Answer:
The diameter of the second pipe is [tex]D_2 = 1.095 D[/tex]
Explanation:
From the question we are told that
The length of the connected pipe is [tex]d = 2L[/tex]
The pressure drop for the first pipe is [tex]\Delta p __{1}} = 1.44* \Delta p__{2}}[/tex]
The diameter of the pipe is [tex]D[/tex]
The rate at which the fluid flows for laminar flow is mathematically represented as
[tex]\r m = \frac{\pi D^4 \Delta p}{128 \mu L}[/tex]
Where L is the length of the pipe
[tex]\mu[/tex] is the dynamic viscosity
[tex]\Delta p[/tex] is the difference in pressure
[tex]\r m[/tex] is the flow rate of the fluid
From the equation of continuity
[tex]\r m_ 1 = \r m_2[/tex]
Where [tex]\r m_1[/tex] is the flow rate in pipe one
[tex]\r m_2[/tex] is the flow rate in pipe two
So
[tex]\frac{\pi D^4 \Delta p_1}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]
Where [tex]D_2[/tex] is the diameter of the second pipe
=> [tex]\frac{\pi D^4 (1.44 \Delta p_2)}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]
=> [tex]1.44 D^4 = D_2 ^4[/tex]
[tex]D_2 =\sqrt[4]{ \frac{D ^4 }{1.44 } }[/tex]
[tex]D_2 = 1.095 D[/tex]
