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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air are
v = 15.06 x 10-6 m2/s
k = 0.0259 W/m-°C
Pr = 0.703
v = 15.06 x 10-6 m2/s
k = 0.0259 W/m-°C
Pr = 0.703
Correct answer is '-62.0 to -61.0'. Can you explain this answer?
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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The ...
The maximum average heat transfer occurs when the air flow is in the direction of the shorter dimension.
The negative sign indicates that heat transfer is towards the freezer.
The negative sign indicates that heat transfer is towards the freezer.
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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The ...
°C. The air temperature is 20°C. What is the rate of heat transfer from the top of the freezer due to convection?
We can use the equation for convective heat transfer:
Q = hA(Ts - Ta)
where Q is the rate of heat transfer, h is the convective heat transfer coefficient, A is the surface area, Ts is the surface temperature, and Ta is the air temperature.
First, we need to find the convective heat transfer coefficient. This depends on the air velocity and other factors, but we can estimate it using the following correlation for natural convection over a flat plate:
h = 5.7 (Grx Pr)^0.25 / (1 + 0.012(Grx Pr)^0.4)^0.2
where Grx is the Grashof number times the length of the plate (in this case, the width of the freezer), and Pr is the Prandtl number for air (about 0.7 at 20°C).
The Grashof number can be approximated as:
Grx = gβ(Ts - Ta) Lx^3 / ν^2
where g is the acceleration due to gravity, β is the coefficient of thermal expansion for air (about 1/Ta), Lx is the length of the plate (in this case, the width of the freezer), and ν is the kinematic viscosity of air (about 15.8 x 10^-6 m^2/s at 20°C).
Plugging in the numbers, we get:
Grx = 9.81 x 1/Ta x 0.3 x 0.9^3 / (15.8 x 10^-6)^2 ≈ 1.8 x 10^11
Pr ≈ 0.7
h = 5.7 (1.8 x 10^11 x 0.7)^0.25 / (1 + 0.012(1.8 x 10^11 x 0.7)^0.4)^0.2 ≈ 6.3 W/(m^2 K)
Now we can calculate the rate of heat transfer:
Q = 6.3 x 0.9 x 1.5 x (10 - 20) = -85.05 W
The negative sign indicates that heat is being lost from the top of the freezer to the air. The rate of heat transfer is about 85 W, which means that the freezer is losing a significant amount of energy due to convection. This could lead to higher energy costs and reduced efficiency. Adding insulation to the top of the freezer or using a fan to increase air circulation could help reduce heat transfer and improve performance.
We can use the equation for convective heat transfer:
Q = hA(Ts - Ta)
where Q is the rate of heat transfer, h is the convective heat transfer coefficient, A is the surface area, Ts is the surface temperature, and Ta is the air temperature.
First, we need to find the convective heat transfer coefficient. This depends on the air velocity and other factors, but we can estimate it using the following correlation for natural convection over a flat plate:
h = 5.7 (Grx Pr)^0.25 / (1 + 0.012(Grx Pr)^0.4)^0.2
where Grx is the Grashof number times the length of the plate (in this case, the width of the freezer), and Pr is the Prandtl number for air (about 0.7 at 20°C).
The Grashof number can be approximated as:
Grx = gβ(Ts - Ta) Lx^3 / ν^2
where g is the acceleration due to gravity, β is the coefficient of thermal expansion for air (about 1/Ta), Lx is the length of the plate (in this case, the width of the freezer), and ν is the kinematic viscosity of air (about 15.8 x 10^-6 m^2/s at 20°C).
Plugging in the numbers, we get:
Grx = 9.81 x 1/Ta x 0.3 x 0.9^3 / (15.8 x 10^-6)^2 ≈ 1.8 x 10^11
Pr ≈ 0.7
h = 5.7 (1.8 x 10^11 x 0.7)^0.25 / (1 + 0.012(1.8 x 10^11 x 0.7)^0.4)^0.2 ≈ 6.3 W/(m^2 K)
Now we can calculate the rate of heat transfer:
Q = 6.3 x 0.9 x 1.5 x (10 - 20) = -85.05 W
The negative sign indicates that heat is being lost from the top of the freezer to the air. The rate of heat transfer is about 85 W, which means that the freezer is losing a significant amount of energy due to convection. This could lead to higher energy costs and reduced efficiency. Adding insulation to the top of the freezer or using a fan to increase air circulation could help reduce heat transfer and improve performance.
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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer?
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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer?.
Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer?.
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Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer?, a detailed solution for Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? has been provided alongside types of Air moving at 0.3 m/s blows over the top of a chest-type freezer. The top of the freezer measures 0.9 m by 1.5 m and is poorly insulted so th a t th e su rface rem ains at 10 ° C . If the temperature of air is 30° C, make calculations for the maximum heat transfer rate by forced convection at the top of the freezer. At a temperature of 20°C, the thermophysical properties of air arev = 15.06 x 10-6 m2/sk = 0.0259 W/m-°CPr = 0.703Correct answer is '-62.0 to -61.0'. Can you explain this answer? theory, EduRev gives you an
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