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Trava [24]
3 years ago
14

6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20

,000-liter cistern has been installed during construction of a new building. The cistern collects water from an HVAC (heating, ventilation, and air-conditioning) system designed to provide 2830 cubic meters of air per minute at 22°C and 50% relative humidity after converting it from ambient conditions (31°C, 70% relative humidity). The collected condensate serves as the source of water for lawn maintenance. Estimate (a) the rate of intake of air at ambient conditions in cubic feet per minute and (b) the hours of operation required to fill the cistern
Engineering
1 answer:
musickatia [10]3 years ago
3 0

<u>Explanation:</u>

At temperature is 33^{\circ} C and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is 33^{\circ} C and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}

Now , we going to calculate the volume,

\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}

The time which is required to fill the cistern can be calculated as follows:

Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)

Now, putting the value in above formula we get,

\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.

Therefore, the hours required to fill the cistern is 4.65 hours.

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8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of
san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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UkoKoshka [18]

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The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
mojhsa [17]

Answer:

A) Upper bound modulus of elasticity; E = 165.6 GPa

B) Lower bound modulus of elasticity; E = 83.09 GPa

Explanation:

A) Formula for upper bound modulus is given as;

E = E_m(1 - V_f) + E_f•V_f

We are given;

E_m = 60 GPa

E_f = 380 GPa

V_f = 33% = 0.33

Thus,

E = 60(1 - 0.33) + 380(0.33)

E = (60 x 0.67) + 125.4

E = 165.6 GPa

B) Formula for lower bound modulus is given as;

E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]

E = 1/[(0.33/380) + ((1 – 0.33)/60)]

E = 1/(0.0008684 + 0.01116667)

E = 1/0.01203507

E = 83.09 GPa

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Answer:

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