Concept:

In a VCRS heat is absorbed in evaporator and rejected in condenser. Therefore, refrigeration effect takes place in evaporator. Also, the throttling process is isenthalpic i.e. enthalpy remains constant.

\({\rm{COP}} = \frac{{{\rm{Refrigeration\;effect}}}}{{{\rm{Work\;input}}}}{\rm{\;}}\)

\({\rm{COP}} = {\rm{\;}}\frac{{{\rm{Refrigerating\;effect}}}}{{{\rm{Work\;input}}}} = {\rm{\;}}\frac{{{{\rm{h}}_1} - {\rm{\;}}{{\rm{h}}_4}}}{{{{\rm{h}}_2} - {\rm{\;}}{{\rm{h}}_1}{\rm{\;}}}}\)

Calculation:

\({\rm{COP}} = {\rm{\;}}\frac{{{\rm{Refrigerating\;effect}}}}{{{\rm{Work\;input}}}} = {\rm{\;}}\frac{{{{\rm{h}}_1} - {\rm{\;}}{{\rm{h}}_4}}}{{{{\rm{h}}_2} - {\rm{\;}}{{\rm{h}}_1}{\rm{\;}}}} = {\rm{\;}}\frac{{240 - 150}}{{300 - 240}} = 1.5\)

Concept:

\({\rm{Condenser\;rating\;}} = \frac{{load}}{{Temperature\;difference}} = \left( {\frac{{load}}{{{\rm{\Delta }}T}}} \right)\)

Calculation:

When load is 30 kW, ΔT = 48 – 34 = 14

Then,

\({\rm{Rating}} = \frac{{30}}{{14}} = 2.1428\;{\rm{kW}}/{\rm{K}}\)

When load is 10 kW

Then

\({\rm{\Delta T}} = \frac{{10}}{{2.1428}} = 4.67\;K\)

∴ Outdoor temp. + ΔT = Condensing temperature

∴ Condensing temperature = 22.67°C

Concept:

\({\rm{Relative\;humidity\;}}\left( \phi \right) = \frac{{{P_v}}}{{{P_{sat}}}}\)

\({\rm{Humidity\;ratio}} = 0.622\frac{{{P_v}}}{{{P_{total}} - {P_v}}}\)

Calculation:

\(\phi = \frac{{0.01}}{{0.02}} \times 100 = 50\% \)

And

\(\omega = 0.622 \times \frac{{0.01}}{{1.01 - 0.01}}\)

∴ ω = 0.00622 kg/kg of dry air

Concept:

The efficiency of the heat engine and COP of heat pump and refrigerator are related by the equation given below

\({\left( {COP} \right)_{HP}} = 1 + {\left( {COP} \right)_R} = \frac{1}{{{\eta _{engine}}}}\)

Calculation:

Given: η_engine = 0.2

\(1 + {\left( {COP} \right)_R} = \frac{1}{{0.2}} = 5\)

∴ (COP)_R = 4

The alternate method to do the problem

\(\begin{array}{l} COP = \frac{{{Q_L}}}{{{W_{in}}}} = \frac{{{Q_L}}}{{{Q_H}\; - \;{Q_L}}}\\ \Rightarrow COP = \frac{{\left( {0.8} \right){Q_H}}}{{\left( {0.2} \right){Q_H}}} = 4 \end{array}\)

Concept:

This numerical can be solved easily by calculating the change in vapour pressure of water and then calculate increase in mass of water vapour.

Calculation:

\({\rm{Relative\;humidity\;}}\left( \phi \right) = \frac{{{P_v}}}{{{P_{sat}}}}\)

\(0.5 = \frac{{{P_{{v_1}}}}}{{1.36}}\)

\(\therefore {P_{{v_1}}} = 0.68\;kPa\) 

This is initial vapour pressure.

Now,

Final vapour pressure will be equal to saturated pressure

Therefore, \({P_{{v_2}}} = 1.36\;kPa\)

So, increase in vapour pressure of water

\( = {P_{{v_2}}} - {P_{{v_1}}} = 1.36 - 0.68\) 

ΔP_v = 0.68 kPa

This increase in vapour pressure is due to water evaporated.

Thus,

\(m = \frac{{\left( {{\rm{\Delta }}{P_v}} \right)V}}{{RT}}\)

\(m = \frac{{0.68 \times 80}}{{0.4618 \times 283}}\;\;\;\;\;\;\;\;\;\;\;\left( {R = \frac{{8.314}}{{18}}} \right)\)

 m = 0.4162 kg

Concept:

Apply the energy balance equation to the subcooling exchanger.

\(h_3' - {h_3} = {h_1} - h_1'\)

Calculation:

Given:

\(h_3' = 78\) kJ/kg, \(h_1' = 182\) kJ/kg, h₂ = 230 kJ/kg, h₃ = 68 kJ/kg = h₄

Now,

Applying the energy balance equation to the subcooling exchanger.

\(h_3' - {h_3} = {h_1} - h_1'\)

78 – 68 = h₁ – 182

∴ h₁ = 192 kJ/kg

Now,

\(COP = \frac{{{h_1}' - {h_4}}}{{{h_2} - {h_1}}} = \frac{{182 - 68}}{{230 - 192}} \)

∴ COP = 3

Concept:

Relative humidity: the amount of water vapour present in air expressed as a percentage of the amount needed for saturation at the same temperature.

Relative humidity = \(\phi= {\rm{\;}}\frac{{{\rm{Partial\;pressure\;of\;vapour}}}}{{{\rm{Saturation\;pressure}}}} = {\rm{\;}}\frac{{{{\rm{P}}_{\rm{V}}}}}{{{{\rm{P}}_{\rm{S}}}}}\) 

Using ideal gas equation

PV = mRT

V/m = RT/P where R= R_u/M

Specific volume = RT/P where R = gas constant

R_air = 287J/kgK ; R_vapour = 461.88 J/kgK

Calculation:

ϕ = 0.70, P_s = 0.0317 bar

P_v = ϕ P_s

∴ Pv= 0.02219 bar

P_air = P_t - P_v = 1 – 0.02219 = 0.97781 bar.

Now,

Applying perfect gas equation

For air:

\({V_a} = \frac{{{R_a}{T_a}}}{{{P_a}}} = \frac{{287 \times 298}}{{0.97781 \times {{10}^5}}}\)

∴ V_a = 0.8746 m³/kg

For vapour:

\({V_v} = \frac{{{R_v}{T_v}}}{{{P_v}}} = \frac{{\frac{{8314}}{{18}} \times 298}}{{0.02219 \times {{10}^5}}}\)

Concept:

When the refrigerant leaves the compressor, it is at superheated state as shown in the figure.

Enthalpy = mc_p∆T kJ

In the question COP of system is given,

\(COP = \;\frac{{{h_1} - \;{h_4}}}{{{h_2} - \;{h_1}}}\)

Using this h₂ can be found out and using that we can then apply

h₂ - h₂' = c_p ∆T

Hence, required temperature can be found out.

Calculation:

Given: COP = 6.6, h₁ = 188.54 kJ/kg, C_P = 0.625 kJ/kgK

h₃ = h₄ = h_f = 68.44 kJ/kg and h₂’ = h_g = 200.44 kJ/kg

Now,

COP = (h₁ – h₄)/(h₂ – h₁)

6.6 = (188.54 – 68.44)/(h₂ – 188.54)

∴ h₂ = 206.73 kJ/kg

Now,

h₂ – h₂' = 206.73 – 200.44 = 6.2969 kJ/kg

6.2969 = 0.625 (T₂ - 30)

∴ T₂ = 40.07° C 

Calculation:

T₂ = 28°C (saturation, vapour), ∴ h₂ = 265.68 kJ/kg, S₂ = 0.91948 kJ/kg k, T₁ = -10°C

Now,

S₁ = S₂

(S_f + x S_fg)₁ = S₂

0.15504 + x₁ (0.93766 – 0.15504) = 0.91948

∴ x₁ = 0.9767

h₁ = (hf + x₁ hf_g) = 38.55 + 0.9767 (244.51 – 38.55)

h₁ = 239.71 kJ/kg 

T₃ = 28° (saturated, liquid)

h₃ = 90.69 kJ/kg   S₃ = S₄ = 0.33846 kJ/kg k

S₃ = S₄

S₃ = (S_f + x₄ Sf_g)₄

0.33846 = (0.15504 + x₄ (0.93766 – 0.15504))

∴ x₄ = 0.2344

Now,

∴ h₄ = (Sf + x₄ hf_g)₄

h₄ = 38.55 + 0.2344 (244.51 – 38.55)

∴ h₄ = 86.82 kJ/kg

Work input to compressor = (h₂ – h₁)

Work input to compressor = 265.68 – 239.71

∴ Work input to compressor = 25.97 kJ/kg

Now,

Work developed by turbine = (h₃ – h₄)

Work developed by turbine = 90.69 – 86.82

∴ Work developed by turbine = 3.87 kJ/kg

Explanation:

Given:

T₁ = 10°C = 283 K, T₃ = 30°C = 303 K

P₂ = P₃ = 9 bar = 9 × 10⁵ Pa

P₄ = P₁ = 1.5 bar = 1.5 × 10⁵ Pa

Refrigeration effect = 1.5 TR = 1.5 × 3.5 = 5.25 kW

Calculation:

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\(\frac{{{T_2}}}{{283}} = {\left( 6 \right)^{\frac{{1.4 - 1}}{{1.4}}}}\)

∴ T₂ = 472.18 K

\(\frac{{{T_3}}}{{{T_4}}} = {\left( {\frac{{{P_3}}}{{{P_4}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\(\frac{{303}}{{{T_4}}} = {\left( 6 \right)^{\frac{{0.4}}{{1.4}}}}\)

∴ T₄ = 181.59 K

Now,

Refrigeration capacity = ṁ (Refrigeration effect)

5.25 kW = ṁ C_p (T₁ – T₄)

5.25 × 10³ = ṁ × 1.005 × 10³ × (283 – 181.59)

ṁ = 0.051 kg/s

P₁V̇₁ = ṁ R T₁

(1.5 × 10⁵) V̇₁ = (0.051) (287)(283)

∴ V̇1 = 0.0276 m³/s