A 5.7 diameter horizontal pipe gradually narrows to 3.6 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 32.5 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

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A 5.7 diameter horizontal pipe gradually narrows to 3.6 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 32.5 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

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Answer 1 · 2018-02-01T07:01:44

The flow rate is $= 2.88 m^{3} s^{- 1}$ $=2.88m^3s^-1$

Explanation:

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Apply Bernoulli's Principle

$P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g z_{2}$ $P_1+1/2rhov_1^2+rhogz_1=P_2+1/2rhov_2^2+rhogz_2$

Since the Pipe is horizontal, $z_{1} = z_{2}$ $z_1=z_2$

So,

$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ $P_1+1/2rhov_1^2=P_2+1/2rhov_2^2$

The flow rate is constant

$Q = A_{1} \cdot v_{1} = A_{2} \cdot v_{2}$ $Q=A_1*v_1=A_2*v_2$

Where $v_{1}$ $v_1$ and $v_{2}$ $v_2$ are the velocities of water in the pipe.

Where $A_{1}, A_{2}$ $A_1, A_2$ are the cross sectional areas of the pipe

$A_{1} = π \frac{d_{1}^{2}}{4} = π \cdot \frac{{(5.7 \cdot 10^{- 2})}^{2}}{4}$ $A_1=pid_1^2/4=pi*(5.7*10^-2)^2/4$

$A_{2} = π \frac{d_{2}^{2}}{4} = π \cdot \frac{{(3.6 \cdot 10^{- 2})}^{2}}{4}$ $A_2=pid_2^2/4=pi*(3.6*10^-2)^2/4$

Therefore,

$v_{1} = \frac{A_{2}}{A_{1}} \cdot v_{2}$ $v_1=A_2/A_1*v_2$

$v_{1} = \frac{π \cdot \frac{{(3.6 \cdot 10^{- 2})}^{2}}{4}}{π \cdot \frac{{(5.7 \cdot 10^{- 2})}^{2}}{4}} \cdot v_{2}$ $v_1=(pi*(3.6*10^-2)^2/4)/(pi*(5.7*10^-2)^2/4)*v_2$

$v_{1} = {(\frac{3.6}{5.7})}^{2} v_{2}$ $v_1=(3.6/5.7)^2v_2$

$v_{1} = 0.4 v_{2}$ $v_1=0.4v_2$

The pressures are

$P_{1} = 32.5 k P a$ $P_1=32.5kPa$

$P_{2} = 24 k P a$ $P_2=24kPa$

And the density of water is $ρ = 1000 k g m^{- 3}$ $rho=1000kgm^-3$

Therefore,

$32.5 \cdot 10^{3} + \frac{1}{2} \cdot 1000 \cdot 0.4 v_{2} = 24 \cdot 10^{3} + \frac{1}{2} \cdot 1000 \cdot v_{2}$ $32.5*10^3+1/2*1000*0.4v_2=24*10^3+1/2*1000*v_2$

$v_{2} (\frac{1}{2} \cdot 1000 - \frac{1}{2} \cdot 1000 \cdot 0.4) = (32.5 - 24) \cdot 10^{3}$ $v_2(1/2*1000-1/2*1000*0.4)=(32.5-24)*10^3$

$v_{2} \cdot \frac{1}{2} \cdot 1000 \cdot 0.6 = 8.5 \cdot 1000$ $v_2*1/2*1000*0.6=8.5*1000$

$v_{2} = \frac{2 \cdot 8.5}{0.6} = 28.33 m s^{- 1}$ $v_2=(2*8.5)/0.6=28.33ms^-1$

Finally,

The flow rate is $Q = A_{2} v_{2} = π \cdot \frac{{(3.6 \cdot 10^{- 2})}^{2}}{4} \cdot 28.33$ $Q=A_2v_2=pi*(3.6*10^-2)^2/4*28.33$

$= 2.88 m^{3} s^{- 1}$ $=2.88m^3s^-1$

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A 5.7 diameter horizontal pipe gradually narrows to 3.6 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 32.5 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

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