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ELGuapo6746
29.07.2020 •
Physics
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 second 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.
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Ответ:
The pressure at point 2 is![P_2 = 254.01 kPa](/tpl/images/0714/6090/3386a.png)
Explanation:
From the question we are told that
The speed at point 1 is![v_1 = 3.57 \ m/s](/tpl/images/0714/6090/8ac60.png)
The gauge pressure at point 1 is![P_1 = 68.7kPa = 68.7*10^{3}\ Pa](/tpl/images/0714/6090/9e07e.png)
The density of water is![\rho = 1000 \ kg/m^3](/tpl/images/0714/6090/e6ba8.png)
Let the height at point 1 be
then the height at point two will be
Let the diameter at point 1 be
then the diameter at point two will be
Now the continuity equation is mathematically represented as
Here
are the area at point 1 and 2
Now given that the are is directly proportional to the square of the diameter [i.e
]
which can represent as
=>![A = c d^2](/tpl/images/0714/6090/380a0.png)
where c is a constant
so![\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}](/tpl/images/0714/6090/99153.png)
=>![\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}](/tpl/images/0714/6090/eb0d7.png)
=>![A_2 = 4 A_1](/tpl/images/0714/6090/7e3b9.png)
Now from the continuity equation
=>![v_2 = \frac{v_1}{4}](/tpl/images/0714/6090/e146c.png)
=>![v_2 = \frac{3.57}{4}](/tpl/images/0714/6090/82470.png)
Generally the Bernoulli equation is mathematically represented as
So
=>![P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )](/tpl/images/0714/6090/65868.png)
substituting values
Ответ:
The answer would be A because he did some bad stuff
Explanation: