Example 8
A vapour compression refrigeration machine, with Freon-12 as refrigerant, has a capacity of 20 tonne of refrigeration operating between - 28°C and 26°C. The refrigerant is subcooled by 4°C before entering the expansion valve and the vapour is superheated by 5°C before leaving the evaporator. The machine has a six-cylinder single-acting compressor with stroke equal to 1.25 times the bore. It has a clearance of 3% of the stroke volume. Determine :
1. Theoretical power required;
2. C.O.P.,
3. Volumetric efficiency; and
4. Bore and stroke of cylinder.
The speed of compressor is 1000 r.p.m.
The following properties of Freon-12 may be used :

Specific heat of liquid refrigerant = 0.963 kJ/kg K and specific heat of superheated vapour = 0.615 kJ/kg K.
Solution
Given :
12 TR ;$T_1’=28º C = -28+273 = 245 K ;T_2’=T_3’=26º C = 26+273 = 299 K ;$
$T_3’-T_3=4º C orT_3=22º C = 22+273 = 295 K ;$
$T_1-T_1’=5º C or T_1=23º C = -23+273= 250 K ;$
$ 1.25 D ; Clearance volume = 3$%$ Stroke volume ; N= 1000 r.p.m. ;$
$v_1’=0.1475 m^3/kg ;v_2’=0.0262 m^3/kg ; h_{f1}=10.64 kJ/kg ;h_{f3}’=60.67 kJ/kg ;$
$h_1’=175.11 kJ/kg ;h_2’=198.1 kJ/kg ;s_{f1}=0.0444 kJ/kg K ;s_{f3}=0.271 kJ/kg K ;$
$s_1’=0.7153 kJ/kg K ; s_2’=0.6865 kJ/kg K ;c_{pl}=0.963 kJ/kg K ;c_{pv}=0.615 kJ/kg K$
The T-s and p-h diagrams are shown in (a) and (b) respectively.
1. Theoretical power required
First of all, let us find the temperature of superheated vapour at point 2 $(T_2)$.
We know that entropy at point 1,
$s_1=s_1’+2.3c_{pv}\log\frac{T_1}{T_1’}=0.7153+2.3\times 0.615\log\frac{250}{245}$
$=0.7277$ ........(i)
and entropy at point 2,
$s_2=s_2’+2.3c_{pv}\log\frac{T_2}{T_2’}=0.6865+2.3\times 0.615\log\frac{T_2}{299}$
$=0.6865+1.4145\log\frac{T_2}{299}$ ..........(ii)

Since the entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii ),
$0.277=0.6865+1.4145\log\frac{T_2}{299}$
$\log\frac{T_2}{299}=\frac{0.7277-0.6825}{1.1415}=0.0291$
$\frac{T_2}{299}=1.0693$ ..........(Taking antilog of 0.0291)
Therefore $T_2=299\times 1.0693=319.7 K$
We know that enthalpy at point 1,
$h_1=h_1’+c_{pv}(T_1-T_1’)=175.1+0.615(250-245)=178.18 kJ/kg$
Enthalpy at point 2,
$h_2=h_2’+c_{pv}(T_2-T_2’)=198.11+0.615(319.17-299)=210.84 kJ/kg$

and enthalpy of liquid refrigerant at point 3,

$h_{f3}=h_{f3}’-c_{pl}(T_3’-T_3)=60.67-0.963(299-295)$
=64.52 kJ/kg

We know that heat extracted or refrigerating effect per kg of the refrigerant, $R_E=h_1-h_{f3}=178.18-64.52=113.66 kJ/kg$

and refrigerating capacity of the system, $Q=12 TR=12\times 210=2520 kJ/kg$ .......(Given)

Therefore Mass flow of the refrigerant, $m_R=\frac{Q}{R_E}=\frac{2520}{113.66}=22.17 kg/min$

Work done during compression of the refrigerant $=m_R(h_2-h_1)=22.17(210.84-178.18)=724 kJ/min$

Therefore Theoretical power required
=720/60=12.07 kW (Ans)
2. C.O.P.
We know that C.O.P.

$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{178.18-64.52}{210.84-178.18}=3.48 $ (Ans)

3. Volumetric efficiency
Let $v_2$ = Specific volume at point 2, and
C = Clearance = 3% = 0.03. … (Given)
First of all, us find the specific volume at suction to the compressor, i.e. at point 1. Applying Charles' law to point 1 and 1',
$\frac{v_1}{T_1}=\frac{v_1’}{T_1’} or v_1=v_1’\times\frac{T_1}{T_1’}$

$=0.1475\times\frac{250}{245}=0.1505 m^3/kg$
Again applying Charles' law to points 2 and 2',
$\frac{v_2}{T_2}=\frac{v_2’}{T_2’} or v_2=v_2’\times\frac{T_2}{T_2’}$

$=0.0262\times\frac{319.7}{299}=0.028 m^3/kg$
We know that volumetric efficiency,
$\eta_V=1+C-C(\frac{v_1}{v_2})=1+0.03-0.03(\frac{0.1505}{0.028}) $

$=0.87 $or 87 % (Ans)
4. Bore and stroke of cylinder
L = Length of cylinder = 1.25 d, and … (Given)
N = Speed of compressor = 1000 r.p.m. … (Given)
We know that theoretical suction volume or piston displacement per minute
$m_R\times v_1\times\frac{1}{\eta_V}=22.17\times 0.1505\times\frac{1}{0.87}=3.84 m^3/min$

Since the machine has six cylinder, single acting compressor, therefore, theoretical suction volume or piston displacement per cylinder per minute
$=\frac{3.84}{6}=0.64 m^3/min$ ............(iii)

We also know that suction volume or piston displacement per minute
= Piston area $\times $ Stroke $\times$ R.P.M. =$\frac{\pi}{4}\times D^2\times L\times N=\frac{\pi}{4}\times 1.5D^2\times D\times1000$
$=982 D^3 m^3/min$ .........(iv)
Equating equations (iii) and (iv),

$D^3=\frac{0.64}{982}=0.000652$
$D=0.0867 m=86.7 mm$(Ans)
$L=1.25\times 86.7=108.4 mm$ (Ans)

Example 9
A food storage locker requires a refrigeration capacity of 12 TR and works between the evaporating temperature of - 8°C and condensing temperature of 30°C. The refrigerant R-12 is sub cooled by 5°C before entry to expansion valve and the vapour is superheated to - 2°C before leaving the evaporator coils. Assuming a two cylinder, single acting compressor operating at 1000 r.p.m. with stroke equal to 1.5 times the bore, determine:
1. Coefficient of performance;
2. Theoretical power per tonne of refrigeration; and
3. Bore and stroke of compressor when (a) there is no clearance; and (b) there is a clearance of 2%.
Use the following data for R-12

The specific heat of liquid R-12 is 1.235 kJ/kg K, and of vapour R-12 is 0.733 kJ/kg K.
Solution
Given:
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,

$s_1=s_1’+2.3c_{pv}\log\frac{T_1}{T_1’}$

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

And entropy at point 2,

$s_2=s_2’+2.3c_{pv}\log\frac{T_2}{T_2’}$

$=0.6853+2.3\times0.733\log\frac{T_2}{303}=0.7171$ .............(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,
$h_1=h_1’+c_{pv}(T_1-T_1’)$

$=184.07+0.733(271-265)=188.47 kJ/kg$
Enthalpy at point 2,
$h_2=h_2’+c_{pv}(T_2-T_2’)$

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

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),
$D^3=\frac{0.785}{1178.25}=0.000666$
$=D=0.087m=87 mm$ (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}) $

=1.02-0.066=0.954
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),

$1178.25D^3=0.8236$

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)