Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?
1 answer:
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Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
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