Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 1 (Q.No. 41)
41.
The equivalent diameter for pressure drop is __________ that for heat transfer.
Discussion:
3 comments Page 1 of 1.
Surya said:
9 years ago
Let's take D1 = 2cm and D2 = 4cm.
CASE 1 for pressure drop;
De = 4 * crossectional area/wetted perimeter.
=> 4 * (pi/4) * (D2^2-D1^2) /pi* (D1 + D2).
=> 2cm.
CASE 2 for heat transfer;
De = 4 * (pi/4) * (D2^2 - D1^2) /pi * D1.
=> 6cm.
So De is more for heat transfer.
CASE 1 for pressure drop;
De = 4 * crossectional area/wetted perimeter.
=> 4 * (pi/4) * (D2^2-D1^2) /pi* (D1 + D2).
=> 2cm.
CASE 2 for heat transfer;
De = 4 * (pi/4) * (D2^2 - D1^2) /pi * D1.
=> 6cm.
So De is more for heat transfer.
(1)
Shubham said:
10 years ago
Equivalent Diameter in Case of Pressure Drop is D2-D1 whereas in case of Heat Transfer is D2^2-D1^2/D1.
(1)
Prasanth said:
1 decade ago
Please explain any one?
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