}{4}\times 1.5D^2\times D\times1000$
$=1178.25 D^3 m^3/min$ ..............(iv)
Equating equations (iii) and (iv),
$=D=0.087m=87 mm$ (Ans)
and $L=1.5D=1.5\times87=130.5 mm$ (Ans)

(b) When there is a clearance of 2%
Let $v_2$= Specific volume at point 2, and
C =Clearance = 2% = 0.02 ..........(Given)
Applying charles law to point 2 and 2'
$\frac{v_2}{T_2}=\frac{v_2’}{T_2’} or v_2=v_2’\times\frac{T_2}{T_2’}$

$=0.0235\times\frac{316.41}{303}=0.0245 m^3/kg$
We know that volumetric efficiency of the compressor,
$\eta_V=1+C-C(\frac{v_1}{v_2})=1+0.02-0.02(\frac{0.081}{0.0245}) $

Therefore Piston displacement per cylinder per min
$ \frac{m_R\times v_1}{2}\times \frac{1}{\eta_V}=\frac{19.4\times0.081}{2}\times\frac{1}{0.954}=0.8236 m^3/min$ ............(v)
Equating equations (iv) and (v),


Or $D^3=\frac{0.8236}{1178.25}=0.00070$

Therefore D=0.0887 m=88.7 mmm (Ans)

And L=1.5D=1.5$\times$8807=1333mm (Ans)

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