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charle [14.2K]
3 years ago
8

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

ciency of 90 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at 200 m depth to be exposed to the atmosphere.
Engineering
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine \rho =1050kg/m^3

Rate at which brine is pumped V=0.3m^3/sec

So mass of the pumped per second

Mass = volume × density = 1050\times 0.3=315 kg/sec

Acceleration due to gravity g=9.8m/sec^2

Depth h = 200 m

So work done W=mgh=315\times 9.8\times 200=617400J

Efficiency is given \eta =0.9

We have to fond the input power

So input power =\frac{617400}{0.9}=686000J

So input power of the geothermal power will be 686000 J

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If you measure 0.7 V across a diode, the diode is probably made of
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Answer:

Made of Silicon.

Explanation:

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Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
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The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

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  • <u>O₂: 0.5 kmol/h</u>

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