250+ TOP MCQs on Psychrometry – Heating and Cooling Factor and Answers

Air-Conditioning online test on “Psychrometry – Heating and Cooling Factor”.

1. What is the By-pass factor for heating coil, if td₁ = 19°C, td₂ = 25°C and td₃ = 37°C?
a) 0.75
b) 0.4
c) 0.7
d) 0.1

Answer: c
Clarification: By-pass factor is the amount of air by-passed in the process. So, for heating coil,
BPF = Temperature difference between coil and exit / Temperature difference between coil and entry
= td₃ – td₂ / td₃ – td₁
= 37 – 25 / 37 – 19
= 0.70588 = 0.7.

2. What is the By-pass factor for cooling coil, if td₁ = 49°C, td₂ = 36°C and td₃ = 30°C?
a) 0.315
b) 0.31
c) 0.320
d) 0.3

Answer: a
Clarification: By-pass factor is the amount of air by-passed in the process. So, for the cooling coil,
BPF = Temperature difference between exit and coil / Temperature difference between entry and coil
= td₂ – td₃ / td₁ – td₃
= 36 – 30 / 49 – 30
= 0.315.

3. If the value of BPF for one row of the coil is one, then what is the value of BPF for 3872 similar rows?
a) 3872
b) 3871
c) 3873
d) 1

Answer: d
Clarification: BPF for multiple similar rows of the coil is the power of one BPF to the number of rows. i.e. (y)ⁿ
So, (1)³⁸⁷² = 1, for any value of n, the value comes one only.

4. If the value of BPF for one row of the coil is 0.4, then what is the value of BPF for 12 similar rows?
a) 0.00000167
b) 0.0000167
c) 0.00167
d) 0.000167

Answer: b
Clarification: BPF for multiple similar rows of the coil is the power of one BPF to the number of rows. i.e. (y)ⁿ
So, (0.4)¹² = 0.0000167.

5. What is the value of sensible heat given out by the coil, if U = 198 W/m² K, A_c = 11 m² and t_m = 0°C?
a) 1
b) 0
c) 594594
d) 543636

Answer: b
Clarification: Sensible heat given out by the coil is the product of the overall heat transfer coefficient, the surface area of the coil, and logarithmic mean temperature difference.
So, Sensible heat = U A_c t_m = 198 x 11 x (273 + 0) = 594594.

6. What is the value of BPF, if U = 1.9 W/m² K, A_c = 1.1 m² and m_a = 0.97 kg?
a) 0.14
b) 0.12
c) 0.13
d) 0.15

Answer: b
Clarification: BPF = e^{-[U A_c / 1.022 m_a]} = e^{-[1.9 x 1.1/ 1.022 x 0.97]} = e^-[2.1082] = 0.12.

7. What is the value of logarithmic mean temperature if td₁ = 300 K, td₂ = 400 K and BPF = 0.75?
a) 70°C
b) 79°C
c) 85°C
d) 75°C

Answer: d
Clarification: Logarithmic mean temperature difference for the given arrangement is,
T_m = td₂ – td₁ / log_e [td₃ – td₁ / td₃ – td₂]
As, BPF for the coil = [td₃ – td₁ / td₃ – td₂]
T_m = td₂ – td₁ / log_e [BPF]
= 400 – 300 / ln (1/0.75) = 100 / 0.287 = 348.43 K = 75.43°C = 75°C.

8. What is the efficiency of the coil, if BPF = 0.6913?
a) 1.6913
b) 0.4467
c) 0.3087
d) 1.4467

Answer: c
Clarification: As the by-pass factor is the inefficiency, so the contact factor or efficiency of the coil is given by 1 – BPF = 1 – 0.6913 = 0.3087.

9. What is the contact factor for heating coil, if td₁ = 11°C, td₂ = 32°C and td₃ = 47°C?
a) 0.5833
b) 0.4833
c) 0.7833
d) 0.1833

Answer: a
Clarification: By-pass factor is the amount of air by-passed in the process. So, for heating coil,
BPF = Temperature difference between coil and exit / Temperature difference between coil and entry
= td₃ – td₂ / td₃ – td₁
= 47 – 32 / 47 – 11
= 0.4167
η_H = 1 – BPF = 1 – 0.4167 = 0.5833.

10. What is the contact factor for cooling coil, if td₁ = 44°C, td₂ = 26°C and td₃ = 19°C?
a) 0.62
b) 0.72
c) 0.78
d) 0.18

Answer: b
Clarification: By-pass factor is the amount of air by-passed in the process. So, for the cooling coil,
BPF = Temperature difference between coil and exit / Temperature difference between coil and entry
= td₂ – td₃ / td₁ – td₃
= 26 – 19 / 44 – 19
= 0.28
η_C = 1 – BPF = 1 – 0.28 = 0.72.

11. For given conditions, td₁ = 288 K, td₂ = 301 K and td₃ = 314 K, the value of by-pass factor and contact factor is equal.
a) True
b) False

Answer: a
Clarification: BPF = 314 – 301 / 314 – 288 = 13 / 26 = 0.5 and η = 1 – BPF = 1 – 0.5 = 0.5
So, for given set of values the value of contact factor and by-pass factor is equal.

12. Higher the value of contact factor, poor is the performance of the coil.
a) True
b) False

Answer: b
Clarification: The contact factor is the efficiency of the coil. So, if the value of it is higher, then better is the performance and vice-versa. It is the opposite of BPF.