# Air-Conditioning Questions and Answers – Psychrometry – Heating and Cooling Factor – 1

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This set of Air-Conditioning Multiple Choice Questions & Answers (MCQs) focuses on “Psychrometry – Heating and Cooling Factor – 1”.

1. What is the By-pass factor for heating coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td3 – td2 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2

Explanation: 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.

2. What is the By-pass factor for cooling coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td3 – td2 / td3 – td1
c) td3 – td2 / td2 – td1
d) td2 – td3 / td1 – td3

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

3. If the value of BPF for one row of a coil is y then what is the value of BPF for n similar rows?
a) n / y
b) n + y
c) (y)n
d) n x y

Explanation: BPF for multiple similar rows of a coil is the power of one BPF to the number of rows. i.e. (y)n.

4. What is the value of sensible heat given out by the coil?
a) U Ac tm
b) U Ac
c) U tm
d) U Ac2 tm

Explanation: Sensible heat given out by the coil is the product of overall heat transfer coefficient, the surface area of the coil and logarithmic mean temperature difference.
So, Sensible heat = U Ac tm.

5. What is the formula of logarithmic mean temperature in terms of By-pass factor?
a) Tm = td2 – td1 / loge [1/BPF]
b) Tm = td2 – td1 / log10 [BPF]
c) Tm = td2 – td1 / loge [BPF]
d) Tm = td2 – td3 / loge [BPF]

Explanation: Logarithmic mean temperature difference for the given arrangement is,
Tm = td2 – td1 / loge [td3 – td1 / td3 – td2]
As, BPF for the coil = [td3 – td1 / td3 – td2]
Tm = td2 – td1 / loge [BPF].

6. What is the efficiency of the coil?
a) 1 + BPF
b) 1 / BPF
c) BPF
d) 1 – BPF

Explanation: As the by-pass factor is the inefficiency, so the contact factor or efficiency of the coil is given by 1 – BPF.

7. What is the contact factor for heating coil, if td1 = temperature at entry, t2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td2 – td1 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for heating coil,
ηH = 1 – BPF
= 1 – [td3 – td2 / td3 – td1]
= td2 – td1 / td3 – td1.

8. What is the contact factor for cooling coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td1 – td2 / td1 – td3
b) td2 – td1 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for cooling coil,
ηC = 1 – BPF
= 1 – [td2 – td3 / td1 – td3]
= td1 – td2 / td1 – td3.

9. A coil with low BPF has better performance.
a) True
b) False

Explanation: As BPF is the amount of air bypassed in the process and is the inefficiency of the coil. So, lower the value of BPF then better is the performance and vice-versa.

10. The value of BPF for the heating and the cooling coil is different under the same temperature conditions.
a) True
b) False

Explanation: BPF for heating coil is, BPF = td3 – td2 / td3 – td1
BPF for cooling coil is, BPF = td2 – td3 / td1 – td3
As the numerator and denominator are reversed in the cooling coil than the heating coil. So, the value we get for either of the coils is the same.

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