# 17
2 answers:
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Answer:
a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr
Explanation:
Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2
Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr
1W = 3.41 BTU/hr
Given parameters:
thickness, t = 7.5mm = 7.5/1000 = 0.0075m
Temperatures 150 C = 150 + 273 = 423 K
50 C = 50 + 273 = 323 K
Temperature difference, T = 423 - 323 = 100 K
We are assuming steady heat flow;
a.) Heat flux, Q" = kT/t
K= thermal conductivity of the material
The thermal conductivity of brass, k = 109.0 W/m.K
Heat flux,
b.) Area of sheet, A = 0.5m2
Heat loss, Q = kAT/t
Heat loss,
Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr
c.) Material is now given as soda lime glass.
Thermal conductivity of soda lime glass, k is approximately 1W/m.K
Heat loss,
Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr
d.) Thickness, t is given as 15mm = 15/1000 = 0.015m
Heat loss,
Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr
Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
