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A large concrete slab 1 m thick has one dimensional temperature distribution:
T = 4 – 10x + 20x2 + 10x3
Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?
T = 4 – 10x + 20x2 + 10x3
Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?
- a)0.1°C/h
- b)0.2°C/h
- c)0.3°C/h
- d)0.4°C/h
Correct answer is option 'B'. Can you explain this answer?
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A large concrete slab 1 m thick has one dimensional temperature distri...
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A large concrete slab 1 m thick has one dimensional temperature distri...
Given:
- Thickness of the concrete slab = 1 m
- Temperature distribution: T = 4 + 10x + 20x^2 + 10x^3
- Thermal diffusivity of the slab material = 2 x 10^-3 m^2/hr
To find:
Rate of change of temperature at the other face of the wall
Solution:
The rate of change of temperature with respect to distance x can be calculated by taking the derivative of the temperature distribution equation.
Differentiating the temperature equation:
dT/dx = 10 + 40x + 30x^2
Rate of change of temperature at the other face:
To find the rate of change of temperature at the other face of the wall, we need to substitute the distance x with the thickness of the slab.
dT/dx = 10 + 40x + 30x^2
dT/dx = 10 + 40(1) + 30(1)^2
dT/dx = 10 + 40 + 30
dT/dx = 80
The rate of change of temperature at the other face of the wall is 80°C/hr.
Conversion from m^2/hr to C/hr:
Since the thermal diffusivity is given in m^2/hr, we need to convert it to C/hr using the following relation:
Thermal diffusivity = k / (ρ * c)
Where:
k = thermal conductivity (in m^2/hr)
ρ = density of the material (in kg/m^3)
c = specific heat capacity (in J/kg°C)
Assuming the density and specific heat capacity of the concrete slab material remain constant, we can write:
k = (ρ * c * α)
Where α is the thermal diffusivity.
Now, we can substitute the given thermal diffusivity and solve for k:
k = (ρ * c * α)
k = (ρ * c * 2 x 10^-3)
Rate of change of temperature at the other face:
Now that we have the value of k, we can calculate the rate of change of temperature at the other face of the wall using Fourier's Law of Heat Conduction:
Q = k * A * (dT/dx)
Where:
Q = rate of heat transfer (in W)
k = thermal conductivity (in W/m°C)
A = cross-sectional area (in m^2)
(dT/dx) = rate of change of temperature (in °C/m)
Since we want the rate of change of temperature in °C/hr, we need to convert the units of Q and (dT/dx) accordingly.
Q = k * A * (dT/dx)
Q = (ρ * c * 2 x 10^-3) * A * (80/3600) [Converting units from °C/m to °C/hr]
Q = (ρ * c * 2 x 10^-3) * A * (80/3600)
Q = (ρ * c * 2 x 10^-3) * A * (2/9)
Q = (ρ * c * A * 2 x 10^-3) * (2/9)
Since the slab is large and the cross-sectional area is not given
- Thickness of the concrete slab = 1 m
- Temperature distribution: T = 4 + 10x + 20x^2 + 10x^3
- Thermal diffusivity of the slab material = 2 x 10^-3 m^2/hr
To find:
Rate of change of temperature at the other face of the wall
Solution:
The rate of change of temperature with respect to distance x can be calculated by taking the derivative of the temperature distribution equation.
Differentiating the temperature equation:
dT/dx = 10 + 40x + 30x^2
Rate of change of temperature at the other face:
To find the rate of change of temperature at the other face of the wall, we need to substitute the distance x with the thickness of the slab.
dT/dx = 10 + 40x + 30x^2
dT/dx = 10 + 40(1) + 30(1)^2
dT/dx = 10 + 40 + 30
dT/dx = 80
The rate of change of temperature at the other face of the wall is 80°C/hr.
Conversion from m^2/hr to C/hr:
Since the thermal diffusivity is given in m^2/hr, we need to convert it to C/hr using the following relation:
Thermal diffusivity = k / (ρ * c)
Where:
k = thermal conductivity (in m^2/hr)
ρ = density of the material (in kg/m^3)
c = specific heat capacity (in J/kg°C)
Assuming the density and specific heat capacity of the concrete slab material remain constant, we can write:
k = (ρ * c * α)
Where α is the thermal diffusivity.
Now, we can substitute the given thermal diffusivity and solve for k:
k = (ρ * c * α)
k = (ρ * c * 2 x 10^-3)
Rate of change of temperature at the other face:
Now that we have the value of k, we can calculate the rate of change of temperature at the other face of the wall using Fourier's Law of Heat Conduction:
Q = k * A * (dT/dx)
Where:
Q = rate of heat transfer (in W)
k = thermal conductivity (in W/m°C)
A = cross-sectional area (in m^2)
(dT/dx) = rate of change of temperature (in °C/m)
Since we want the rate of change of temperature in °C/hr, we need to convert the units of Q and (dT/dx) accordingly.
Q = k * A * (dT/dx)
Q = (ρ * c * 2 x 10^-3) * A * (80/3600) [Converting units from °C/m to °C/hr]
Q = (ρ * c * 2 x 10^-3) * A * (80/3600)
Q = (ρ * c * 2 x 10^-3) * A * (2/9)
Q = (ρ * c * A * 2 x 10^-3) * (2/9)
Since the slab is large and the cross-sectional area is not given
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A large concrete slab 1 m thick has one dimensional temperature distri...
From generalised 1 d heat heat eqn. 100*2/1000
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A large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. Can you explain this answer?
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A large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. 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 large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. 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 large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. Can you explain this answer?.
A large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. 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 large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. 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 large concrete slab 1 m thick has one dimensional temperature distribution:T = 4 – 10x + 20x2 + 10x3Where T is temperature and x is distance from one face towards other face of wall. If the slab material has thermal diffusivity of 2 × 10-3 m2/hr,what is the rate of change of temperature at the other face of the wall?a)0.1°C/hb)0.2°C/hc)0.3°C/hd)0.4°C/hCorrect answer is option 'B'. Can you explain this answer?.
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