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 td_{1} = temperature at entry, td_{2} = temperature at exit and td_{3} = coil temperature?

a) td_{3} – td_{1} / td_{3} – td_{1}

b) td_{3} – td_{2} / td_{3} – td_{1}

c) td_{3} – td_{2} / td_{2} – td_{1}

d) td_{3} – td_{2} / td_{1} – td_{2}

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

= td

_{3}– td

_{2}/ td

_{3}– td

_{1}.

2. What is the By-pass factor for cooling coil, if td_{1} = temperature at entry, td_{2} = temperature at exit and td_{3} = coil temperature?

a) td_{3} – td_{1} / td_{3} – td_{1}

b) td_{3} – td_{2} / td_{3} – td_{1}

c) td_{3} – td_{2} / td_{2} – td_{1}

d) td_{2} – td_{3} / td_{1} – td_{3}

View Answer

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

= td

_{2}– td

_{3}/ td

_{1}– td

_{3}.

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

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 A_{c} t_{m}

b) U A_{c}

c) U t_{m}

d) U A_{c}^{2} t_{m}

View Answer

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 A

_{c}t

_{m}.

5. What is the formula of logarithmic mean temperature in terms of By-pass factor?

a) T_{m} = td_{2} – td_{1} / log_{e} [1/BPF]

b) T_{m} = td_{2} – td_{1} / log_{10} [BPF]

c) T_{m} = td_{2} – td_{1} / log_{e} [BPF]

d) T_{m} = td_{2} – td_{3} / log_{e} [BPF]

View Answer

Explanation: Logarithmic mean temperature difference for the given arrangement is,

T

_{m}= td

_{2}– td

_{1}/ log

_{e}[td

_{3}– td

_{1}/ td

_{3}– td

_{2}]

As, BPF for the coil = [td

_{3}– td

_{1}/ td

_{3}– td

_{2}]

T

_{m}= td

_{2}– td

_{1}/ log

_{e}[BPF].

6. What is the efficiency of the coil?

a) 1 + BPF

b) 1 / BPF

c) BPF

d) 1 – BPF

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.

7. What is the contact factor for heating coil, if td_{1} = temperature at entry, t_{2} = temperature at exit and td_{3} = coil temperature?

a) td_{3} – td_{1} / td_{3} – td_{1}

b) td_{2} – td_{1} / td_{3} – td_{1}

c) td_{3} – td_{2} / td_{2} – td_{1}

d) td_{3} – td_{2} / td_{1} – td_{2}

View Answer

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for heating coil,

η

_{H}= 1 – BPF

= 1 – [td

_{3}– td

_{2}/ td

_{3}– td

_{1}]

= td

_{2}– td

_{1}/ td

_{3}– td

_{1}.

8. What is the contact factor for cooling coil, if td_{1} = temperature at entry, td_{2} = temperature at exit and td_{3} = coil temperature?

a) td_{1} – td_{2} / td_{1} – td_{3}

b) td_{2} – td_{1} / td_{3} – td_{1}

c) td_{3} – td_{2} / td_{2} – td_{1}

d) td_{3} – td_{2} / td_{1} – td_{2}

View Answer

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for cooling coil,

η

_{C}= 1 – BPF

= 1 – [td

_{2}– td

_{3}/ td

_{1}– td

_{3}]

= td

_{1}– td

_{2}/ td

_{1}– td

_{3}.

9. A coil with low BPF has better performance.

a) True

b) False

View Answer

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

View Answer

Explanation: BPF for heating coil is, BPF = td

_{3}– td

_{2}/ td

_{3}– td

_{1}

BPF for cooling coil is, BPF = td

_{2}– td

_{3}/ td

_{1}– td

_{3}

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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