# Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion

Discussion Forum : Heat Transfer, Refrigeration and Air Conditioning - Section 5 (Q.No. 9)
9.
The atmospheric air at dry bulb temperature of 15°C enters a heating coil maintained at 40°C. The air leaves the heating coil at 25°C. The by-pass factor of the heating coil is
0.376
0.4
0.6
0.67
Explanation:
No answer description is available. Let's discuss.
Discussion:
5 comments Page 1 of 1.

Arun said:   4 years ago
@Sanjay.

You are correct. As per the question, this is heating coil so the answer is option B.

By pass factor Basic Fanda output/input = Effieiency = BPF.

Debashis said:   6 years ago
You are right @Swapnil.

Swapnil said:   8 years ago
By pass factor= (tb2 - tb3)/(tb1 - tb3).

So,
tb1 = ambient air temperature = 15,
tb2 = temperater of air leaving the coil's = 25.
tb3 = heating coil temperature = 40.

By solving.

(25 - 40)/(15 - 40) = 0.6.

Sanjay kumar said:   1 decade ago
By-pass factor of the heating coil is = x,

x=((td3-td2)/(td3-td1)).

Where,

td3= DBT of the heating coil.
td1= DBT of air entering the heating coil.
td2= DBT of air leaving the heating coil.