- #1

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let θ = the acceptance angle;

let a' = the radius of the base;

let a = the radius of the aperture, where a=a'/sin(θ);

let L = the height of the CPE from base to aperture, where L=(a+a')/tan(θ);

let T

_{1}be the temperature of a';

let T

_{2}be the temperature of a;

where T

_{2}>T

_{1};

y(Φ)=((2a'(1+sin(θ)sin(Φ-θ))/(1-cos(Φ))-a, gives us the radius at Φ;

z(Φ)=((2a'(1+sin(θ)cos(Φ-θ))/(1-cos(Φ)), gives us the height at Φ;

Solving z for Φ gives us,

Φ(z)=2cot

^{-1}((sec(θ)sqrt(zcos(θ)a(sin(θ)+1)+a(sin(θ)+1)

^{2})-tan(θ)a(sin(θ)+1))/a(sin(θ)+1));

A(x)=π•(y(Φ(x)))

^{2};

Q'•dx/A(x)=-k*dt;

Integrate the left side from 0 to L. Integrate the right side from T1 to T2.

The calculus gets very hairy at the function composition (y ° Φ)(x). It's even too dense for Wolfram Alpha, which times out, given such a monstrosity. It looks to me like a differential equation, but that looks almost as hairy. Is there a better integral technique to address this mess?