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Water is flowing through a horizontal pipe of constant diameter and the flow is laminar. If the diameter of the pipe is increased by 50% keeping the volume flow rate constant, then the pressure drop in the pipe due to friction will decrease by
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- 33%
- 50%
- 70%
- 80%
Correct Option: D
| ∆ P = | ||
| ∆² |
| Q = AV = | d²V | |
| 4 |
| V = | ||
| πd² |
| ∆P = | ||
| πD4 |
Above equation shows that
| ∆P ∝ | ||
| D4 |
| ∆P1 = | = | where D1 = D | ||
| D41 | D4 |
| & ∆P2 = | = | where D2 = 1.5D | ||
| D42 | (d42) |
% change in pressure drop
| % = | × 100 | |
| ∆P1 |
= 80.24% ≃ 80%
