hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp⁡(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ??hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp⁡(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ?? hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp⁡(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ?? finite difference method, temperature distribution, annular pellet, conductivity, cladding, heat generation, homework, temperature, external cooling, uo2 (fuel), matlab, plot, matrix, graph, function, differential equations, matrices, iteration MATLAB Answers — New Questions

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