Business: A plane wall of thickness 8 cm experiences uniform volumetric heat generation ( q ). One surface of the wall is insulated, and the other surface is exposed to a fluid at T(infinite) = 20 degree with convection heat transfer characterized by h=1000W/m2.K. The initial temperature distribution on the wall is given as: T (x,0) = 300-0.1*10^4*X^2 , where x is in meters. Suddenly, the volumetric heat generation is deactivated, while convection heat transfer continues to occur at x=L. If the wall properties

I believe the the answer is b.social security number....

A plane wall of thickness 8 cm experiences uniform

Ответ:

hannahkharel2

18.01.2022

Business

Part a: the magnitude of the volumetric heat generation rate is $21.6 \times 10^6 W/m^3$

Part b: the convective heat flux is 600 W/m^2 .

Part c: the amount of energy removed per unit area is $1.2264 \times 10^7 J/m^2$

Explanation:

Part a

As per the heat equation

$\frac{d}{dx}(\frac{dT}{dx})+\frac{\dot{q}}{k}=0\\$

Here

T is given as $300-0.1\times 10^4x^2$ where x is in meters

k is given as 108 W/mK

So rearranging and solving for q' gives

$\dot{q}=-k\frac{d}{dx}(\frac{dT}{dx})\\\dot{q}=-108\frac{d}{dx}(\frac{d(300-0.1\times 10^4 x^2)}{dx})\dot{q}=-108\frac{d}{dx}(2\times (300-0.1\times 10^4 x))\\\dot{q}=-108\times 2\times (300-0.1\times 10^4 )\\\dot{q}=21.6 \times 10^6 W/m^3$

So the magnitude of the volumetric heat generation rate is $21.6 \times 10^6 W/m^3$

Part b

$q''_{conv}=h(T(L,0)-T_{\infty})$

Here

h is given as 1000W/m2K

T(L,0) is found by putting x=L in equation of T. L is given as 8cm=0.08m.

T_inf is the infinite temperature given as 20C or 293K.

So by substituting values in the equation.

$q''_{conv}=h(T(L,0)-T_{\infty})\\q''_{conv}=1000((300-0.1\times 10^4 (L^2)-293)\\q''_{conv}=1000((300-0.1\times 10^4 (0.08^2)-293)\\q''_{conv}=1000((300-0.1\times 10^4 (0.0064)-293)\\q''_{conv}=1000((300-6.4-293))\\q''_{conv}=600 W/m^2$

So the convective heat flux is 600 W/m^2 .

Part c:

For energy balance

$\dot{E_{in}}-\dot{E_{out}}=\dot{E_{st}}$

Here as

$\dot{E_{in}}$ is zero

$-\dot{E_{out}}=\dot{E_{st}}$

It is given as

$-\dot{E_{out}}=\dot{E_{st}}=\rho c_p(T_{final}-T_{\infty})-\rho c_p\int\limits^{x=L}_{x=0} ({T_{x,0}-T_{\infty}) \, dx$

Here as T_final and T_inf will be same so

Also ρ is given as 7000 kg/m3

cp=450 J/kg.K

$\dot{E_{out}}=\rho c_p\int\limits^{x=L}_{x=0} {(T_{x,0}-T_{\infty}) \, dx}\\\dot{E_{out}}=7000 \times 450\int\limits^{x=L}_{x=0} {(300-0.1\times 10^4 x^2-293) \, dx}\\\dot{E_{out}}=3150000[300x-0.1\times 10^4 \frac{x^3}{3}-293x]_0^L\\\dot{E_{out}}=3150000[300(0.08)-0.1\times 10^4 \frac{(0.08)^3}{3}-293(0.08)]\\\dot{E_{out}}=3150000(0.3893)\\\dot{E_{out}}=1.2264 \times 10^7 J/m^2$

So the amount of energy removed per unit area is $1.2264 \times 10^7 J/m^2$

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