$α = 15˚ \\
P=10kw \\
N = 600 rpm \\
Pn = 100 KN/m^2 \\
Width l = (Mean Radius)/2 \\
= \dfrac{\dfrac{(R_1+R_2)}{2}}{2} \\
= \dfrac{R_1+R_2}{4} \\
µ = 0.25 \\
P=\dfrac{2πNT}{60}=10 \times 10^3=\dfrac{2 \times π \times 600 \times T}{60} \\
\therefore,T=159.15 N-m \\
=159.15 \times 10^3 N-mm \\
T = \dfrac12 \times \dfrac{u.w(R1+R2)}{sin∝} -(I) Since,\dfrac{(R_1+R_2)}{2}=Rm
\dfrac{u.w.Rm}{\sin∝} \\
P=\dfrac{w}{π[R_1^2-R_2^2]}=\dfrac{w}{π[R_1+R_2][R_1-R_2]} \\
=\dfrac{W}{π \dfrac{[R_1+R_2]}{2}×2×[R_1-R_2]} \\
=\dfrac{W}{πRm×2×[R_1-R_2]} \\
=\dfrac{w}{π[l.\sin α]×Rm×2} \ \ \ \ \ \ \ Since,R_1-R_2=l \sin α \\
w= ρ×π×l \sinα×Km×2 \\
\text{Substituting the value of w in equation (I)} \\
T=\dfrac{1×μ×P×π×l \sinα×Rm×2Rm}{ \sinα} \\
=2×μ×ρ×πl×Rm^2 \\
159.15 = 2 × 0.25 ×π× 100 × 10^3×Rm^2×0.5×2×R_m \\
\therefore,
Rm^3 = 1.0131 × 10^{-3} \\
Rm = 0.1004m \\
Rm = 100.4mm \\
Dm = 2 × Rm \\
Dm = 200.87mm \\
l = 0.5 Dm \\
= 0.5 × 200.87 \\
= 100.43 mm \\
R1 + R2 = 2Rm \\
R1 + R2 = 2 × 100.43 \\
R1 + R2 = 200.87 \\
R1 – R2 = l \sin α \\
= 100.43 × sin 15˚ \\
R1 – R2 = 25.993 mm$
Solving $R_1 \& R_2$ we get
$2R_1 = 226.86 \\
R_1 = 113.43mm \\
And R_2 = 87.4385 mm$
$P_{\max} = c/R_2 \\
\therefore, c = P_{\max} \times R_2 \\
=100 \times 10^3 \times 87.4385/1000 \\
\therefore, c = 8743 N/m \\
\text{Axial load m clutch} \\
w = 2 \times π c (R_1 – R_2) \\
= 2 \times π \times 8743 \times 25.983 \\
W = 1427 N$