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
Discussion:
5 comments Page 1 of 1.
Arun said:
5 years ago
@Sanjay.
You are correct. As per the question, this is heating coil so the answer is option B.
You are correct. As per the question, this is heating coil so the answer is option B.
Shivaprasad said:
6 years ago
By pass factor Basic Fanda output/input = Effieiency = BPF.
Debashis said:
7 years ago
You are right @Swapnil.
Swapnil said:
9 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.
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.
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.
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