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Homework Statement
A pipe 30 cm in diameter and 420 m long has a wall thickness of 1 cm. The pipe is commercial steel (ks = 0.045 mm) and carries water from a reservoir to an elevation 100 m below the bottom of the reservoir and discharges into the air. A rotary valve is installed at the downstream end. Minor losses include an entry (kL = 0.5) and rotary valve (kL = 10). Calculate the maximum water hammer pressure that can be expected on the valve if it closes in 0.5 sec. Also determine the total pressure the pipeline will be exposed to during the water hammer phenomenon. Assume: K = 2.2 x 109 N/m2, E = 1.9 x 1011 N/m2
Homework Equations
I am having trouble with determining the water pipe velocity and pipe pressure.
The Bernoulli:
$$ \frac{P_1}{2g} + \frac{v_1^2}{2g}+h_1 = \frac{P_2}{2g} + \frac{v_2^2}{2g}+h_2+ \frac{v^2}{2g}(\lambda \cdot L/D + \sum K_L) $$
The Attempt at a Solution
I am having a really hard with this problem. I think we need more details about the reservoir depth to solve the problem.
Applying the Bernoulli equation from the bottom of the reservoir to the exit of the pipe:
$$ \frac{P_{pipe}}{2g} + \frac{v_{pipe}^2}{2g}+h_1 = \frac{P_{exit}}{2g} + \frac{v_{exit}^2}{2g}+h_2+ \frac{v_{pipe}^2}{2g}(\lambda \cdot L/D + \sum K_L) $$
Exit pressure is atmospheric, and velocity in pipe and at exit are equal.
$$\frac{P_{pipe}}{2g} + \frac{v_{pipe}^2}{2g} = \frac{v_{exit}^2}{2g} - 100 + \frac{v_{pipe}^2}{2g}(\lambda \cdot 450/0.3 + 10.5) $$
So I get this equation with two unknowns velocity ( and lambda depends on v ...) and pipe pressure. Can some one please help - this is so frustrating to solve...
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