
The form of the heat diffusion equation for each case, and to write the equation for the initial condition and the boundary conditions that are applied at
The equation for case (a) are, the heat equation is
Given:
The given diagram is shown in Figure 1.
Figure 1
Formula Used:
The expression for the governing differential equation for the one-dimensional heat transfer without any heat generation is given by,
Here,
The expression for the slope of the temperature distribution is given by,
Calculation:
Case (a):
The temperature distribution at steady-state conditions is constant due to which the thermal energy generation does not take place. Therefore the dimensional heat conduction equation for case (a) is calculated as,
The initial condition of temperature for the plain wall is evaluated as,
Here,
The boundary condition equation for the temperature distribution at condition of
Here,
The temperature remains the same at a time when step is greater than zero, thus the temperature at the surface remains constant. Therefore the boundary condition equation of temperature distribution at condition of
Here,
Case (b):
The temperature distribution at the steady-state condition is constant due to which the thermal energy generation does not take place, therefore the one-dimensional heat conduction equation is given by,
The initial condition of temperature distribution in a plane wall at
The initial temperature is the same throughout the solid plane, therefore, the boundary condition equation of temperature at condition of
The temperature gradient at the surface decreases with the time step and the thermal energy generation leads to further fall in temperature. Thus the boundary condition equation of temperature distribution at condition of
Case (c)
The temperature distribution at steady-state conditions is parabolic due to which the thermal energy generation takes place since the temperature increase with time due to large temperature the thermal energy becomes positive. Therefore, the one-dimensional heat conduction equation for the Cartesian coordinate in
The initial temperature is uniformly distributed across the solid and the initial condition temperature distribution for
The boundary condition equation of temperature distribution at condition of
The boundary condition equation of temperature distribution at condition of
Case (d)
The temperature distribution at steady-state conditions is parabolic due to which the thermal energy generation takes place. Therefore the one-dimensional heat conduction equation for the Cartesian coordinate in
The initial temperature is uniformly distributed across the solid and the initial condition temperature distribution for
The boundary condition equation of temperature distribution at condition of
The temperature gradient at the surface decreases with the time step and the thermal energy generation leads to further fall in temperature. Thus the boundary condition equation of temperature distribution at condition of
Conclusion:
Therefore, the equation for case (a) is, the heat equation is

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Chapter 2 Solutions
Fundamentals of Heat and Mass Transfer
- Experiment تكنولوجيا السيارات - Internal Forced convenction Heat transfer Air Flow through Rectangular Duct. objective: Study the convection heat transfer of air flow through rectangular duct. Valve Th Top Dead Centre Exhaust Valve Class CP. N; ~ RIVavg Ti K 2.11 Te To 18.8 21.3 45.8 Nath Ne Pre Calculations:. Q = m cp (Te-Ti) m: Varg Ac Acca*b Q=hexp As (Ts-Tm) 2 2.61 18.5 20.846.3 Tm = Te-Ti = 25 AS-PL = (a+b)*2*L Nu exp= Re-Vavy D heep Dh k 2ab a+b Nu Dh the- (TS-Tm) Ts. Tmy Name / Nu exp Naxe بب ارتدان العشريarrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findarrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of thearrow_forward
- Procedure: 1- Cartesian system, 2D3D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D ∑Fx=0 ∑Fy=0 ∑Fz=0 ∑Mx=0 ∑My=0 ΣMz=0 2D ΣFx=0 ΣFy=0 ΣMz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of thearrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of thearrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forward
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- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning
