Chemical Engineering Exam > Chemical Engineering Questions > The diagram shows heat conduction through a p...
Start Learning for Free
The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2 degree and radiation factor is 0.9
- a)– 1052.4 degree celsius
- b)– 2052.4 degree celsius
- c)– 3052.4 degree celsius
- d)– 4052.4 degree celsius
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The diagram shows heat conduction through a plane wall. The surface te...
Heat conducted through the plate = convection heat losses + radiation heat losses. So, d t /d x = – 13155/12.5 = – 1052.4 degree Celsius.
Most Upvoted Answer
The diagram shows heat conduction through a plane wall. The surface te...
To find the temperature gradient, we can use the formula:
Temperature gradient = (surface temperature - surrounding temperature) / distance
Since we are not given the distance, we cannot calculate the exact temperature gradient. However, we can calculate the rate of heat transfer using the given values.
The rate of heat transfer by conduction can be calculated using Fourier's Law:
Q_conduction = k * A * (surface temperature - surrounding temperature) / distance
where:
Q_conduction is the rate of heat transfer by conduction
k is the thermal conductivity (12.5 W/m degree)
A is the cross-sectional area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature - surrounding temperature) is the temperature difference
distance is the distance through which heat is being transferred (not given)
The rate of heat transfer by convection can be calculated using Newton's Law of Cooling:
Q_convection = h * A * (surface temperature - surrounding temperature)
where:
Q_convection is the rate of heat transfer by convection
h is the convective coefficient (80 W/m2degree)
A is the surface area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature - surrounding temperature) is the temperature difference
The rate of heat transfer by radiation can be calculated using the Stefan-Boltzmann Law:
Q_radiation = ε * σ * A * (surface temperature^4 - surrounding temperature^4)
where:
Q_radiation is the rate of heat transfer by radiation
ε is the radiation factor (0.9)
σ is the Stefan-Boltzmann constant (5.67 * 10^-8 W/m^2K^4)
A is the surface area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature^4 - surrounding temperature^4) is the temperature difference
The total rate of heat transfer is the sum of the rates of heat transfer by conduction, convection, and radiation:
Q_total = Q_conduction + Q_convection + Q_radiation
Substituting the given values into the equations and solving for Q_total will give us the rate of heat transfer. However, we cannot calculate the exact temperature gradient without knowing the distance.
Temperature gradient = (surface temperature - surrounding temperature) / distance
Since we are not given the distance, we cannot calculate the exact temperature gradient. However, we can calculate the rate of heat transfer using the given values.
The rate of heat transfer by conduction can be calculated using Fourier's Law:
Q_conduction = k * A * (surface temperature - surrounding temperature) / distance
where:
Q_conduction is the rate of heat transfer by conduction
k is the thermal conductivity (12.5 W/m degree)
A is the cross-sectional area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature - surrounding temperature) is the temperature difference
distance is the distance through which heat is being transferred (not given)
The rate of heat transfer by convection can be calculated using Newton's Law of Cooling:
Q_convection = h * A * (surface temperature - surrounding temperature)
where:
Q_convection is the rate of heat transfer by convection
h is the convective coefficient (80 W/m2degree)
A is the surface area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature - surrounding temperature) is the temperature difference
The rate of heat transfer by radiation can be calculated using the Stefan-Boltzmann Law:
Q_radiation = ε * σ * A * (surface temperature^4 - surrounding temperature^4)
where:
Q_radiation is the rate of heat transfer by radiation
ε is the radiation factor (0.9)
σ is the Stefan-Boltzmann constant (5.67 * 10^-8 W/m^2K^4)
A is the surface area through which heat is being transferred (assumed to be 1 m^2 for simplicity)
(surface temperature^4 - surrounding temperature^4) is the temperature difference
The total rate of heat transfer is the sum of the rates of heat transfer by conduction, convection, and radiation:
Q_total = Q_conduction + Q_convection + Q_radiation
Substituting the given values into the equations and solving for Q_total will give us the rate of heat transfer. However, we cannot calculate the exact temperature gradient without knowing the distance.
Free Test
FREE
| Start Free Test |
Community Answer
The diagram shows heat conduction through a plane wall. The surface te...
To find the temperature gradient, we can use the equation for heat transfer through conduction:
Q = k * A * ΔT / L
Where:
Q is the heat transfer rate (in watts)
k is the thermal conductivity of the material (in watts per meter degree)
A is the surface area (in square meters)
ΔT is the temperature difference (in degrees)
L is the thickness of the material (in meters)
In this case, we are given:
k = 12.5 W/m degree
A = not given (assume 1 square meter for simplicity)
ΔT = 475 K - 335 K = 140 K
L = not given (assume 1 meter for simplicity)
Let's plug in the values and solve for the temperature gradient:
Q = k * A * ΔT / L
Q = 12.5 W/m degree * 1 m^2 * 140 K / 1 m
Q = 12.5 * 140 W/degree
Q = 1750 W
Now, we can use the equation for heat transfer through convection:
Q = h * A * (Ts - Ta)
Where:
Q is the heat transfer rate (in watts)
h is the convective coefficient (in watts per square meter degree)
A is the surface area (in square meters)
Ts is the surface temperature (in degrees)
Ta is the ambient temperature (in degrees)
In this case, we are given:
h = 80 W/m^2 degree
A = not given (assume 1 square meter for simplicity)
Ts = 475 K
Ta = 335 K
Let's plug in the values and solve for the temperature gradient:
Q = h * A * (Ts - Ta)
1750 W = 80 W/m^2 degree * 1 m^2 * (475 K - 335 K)
1750 W = 80 W/degree * 140 K
1750 W = 11200 W
Since the heat transfer rate is the same for both conduction and convection, we can set the two equations equal to each other:
k * A * ΔT / L = h * A * (Ts - Ta)
We can cancel out the surface area (A) from both sides of the equation:
k * ΔT / L = h * (Ts - Ta)
Now, we can solve for the temperature gradient (ΔT / L):
ΔT / L = h * (Ts - Ta) / k
ΔT / L = 80 W/m^2 degree * (475 K - 335 K) / 12.5 W/m degree
ΔT / L = 80 W/m^2 degree * 140 K / 12.5 W/m degree
ΔT / L = 11200 W / 12.5 W
ΔT / L = 896 degree/m
Therefore, the temperature gradient is 896 degree/m.
Q = k * A * ΔT / L
Where:
Q is the heat transfer rate (in watts)
k is the thermal conductivity of the material (in watts per meter degree)
A is the surface area (in square meters)
ΔT is the temperature difference (in degrees)
L is the thickness of the material (in meters)
In this case, we are given:
k = 12.5 W/m degree
A = not given (assume 1 square meter for simplicity)
ΔT = 475 K - 335 K = 140 K
L = not given (assume 1 meter for simplicity)
Let's plug in the values and solve for the temperature gradient:
Q = k * A * ΔT / L
Q = 12.5 W/m degree * 1 m^2 * 140 K / 1 m
Q = 12.5 * 140 W/degree
Q = 1750 W
Now, we can use the equation for heat transfer through convection:
Q = h * A * (Ts - Ta)
Where:
Q is the heat transfer rate (in watts)
h is the convective coefficient (in watts per square meter degree)
A is the surface area (in square meters)
Ts is the surface temperature (in degrees)
Ta is the ambient temperature (in degrees)
In this case, we are given:
h = 80 W/m^2 degree
A = not given (assume 1 square meter for simplicity)
Ts = 475 K
Ta = 335 K
Let's plug in the values and solve for the temperature gradient:
Q = h * A * (Ts - Ta)
1750 W = 80 W/m^2 degree * 1 m^2 * (475 K - 335 K)
1750 W = 80 W/degree * 140 K
1750 W = 11200 W
Since the heat transfer rate is the same for both conduction and convection, we can set the two equations equal to each other:
k * A * ΔT / L = h * A * (Ts - Ta)
We can cancel out the surface area (A) from both sides of the equation:
k * ΔT / L = h * (Ts - Ta)
Now, we can solve for the temperature gradient (ΔT / L):
ΔT / L = h * (Ts - Ta) / k
ΔT / L = 80 W/m^2 degree * (475 K - 335 K) / 12.5 W/m degree
ΔT / L = 80 W/m^2 degree * 140 K / 12.5 W/m degree
ΔT / L = 11200 W / 12.5 W
ΔT / L = 896 degree/m
Therefore, the temperature gradient is 896 degree/m.
|
Explore Courses for Chemical Engineering exam
|
|
Similar Chemical Engineering Doubts
The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer?
Question Description
The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer?.
The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
Download more important topics, notes, lectures and mock test series for Chemical Engineering Exam by signing up for free.
Here you can find the meaning of The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m2degree and radiation factor is 0.9a)– 1052.4 degree celsiusb)– 2052.4 degree celsiusc)– 3052.4 degree celsiusd)– 4052.4 degree celsiusCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
|
|
Explore Courses for Chemical Engineering exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup