Refrigeration and Air Conditioning Test 1

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Refrigeration and Air Conditioning Test 1

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Weekly Quiz Competition

Question 1
1 / -0

The enthalpy of working fluid at the condenser inlet in a VCRS was found to be 300 kJ/kg and that at evaporator outlet was 240 kJ/kg. If the enthalpy at exit of condenser was 150 kJ/kg, then calculate the COP of this system.

A
1.5-1.5

B
1.54-1.51

C
1.57-1.56

D
1.59-1.52

Solution

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:
\({\rm{COP}} = {\rm{\;}}\frac{{{\rm{Refrigerating\;effect}}}}{{{\rm{Work\;input}}}} = {\rm{\;}}\frac{{{{\rm{h}}_1} - {\rm{\;}}{{\rm{h}}_4}}}{{{{\rm{h}}_2} - {\rm{\;}}{{\rm{h}}_1}{\rm{\;}}}} = {\rm{\;}}\frac{{240 - 150}}{{300 - 240}} = 1.5\)
Question 2
1 / -0

When the max outdoor temperature is 34°C, then the condensing temperature is 48°C and the heat rejected is 30 kW, what is the condensing temperature (in °C) when the load is reduced to 10 kW and outdoor temperature is 18°C?

A
22-30

B
224-301

C
227-306

D
229-302

Solution

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
Question 3
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Moist air exists at the pressure of 1.01 bar. The partial pressure and saturation pressure of water vapor are 0.01 bar and 0.02 bar respectively. What are relative humidity and humidity ratio of moist air, respectively?

A
25% and 0.0622 kg/kg dry air

B
50% and 0.0622 kg/kg dry air

C
25% and 0.00622 kg/kg dry air

D
50% and 0.00622 kg/kg dry air

Solution

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
Question 4
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A container of water is placed in room of 80 m³ volume. The initial temperature and relative humidity of room is 10°C and 50 % respectively. The saturation pressure at 10°C is 1.35 kPa, then the amount of water (in grams) which will get evaporated is

A
0.416

B
0.392

C
416.2

D
392

Solution

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
∴ m = 416.2 grams
Question 5
1 / -0

The relative humidity of a mixture of air and water vapour at 298 K and 1 bar is 70%. Then the specific volume (m³/kg) of air and water vapour are _____ and _____ respectively, if P_s = 0.0317 bar

A
0.8124, 56.0034

B
0.546, 45.767

C
0.8746, 62.0292

D
0.7346, 60.3244

Solution

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:
\({V_v} = \frac{{{R_v}{T_v}}}{{{P_v}}} = \frac{{\frac{{8314}}{{18}} \times 298}}{{0.02219 \times {{10}^5}}}\)
∴ V_v = 62.0292 m³/kg