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Solve the one-dimensional heat conduction problem 6 using the Rayleigh-Ritz method. For the heat conduction problem, the total potential can be defined as
Use the approximate solution
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- The 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_forwardThe wall (thickness L) of a furnace is comprised of brick material (thermal conductivity, k = 0.2 Wm¯' K'). Given that the atmospheric temperature is 0°C at both sides of wall, the density (p) and heat capacity (c) of the brick material are 1.6 gm cm³ and 5.0 J kg K¯l respectively. du Solve pc = k- subject to initial conditions as u(x,0) = x²(L – x). ốt Consider the case 2=- p² only.arrow_forward1. The general form of linear second-order differential equation can be written in the form: و بار / كلية الهندسة Q4)/ grap dy q(x)y = r(x) d'y +p(x) dx dy b. dx - F(x)y = F(x) x2 dy dx - xy = C. d. r2 d?y dx2 -f(x)y = F(x) 2431)(5-1) 3 (3-21)2 a. (221 -91i) / 169 b. (21 + 52i)/ 13 c. (-90+220i)/169 d. (-7+17i)/ 13 2. Simplify: الحدار المك المراغة 3. If the roots of second order differential equation is complex conjugate, then the gene contain: a. sinusoidal functions and exponentials b. constant and two exponentials c. two constants and two exponentials d. two constants and one exponential 5 4. The order and degree of the differential: 3(3 - + 4y = sinx* are: d²y a. First-order, First-degree- b. First-order, second-degree Second -order, First -degree d. Second -order, second-degree dx2 lo - 2i tisi. 8- 12i 5. The particular solution of (D² + 4)y = cos 2x is equal to: a. sin 2x b. cos 2x 13+159 C. 4 cos 2x d. 4 sin 2x 5-12 lo Best wishes الامتحانية د. مازن ياسین عبود رئيس القسم بن فاضل…arrow_forward
- Consider the following linear equations,arrow_forwardFind the two-dimensional temperature distribution T(x,y) and midplane temperature T(B/2,W/2) under steady state condition. The density, conductivity and specific heat of the material are p=(1200*32)kg/mº, k=400 W/m.K, and cp=2500 J/kg.K, respectively. A uniform heat flux 9" =1000 W/m² is applied to the upper surface. The right and left surfaces are also kept at 0°C. Bottom surface is insulated. 9" (W/m) T=0°C T=0°C W=(10*32)cm B=(30*32)cmarrow_forward3.10 By neglecting lateral temperature variation in the analysis of fins, h,T. 木 H two-dimensional conduction is modeled as a one-dimensional H problem. То examine this T, h,T. approximation, consider a semi- infinite plate of thickness 2H. The base is maintained at uniform temperature T,. The plate exchanges heat by convection at its semi- infinite surfaces. The heat transfer coefficient is h and the ambient temperature is T.. Determine the heat transfer rate at the base.arrow_forward
- The amount of heat conducted through a wall of length r is given by Fourier's Law:. CONDUCTION RATE EQUATION T FOURIER'S LAW q, = -k A dT dx T, >T, where q, is the heat flux, k is a proportionality factor, Ais the wall's cross-sectional area, and 4 is the temperature gradient throughout the wall. Our friend Matt Labb wants to find T2 (temperature of the wall's rightmost edge) given q,, k, and A. Is this possible? If so, briefly explain how to find Tp. If not, briefly explain why.arrow_forwardConsider a wall of thickness 50 mm and thermal conductivity 14 W/m.K, the left side (x-0) is insulated. Heat generation (q,) is present within the wall and the one dimensional steady-state temperature distribution is given by T(x) = ax +bx+c [°CJ, where c 200 °C, a = -1144 °C/m is the heat fluxes at the right side, x L, (kW/m)? b= needs to he determined, and x is in meters. What 9, K 4L) Insulationarrow_forwardUsing an appropriate method, find the general solution for the following differential equation, dy +2= cos 2x . dxarrow_forward
- Help mearrow_forwardunder steady-state conditions. If you are given T1 = 200 °C and T2 = 164 °C, determine: a) the conduction heat flux, q,.cond, in m2 W from x = 0 to x = L b) if the dimensions of the triangle ares 15 mm and h 13 mm, calculate the heat transfer due to convection, q,y, in W at x = L Finsulation T2 T T = 20°C h = 500 W/m2.K Triangular Prism x L x 0 L= 50 mm k = 100 W/m-Karrow_forwardLet's assume that the outdoor temperature in your region was 1 C on 26.12.2002. Let's assume that you use a 2088 W heater in the room in order to keep the indoor temperature of the room at 20 ° C. In the meantime, a 68 W light bulb for lighting, a computer you use to solve this question and load it into the system (let's assume it consumes 217 W of energy), you and your two friends (three people in total) are in the room to assist you in solving the questions. A person radiates 45 J of heat per second to his environment. When you consider all these conditions, calculate the exergy destruction caused by the heat loss from the exterior wall of your room.arrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning