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galina1969 [7]
3 years ago
8

How is the air delivery temperature controlled during A/C operation?

Engineering
1 answer:
shepuryov [24]3 years ago
3 0
A IS THE CORRECT ANSWER
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1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet
Setler79 [48]

Answer:d

Explanation:

(d) chain drive belt drive gear drive

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3 years ago
Select the correct answer.
Lilit [14]

Answer:

D

Explanation:

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A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
PIT_PIT [208]

Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0
3 years ago
A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
Mazyrski [523]

Answer:

P > 142.5 N  (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If  Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

Then  we can get F (the force of friction) as follows

F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F  > 0

P  > 142.5 N (→)

the motion sliding

6 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
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