consider coaxial, parallel, black disks separated a distance of 0.20 m. the lower disk of diameter 0.40 m is maintained at 500 k and the surroundings are at 310 k. 0.40 m 0.20 m 0.20 m heater 500 k what temperature will the upper disk of diameter 0.20 m achieve if electrical power of 15 w is supplied to the heater on the back side of the disk?

Let's now think about the net rate of radiation from the upper disk. If we take a look at the given figure we will notice that the heat is transferred by radiation from surface A1 to both lower disk of surface A2and the surroundings. So, we will write the net rate of heat transfer from A1as:

Q˙1=Q˙12+Q˙1,surr

where: σ=5.67⋅10−8 W m2K2σ=5.67⋅10−8 m2K2 W.

According to eq. (2), the unknown values that have to be determined in order to calculate T2 are:

Surface area A1A1;

The view factors F12F12 and F1,surrF1,surr.

The surface area A1A1 is the surface of the disk, calculated using the well-known equation:

A1=D12π4=0.22π4=0.0314 m2

This was easy. Let's move on to a little bit more difficult part, which is determining the view factors.

We will now calculate the view factor F12F12. Since surfaces A1A1 and A2A2 are coaxial parallel disks, the view factor will be determined using the following expression:

Remember that distance between the disks is given to be:

L=0.2 m

Therefore, we will calculate SS as:

S=1+1+(0.20.2)2(0.10.2)2=1+1+0.520.52=9

Now, let's substitute values to eq. (3) and determine the view factor from the upper to the lower disk:

F12=21⎝

⎛9−[92−4(0.10.2)2]1/2⎠

⎞=0.47

Next, by applying the summation rule for surfaceA1A1, F1, surrF1, surr will be determined:

F11+F12+F1, surr=1F1, surr=1−F11−F12=1−0−0.47=0.53

Note: The disk is a plate surface that can not see itself. Therefore, the view factor from the plate surface to itself is equal to zero: F11=0F11=0.

Now, that we have all the values we need, by substituting them into eq. (2), it follows:

17.5=1⋅5.67⋅10−8⋅0.0314⋅[0.47⋅(T14−5004)+0.53⋅(T14−3004)]17.5=0.178⋅10−8⋅[0.47⋅(T14−5004)+0.53⋅(T14−3004)]

17.517.5=1⋅5.67⋅10−8⋅0.0314⋅[0.47⋅(T14−5004)+0.53⋅(T14−3004)]=0.178⋅10−8⋅[0.47⋅(T14−5004)+0.53⋅(T14−3004)]

Finally, if we rearrange the previous equation, we will calculate the temperature of the upper disk:

0.08366⋅10−8⋅T14+0.09434⋅10−8⋅T14=17.5+49.94+7.640.178⋅10−8⋅T14=75.08T14=75.080.178⋅10−8T1=421.8⋅1084=453.2 K

0.08366⋅10−8⋅T14+0.09434⋅10−8⋅T140.178⋅10−8⋅T14T14T1=17.5+49.94+7.64=75.08=0.178⋅10−875.08=4421.8⋅108

=453.2 K

Therefore, the required value of T1 is obtained and our problem is solved.

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