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#### Concept explainers

Assume steady-state, one-dimensional heat conduction through the axisymmetric shape shown below.

Assuming constant properties and no internal heat generation, sketch the temperature distribution on

To sketch: The temperature distribution on T-x coordinates and explain the shape.

### Explanation of Solution

Draw the axisymmetric shape as shown below:

Write the expression as per Fourier law.

Here,

Since the heat transfer through the body remains constant and thermal conductivity of the body remains constant, the Fourier law can be explained as:

The above expression indicated that temperature and thickness is inversely proportional to one another.

On T-x curve the independent variable is x and the dependent variable is T .

Draw the T-x curve for axisymmetric shape as shown below:

The T-xcurve for axisymmetric shape is shown above and the slope

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# Chapter 2 Solutions

Fundamentals of Heat and Mass Transfer

- #4: A rod of length L. coincides with the interval [0, L] on the x-axis. Let u(x, t) be the temperature. Consider the following conditions. (A) The left end is held at temperature 0°. (B) The right end is insulated. (C) There is heat transfer from the lateral surface of the rod into the surrounding medium, which is held at temperature 0° (D) The left end is insulated. (E) The initial temperature is 0° throughout. (F) The right end is held at temperature 0°. (G) There is heat transfer from the right end into the surrounding medium, which is held at a constant temperature of 0°. (H) There is heat transfer from the left end into the surrounding medium, which is held at a constant temperature of 0°. In each part below, determine which of the above conditions corresponds to the given initial or boundary condition for the heat equation. (a) u(x, 0) = 0 (b) u(0, 1) = 0 (c) du (d) ou x=L ox|x=0 = -hu(L, 1) = hu(0, 1)
*arrow_forward*Derive an expression for the temperature distribution within a sphere that has inner radius r, where the temperature T, and outer radius r, where the temperature T,. Assume the heat source within the wall of sphere is q' and the heat conductivity is k. also assume one-dimensional heat transfer (r - direction)*arrow_forward*The thermal conductivity of a specific material is highly dependent on a number of factors. These include the temperature gradient, the properties of the material, and the path length that the heat follows. My question is how to add thermal conductivity? for example SS316 material has thermal conductivity=15W/mk if Gyroid shape is created with SS316 then what will be the total thermal conductivity?*arrow_forward* - Question 5:Assume steady-state, one-dimensional heat conduction through the symmetric shape shown in Figure 1.Assuming that there is no internal heat generation, derive an expression for the thermal conductivity k(x) for these conditions: A(x) = (1 -x), T(x) = 300(1 - 2x -x3),and q = 6000 W, where A is in square meters, T in kelvins, and x in meters. Consider x= 0 and 1
*arrow_forward*for the following differential equation, determine (a) order, (b)degree, (c) linearity, (d) unknown function, and (e) independent variables*arrow_forward*Which one of these is false about heat conduction equation? In plane coordinates it is linear but in cylindrical ones it is logarithmic The equation may be solved in cartesian or polar coordinates None of the given We have assumed that the flow is at least isothermal with adiabatic flow lines The equation allows us to determine how the temperature changes inside the material*arrow_forward* - After a thorough derivation by Doraemon to establish an equation for cylindrical fuel rod of a nuclear reactor. Here he was able to come up an equation of heat generated internally as shown below. 96 = 9. where qG is the local rate of heat generation per unit volume at radius r, ro is the outside radius, and qo is the rate of heat generation per unit volume at the centre line. Calculate the temperature drop from the centre line to the surface for a 2.5 cm outer diameter rod having k = 25 W/m K, if the rate of heat removal from the surface is 1650 kW/m2 A 619 °C 719 °C C) 819 °C 919 °C E 1019 °C F None of these
*arrow_forward*Consider 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_forward*Please provide accurate answer with proper steps The wall of the furnace is 30.48 mm thick and is insulated from outside. Thermal conductivity of the wall material is 0.1 W/m K and the insulation material is 0.01 W/m K. The furnace operates at 650 0C and the ambient temperature is 30 0 Allowable temperature on the outer side of the insulation is 1000C. Determine the overall heat transfer by conduction per unit area occurring across a furnace wall made from clay. If the air side heat transfer coefficient is 0.4 W/m2 K, calculate the minimum insulation thickness requirement.*arrow_forward* - After a thorough derivation by Doraemon to establish an equation for cylindrical fuel rod of a nuclear reactor. Here he was able to come up an equation of heat generated internally as shown below. 9G = 9. where qG is the local rate of heat generation per unit volume at radius r, ro is the outside radius, and qo is the rate of heat generation per unit volume at the centre line. Calculate the temperature drop from the centre line to the surface for a 2.5 cm outer diameter rod having k = 25 W/m K, if the rate of heat removal from the surface is 1650 kW/m² А) 619°C В 719 °C C) 819 °C D) 919 °C E 1019 °C F None of these
*arrow_forward*The class is Thermodynamics and Kinetics of Materials, the section is on surface energy using "Phase Transfromations in Metals and Alloys" by Porter, Easterling, and Sharif. The first image shows the problem. My understanding from Eq 3.1 is that the free energy of the 2 dimensional rectangle (with no bulk and two surfaces) would be G=2AY (Y being gamma [free energy per unit area] in the text). The second image is the pertinant section. The problem gives two length variables, each with an assigned free energy (Y) and says the area is constant. The solution claims the Gibbs Energy is G=2(l1Y1+l2Y2) which I don't understand if the area should be the product of the lengths, not the sum.*arrow_forward*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_forward*

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- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning