8. Consider laminar viscous flow over a flat plate. We will earn later in the course that the skin friction coefficient Cf is gi
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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
The total amount of daily heat transfer is 1382.38 M w.
The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.
<u>Explanation:</u>
Given data,
= 10° C
= 250 w/
k
Pipe length = 20 m
Inner diameter = 6 cm,
= 3 cm
Outer diameter = 8 cm,
= 4 cm
The thickness of insulation is 4 cm.
=
+ 4
= 4+4
= 8 cm
is the heat transfer coefficient of convection inside,
is the heat transfer coefficient of convection outside.
The heat transfer rate between ambient and steam is
watt
= watt
= watt
q = 15999.86 watt
The total amount of daily heat transfer = 15999.86 × 86400
= 1382.387904 watt
= 1382.38 M w
The total amount of daily heat transfer is 1382.38 M w.
b) The temperature on the outside surface of the gypsum plaster insulation.
q =
15999.86
- 10 = 7.96
= 17.96 ° C.