Concept explainers
In the method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction, the separation constant
To show: The value of
Explanation of Solution
From the textbook,
Constant of separation is given as,
The distribution of temperature can be given as
Now,
The solution of equation (1) and (2) can be expressed as,
Therefore,
We obtain the value of C8 or C5 is zero. So, it will generate a trivial solution that does not depend on the value of x and y.
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Chapter 4 Solutions
Fundamentals of Heat and Mass Transfer
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- Drive an expression for heat transfer and temperature distribution for steady state one dimensional heat conduction in a plan wall. The temperature is maintained at a temperature Ti at x=0, while the other face X-L is maintained at temperature T2, the thickness of the wall may be taken as L and the energy equation is given by: d²T/dx² = 0. : Sketch a simple diagram for the temperature distribution in plane wall for a steady state one dimensional heat conduction, with heat generation. The surface temperature of the walls Ti and T2, for the cases Ti>T2, T1-T2, and T2>T1. The thickness of the wall may be taken as 2Larrow_forward4.41 Solve the steady-state, 2-D heat conduction equation in the unit square, 0arrow_forwardA plane wall of thickness 8cm and thermal conductivity k=5W/mK experiences uniform volumetric heat generation, while convection heat transfer occurs at both of its surfaces (x= -L, x= + L), each of which is exposed to a fluid of temperature T∞ = 20˚C. The origin of the x-coordinate is at the midplane of the wall. Under steady-state conditions, the temperature distribution in the wall is of the form T(˚C) = a + bx - cx^2, where x is in meters, a =86˚C, b = -500˚C/m, and c=4459. 1) Heat Flux Entering the wall is ? 2) Temperature at the left face is /arrow_forwardConsider a solid sphere of radius R with a fixed surface temperature, TR. Heat is generated within the solid at a rate per unit volume given by q = ₁ + ₂r; where ₁ and ₂ are constants. (a) Assuming constant thermal conductivity, use the conduction equation to derive an expression for the steady-state temperature profile, T(r), in the sphere. (b) Calculate the temperature at the center of the sphere for the following parameter values: R=3 m 1₁-20 W/m³ TR-20 °C k-0.5 W/(m K) ₂-10 W/m³arrow_forwardDrive an expression for heat transfer and temperature distribution for steady state one dimensional heat conduction in a plan wall. The temperature is maintained at a temperature Ti at x=0, while the other face X-L is maintained at temperature T2, the thickness of the wall may be taken as L and the energy equation is given by: d²T/dx²=0.arrow_forwardProblem: Conduction related Uniform internal heat generation at q =6.0×10^7 W/m3 is occurring in a cylindrical nuclear reactor fuel rod of 60-mm diameter, and under steady-state conditions the temperature distribution is of the form T\left(r\right)=a+br^2T(r)=a+br2, where T is in degrees Celsius and r is in meters, while a = 900°C and b = -5.26 × 10^5 °C/m^2. The fuel rod properties are k = 30 W/m · K, ρ = 1,100 kg/m^3, and cp = 800 J/kg · K. (a) What is the rate of heat transfer per unit length of the rod at r = 0 (the centerline) and at r = 30 mm (the surface)? (b) If the reactor power level is suddenly increased to q2dot = 10^8 W/m^3, what is the initial time rate of temperature change at r = 0 and r = 30 mm?arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning