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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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A commercial refrigerator with refrigerant -134a as the working fluid is used to keep the refrigerated space at -30C by rejectin
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Answer:

a) 0.487

b) refrigeration load = 5.46w

c) cop = 2.24

d)ref load max = 12.43kw

Explanation:

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The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
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Answer:

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

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3 years ago
What parts of a ship have plastic? And what type of plastic?
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3 years ago
A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging
dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

\alpha = \frac{k}{\rho c}

           = \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})

\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]

0.909 = erf{\frac{3.8245}{\sqrt{t}}}

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]

0.08073 = \frac{3.8245}{\sqrt{t}}

solving fot t we get

t = 2244.3 sec

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3 years ago
​What should be a concern as a weldment becomes larger as more parts are added?
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