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USPshnik [31]
3 years ago
14

A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a

t a constate rate of 0.05 cubic meters per second, the drawdowns at distances of 50 and 150 meters from the well were observed to be three and 1.2 meters respectively. Computer the hydraulic conductivity of the aquifer and the radius of influence of the pumping well.

Engineering
1 answer:
GarryVolchara [31]3 years ago
3 0

Answer:

Explanation:

Find attach the solution

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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Data;

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  • T = 288.2K
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<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

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Which of the following explains why material properties present challenges for engineers?
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