__Concept__:

The power transmitted by the shaft is given by,

\(P = T.\omega = \frac{{2\pi NT}}{{60}}\)

Torsion equation,

\(\frac{T}{{{J}}} = \frac{\tau }{{{r}}}=\frac{{Gθ }}{l}\;\)

**Calculation:**

**Given:**

P = 30 kW, N = 700 rpm, θ = 1^{° }, G = 80 GPa and d_{i} = 0.7 d_{o}

from **\(P = T.\omega = \frac{{2\pi NT}}{{60}}\)**

\(30 \times 1000 = T \times \frac{{2π \times 700}}{{60}}\)

∴ T = 409.256 N – m

From **Torsion equation**,

**\(\frac{T}{{{J}}} = \frac{{Gθ }}{l}\;\)**

L = 1 m, θ = 1° = π/180 rad

\(\frac{{409.256}}{{\frac{π }{{32}}\left( {1 - {{0.7}^4}} \right)d_0^4}} = \frac{{80 \times {{10}^9}}}{1} \times \frac{π }{{180}}\)

on Solving, we get

**d**_{0} = 44.52 mm