This set of Air-Conditioning Multiple Choice Questions & Answers (MCQs) focuses on “Psychrometry – Adiabatic Mixing of Two Streams”.

1. What is the formula for m_{1} / m_{2} in terms of enthalpies for adiabatic mixing of two streams?

a) h_{1} – h_{2} / h_{1} – h_{3}

b) h_{3} – h_{2} / h_{1} – h_{3}

c) h_{3} – h_{2} / h_{2} – h_{3}

d) h_{3} – h_{2} / h_{1} – h_{2}

View Answer

Explanation: For energy balance, m

_{1}h

_{1}+ m

_{2}h

_{2}= m

_{3}h

_{3}

m

_{3}= m

_{1}+ m

_{2}

So, m

_{1}h

_{1}+ m

_{2}h

_{2}= (m

_{1}+ m

_{2}) h

_{3}

By solving we get,

m

_{1}/ m

_{2}= h

_{3}– h

_{2}/ h

_{1}– h

_{3}.

2. What is the formula for m_{1} / m_{2} in terms of specific humidity for adiabatic mixing of two streams?

a) W_{2} – W_{1} / W_{1} – W_{3}

b) W_{3} – W_{1} / W_{1} – W_{3}

c) W_{3} – W_{2} / W_{1} – W_{3}

d) W_{3} – W_{2} / W_{2} – W_{3}

View Answer

Explanation: For energy balance, m

_{1}W

_{1}+ m

_{2}W

_{2}= m

_{3}W

_{3}

m

_{3}= m

_{1}+ m

_{2}

So, m

_{1}W

_{1}+ m

_{2}W

_{2}= (m

_{1}+ m

_{2}) W

_{3}

By solving we get,

m

_{1}/ m

_{2}= W

_{3}– W

_{2}/ W

_{1}– W

_{3}.

3. Which of the following is true for the adiabatic mixing of two streams?

a) m_{3} = m_{3} – m_{1}

b) m_{2} = m_{1} + m_{3}

c) m_{1} = m_{3} + m_{2}

d) m_{3} = m_{1} – m_{2}

View Answer

Explanation: As two streams are mixed, then the result of it is the summation of mixed masses, and using the energy balance ratio of masses can be obtained. So, m

_{3}= m

_{1}+ m

_{2}.

4. When the adiabatic mixing is carried out, the air having ______ enthalpies and ____________ specific humidities are mixed.

a) similar, similar

b) different, similar

c) similar, different

d) different, different

View Answer

Explanation: The two air streams getting mixed adiabatically have different enthalpies and different specific humidities to get the final condition of air.

5. What is the value of m_{1} / m_{2} if h_{1} = 81 kJ/ kg of dry air, h_{2} = 46 kJ/ kg of dry air and h_{3} = 58 kJ/ kg of dry air?

a) 1

b) 0

c) 0.51

d) 0.52

View Answer

Explanation: As we know, m

_{1}/ m

_{2}= h

_{3}– h

_{2}/ h

_{1}– h

_{3}

= 58 – 46 / 81 – 58

= 0.5217 = 0.52.

6. What is the value of m_{1} / m_{2} if W_{1} = 0.0157 kg/ kg of dry air, W_{2} = 0.0084 kg / kg of dry air and W_{3} = 0.0103 kJ/ kg of dry air?

a) 0.31

b) 0.28

c) 0.35

d) 0.52

View Answer

Explanation: As we know, m

_{1}/ m

_{2}= W

_{3}– W

_{2}/ W

_{1}– W

_{3}

= 0.0103 – 0.0084 / 0.0157 – 0.0103

= 0.0019 / 0.0054 = 0.3518 = 0.35.

7. What is the value of one of the mass before mixing if the other mass is 1.98 kg, and the final is 4.22 kg?

a) 2.24

b) 1.98

c) 4.22

d) 6.20

View Answer

Explanation: As two streams are mixed, then the result of it is the summation of mixed masses, and using the energy balance ratio of masses can be obtained. So, m

_{3}= m

_{1}+ m

_{2}

If, m

_{1}= 1.98 and m

_{3}= 4.22 then, m

_{2}= 4.22 – 1.98 = 2.24 kg.

8. What is the value of final enthalpy if the enthalpies before mixing were 50 and 28 kJ / kg of dry air respectively and m_{2} / m_{1} = 2?

a) 38.67

b) 48.67

c) 40.05

d) 52

View Answer

Explanation: For energy balance, m

_{1}h

_{1}+ m

_{2}h

_{2}= m

_{3}h

_{3}

m

_{3}= m

_{1}+ m

_{2}

So, m

_{1}h

_{1}+ m

_{2}h

_{2}= (m

_{1}+ m

_{2}) h

_{3}

m

_{1}/ m

_{2}= h

_{3}– h

_{2}/ h

_{1}– h

_{3}

0.5 = h

_{3}– 28 / 50 – h

_{3}

50 – h

_{3}= 2h

_{3}– 56

3h

_{3}= 50 + 56 = 116

h

_{3}= 38.67.

9. Even a lesser quantity of steam may result in the formation of fog.

a) True

b) False

View Answer

Explanation: Fog also results when steam or very fine water spray is injected into the air in a greater quantity than required to do the saturation of the air. So, even less quantity of steam without mixing properly can result in fog.

10. Fog can be cleared by cooling.

a) True

b) False

View Answer

Explanation: Fog can be cleared by heating or mixing it with unsaturated air or mechanically separating water droplets from the air.

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