Question 7 Critical Mass and Geometry (10) a) Calculate the...

Question 7 Critical Mass and Geometry (10) a) Calculate the geometric buckling ($B_g^2$) for 1 m³ reactor with the following geometry - (neglecting the extrapolation length). 1. Cubic 2. Cylindrical with height-to-diameter ratio 1:1 3. Spherical For reactors composed of the same homogeneous material, which geometry in part a) has the smallest critical mass, and which one has the largest and why?

Transcript

00:01 In the question we have to determine the height to diameter ratio of a bare cylindrical reactor.

00:06 So smallest critical mass means that smallest surface area so that means with the volume of reactor being the same we wish to minimize the surface area of cylindrical reactor and determine height to diameter ratio.

00:25 So we will let the volume of cylinder be v.

00:35 So we know that volume of cylinder is pi r square h where r and h are the radius and height of the cylinder.

00:45 So from here h will be equal to v upon pi r square.

00:49 So let us take this as equation 1.

00:52 Also let the surface area of cylinder be s.

00:54 So we know that the surface area of cylinder is 2 pi r h plus 2 pi r square.

01:05 So from here we know the value of h so we will put the value of h in this area and we will in this surface area and we will s of r will be equal to 2 pi r on taking the value of h that is v upon pi r square plus 2 pi r square.

01:26 So from here 2 pi will get cancelled by pi and we will have 2 v upon r plus 2 pi r square.

01:36 So that was s of r.

01:38 So s of r will be equal to 2 v upon r plus 2 pi r square.

01:46 Let us take this as equation 2.

01:48 Now we will differentiate equation 2.

01:52 Differentiating equation 2 with respect to r we will get s dash r will be equal to minus 2 v upon r square plus 4 pi r.

02:13 So if we will put s dash of r is equal to 0 that means we will have minus 2 v upon r square plus 4 pi r is equal to 0.

02:28 So from here 4 pi r will be equal to 2 v upon r square.

02:34 So on cross multiplicating we will get 4 pi r cube is equal to 2 v.

02:42 So from here r cube will be equal to 2 v upon 4 pi...

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