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denis-greek [22]
3 years ago
10

A room in the lower level of a cruise ship has a 30 cm diameter circular window If the midpoint of the window is 4 m below the w

ater surface, determine the hydrostatic force in Newtons acting on the window. Answer 2840N.
Engineering
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

F = 2840.3 N

Explanation:

Given:

- Diameter of window D = 0.3 m

- Midpoint of window from sea level h = 4 m

- Specific gravity of sea water S.G = 1.024

- Density of water p = 1000 kg/m^3

Find:

The hydro-static force F_r acting on the mid-point of the window.

Solution:

- The average pressure P acting on the midpoint of the window:

                               P = S.G p*g*h

                               P = 1.024*1000*9.81*4

                               P = 40181.76

- The hydro-static force F_r acting on the mid-point of the window:

                               F = P*A = P*pi*D^2 / 4

                               F = 40181.76*pi*0.3^2 / 4

                               F = 2840.3 N

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Answer:

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4 0
4 years ago
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blagie [28]

Answer:

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3 years ago
The Cv factor for a valve is 48. Compute the head loss when 30 GPM of water passes through the valve.
dlinn [17]

Answer:

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Explanation:

Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.

Given:

Factor cv is 48.

Flow rate of water is 30 GPM.

GPM means gallon per minute.

Calculation:

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Expression for head loss for the water is given as follows:

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Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.

Step2

Substitute 48 for cv and 30 for Q in above equation as follows:

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h = 0.390625 psi.

Thus, the head loss in Psi is 0.390625 psi.

 

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