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3 years ago

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water at 20c discharges from a stroke tank through a 150m length of horizontal pipe. The pipe is smooth and has a diameter of 75

mm. The entrance of the pipe is square edged. What is depth of water needed to produce a volumetric flow rate of .1m3/s

1 answer:

inessss [21]3 years ago

7 0

Answer:

The dept of the water needed is 26.11m

Explanation:

Given the following parameters:

Length of the pipe = $150m$

Diameter of the pipe = $75mm = 0.75m$

Volumetric flow rate $1m^{3}/s$

Knowing that:

Volumetric flowrate $= area (A) * Velocity(V)$

==> $0.1 = A*V$ *Knowing that $V = \sqrt{2gh}$

==> $0.1 = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}$

==> $h = 26.11m$

Hence, the dept of water needed to produce a volumetric flow rate of $1m^{3}/s$ is 26.11m.

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Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

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ANSWERS:

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Explanation:

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$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

b)

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

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