This set of Air-Conditioning online test focuses on “Psychrometry – Heating and Cooling Factor – 2”.
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
View Answer
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
= 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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, Ac = 11 m2 and tm = 0°C?
a) 1
b) 0
c) 594594
d) 543636
View Answer
Explanation: 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 Ac tm = 198 x 11 x (273 + 0) = 594594.
6. What is the value of BPF, if U = 1.9 W/m2 K, Ac = 1.1 m2 and ma = 0.97 kg?
a) 0.14
b) 0.12
c) 0.13
d) 0.15
View Answer
Explanation: BPF = e-[U Ac / 1.022 ma] = 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
View Answer
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]
= 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
View Answer
Explanation: 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
View Answer
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
= 47 – 32 / 47 – 11
= 0.4167
ηH = 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
View Answer
Explanation: 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
ηC = 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
View Answer
Explanation: 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
View Answer
Explanation: 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.
Sanfoundry Global Education & Learning Series – Refrigeration & Air Conditioning.
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