Air-Conditioning Multiple Choice Questions & Answers (MCQs) 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}

Answer: b

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

= 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}

Answer: d

Clarification: 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

Answer: c

Clarification: 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}

Answer: a

Clarification: 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]

Answer: c

Clarification: 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

Answer: d

Clarification: 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}

Answer: b

Clarification: 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}

Answer: a

Clarification: 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

Answer: a

Clarification: 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

Answer: b

Clarification: 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.