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Trava [24]
2 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]2 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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Use superpositions find​
Sunny_sXe [5.5K]

Answer:

no

Explanation:

no

3 0
3 years ago
Your study space does not need to be quiet as long as you can ignore any noise coming from the space true or false?
Makovka662 [10]

Answer:

False

Explanation:

When you're studying, you need to make sure that you can focus properly. This means that you shouldn't be hungry or too full and that you should be well-rested, in a quiet room with good lighting and no distractions. Noise is never good when you need to memorize something. Some people can partially ignore it as long as it isn't too loud, but it will begin to bother them eventually. That's why it's better to study in a quiet room.

3 0
2 years ago
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A ________ can be installed in a cast-iron block to repair a worn or cracked cylinder. Question 24 options:
VLD [36.1K]

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Cylinder sleeve

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7 0
2 years ago
Determine the following parameters for the water having quality x=0.7 at 200 kPa:
ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

   Temperature = 120.21°C

c). Specific volume

  $v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

  $v_x=v_f+x(v_g-v_f)$

       $=0.001061+0.7(0.88578-0.001061)$

       $=0.62036 \ m^3/kg$

d). Specific energy (u_x)

    $u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

   $u_x=504.5 + 0.7(2024.6)$

         $=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

   At $h_f = 504.71, \ \ h_{fg} = 2201.6$

   h_x=504.71+(0.7\times 2201.6)

        $= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

  $H=mh_x$

       $= 0.5 \times 2045.83$

       = 1022.91 kJ

7 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
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