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
View Answer

Answer: b
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.
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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
View Answer

Answer: d
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
View Answer

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

Answer: a
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]
View Answer

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

Answer: d
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
View Answer

Answer: b
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.
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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
View Answer

Answer: a
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
View Answer

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

Answer: b
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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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter