Question

A refrigerator is operating between a high pressure of 0.7 MPa and low pressure of 0.1 MPa. The Isentropic efficiency of the compressor is 0.8. The working fluid of the refrigeration cycle is R134a flowing at 0.1 kg/s. The specific heat of air is 1.005 kJ/kgk and density of air is 1.2 kg/m. 

Calculate 

a. The compressor exit temperature. 

b. The compressor power input. 

c. The evaporator cooling load. 

d. The coefficient of performance.

e. The steam quality at the evaporator inlet.

f. If the cooling load obtained in part is used to cool a building space for 30 minutes causing drop of 20°C in the space temperature calculate the volume of the space to be cooled. 

Answer 1

A)

Since, which refrigeration cycle is to be followed is not mentioned. we will consider Vapour Compression Refrigeration System (VCRS) for this case.

The following is the T-S chart for the current refrigeration cycle.

Ideal compression process Condenser Throttle valve Actual compression process 4 Evaporator VCRS

Since the properties of R-134a is not given to us in the question, I shall be using the tables which I have with me, for solving this sum.

The data is as follows:

Pressure (MPa)	T(saturation) (C)	hf (kJ/kg)	hg (kJ/kg)	sf (kJ/kgK)	sg (kJ/kgK)
0.7	26.69	88.82	265	0.3323	0.9199
0.1	-26.37	17.28	234.4	0.07188	0.9518

The compressor inlet enthalpy h1 = hg (at 0.1 MPa) = 234.4 kJ/kg

It is said that the compressor is has an isentropic efficiency of 0.8.

If the compressor was supposed to be ideal, then the exist state would have been (2') (as marked in the image)

In such a case, the s1 = s2'.

• s1 = sg (at 0.1 MPa) = 0.9518 kJ/kgK

Now, s1 = s2' = 0.9518 kJ/kgK

But $s2' = sg+C.ln\frac{T_{final}}{T_{initial}}$

where

• sg (at 0.7 MPa) = 0.9199 kJ/kgK
• C = Specific heat of R-134a = 1.036 kJ/kgK
• T(final) = T2' = ?
• T(initial) = Saturation temperature at 0.7 MPa = 26.69 C = 299.69 K

Thus, $s2' = sg+C.ln\frac{T_{final}}{T_{initial}}$

$0.9518 = 0.9199+1.036\times ln\frac{T_{2'}}{299.69}$

T2' = 309.1 K

Now, the enthalpy h2' = hg + C x (T2' - T(sat))

where

• hg (at 0.7 MPa) = 265 kJ/kg
• C = Specific heat of R-134a = 1.036 kJ/kgK
• T2' = 309.1 K
• T(sat) = Saturation temperature at 0.7 MPa = 26.69 C = 299.69 K

h2' = hg + C x (T2' - T(sat))

h2' = 265 + 1.036 x (309.1 - 299.69) => h2' = 274.75 kJ/kg

Now, since it is given that the isentropic efficiency = 0.8

$\eta _{isen}=\frac{h2'-h1}{h2-h1}$

We have all the required values, thus,

$\eta _{isen}=\frac{h2'-h1}{h2-h1}$

$0.8=\frac{274.75-234.4}{h2-234.4}=>\mathbf{h2 = 284.84\: kJ/kg}$

Now, finally, h2 = hg + C x (T2 - T(sat))

where

• hg (at 0.7 MPa) = 265 kJ/kg
• C = Specific heat of R-134a = 1.036 kJ/kgK
• T2 = Exit temperature of compressor = ? (our answer)
• T(sat) = Saturation temperature at 0.7 MPa = 26.69 C = 299.69 K

h2 = hg + C x (T2 - T(sat))

284.84 = 265 + 1.036 x (T2 - 299.69) => Exit temperature = T2 = 318.834 K

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B)

Compressor Power input = m(refrigerant) x (h2 - h1)

where

• m(refrigerant) = 0.1 kg/s
• h2 = 284.84 kJ/kg
• h1 = 234.4 kJ/kg

Compressor Power input = m(refrigerant) x (h2 - h1) = 0.1 x (284.84 - 234.4)

Compressor Power input = 5.044 kW

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C)

Evaporator cooling load = m(refrigerant) x (h1 - h4)

where

• m(refrigerant) = 0.1 kg/s
• h4 = h3 (because of isenthalpic expansion in throttle valve) = hf (at 0.7 MPa) = 88.82 kJ/kg
• h1 = 234.4 kJ/kg

Evaporator cooling load = m(refrigerant) x (h1 - h4) = 0.1 x (234.4 - 88.82)

Evaporator cooling load = 14.558 kW

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D)

Coefficient of Performance = $\mathbf{COP}=\frac{Refrigeration\: \: load}{Compressor\: \: work}=\frac{14.558}{5.044}=\mathbf{2.886 }$

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E)

We can get the steam quality by equating the enthalpy at the condenser outlet (h3) and enthalpy at the evaporator inlet (h4) because of isenthalpic expansion, these two values are the same.

h4 = h3 (because of isenthalpic expansion in throttle valve) = hf (at 0.7 MPa) = 88.82 kJ/kg

But, $h4+hf+x\times (hg-hf)$

where

• hf = hf (at 0.1 MPa) = 17.28 kJ/kg
• hg = hg (at 0.1 MPa) = 234.4 kJ/kg

Therefore, $h4+hf+x\times (hg-hf)$

$88.82=17.28+x\times (234.4-17.28)=>\mathbf{Quality = x = 0.3295}$

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F)

The cooling load obtained in (C) is 14.558 kW.

We are using this cooling load for 30 mins ( = 30 x 60 = 1800 seconds).

Thus, net cooling energy is 14.558 x 1800 = 26204.4 kJ

This causes the temperature of the room to fall by 20 C.

Net cooling energy = m x Cp x Temp. Change

where

• m = mass of air in the room = ?
• Cp = 1.005 kJ/kgK (given)
• Temp. Change = 20 C

Net cooling energy = m x Cp x Temp. Change

26204.4 = m x 1.005 x 20 => mass of air in the room = m = 1303.7014 kg

Now, we know, $Volume=\frac{m}{Density}$

where

• m = mass of air in the room = 1303.7014 kg
• Density = 1.2 kg/m3 (given)

$Volume=\frac{m}{Density}=\frac{1303.714}{1.2}=>\mathbf{Volume=1086.417 m^{3}}$

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