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Nesterboy [21]
3 years ago
8

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
Engineering
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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Answer:

A) About 9.273 \times 10^{-9} newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) F_g=\dfrac{GM_1M_2}{r^2} , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

3 0
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BabaBlast [244]

Answer:

a)

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b)

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c)

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Answer:

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