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A horizontal heated plate at 200 degree Celsius and facing upwards has been placed in still air at 20 degree Celsius. If the plate measures 1.25 m by 1 m, make calculations for the heat loss by natural convection. The convective film coefficient for free convection is given by the following empirical relation
h = 0.32 (α) 0.25 W/m2 K
Where α is the mean film temperature in degrees kelvin
h = 0.32 (α) 0.25 W/m2 K
Where α is the mean film temperature in degrees kelvin
- a)6006 W
- b)5006 W
- c)4006 W
- d)3006 W
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A horizontal heated plate at 200 degree Celsius and facing upwards has...
α = 273 + 200 + 20/2 = 383 degree Celsius. Therefore, rate of heat loss = h A d t = 3006 W.
Most Upvoted Answer
A horizontal heated plate at 200 degree Celsius and facing upwards has...
(Ts-T∞)/L)^0.25
where h is the convective film coefficient in W/(m^2*K), Ts is the surface temperature of the plate in K, T∞ is the ambient temperature in K, and L is the characteristic length in meters (in this case, the width of the plate).
First, we need to convert the temperatures to Kelvin:
Ts = 200 + 273 = 473 K
T∞ = 20 + 273 = 293 K
The characteristic length is the width of the plate, so L = 1 m.
Plugging these values into the equation for the convective film coefficient, we get:
h = 0.32 ((473-293)/1)^0.25 = 21.5 W/(m^2*K)
Next, we can use Newton's Law of Cooling to calculate the heat loss from the plate:
q = hA(Ts-T∞)
where q is the heat loss in watts, A is the surface area of the plate in square meters (in this case, 1.25 m by 1 m), and Ts and T∞ are in Kelvin.
A = 1.25 * 1 = 1.25 m^2
Plugging in the values, we get:
q = 21.5 * 1.25 * (473-293) = 526 W
Therefore, the heat loss by natural convection from the heated plate is 526 W.
where h is the convective film coefficient in W/(m^2*K), Ts is the surface temperature of the plate in K, T∞ is the ambient temperature in K, and L is the characteristic length in meters (in this case, the width of the plate).
First, we need to convert the temperatures to Kelvin:
Ts = 200 + 273 = 473 K
T∞ = 20 + 273 = 293 K
The characteristic length is the width of the plate, so L = 1 m.
Plugging these values into the equation for the convective film coefficient, we get:
h = 0.32 ((473-293)/1)^0.25 = 21.5 W/(m^2*K)
Next, we can use Newton's Law of Cooling to calculate the heat loss from the plate:
q = hA(Ts-T∞)
where q is the heat loss in watts, A is the surface area of the plate in square meters (in this case, 1.25 m by 1 m), and Ts and T∞ are in Kelvin.
A = 1.25 * 1 = 1.25 m^2
Plugging in the values, we get:
q = 21.5 * 1.25 * (473-293) = 526 W
Therefore, the heat loss by natural convection from the heated plate is 526 W.
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A horizontal heated plate at 200 degree Celsius and facing upwards has been placed in still air at 20 degree Celsius. If the plate measures 1.25 m by 1 m, make calculations for the heat loss by natural convection. The convective film coefficient for free convection is given by the following empirical relationh = 0.32 (α)0.25W/m2KWhere α is the mean film temperature in degrees kelvina)6006 Wb)5006 Wc)4006 Wd)3006 WCorrect answer is option 'A'. Can you explain this answer?
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