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During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:
Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°C
- a)36
- b)38
Correct answer is between '36,38'. Can you explain this answer?
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Verified Answer
During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm he...
For a cylindrical piece, the characteristic linear dimension is
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= 4.41 X 10-3m
Biot number Bi
= 0.00882 < />
Since Bi < 0.1,the="" lumped="" parameter="" model="" can="" be="" adopted.="" />
= 0.004814
= loge4.8 = 1.5686
∴ Time required for heating,t
Let T be the temperature attained when the pieces are taken out from the furnace after 280 seconds. Then
∴ Shortfall in temperature
= 600 - 563 = 37°C
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During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm he...
Solution:
Given data:
Diameter of cylindrical piece, D = 25 mm
Height of cylindrical piece, L = 30 mm
Temperature of cylindrical piece, T1 = 30°C
Temperature to be achieved, T2 = 600°C
Convection coefficient, h = 80 W/m2-°C
Time taken for heat treatment, t = 280 seconds
Density of cylindrical piece, ρ = 7850 kg/m3
Specific heat of cylindrical piece, Cp = 480 J/kg-K
Thermal conductivity of cylindrical piece, k = 40 W/m-°C
Step 1: Calculation of surface area and volume of cylindrical piece
Surface area of cylindrical piece, A = πDL + πD2/4 = (3.14)(0.025)(0.03) + (3.14)(0.025)2/4 = 0.00236 m2
Volume of cylindrical piece, V = πD2L/4 = (3.14)(0.025)2(0.03)/4 = 1.47 x 10-5 m3
Step 2: Calculation of heat transfer rate
Heat transfer rate, Q = hA(T2 - T1) = (80)(0.00236)(600 - 30) = 111.2 W
Step 3: Calculation of temperature distribution inside the cylindrical piece
Assuming one-dimensional heat transfer, the temperature distribution inside the cylindrical piece can be given by:
T(x, t) = T1 + (T2 - T1)erf(x/2√(kt))
Where,
erf is the error function
x is the distance from the center of the cylinder
t is the time
Using this equation, we can calculate the temperature at the center of the cylindrical piece at the end of the heat treatment (t = 280 s):
T(0, 280) = 30 + (600 - 30)erf(0/2√(40(7850)(480))) = 590.6°C
Step 4: Calculation of shortfall in temperature
Shortfall in temperature, ΔT = T2 - T(0, t) = 600 - 590.6 = 9.4°C
Therefore, the shortfall in temperature is 9.4°C.
Answer: 9.4°C (rounded off to one decimal place)
Note: The error function (erf) is a special function used in mathematics to describe the probability of a Gaussian distribution. It can be calculated using special software or tables.
Given data:
Diameter of cylindrical piece, D = 25 mm
Height of cylindrical piece, L = 30 mm
Temperature of cylindrical piece, T1 = 30°C
Temperature to be achieved, T2 = 600°C
Convection coefficient, h = 80 W/m2-°C
Time taken for heat treatment, t = 280 seconds
Density of cylindrical piece, ρ = 7850 kg/m3
Specific heat of cylindrical piece, Cp = 480 J/kg-K
Thermal conductivity of cylindrical piece, k = 40 W/m-°C
Step 1: Calculation of surface area and volume of cylindrical piece
Surface area of cylindrical piece, A = πDL + πD2/4 = (3.14)(0.025)(0.03) + (3.14)(0.025)2/4 = 0.00236 m2
Volume of cylindrical piece, V = πD2L/4 = (3.14)(0.025)2(0.03)/4 = 1.47 x 10-5 m3
Step 2: Calculation of heat transfer rate
Heat transfer rate, Q = hA(T2 - T1) = (80)(0.00236)(600 - 30) = 111.2 W
Step 3: Calculation of temperature distribution inside the cylindrical piece
Assuming one-dimensional heat transfer, the temperature distribution inside the cylindrical piece can be given by:
T(x, t) = T1 + (T2 - T1)erf(x/2√(kt))
Where,
erf is the error function
x is the distance from the center of the cylinder
t is the time
Using this equation, we can calculate the temperature at the center of the cylindrical piece at the end of the heat treatment (t = 280 s):
T(0, 280) = 30 + (600 - 30)erf(0/2√(40(7850)(480))) = 590.6°C
Step 4: Calculation of shortfall in temperature
Shortfall in temperature, ΔT = T2 - T(0, t) = 600 - 590.6 = 9.4°C
Therefore, the shortfall in temperature is 9.4°C.
Answer: 9.4°C (rounded off to one decimal place)
Note: The error function (erf) is a special function used in mathematics to describe the probability of a Gaussian distribution. It can be calculated using special software or tables.
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During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer?
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During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. 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 During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. 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 During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer?.
During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. 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 During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. 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 During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer?.
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During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer?, a detailed solution for During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer? has been provided alongside types of During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30°C are placed in a furnace at 750°C with convection coefficient 80 W/m2-°C. It is required to heat the pieces to 600°C before pieces are taken out. What will be the shortfall in temperature if the pieces are taken out from the furnace after 280 seconds? Assume the following property values:Density 7850 kg/m3 ; specific heat 480 J/kg-K; Conductivity 40 W/m-°Ca)36b)38Correct answer is between '36,38'. Can you explain this answer? theory, EduRev gives you an
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