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3 years ago

The themes around which social sciences texts are organized boost understanding by

Engineering

1 answer:

Nadya [2.5K]3 years ago

7 0

Answer:

Facilitating an immensely focused setting where one does not get carried away
Creates an easier means to develop essays in relation to targeted themes
Developing "umbrellas" that one can create sub groups that a reader comes across
Instilling a directive path for vocabulary reading

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

kicyunya [14]

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.

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

$\bigtriangleup p=3217.68$ kpa.

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

6 0

3 years ago

A certain heat pump produces 200 kW of heating for a 293 K heated zone while only using 75 kW of power and a heat source at 273

vodka [1.7K]

Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

8 0

3 years ago

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

damaskus [11]

Answer:

$P_2-P_1=27209h$

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

$P_1+p_1gh_1=p_2_gh_2+P_2$