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ddd [48]
2 years ago
8

1. An ideal gas undergoes a process according to PV" = k. Initially, the temperature and volume are 2 bar and 0.1m?, respectivel

y. After the process, the volume doubles. Determine the work done during the process. Take n = 1.4 ​
Engineering
1 answer:
ella [17]2 years ago
8 0
Tryna boost my score for college stuff could you give me the brainiest and a thanks? Hope you find your answer your looking for!
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Which element of Miranda's character is best illustrated by this excerpt?
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B. She is compassionate

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Explain the LWD process why is it important in drilling operations?
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Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.

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Drag each item to show if it is an element or not an element.
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5 0
3 years ago
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.

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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3 years ago
Steam enters an adiabatic turbine at 400◦C, 2 MPa pressure. The turbine has an isentropic efficiency of 0.9. The exit pressure i
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