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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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The drag force, Fd, imposed by the surrounding air on a
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a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

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<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

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3 years ago
N DevOps, high levels of automation are expected, which increases productivity. Which fact illustrates this productivity increas
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Answer:

Less intervention of humans.

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