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problems on effect of liquid sub cooling
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Illustrative examples
A. Effect of liquid sub-cooling

Example 1
A vapour compression refrigerator uses R-12 as refrigerant and the liquid evaporates in the evaporator at - 15°C. The temperature of this refrigerant at the delivery from the compressor is 15 oC when the vapour is condensed at 10°C. Find the co-efficent of performance if
(i) there is no undercooling, and
(ii) (ii) the liquid is cooled by 5°C before expansion by throttling.
(iii) Take specific heat at constant pressure for the superheated vapour as 0.64 kJ/kg K and that for liquid as 0.94 kJ/kg K. The other properties of refrigerant are as follows

Solution
Given :
$T_1=T_4=-15°C = -15+273 = 258 K ;T_2=15°C = 15+273 = 288 K ;T_2’=10°C = 10+273 = 283 K $; $ c_{pv}=0.64 kJ/kg K ;c_{p1}=0.94 kJ/kg K ;h{f1}=22.3 kJ/kg ;h_{f3}=45.4 kJ/kg ;$
$ h_1’=180.88 kJ/kg ; h_2’=19.76 kJ/kg ;s_{f1}=0.0904 kJ/kg ;s_{f3}=0.1750 kJ/kg K ;$
$s_{g1}=0.7051 kJ/kg K ;s_2’=0.6921 kJ/kg K$

(i) Coefficient of performance if there is no undercooling
The T-s and p-h diagrams, when there is no undercooling, are shown in Fig a and b respectively.
Let $x_1$=Dryness fraction of the refrigerant at point 1.
We know that entropy at point 1,

$s_1=s_{f1}+x s_{fg1}=s_{f1}+x(s_{g1}-s_{f1})$ .............$(s_{g1}=s_{f1}+s_{fg1})$

=$0.0904+x_1(0.7051-0.0904)$

=0.0904+0.6147x_1 ..........(i)
and entropy at point 2,
$s_2=s_2’+2.3c_{pv}\log\frac{T_2}{T_2’}$

$=0.6921+2.3\times \log\frac{288}{283} =0.6921+2.3\times 0.46\times 0.0077=0.7034$ ..........(ii)
Since the entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii),
$0.0904+0.6147x_1=0.7034 or x_1=0.997$

We know that the enthalpy at point 1, $h_1=h_{f1}+xh_{fg1}=h_{f1}+x(h_{g1}-h_{f1})$

$22.3+0.997(18-22.3)=180.4 kJ/kg$ ...........($h_{g1}=h_1’$)
and enthalpy at point 2, $h_2=h_2’+c_{pv}(T_2-T_2’)$
$=191.76+0.64(288-22.3)=194.96 kJ/kg$

Therefore C.O.P=$\frac{h_1-h_{f3}’}{h_2-h_1}=\frac{180.4-45.4}{194.96-180.4}=9.27$ (Ans)
(ii) Coefficient of performance when there is an undercooling of 5 °C

The T-s and p-h diagrams, when there is an undercooling of 5 oC, are shown in below (a) and (b) respectively.
We know that enthalpy of liquid refrigerant at point 3

$h_{f3}=h_{f3’}-c_{p1}\times $degree of undercooling $=45.4-0.94\times5=40.7 kJ/kg$

Therefore C.O.P=$\frac{h_1-h_{f3}’}{h_2-h_1}=\frac{180.4-40.7}{194.96-180.4}=9.59$ (Ans)


Example 2
Simple NH3 vapour compression. system has compressor with piston displacement of 2 M3/Min, a condenser pressure of 12 bar and evaporator pressure of 2.5 bar. The liquid is sub-cooled to 20°C by soldering the liquid line to suction line. The temperature of vapour leaving. the compressor is 100°C, heat rejected to compressor cooling water is 5000 kJ/hour, and volumetric efficiency of compressor is 0.8. Compute :
(i) Capacity ;
(ii) Indicated power ; and
(iii) C.O.P. of the system.

Solution
Given :
$v_p=2 m^3/min;p_2=p_2’=p_3’=p_3=12 bar ;p_1=p_4=2.5 bar ;$
$T_3=20°C = 20+273 = 293 K ;T_2=100°C= 100+273 = 373 K ;\eta_V=0.8$

1.Capacity of the system
The T-s and p-h diagrams are shown in a and b respectively.
From p-h diagram, we find that the evaporating temperature corresponding to 2.5 bar is

$T_1=T_4=-14°C =-14+273=259 K$
Condensing temperature corresponding to 12 bar is

$T_2’=T_2=30°C=30+273=303 K$
Specific volume of dry saturated vapour at 2.5 bar ( i.e. at point 1 ),
$v_1=0.49 m^3/kg$
Enthalpy of dry saturated vapour at point 1,
$h_1=1428 kJ/kg$
Enthalpy of superheated vapour at point 2,
$h_2=$1630 kJ/kg

and enthalpy of sub-cooled liquid at 200 C at point at 3
$h_{f3}=h_4=270 kJ/kg$

Let $m_R$= Mass flow of the refrigerant in kg/min.
We know that piston displacement,
$v_p=\frac{m_R\times V_1}{\eta_V}$ or $m_R=\frac{v_p\times\eta_V}{v_1}=\frac{2\times0.8}{0.49}=3.265 kg/min$

We know that refrigerating effect per kg of refrigerant
$h_1-h_{f3} =1428-270=1158 kJ/kg$

and total refrigerating effect
$=m_R(h_1-h_{f3}=3.265(1428-270)=3781 kJ/min$

Therefore capacity of the system
$=\frac{3781}{210}=18 TR$ (Ans)

2.Indicated power, of the system
We know that work done during compression of the refrigerant
$=m_R(h_2-h_1)=3.265(1630-1428)=659.53 kJ/min$

Heat rejected to compressor cooling water
$=5000kJ/J=\frac{5000}{60}=83.33 kJ/min$ .......(Given)

Therefore total work done by the system
$=659.53+83.33=742.86 kJ/min$

and indicated power of the system
$=\frac{742.86}{60}=12.38 kW$ (Ans)

3.C.O.P. of the system
We know that C.O.P. of the system

$\frac{Total\space refrigerating \space effect}{Total \space work done}=\frac{3781}{742.86}=5.1 $ (Ans)


Example 3
A vapour compression refrigerator uses methyl chloride (R-40) and operates between pressure limits of 177.4 kPa and 967.5 kPa. At entry to the compressor, the methyl chloride is dry saturated and after compression has a temperature of 102°C. The compressor has a bore and stroke of 75 mm and runs at 480 r.p.m with a volumetric efficiency of 80%. The temperature of the liquid refrigerant as it leaves the condenser is 35°C and its specific heat capacity is 1.624 kJ/kg K. The specific heat capacity of the superheated vapour may be assumed to be constant. Determine : 1. refrigerator C.O. P.; 2. mass flow rate of refrigerant; and 3. cooling water required by the condenser if its temperature rise is limited to 12°C. Specific heat capacity of water= 4.187 kJ/kg K. The relevant properties of methyl chloride are as follows:

Solution
Given :
$p_1=p_4=177.4 kPa ;p_2=p_3=967.5 kPa ;T_2=102°C= 102+273 = 375 K ; $
$ 75 mm = 0.075 m ; 480 r.p.m ; \eta_V=80$%$ = 0.8 ;$
$T_3=35°C = 35+273 = 308 K ;c_{p1}=c_{pv}=1.624 kJ/kg K ;c_{pw}=4.187 kJ/kg K;$
$T_1=T_4=-10°C = -10+273 =263 K ; T_2’=T_3’=45°C = 45+273 = 318 K ;v_1=0.233 m3/kg ;$
$v_2’=0.046 m3/kg ;h_{f1}=45.38 kJ/kg ;h_{f3}’=132.98 kJ/kg ;h_1=460.76 kJ/kg ;h_2’483.6 kJ/kg ;$
$s_{f1}=0.183 kJ/kg K ; s_{f3}=0.485 kJ/kg K ;s_1=s_2=1.762 kJ/kg K;s_2’=1.587 kJ/kg K.$
The T-s and p-h diagrams are shown in (a) and (b) respectively.

1. Refrigerator C.O.P.
We know that enthalpy at point 2, $h_2=h_2’+c_{pv}(T_2-T_2’)$
$=483.6+1.624(375-318)=576.2 kJ/kg$
and enthalpy of liquid refrigerant at point 3,
$h_{f3}=h_{f3}’-c_{pv}(T_3’-T_3)$

=$132.98-1.624(318-308)=116.74 kJ/kg$

We know that refrigerator C.O.P.
$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{460.76-116074}{576.2-460.76}=2.98 $ (Ans)
2. Mass flow rate of refrigerant
Let $m_R$= Mass flow rate of refrigerant in kg/min.
We know that suction volume or piston displacement per minute,
$=piston area\times stroke\times R.P.M$
$=\frac{\pi}{4}(0.075)_2\times 480=0.16 m^3/min$ ................(i)
We also know that piston displacement per minute
$=m_R\times v_1\times\frac{1}{\eta_V}=m_R\times0.233\times\frac{1}{0.8}=0.29 m_R$ .....(ii)
Equating equations (i) and (ii),
$m_R=\frac{0.16}{0.29}=0.55 kg/min$
3. Cooling water required by the condenser
Let $m_W=$ Cooling water required by the condenser in kg/min.
We know that heat given out by the refrigerant in the condenser

$=m_R(h_2-h_{f3}=0.55(576.2-116.74)=252.7 kJ/min$ ............(iii)

and heat taken by water in the condenser
$=m_W\times c_{pw}\times $rise in temperature

$=m_W\times 4.187\times 12=50.244 m_W$ ...........(iv)
Equating equations (iii) and (iv),

$m_w=\frac{252.7}{50.244}=5.03 kg/min$ (Ans)


Example 4
A commercial refrigerator operates with R-12 between 1.2368 bar and 13.672 bar. The vapour is dry and saturated at the compressor inlet. Assuming isentropic compression, determine the theoretical C.O.P. of the plant. The isentropic discharge temperature is 64.86°C. If the actual C.O.P. of the plant is 80% of the theoretical, calculate the power required to run the compressor to obtain a refrigerating capacity of 1 TR. If the liquid is sub-cooled through 10°C after condensation, calculate the power required the properties of R-12 are given below:

Assume specific heat of liquid to be 1.055 kJ/kg K.
Solution
Given :
$p_1=p_4=1.2368 bar ;p_2=p_3=13.672 bar ;T_2=64.86°C = 64.86+273 = 337.86 K ; $
$(C.O.P)_{actual}=80$%$(C.O.P)_{th} ; 1 TR ;T_1=-25°C = -25+273 = 248 K ;$
$T_2’=55°C = 55+273 = 328 K ;h_{f1}=13.33 kJ/kg ;h_1=176.48 kJ/kg ;s_{f1}=0.0552 kJ/kg ;$
$s_1=s_2=0.7126 kJ/kg K ;h_{f3}’=90.28 kJ/kg ;h_2’=207.95 kJ/kg ;s_{f3}=0.3197 kJ/kg ;$
$s_2’= 0.6774 kJ/kg K ;h_2=220.6 kJ/kg ;c_{p1}=1.055 kJ/kg K.$

The T-s and p-h diagrams are shown in a and b respectively.
Theoretical C.O.P. of the plant

We know that theoretical C.O.P. of the plant
$=\frac{h_1-h_{f3}’}{h_2-h_1}=\frac{16048-90.88}{220.6-176.48}=1.95 $ (Ans)
Power required to run the compressor
Since the actual C.O.P., of the plant is 80% of the theoretical, therefore
$(C.O.P)_{actual}=0.8\times 1.95=1.5$
refrigerating effect produced per kg of refrigerant

$=W_{actual}\times (C.O.P)_{actual}$
$=44.12\times 1.56=68.83 kJ/kg$

We know that refrigerating capacity
1TR=210 kJ/min
Therefore mass flow of refrigerant,
$m_R=\frac{210}{68.83}=3.05 kg/min$
and work done during compression of the refrigerant

$=m_R(h_2-h_1)=3.05(2220.6-176048)=134.57 kJ/min$

Therefore power required to run the compressor =$\frac{134.57}{60}=2024 kW$ (Ans)
Power required if the liquid is subcooled through 10 ºC

The T-s and p-h diagrams with sub cooling of liquid are shown in a and b respectively. We know that enthalpy of liquid refrigerant at point 3,
$h_{f3}=h_{f3}’-c_{p1}\times $degree of subcooling

$=90.28-1.055\times10=79.73 kJ/kg$
Theoretical
(C.O.P)=$\frac{h_1-h_{f3}}{h_2-h_1}=\frac{176.48-79.73}{220.6-176.48}=2.2$

And $(C.O.P)_{actual}=0.8\times 2.2 =1.76$
Actual work done by the compressor,

$W_{actual}=h_2-h_1=220.6-176.48=44.12 kJ/kg$
Therefore net refrigeration effect produced per kg of refrigerant
$=w_{actual}\times (C.O.P)_{actual}$
$=44.12\times 1.76=77.65 kJ/kg$

We know that refrigerating capacity =1 TR=210 kJ/min Therefore mass flow of the refrigerant, $m_R=\frac{210}{77.65}=2.7 kg/min$
and work done during compression of the refrigerant
$=m_R(h_2-h_1)=2.7(220.6-176.48)=119.12 KJ/min$

Power required $=\frac{119.12}{60}=1.985 kW$ (Ans)


Example 5 A vapour compression refrigerator works between the pressures 4.93 bar and 1.86 bar. The vapour is superheated at the end of compression, its temperature being 25°C. The liquid is cooled to 9°C before throttling. The vapour is 95% dry before compression. Using the data given below, calculate the coefficient of performance and refrigerating effect per kg of the working substance circulated

The specific heat at constant pressure for he superheated vapour is 0.645 kJ/kg K and for the liquid is 0.963 kJ/kg K. Solution
Given :
$p_2=p_3=4.93 bar ;p_1=p_4=1.86 bar ;T_2=25º C = 25+273 = 298 K ;$
$T_3= 9º C = 9+273 = 282 K ;x_1=95$%$= 0.95 ; T_3’=T_2’=14.45º C = 14.45+273 = 287.45 K ;$
$T_1=T_4=-15º C = -15+273 = 258 K ;h_{f1}=21.67 kJ/kg ;h_{f3}’=h{f2}’=49.07 kJ/kg ;$
$h_{fg1}=161.41 kJ/kg ;h_{fg2}’=147.8 kJ/kg ;c_{pv}=0.645 kJ/kg K ;c_{pl}=0.963 kJ/kg K$
The T-s and p-h diagrams are shown in a and b respectively.

Coefficient of performance
We know that enthalpy at point 1,
$h_1=h_{f1}+x_1h_{fg1}=21.67+0.95\times161.41=175 kJ/kg$

Similarly, enthalpy at point 2,
$h_2=h_2’+c_{pv}\times degree of superheat $

=$(h_{f2}’+h_{fg2}’=c_{pv}(T_2-T_2’)$

$=(490.7+147.8)+0.645(298-287.45)=203.67 kJ/kg$

and enthalpy of liquid refrigerant at point 3,
$h_{f3}=h_{f3}’-c_{p1}\times degree of cooling$

$=h_{f3}’-c_{p1}(T_3’-T_3)$

=$49.07-0.963(287.45-282)=43.82 kJ/kg$

Therefore coefficient of performance,

$(C.O.P)=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{175-43.82}{203.67-175}=4.57$ (Ans)

Refrigerating effect per kg of the working substance

We know that the refrigerating effect
$=h_1-h_{f3}=175-43.82=131.18 kJ/kg$ (Ans)


Example 6
In a 15 TR ammonia refrigeration plant, the condensing temperature is 25° C and evaporating temperature -10° C. The refrigerant ammonia is sub-cooled by 5° C before passing through the throttle valve. The vapour leaving the evaporator is 0.97 dry. Find

  1. coefficient of performance ; and
  2. power required.
    Use the following properties of ammonia

Solution
Given :
15 TR ; $T_2’=T_3’= 25º C = 25+273 = 298 K ;T_1=T_4=-10º C = -10+273 = 263 K ;$
$T_3=25-5 = 20º C = 20+273 = 293 K ;x_1=0.97 ;h_{f3}’=298.9 kJ/kg ;$
$h_2’=1465.84 kJ/kg ;s_{f3}’=1.1242 kJ/kg K ; s_{g2}’==s_2’=5.0391 kJ/kg K ;$
$c_{p1}=4.6 kJ/kg K ;c_{pv}=2.8 kJ/kg K ;h_{f1}=135.37 kJ/kg ;h_1’=1433.05 kJ/kg ;$
$s_{f1}=0.5443 kJ/kg ;s_{g1}=5.4770 kJ/kg K$
The T-s and p-h diagrams are shown in a and b respectively. First of all, let us find the temperature of refrigerant at point 2 $(T_2)$.

We know that entropy at point l,
$s_1=s_{f1}+x_1s_{fg1}=s_{f1}+x(s_{g1}-s_{f1}) ...........(s_{g1}=s_{f1}+s_{fg1})$
$=0.5443+0.97(5.4770-0.5443)=5.329 kJ/kg K$ .......(i)
and entropy at point 2, $s_2=s_2’+2.3 c_{pv}\log\frac{T_2}{T_2’}=5.0391+2.3\times \log\frac{T_2}{298}$ .........(ii) $=50.0391+6.44\log\frac{T_2}{298}$ ........(ii)
Since entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii),
$5.329=50.0391+6.44\log\frac{T_2}{298}$
$\log\frac{T_2}{298}=\frac{5.329-5.0391}{6.44}=0.445$

Or $\frac{T_2}{298}=1.109$ .........(Taking antilog of 0.0405)
$T_2=298\times 1.09=330 K$
1. Coefficient of performance
We know that enthalpy at point 1
$h_1=h_{f1}+x_1h_{fg1}=h_{f1}+x_1(h_{g1}-h_{f1}) ........(h_{g1}=h_{f1}+h_{fg1})$
$=135.37+0.97(1433.05-135.37)=1394.12 kJ/kg ...........(h_{g1}-h_1’)$

Enthalpy at point 2, $h_2=h_2’+c_{pv}\times degree pf superheat=h_2’+c_{pv}(T_2-T_2’)$
$=1465.84+2.8(3330-298)=1555.44 kJ/kg$

and enthalpy of liquid refrigerant at point 3,
$h_{f3}=h_{f3}’-c_{pl}\times degree of undercooling $

$=298.9-4.6\times5=275.9 kJ/kg$
Therefore Coefficient of performance,

C.O.P=$\frac{h_1-h_{f3}}{h_2-h_1}=\frac{1394.12-275.9}{1555.44-1394.12}=6.93 $ (Ans)

2. Power required
We know that the heat extracted or refrigerating effect produced per kg of refrigerant
$R_E=h_1-h_{f3}=1394.12-275.9=1118.22 kJ/kg $

and refrigerating capacity of the system,
Q=15 TR=15$\times $210=3150 kJ/min ..........(Given)

Therefore Mass flow of the refrigerant,
$m_R=\frac{Q}{R_E}=\frac{3150}{1118.22}=2.81 kJ/min$

Work done during compression of the refrigerant
$m_R(h_2-h_1)=2.81(1555.44-1394.12)=453.3 kj/min$
Therefore Power required =453.3/60=7.55kW (Ans)


Example 7
The following data refers to a 20 TR ice plant using ammonia as refrigerant : The temperature of water entering and leaving the condenser are 20°C and 27°C and temperature of brine in the evaporator is -15°C.Before entering the expansion valve, ammonia is cooled to 20°C and the ammonia enters the compressor dry saturated.Calculate for one tonne of refrigeration the power expended, the amount of cooling water in the condenser and the coefficient of performance of the plant. Use the properties given in the table below:

Solution
Given :
20 TR ;$T_{w2}=20º C = 20+273 = 293 K;T_{W1}=27º C = 27+273 = 300 K $
$;T_2’=T_3’=25 ºC = 25+273 = 298 K ;T_1=T_4=-15 ºC = -15+273 = 258 K ;T_3=20ºC = 293 K ;$
$h_{f1}=112.34 kJ/kg ;h_{f3}’= 298.90 kJ/kg ;h_1=1426.54 kJ/kg ;h_2’=1465.84 kJ/kg;$
$s_{f1}=0.4572 kJ/kg K ;s_{f3}=1.1242 kJ/kg K ;s_1=s_2=5.5490 kJ/kg K s_2’=5.0391 kJ/kg K ;$
$c_{pl4}=4.396 kJ/kg K ;c_{pl3}=4.06 kJ/kg K ;c_{pv1}=2.303 kJ/kg K ;c_{pv2}’= 2.805 kJ/kg K $
The T-s and p-h diagrams are shown in a and b respectively. First of all, let us find the temperature of refrigerant at point 2 ($T_2$). We know that entropy at point 2,

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

$5.5490=5.0391+2.3\times 2.805\log\frac{T_2}{298}$

$\log\frac{T_2}{298}=\frac{5.5490-5.0391}{2.3\times 2.805}=0.079$

$\frac{T_2}{298}=1.2$ ..(Taking antilog of 0.079)

Or $T_2=1.2\times 298=357.6 K$ or 84.6 º C

Enthalpy at point 2,

$h_2=h_2’+c_{pv2}’(T_2-T_2’)$

$=1465.84+2.805(357.6-298)=1633.02 kJ/kg$

and enthalpy of liquid refrigerant at point 3,

$h_{f3}=h_{f3}’-c_{pl3}\times$ Degree of under cooling
$h_{f3}’-c_{pl3}(T_3’-T_3)$
$=298.9-4.06(298-293)=275.87 kJ/kg$

We know that heat extracted or refrigerating effect produced per kg of the refrigerant,
$R_E=h_1-h_{f3}=1426.54-275.87=1150.67 kJ/kg $

and capacity of the ice plant,
$Q=20 TR=20\times 210=420 kJ/min$
Therefore Mass flow of the refrigerant,
$m_R=\frac{Q}{R_E}=\frac{4200}{1150.67}=3.65 kg/min$

Work done by the compressor per minute $=m_R(h_2-h_1)=3.65(1633.02-1426.54)=753.65 kJ/min$
Therefore Power expended per TR $=\frac{753.65}{60\times 20}=0.628 kW/TR$ (Ans)

Amount of cooling water in the condenser
Let $m_W$= Amount of cooling water in the condenser.

We know that heat given out by the refrigerant in the condenser $=m_R(h_2-h_{f3})$
$=3.65(1633.02-275.87)=4953.6 kJ/min$ .......(i)
Since the specific heat of water,
$c_W$=4.187 kJ/kg K,
therefore heat taken by water in the condenser $=m_W\times c_W(T_{W1}-T_{W2})$
$=m_W\times 4.187(300-293)=29.3 kJ/min$ ....(ii)

Since the heat given by the refrigerant in the condenser is equal to the heat taken by water in the condenser, therefore equating equations (i) and (ii),
$29.3m_W=4953.6 $ or $m_W=169 kg/min$ (Ans)

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