This question was previously asked in

ISRO LPSC Technical Assistant Mechanical held on 23/02/2020

- 25
- 50
- 75
- 100

Option 1 : 25

Free

CT 1: Growth and Development - 1

46784

10 Questions
10 Marks
10 Mins

**Concept:**

For fully developed laminar flow in the pipe

\(U_{max} =-{1\over 4μ}({\partial p\over \partial x})R^2\)

where U_{max} = Maximum velocity, μ = dynamic viscosity, \({\partial p\over \partial x} = axial~pressure~gradient\), R = radius of the pipe

**Calculation:**

**Given:**

μ = 0.001 Pa-s, \({\partial p\over \partial x} = -10~ Pa/m\), R = 0.1 m, Umax = ?

For fully developed laminar flow in the pipe

\(U_{max} =-{1\over 4μ}({\partial p\over \partial x})R^2\)

\(U_{max} =-{1\over 4\times 0.001}(-10)(0.1)^2\)

**Umax = 25 m/s**

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