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A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.
- a)12
- b)13
Correct answer is between '12,13'. Can you explain this answer?
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A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wi...
Heat conducted to the outside surface of the wire = Heat convected to the air,
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where Tw is the outside surface temperature of the wire. At the centre, T is maximum and
From the heat conduction equation,
From equation 
and 
, C1= 0
and , C1= 0Equation 
reduces to
reduces toFrom Equation 
,
,Substituting Tw in Equation 
, we get the temperature distribution,
, we get the temperature distribution,Putting r = 0, T = Tmax
Imax = 12.19 amperes
Most Upvoted Answer
A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wi...
To find the maximum current that the wire can carry, we need to consider the heat generated in the wire due to the current flow and the heat dissipated by convection to the surrounding air.
1. Heat generated in the wire:
The heat generated in the wire can be calculated using the formula:
Q = I^2*R*t
Where,
Q = Heat generated (in Joules)
I = Current (in Amperes)
R = Resistance of the wire (in Ohms)
t = Time (in seconds)
2. Heat dissipated by convection:
The heat dissipated by convection can be calculated using the formula:
Q = h*A*(T_wire - T_air)*t
Where,
Q = Heat dissipated (in Joules)
h = Convective heat transfer coefficient (in W/m^2K)
A = Surface area of the wire (in m^2)
T_wire = Temperature of the wire (in °C)
T_air = Temperature of the air (in °C)
t = Time (in seconds)
3. Equating the heat generated to the heat dissipated, we have:
I^2*R*t = h*A*(T_wire - T_air)*t
4. Rearranging the equation to solve for the current (I), we get:
I = sqrt((h*A*(T_wire - T_air))/(R*t))
5. Substituting the given values:
- Diameter of the wire = 1 mm = 0.001 m
- Radius of the wire (r) = 0.0005 m
- Length of the wire (L) = unknown
- Specific resistance of the wire (ρ) = 0.037 ohm/m
- Temperature of the air (T_air) = 25°C = 25 + 273 = 298 K
- Maximum temperature of the wire (T_max) = 200°C = 200 + 273 = 473 K
- Convective heat transfer coefficient (h) = 10 W/m^2K
6. Calculating the surface area of the wire (A):
A = π*r^2 = π*(0.0005)^2
7. Calculating the resistance of the wire (R):
R = (ρ*L)/A
We can rearrange this equation to solve for the length of the wire (L):
L = (R*A)/ρ
8. Substituting the values of A and ρ, we can calculate the length of the wire (L).
9. Finally, substituting the calculated values of A, R, h, T_wire, T_air, and L into the formula for current (I), we can calculate the maximum current that the wire can carry.
The correct answer lies between '12' and '13' Amperes.
1. Heat generated in the wire:
The heat generated in the wire can be calculated using the formula:
Q = I^2*R*t
Where,
Q = Heat generated (in Joules)
I = Current (in Amperes)
R = Resistance of the wire (in Ohms)
t = Time (in seconds)
2. Heat dissipated by convection:
The heat dissipated by convection can be calculated using the formula:
Q = h*A*(T_wire - T_air)*t
Where,
Q = Heat dissipated (in Joules)
h = Convective heat transfer coefficient (in W/m^2K)
A = Surface area of the wire (in m^2)
T_wire = Temperature of the wire (in °C)
T_air = Temperature of the air (in °C)
t = Time (in seconds)
3. Equating the heat generated to the heat dissipated, we have:
I^2*R*t = h*A*(T_wire - T_air)*t
4. Rearranging the equation to solve for the current (I), we get:
I = sqrt((h*A*(T_wire - T_air))/(R*t))
5. Substituting the given values:
- Diameter of the wire = 1 mm = 0.001 m
- Radius of the wire (r) = 0.0005 m
- Length of the wire (L) = unknown
- Specific resistance of the wire (ρ) = 0.037 ohm/m
- Temperature of the air (T_air) = 25°C = 25 + 273 = 298 K
- Maximum temperature of the wire (T_max) = 200°C = 200 + 273 = 473 K
- Convective heat transfer coefficient (h) = 10 W/m^2K
6. Calculating the surface area of the wire (A):
A = π*r^2 = π*(0.0005)^2
7. Calculating the resistance of the wire (R):
R = (ρ*L)/A
We can rearrange this equation to solve for the length of the wire (L):
L = (R*A)/ρ
8. Substituting the values of A and ρ, we can calculate the length of the wire (L).
9. Finally, substituting the calculated values of A, R, h, T_wire, T_air, and L into the formula for current (I), we can calculate the maximum current that the wire can carry.
The correct answer lies between '12' and '13' Amperes.
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A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. Can you explain this answer?
Question Description
A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. 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 A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. 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 A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. Can you explain this answer?.
A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. 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 A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. 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 A 1 mm dia aluminium (k = 204 W/m-K) wire can carry a current till wire temperature does not exceed Tmax = 200°C when suspended in air at 25°C where h = 10 W/m2K. For the wire, the specific resistance is 0.037 ohm/m. The maximum current the wire can carry is __________A.a)12b)13Correct answer is between '12,13'. Can you explain this answer?.
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