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3 years ago

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Wells drilled by a nonprofit called Water for South Sudan use a pump that can provide up to 5,500 gallons of water per day. Use

this total amount to calculate the following:
How many gallons per day per person could the well produce for a village with a population of 850 if every gallon went to domestic use?

1 answer:

Soloha48 [4]3 years ago

7 0

Answer:

6.47 gallon per person per day

Explanation:

Given

The total amount of water available for domestic use is 5500 gallon

Let us consider that the entire available water goes to domestic use.

The total population of the village is 850.

The amount of water available to each villager is

$\frac{5500}{850}\\6.47$ gallon per person per day

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2 years ago

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2 years ago

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil (cp = 2000 J/kg. K) f

romanna [79]

Answer:

A) $\epsilon = 0.67122$

B) $T_{h, out} = 44.795\ degree\ C$

Explanation:

Heat capacity of oil $C_h =m_h C_{ph} = 0.02 \times 2 = 0.04$ Kw/K

Heat capacity of air $C_C = m_C C_{pc} = 0.2 \times 1 = 0.2$ Kw/K

therefore $C_{min = C_h$

and $C_{max} = C_C$

we know that capacity ratio is

$c = \frac{C_{min}}{C_{max}} = 0.2$

$NTU = \frac{U A_s}{DC_{min}} = \frac{0.05 \times 1}{0.04} 1.25$

effectiveness is given as

$\epsilon = 1 -e^{\frac{NTU^0.22}{c} [ exp (-cNTU^{0.78}) -1]}$

$\epsilon = 1 -e^{\frac{1.25^0.22}{0.2} [ exp (-0.2\times 1.25^{0.78}) -1]}$

$\epsilon = 0.67122$

we knwo that actual heat Q is given

$Q = \epsilon \times Q_{max}$

$Q_H = \epsilon \times Q_{max}$

$Q_H = C_h (T_{h, in} -T_{h, out})$

$Q_H = \epsilon C_{min} (T_{h, in} - T_{c, in})$

$T_{h, out} = T_{h, in} - \epsilon (T_{h, in} - T_{c, in})$

$T_{h, out} = 75 - 0.67122(72 - 30)$

$T_{h, out} = 44.795\ degree\ C$

6 0

3 years ago