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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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1)A wheel is used to turn a valve stem on a water valve. If the wheel radius is 1 foot and the stem, (axle), radius is .5 inches
Novay_Z [31]

Answer:

1.) 2.4

2.) 112 lbs

3.) 7.85 inches

4.) 6 lbs

5.) 2 lbs

6.) 67%

Explanation:

Given that

Radius of the wheel R = 1 foot

1 foot = 12 inches.

Radius of the axle r = 5 inches

1.) The mechanical advantage MA is :

MA = R/r = 12/5 = 2.4

2.) How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1?

MA = Load / effort

Where effort = 80 lbs

Substitute MA and effort into the formula

2.4 = Load / 80

Cross multiply

Load = 2.4 × 80 = 192 lbs

The resistance force to be overcome will be

Force = load - effort

Resistance force = 192 - 80 = 112 lbs

3) What is the linear distance traveled when a 2.5' diameter wheel makes one revolution

One revolution = 2π

Radius = 2.5 /2 = 1.25 inches

Linear distance S = angular distance Ø × radius r

S = Ør

S = 2π × 1.25

S = 7.85 inches

4. ) given that

Wheel radius R = 4

Axle radius r = 1

MA = 4/1 = 4

MA = Load / effort

4 = 24/ effort

Effort = 24/4 = 6 lbs

5.) 6 - 4 = 2lb

6.) Efficiency = MA / VR × 100

Efficiency = 4 / 6 × 100

Efficiency = 67%

3 0
3 years ago
Engineered lumber should not be used for
Dimas [21]

Answer:

Composite panel garage doors

Explanation:

8 0
2 years ago
What parts of a ship have plastic? And what type of plastic?
goblinko [34]

Answer:

I know ship has stream line body

3 0
3 years ago
1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and
belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = \frac{wet mass - dry mass }{wet mass} = \frac{0.525 - 0.49}{0.525} = 0.07 or 7%

water content =  \frac{wet mass - dry mass}{wet mass} = 7%

8 0
3 years ago
Read 2 more answers
A 10 hp motor is used to raise a 1000 Newton weight at a vertical distance of 5 meters. What is the work the motor performs?
Aleksandr [31]

The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.

Define work. Explain the rate of doing work.

Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.

Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.

Solution Explained:

Given,

Weight = 1000N and distance = 5m

A/Q, the work here is done in lifting then

Work = (weight) × (distance moved)

         = 1000 X 5

         = 5000Nm or 5000J = 5kJ

Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

To learn more about work, use the link given
brainly.com/question/25573309
#SPJ9

<u />

4 0
1 year ago
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