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Sunny_sXe [5.5K]
3 years ago
9

9. How is the Air Delivery temperature controlled during A/C operation?

Engineering
1 answer:
Rzqust [24]3 years ago
6 0

Answer:

i believe the answer is a but i could be wrong

Explanation:

i hope it helps

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Identify SIX (6) objectives of maintenance.<br>​
Rasek [7]

Answer:

to optimize the reliability of equipment and infrastructure;

- to ensure that equipment and infrastructure are always in good condition;

- to carry out prompt emergency repair of equipment and infrastructure so as to secure the best possible availability for production;

- to enhance, through modifications, extensions, or new low-cost items, the productivity of existing equipment or production capacity;

- to ensure the operation of equipment for production and for the distribution of energy and fluids;

- to improve operational safety;

- to train personnel in specific maintenance skills;

- to advise on the acquisition, installation and operation of machinery;

- to contribute to finished product quality;

- to ensure environmental protection.

Explanation:

pick whichever you want

8 0
3 years ago
Question 9 of 25
mafiozo [28]

Answer:

D

Explanation:

took test failed question D is the right answer

3 0
3 years ago
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0
3 years ago
A growing trend in urban design is the concept of a rooftop garden. If every building in a city were to install a rooftop garden
vlabodo [156]

Answer:The Urban heat island temperature will be REDUCED.

Two Impacts of Rooftop gardens

1) provision of shade against Sunlight.

2) It helps to purify the air around the building.

Explanation: Rooftop gardens are gardens made on top of the roofs of buildings, it is a Green initiative aimed at helping to improve the overall Environment.

Rooftop gardens have several significant benefits which includes

Reduction of the surrounding temperatures and the Urban heat Island temperatures.

Rooftop gardens helps to shade the roof from the direct impacts of harsh weather conditions.

Generally, plants are known as air purifiers as they remove the excess Carbondioxide around the environment through photosynthesis, and they also help to release water vapor which will help to improve the humidity of the environment.

5 0
3 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
3 years ago
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