Concept explainers
An electrical cable, experiencing uniform volumetric generation
(a) Derive the explicit, finite-difference equations for an interior node (m, n ), the center node
(b) Obtain the stability criterion for each of the finite-difference equations. Identify the most restrictive criterion.
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Introduction to Heat Transfer
- (a) Consider nodal configuration shown below. (a) Derive the finite-difference equations under steady-state conditions if the boundary is insulated. (b) Find the value of Tm,n if you know that Tm, n+1= 12 °C, Tm, n-1 = 8 °C, Tm-1, n = 10 °C, Ax = Ay = 10 mm, and k = = W 3 m. k . Ay m-1, n m, n | Δx=" m, n+1 m, n-1 The side insulatedarrow_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_forward(a) Consider nodal configuration shown below. (a) Derive the finite-difference equations under steady-state conditions if the boundary is insulated. (b) Find the value of Tm,n if you know that Tm, n+1= 12 °C, Tm, n-1 = 8 °C, Tm-1, n = 10 °C, Ax = Ay = 10 mm, and k = W 3 m. k Ay m-1, n 11- m2, 11 m, n+1 m, n-1 The side insulatedarrow_forward
- Q1 Passage of an electric current through a long conducting rod of radius r; and thermal conductivity k, results in uniform volumetric heating at a rate of ġ. The conduct- ing rod is wrapped in an electrically nonconducting cladding material of outer radius r, and thermal conduc- tivity k, and convection cooling is provided by an adjoining fluid. Conducting rod, ġ, k, 11 To Čladding, ke For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express ap- propriate boundary conditions for the solution of these equations.arrow_forwardA 1-D conduction heat transfer problem with internal energy generation is governed by the following equation: +-= dx2 =0 W where è = 5E5 and k = 32 If you are given the following node diagram with a spacing of Ax = .02m and know that m-K T = 611K and T, = 600K, write the general equation for these internal nodes in finite difference form and determine the temperature at nodes 3 and 4. Insulated Ar , T For the answer window, enter the temperature at node 4 in Kelvin (K). Your Answer: EN SORN Answer units Pri qu) 232 PM 4/27/2022 99+ 66°F Sunny a . 20 ENLARGED oW TEXTURE PRT SCR IOS DEL F8 F10 F12 BACKSPACE num - %3D LOCK HOME PGUP 170arrow_forward(a) Consider nodal configuration shown below. Derive the finite-difference equations under steady-state conditions for the following situations. (a) The boundary is insulated. (b) The boundary is subjected to a constant heat flux. m, n+1 Ay Im, n The side insulated m-1, n I I Ax- m, n-1arrow_forward
- The steady state temperature3/3 distribution in a one dimensional wall of thermal conductivity 50W/mK and thickness 5cm is observed to be T("C) = a + bx2, where a =200°C, b = -2000°C/m2, and x is in meters. Determine the heat generation rate in the wall.arrow_forwardConsider the square channel shown in the sketch operating under steady state condition. The inner surface of the channel is at a uniform temperature of 600 K and the outer surface is at a uniform temperature of 300 K. From a symmetrical elemental of the channel, a two-dimensional grid has been constructed as in the right figure below. The points are spaced by equal distance. Tout = 300 K k = 1 W/m-K T = 600 K (a) The heat transfer from inside to outside is only by conduction across the channel wall. Beginning with properly defined control volumes, derive the finite difference equations for locations 123. You can also use (n, m) to represent row and column. For example, location Dis (3, 3), location is (3,1), and location 3 is (3,5). (hint: I have already put a control volume around this locations with dashed boarder.) (b) Please use excel to construct the tables of temperatures and finite difference. Solve for the temperatures of each locations. Print out the tables in the spread…arrow_forwardWrite the finite difference form of the two dimensional steady state heat conduction equation with internal heat generation at a constant rate ‘g’ for a region 0.03m X 0.03m by using a mesh size ∆x=∆y= 0.01 m for a material having thermal conductivity 25 W/m.K and heat generation rate, 107 W/m3 . All the boundary surfaces are maintained at 10°C. Express the finite difference equations in matrix form for the unknown node temperatures.arrow_forward
- 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_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_forwardThe initial temperature distribution of a 5 cm long stick is given by the following function. The circumference of the rod in question is completely insulated, but both ends are kept at a temperature of 0 °C. Obtain the heat conduction along the rod as a function of time and position ? (x = 1.752 cm²/s for the bar in question) 100 A) T(x1) = 1 Sin ().e(-1,752 (³¹)+(sin().e (-1,752 (²) ₁ + 1 3π TC3 .....) 100 t + ··· ....... 13) T(x,t) = 200 Sin ().e(-1,752 (²t) + (sin (3). e (-1,752 (7) ²) t B) 3/3 t + …............) C) T(x.t) = 200 Sin ().e(-1,752 (²t) (sin().e(-1,752 (7) ²) t – D) T(x,t) = 200 Sin ().e(-1,752 (²)-(sin().e (-1,752 (²7) ²) t E) T(x.t)=(Sin().e(-1,752 (²t)-(sin().e(-1,752 (²) t+ t + ··· .........) t +.... t + ··· .........) …..)arrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning