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2 years ago

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

d point in the line, 18.5 m lower than the first, if the pipe diameter at the second point is twice that at the first. Remember that the density of water is 1000 kg/m3. Please give your answer in units of kPa.

1 answer:

ArbitrLikvidat [17]2 years ago

8 0

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

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