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# 2d Steady Heat Conduction Assignment Sample

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## Introduction - 2d Steady Heat Conduction

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### 1.0 Introduction

The heat conduction takes place between hot steel billet and ceramic table on which that billet is placed. The temperature of billet is about 1500 degree Celsius. The heat conduction takes place between hot billet and table through conduction mode of heat transfer. The aluminum legs are attached to table and there is also heat conduction takes place. The bottoms of legs are insulated with insulation material through which negligible heat transfer takes place. The main problem which is needed to be solved to overheating of floor below insulated material. This requires defining the thermal conductivity of insulating material and reducing the floor temperature to 60-degree Celsius.

### 2.0 Sketch of problem

The above figure describes about sketch of the problem which is needed to be answered with help of the calculation and simulation process. The different mode of heat transfer is shown in above figure. The conduction and convection mode of heat transfer is shown with spline curve and arrow respectively in above figure (Peng et al 2020). The insulation is provided below the aluminum legs at the bottom and hence the main problem which is needed to be solved is reducing the temperature of floor below 60 degrees Celsius. The different nodes with convective heat transfer are shown in the above figure in which heat transfer takes place between ceramic table and surrounding air. A high amount of heat transfer will take place between hot billet and the table where the temperature will be very high. The heat transfer will then process through aluminium legs in which temperature will be also very high due to its high thermal conductivity. The insulation material is provided below the aluminium legs and at the bottom so as to reduce the effect of heat transfer through the conduction process (Dördüncü, 2019). This requires to select the material of insulation so that overall temperature below the insulation materials will be below 60-degree Celsius. This will solve the problem of overheating for the floor below the insulated surface attached to legs of table.

### 3.0 Temperature distribution of table

The temperature distribution of table is analyzed with help of conduction and convection equations. The heat transfer equation of conduction and convection is shown with help of excel calculations.

The heat transfer equation that is used for table is shown below:

 Direction Conduction Convection X axis Qx = -K*Ax*dT/dx Qx = h* Ax * (T1 - T2) Y axis Qy = -K*Ay*dT/dy Qy = h* Ay * (T1 - T2)

The heat transfer equation is shown for conduction and convection equation that takes place through table.

The total heat transfer with conduction process in table is given by:

Q = Qx + Qy

Q = -K*Ax*dT/dx -K*Ay*dT/dy

The heat transfer through convection process in both z and y directions are given by following equation.

Q` = Q`x + Q`y

Q` = h* Ax * (T1 - T2) + h* Ay * (T1 - T2)

K and h in above two equations are thermal conductivity and convective heat transfer coefficient for heat transfer process.

The elemental strip is selected as ?x and ?y for analyzing heat transfer process from table and this will help to analyze the heat transfer process at each specified node in table in x and y direction. The change in temperature between these nodes is shown with help of differential equations dT/dx and dT/dy. This temperature differential in the x and y direction will result into an evaluation of heat transfer in that elemental strip of the element.

The overall heat transfer rate from billet to insulation is shown below:

Q = -K1*Ax*dT/dx -K1*Ay*dT/dy + h1* A * (T2 - Ta) + h2* A * (T2 - Ta) - K2*Ax*dT/dx -K2*Ay*dT/dy - Kin*Ain*dTin/dx

Here h1 = 3.1 W/m^2/K

h2 = 0.7 W/m^2/K

K1 = 27.8 W/m/K

K2 = 240 W/m/K

T1 = 1500 C

T2 = 500 C (Assumed for calculation process for differential equations)

The further calculations are shown in excel in appendix 1 and temperature analyzed for aluminium legs are 344.66 C after solving heat conduction and convection equation. [Refer to appendix 1]

Total heat transferred in aluminium legs with help of calculation process is -2797.56 W and this negative value shows that heat is taken by the body not rejected (Li et al 2019). The further heat equation is solved for insulated material and the unknowns in heat conduction equation for insulated material are heat transfer rate, thermal conductivity of insulation and temperature after insulation process. [Refer to appendix 2]

After 25 iterations it is confirmed that, at 0.038 W/m/K thermal conductivity of insulation, the dropped temperature will be 56.88 C which is below 60 C. The heat transfer through insulation is analyzed as 0.82 W as any insulated materials cannot have totally zero heat transfer rate.

### 4.0 Maximum value of insulation

The maximum value of insulation analyzed is 0.04 W/m/K after carrying out iteration process. There are 25 iterations conducted with help of an excel sheet and approximate value of Kins where temperature reaches near 60 C is 0.04 W/m/K.

The heat transfer rate at 0.04 W/m/K and 60 C temperature is shown with help of conduction equation.

 Qin = Kin * 0.025 * (344.6 - Tin) + Kin* 0.05* (344.6 - Tin)
 Qin = 0.075 * kin * (344.6 - Tin)

Qin = 0.075 * 0.04 * (344.6 - 60)

Qin = 0.8538 W

This heat conduction is very approx to 0.82 W of heat transfer analyzed with help of conduction process and material analyzed for insulation is fiberglass.

### 5.0 Temperature distribution contour plot

The contour plot of temperature distribution is analyzed with help of design and simulation in Ansys. The temperature distribution is shown for temperature acting on table and its distribution over all surfaces.

The above figure shows temperature distribution on table for high temperature of hot billet placed on table surface. The range of temperature acting on the table is from 1020 C to 220 C and this distribution is shown uniformly on all over the body of aluminium legs below the table.

### 6.0 Plot of temperature as a function of height

The graph is drawn for variation of height with temperature that is analyzed with help of contour plot. The maximum temperature analyzed is 1020 C and this is analyzed at top of leg and minimum temperature analyzed is 220 C at the bottom of leg. The data is collected from contour plot to generate graph between these two parameters.

The above figure shows temperature vs length of table leg graph and fall of temperature is clearly analyzed when decreasing length of leg towards surface of insulation.

### 7.0 Total convective heat loss from table

The convective heat loss from table is analyzed as

 Qc = -h1*A * (T2 - Ta) - h2 * A * (T2-Ta) - h2*Ain*(T3-Tins)
 Qc = -3.1*0.1*0.1 * (500 - 30) - 0.7 * 0.1*0.1 * (500 - 30) - 0.7 * 0.075* (344.6 - 60) Qc = -17.86 - 14.94 Qc = -32.80 W

The total heat rejection rate through convection process in table is 32.80 W.

Conclusion

The problem is analyzed for reducing the overall temperature below insulation surface to 60 C with using of fiberglass material whose thermal conductivity is 0.04 W/m/K. This is analyzed with finite differential equation in two dimensions and a finite differential equation for heat conduction and convection process. The contour plot of temperature variation in leg of table with temperature variation graph between length of leg and temperature is shown here. The main heat transfer process which is considered in this problem are conduction and radiation mode of heat transfer.

Reference list

Journals

Alrwashdeh, S.S., 2018. Modelling of Operating Conditions of Conduction Heat Transfer Mode Using Energy 2D Simulation. Int. J. Online Eng.14(9), pp.200-207.

Baral, P. and Bhattarai, N., 2019. Non-linear Steady State Heat Conduction using Element-Free Galerkin Method. In Proceedings of IOE Graduate Conference.

Dördüncü, M., 2019. Peridynamics for the Solution of the Steady State Heat Conduction Problem in Plates with Insulated Cracks. Journal of Aeronautics and Space Technologies12(2), pp.145-155.

Ho?ejowski, L., 2018. An inverse geometric problem in steady state heat conduction–the solution and stability analysis. In MATEC Web of Conferences (Vol. 240, p. 01013). EDP Sciences.

Li, J.P., Fu, Z.J., Chen, W. and Liu, X., 2019. A dual-level method of fundamental solutions in conjunction with kernel-independent fast multipole method for large-scale isotropic heat conduction problems. Adv. Appl. Math. Mech11(2), pp.501-517.

Li, Z., Xiong, S., Sievers, C., Hu, Y., Fan, Z., Wei, N., Bao, H., Chen, S., Donadio, D. and Ala-Nissila, T., 2019. Influence of thermostatting on nonequilibrium molecular dynamics simulations of heat conduction in solids. The Journal of chemical physics151(23), p.234105.

Peng, J.Z., Liu, X., Aubry, N., Chen, Z. and Wu, W.T., 2020. Data-driven modeling of geometry-adaptive steady heat transfer based on convolutional neural networks: heat conduction. arXiv preprint arXiv:2010.03854.

Ranjan, A., Kumar, A., Roy, A.K. and Kumar, K., 2022. Python assisted numerical analysis of heat conduction for an orthotropic material. Advances in Materials and Processing Technologies, pp.1-15.

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