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vovikov84 [41]
3 years ago
7

A horizontal, opaque surface at a steady-state temperature of 80°C is exposed to an airflow having a free stream temperature of

25°C with a convection heat transfer coefficient of 20 W/m2 ·K. The emissive power of the surface is 628 W/m2 , the irradiation is 1380 W/m2 , and the reflectivity is 0.30.
Determine the absorptivity of the surface.

Determine the net radiation heat transfer rate for this surface. Is this heat transfer to the surface or from the surface?

Determine the com-bined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

Engineering
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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8. Air at 25C, 100 kPa and air at 50C, 200 kPa at 1 to 1 volume ratio are mixed inside an adiabatic compressor to 55C, 500 kPa a
quester [9]

Answer:

Workdone, w = 68.935 kJ

Explanation:

m1h1 + m2h2 + Q = m3h3 - Wc

Final mass of mixture,

m3 = m1 + m2 = 5kg

p1V1 = m1R1T1

p2V2 = m2R2T2

Since they are at 1:1,

V1 = V2 and R1 = R2

Comparing equations,

p1/p2 = (m1/m2) x (T1/T2)

m1/m2 = (100/200) x ((50 + 273)/(25+273))

m1/m2 = 0.542

m1 = 0.542m2

m3 = m1 + m2

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m1 = 1.757 kg

Workdone is given as,

Wc = m3h3 - m1h1 - m2h2

h = Cp x T, since air is an ideal gas

Cp = 1.005kJ/kGK

Wc = (5 x 1.005 x (55+273)) - (1.757 x 1.005 x 298) -(3.243 x 1.005 x 323)

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/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a
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Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

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k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

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