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For steady, fully developed flow Inside straight pipe of diameter D, neglecting gravity effects, the pressure drop ∆p over a length L and the wall shear stress τW are related by
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τW = ∆pD 4L -
τW = ∆pD² 4L² -
τW = ∆pD 2L -
τW = 4∆pD D
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Correct Option: A
Shear stress on wall,
τW = | . | δx | 2 |
where | . | δx | L |
& R = | 2 |
∴ τW = | × | = | L | 2 x 2 | 4L |