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A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,
[2005]
- a)8.7
- b)13.9
- c)17.3
- d)27.7
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A small copper ball of 5 mm diameter at 500 K is dropped into an oil b...
Given data
K = 400 W/mK; Ti = 500 K; ρ = 9000 Kg/m3.
K = 400 W/mK; Ti = 500 K; ρ = 9000 Kg/m3.
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A small copper ball of 5 mm diameter at 500 K is dropped into an oil b...
Given data
K = 400 W/mK; Ti = 500 K; ρ = 9000 Kg/m3.
K = 400 W/mK; Ti = 500 K; ρ = 9000 Kg/m3.
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Community Answer
A small copper ball of 5 mm diameter at 500 K is dropped into an oil b...
Given data:
- Diameter of copper ball (D): 5 mm
- Initial temperature of copper ball (T1): 500 K
- Temperature of oil bath (T2): 300 K
- Thermal conductivity of copper (k): 400 W/mK
- Density of copper (ρ): 9000 kg/m3
- Specific heat of copper (c): 385 J/kgK
- Heat transfer coefficient (h): 250 W/m2K
To find: Rate of fall of temperature of the ball at the beginning of cooling (dT/dt)
We can use the lumped analysis approach to solve this problem. The lumped analysis assumes that the entire mass of the object is at the same temperature and that heat transfer occurs only at the surface.
1. Calculation of surface area:
The surface area of the ball can be calculated using the formula:
A = π D^2 / 4
A = π (0.005 m)^2 / 4
A = 0.00001963 m2
2. Calculation of volume:
The volume of the ball can be calculated using the formula:
V = π D^3 / 6
V = π (0.005 m)^3 / 6
V = 0.00000006545 m3
3. Calculation of mass:
The mass of the ball can be calculated using the formula:
m = ρ V
m = 9000 kg/m3 * 0.00000006545 m3
m = 0.000588 kg
4. Calculation of Biot number:
The Biot number (Bi) can be calculated using the formula:
Bi = h L / k
Where L is the characteristic length, which can be taken as the diameter of the ball.
Bi = 250 W/m2K * 0.005 m / 400 W/mK
Bi = 0.003125
5. Calculation of Fourier number:
The Fourier number (Fo) can be calculated using the formula:
Fo = α t / L^2
Where α is the thermal diffusivity, given by α = k / (ρ c).
Fo = (400 W/mK) / (9000 kg/m3 * 385 J/kgK) * t / (0.005 m)^2
Fo = 0.00002177 t
6. Calculation of Nusselt number:
The Nusselt number (Nu) can be calculated using the formula:
Nu = Bi / (4 + 0.065 Bi)
Nu = 0.003125 / (4 + 0.065 * 0.003125)
Nu = 0.000536
7. Calculation of heat transfer coefficient:
The heat transfer coefficient (h) can be calculated using the formula:
h = Nu k / L
h = 0.000536 * 400 W/mK / 0.005 m
h = 42.88 W/m2K
8. Calculation of rate of fall of temperature:
The rate of fall of temperature (dT/dt) can be calculated using the formula:
dT/dt = h A (T2 - T1) / (m c)
dT/dt = 42.88 W/m2K * 0.00001963 m2 * (300 K - 500 K) / (0.000588 kg * 385
- Diameter of copper ball (D): 5 mm
- Initial temperature of copper ball (T1): 500 K
- Temperature of oil bath (T2): 300 K
- Thermal conductivity of copper (k): 400 W/mK
- Density of copper (ρ): 9000 kg/m3
- Specific heat of copper (c): 385 J/kgK
- Heat transfer coefficient (h): 250 W/m2K
To find: Rate of fall of temperature of the ball at the beginning of cooling (dT/dt)
We can use the lumped analysis approach to solve this problem. The lumped analysis assumes that the entire mass of the object is at the same temperature and that heat transfer occurs only at the surface.
1. Calculation of surface area:
The surface area of the ball can be calculated using the formula:
A = π D^2 / 4
A = π (0.005 m)^2 / 4
A = 0.00001963 m2
2. Calculation of volume:
The volume of the ball can be calculated using the formula:
V = π D^3 / 6
V = π (0.005 m)^3 / 6
V = 0.00000006545 m3
3. Calculation of mass:
The mass of the ball can be calculated using the formula:
m = ρ V
m = 9000 kg/m3 * 0.00000006545 m3
m = 0.000588 kg
4. Calculation of Biot number:
The Biot number (Bi) can be calculated using the formula:
Bi = h L / k
Where L is the characteristic length, which can be taken as the diameter of the ball.
Bi = 250 W/m2K * 0.005 m / 400 W/mK
Bi = 0.003125
5. Calculation of Fourier number:
The Fourier number (Fo) can be calculated using the formula:
Fo = α t / L^2
Where α is the thermal diffusivity, given by α = k / (ρ c).
Fo = (400 W/mK) / (9000 kg/m3 * 385 J/kgK) * t / (0.005 m)^2
Fo = 0.00002177 t
6. Calculation of Nusselt number:
The Nusselt number (Nu) can be calculated using the formula:
Nu = Bi / (4 + 0.065 Bi)
Nu = 0.003125 / (4 + 0.065 * 0.003125)
Nu = 0.000536
7. Calculation of heat transfer coefficient:
The heat transfer coefficient (h) can be calculated using the formula:
h = Nu k / L
h = 0.000536 * 400 W/mK / 0.005 m
h = 42.88 W/m2K
8. Calculation of rate of fall of temperature:
The rate of fall of temperature (dT/dt) can be calculated using the formula:
dT/dt = h A (T2 - T1) / (m c)
dT/dt = 42.88 W/m2K * 0.00001963 m2 * (300 K - 500 K) / (0.000588 kg * 385
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A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer?
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A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. 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 A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. 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 A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer?.
A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. 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 A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. 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 A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer?.
Solutions for A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,[2005]a)8.7b)13.9c)17.3d)27.7Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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