WDocument1 - Microsoft Word (Product Activation Failed)FileHomeInsertPage LayoutReferencesMailingsReviewViewa ?A Find -A% Cut-=。青Calibri (Body) - 11A AAaAalAaBbCcDc AaBbCcDc AaBbC AaBbCc AaB AaBbCcLE Copyae ReplacePasteBI Uabe x, x*AabAI NormalI No Spaci. Heading 1ChangeStyles.Heading 2TitleSubtitleWA Select -Format PainterClipboardFontParagraphStylesEditing• 2. 1:.I' 2: 1 : 3:1• 4.I 5.1' 6.1'7I'8: 1 9 ' 10.L· 11: 1 ' 12.'13 · L 14: 1· 15.1A L'17:1 18PARTIAL DIFFERENTIAL EQUATIONSPLEASE ANSWER ALL QUESTIONSConsider the temperature distribution in a 2 x 1 rectangular plate as described in the figure below.Show that the steady-state temperature distribution u (1, y) in the plate is given by200пя (1- у)u (1, 3) = Esinh:(1 +(-1)") sin -sinhn sinhn=12Calculate the steady-sate temperature at the centre of the plate.Ice at 0°-1Ice at 0°Ice at 0°Boiling water at 100°all06:14 PM2022-05-10Page: 1 of 1Words: 7E EA E E E100% -+)

Given : To find : Temperature at center

Consider the temperature distribution in a 2 x 1 rectangular plate as described in the figure belov Show that the steady-state temperature distribution u (r, y) in the plate is given by u (1, y) n sinh # (1 + (-1)") sin "L sinh "x (1 – y) . Calculate the steady-sate temperature at the cœntre of the plate. Ice at 0°- Ice at 0° Ice at 0° Boiling water at 100°

Consider the temperature distribution in a 2 x 1 rectangular plate as described in the figure belov Show that the steady-state temperature distribution u (r, y) in the plate is given by u (1, y) n sinh # (1 + (-1)") sin "L sinh "x (1 – y) . Calculate the steady-sate temperature at the cœntre of the plate. Ice at 0°- Ice at 0° Ice at 0° Boiling water at 100°

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Solve for the temperature distribution of a long, thin circular solid tube with a length of 10 cm (shown in Figure2), using the Finite Difference Method: Parabolic Equation (Explicit Method). The following values of the circular solid tube is as provided: k’= 0.49 cal/ (s · cm · oC), Δt = 0.1 s and Δx = 2 cm. When t = 0, the temperature of the circular solid tube is zero and the boundary conditions are fixed for all times at temperature T, at x = 0 cm is 100oC where as the temperature T, at x =10 is 50oC. From this information it is given that the circular solid rod is made of material with heat capacity, C = 0.2174 cal/(g· oC) and density (ρ) = 2.7 g/cm3. Use Explicit Method to solve this problem. Task: A Visual Basic of the FDM solution for (a) has been initially developed based on the required calculation of task Q2(a). The source code is as provided below (may need some minor tweak): Option Explicit Sub Explicit() Dim i As Integer, j As Integer, np As Integer, ns As Integer…
T1=68 T2=43 Q3:
please help with this im not sure what i did wrong
1. Temperatures are measured at the left-hand face and at a point 4 cm from the left-hand face of the planar wall shown in the figure below. These temperatures are T₁ = 45.3 °C and T* = 21.2 °C. The heat flow through the planar wall is steady and one dimensional. What is the value of T2 at the right-hand surface of the wall? TI T* 4 cm 10 cm T2
1. An infinite parallel-sided slab of non-dimensional thickness L 1, has an initial (at non- dimensional time t= 0) temperature distribution To (x) given in below. Its left and right ends are subsequently (for t> 0) maintained at temperatures T1, and T2, respectively. The initial temperature distribution is given by T.(x) = 250*x*sin(rx) K, and the boundary temperatures for t> 0 are: T = 50K and T2 = 100K. Use 10 intervals in the x- direction, and At = 0.01. (a) Write the equations you'll be using to solve the problem in matrix form, using the implicit method. The coefficient matrix and the right-hand-side vector should show numerical values. Do not write every value, but use the matrix representation given in the presentation slides so you can fit the answer in the width of the page. Your equations should be for solution at time level (n=2), where n = Irepresents time t = 0.
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Problem 4 A cork board (k = 0.039W/m K) 3 cm thick is exposed to air with an average temperature T.. = 30°C with a convection heat transfer coefficient, h = 25 W/m².K. The other surface of the board is held at a constant temperature of 15°C. A volumetric heat generation of 5 W/m³ is occurring inside the wall. Assuming one dimensional conduction, write the governing equation and express the boundary conditions for the wall. (solve the equation is not required) Problom
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Please answer the question with explicit scheme and calculation is done until the second time!
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### Given the resolution of the problem 5.102 below, develop a python program that shows the found temperature distribution PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q = 10³ W/m³, occurs when the element is inserted into the core while the surfaces experience convection (To,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to Fo aAt 5x10 m²/sx0.3s 4x² t=pAt=0.3p(s). =0.375 (0.002m)² Fuel element- x=5x10 m²/s k-30W/m-K Coolant Too-250°C h-1100 W/m²-K رجا T(x,0)=250°C 9-10°W/m³ at +>0 Coolant Too,h L-10mm k+ax=2mm ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q=0, initially; at t> 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal…
A long cylinder of radius, ro = 2.8cm, thermal conductivity k = 225 S mk thermal diffusivity 1.10 kJ and initial temperature, T = 200°C is kgK suddenly exposed to a convective environment at To 65°C with a convection coefficient, a = 6.82x105 m², specific heat, c h=510- W Watch your units on this one! m²K Determine the temperature at the center of the cylinder after t 130s, time, of exposure to the convective environment in degrees Celsius. 97.14C 95.30 198.08 70.94