Question

Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q

Answer 1

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta P = \dfrac{128 \mu L Q}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D = D_A = D_B$

where;

length of the first pipe A $L_A = L$ and the length of the second pipe B $L_B = 2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}$

$\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}$

$\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}$

$\mathbf{Q_1 = 2Q}$

Answer 2

The flow rate in pipe B is 2Q of the flow rate of the pipe A

What is flow rate?

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta p=\dfrac{128\mu LQ}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D=D_A=D_B$

where;

length of the first pipe A  $L_A=L$ and the length of the second pipe B  

$L_B=2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}$

$\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}$

$\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}$

$Q_1=2Q$

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

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