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goldenfox [79]
3 years ago
6

Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph

eric pressure at the surface is the same as in the submarine. (S.G. of sea water =1.025).
Engineering
1 answer:
kicyunya [14]3 years ago
6 0

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

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1 year ago
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
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Answer:

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Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

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- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

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    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

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          up = λ*f = 0.5169*100*10^6

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          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

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          ηc = 6.28*( 1 + j )

         

           

         

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