1. **Problem Statement:** Calculate the module and pitch circle diameters of a pair of helical gears transmitting 25 kW power. Given: - Power, $P = 25$ kW - Speed of driver gear, $N_1 = 1800$ rpm - Speed of driven gear, $N_2 = 600$ rpm - Helix angle, $\beta = 30^\circ$ - Pressure angle, $\phi = 20^\circ$ - Number of teeth on driven gear, $T_2 = 24$ - Allowable stress, $\sigma = 5$ MN/m$^2$ - Face width, $b = 4 \times$ circular pitch 2. **Calculate the number of teeth on the driver gear $T_1$:** Since speeds are inversely proportional to teeth numbers, $$\frac{N_1}{N_2} = \frac{T_2}{T_1} \implies T_1 = T_2 \times \frac{N_2}{N_1} = 24 \times \frac{600}{1800} = 8$$ 3. **Calculate the velocity ratio $i$:** $$i = \frac{N_1}{N_2} = \frac{1800}{600} = 3$$ 4. **Calculate pitch circle diameters $d_1$ and $d_2$:** Pitch circle diameter is related to module $m$ and number of teeth $T$ by $$d = m \times T$$ 5. **Calculate circular pitch $p$:** Circular pitch is related to module by $$p = \pi m$$ 6. **Calculate face width $b$:** Given as 4 times circular pitch, $$b = 4p = 4 \pi m$$ 7. **Calculate tangential velocity $v$ of pinion:** $$v = \frac{\pi d_1 N_1}{60} = \frac{\pi m T_1 N_1}{60}$$ 8. **Calculate tangential force $F_t$ on the gear teeth:** Power transmitted, $$P = F_t v \implies F_t = \frac{P}{v} = \frac{25000}{v}$$ 9. **Calculate normal force $F_n$ on the teeth:** Due to helix angle, $$F_n = \frac{F_t}{\cos \beta}$$ 10. **Calculate module $m$ using allowable stress:** Bending stress formula for helical gears, $$\sigma = \frac{F_n}{b m} \implies m = \sqrt{\frac{F_n}{b \sigma}}$$ Substitute $b = 4 \pi m$ and solve for $m$: $$\sigma = \frac{F_n}{4 \pi m^2} \implies m^2 = \frac{F_n}{4 \pi \sigma} \implies m = \sqrt{\frac{F_n}{4 \pi \sigma}}$$ 11. **Calculate $v$ and $F_t$ numerically:** $$v = \frac{\pi m \times 8 \times 1800}{60} = 240 \pi m$$ $$F_t = \frac{25000}{240 \pi m} = \frac{25000}{753.98 m} = \frac{33.17}{m}$$ 12. **Calculate $F_n$ numerically:** $$F_n = \frac{F_t}{\cos 30^\circ} = \frac{33.17/m}{0.866} = \frac{38.28}{m}$$ 13. **Calculate module $m$ numerically:** $$m = \sqrt{\frac{F_n}{4 \pi \sigma}} = \sqrt{\frac{38.28/m}{4 \pi \times 5 \times 10^6}}$$ Multiply both sides by $m$: $$m^2 = \sqrt{\frac{38.28}{4 \pi \times 5 \times 10^6}}$$ Calculate inside the root: $$4 \pi \times 5 \times 10^6 = 62,831,853$$ $$\frac{38.28}{62,831,853} = 6.09 \times 10^{-7}$$ $$m = \sqrt{6.09 \times 10^{-7}} = 0.00078 \text{ m} = 0.78 \text{ mm}$$ 14. **Calculate pitch circle diameters:** $$d_1 = m \times T_1 = 0.78 \times 8 = 6.24 \text{ mm}$$ $$d_2 = m \times T_2 = 0.78 \times 24 = 18.72 \text{ mm}$$ **Final answers:** - Module, $m = 0.78$ mm - Pitch circle diameter of pinion, $d_1 = 6.24$ mm - Pitch circle diameter of gear, $d_2 = 18.72$ mm