A worm gear box with an effective surface area of 1.5 m² is operating in still air with a heat transfer coefficient of 15 W/m²°C. The temperature rise of the lubricating oil above the atmospheric temperature is limited to 50°C. The worm gears are designated as,
1/30/10/8
The worm shaft is rotating at 1440 rpm and the normal pressure angle is 20°. Calculate the power transmitting capacity based on the thermal considerations.

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Accepted Answer

[latex] ext { Given } A=1.5 m ^{2} \quad k=15 W / m ^{2}{ }^{\circ} C [/latex].

[latex] \left(t-t_{o} ight)=50^{\circ} C \quad z_{1}=1 \quad z_{2}=30 ext { teeth } \quad q=10 [/latex].

m = 8 mm n = 1440 rpm
Step I Efficiency of worm gear drive

[latex] an \gamma=\frac{z_{1}}{q}=\frac{1}{10}=0.1 \quad ext { or } \quad \gamma=5.71^{\circ} [/latex].

[latex] d_{1}=m q=8(10)=80 mm [/latex].

[latex] V_{s}=\frac{\pi d_{1} n_{1}}{60000 \cos \gamma}=\frac{\pi(80)(1440)}{60000 \cos (5.71)}=6.06 m / s [/latex].

From Fig. 20.9, the coefficient of friction is 0.024. From Eq. (20.34),

[latex] \eta=\frac{(\cos \alpha-\mu an \gamma)}{(\cos \alpha+\mu \cot \gamma)} [/latex] (20.34),

[latex] \eta=\frac{(\cos \alpha-\mu an \gamma)}{(\cos \alpha+\mu \cot \gamma)}=\frac{[\cos (20)-0.024(0.1)]}{[\cos (20)+0.024(1 / 0.1)]} [/latex].

= 0.7945.

Step II Power transmitting capacity based on thermal considerations
From Eq. (20.40),

[latex] kW =\frac{k\left(t-t_{o} ight) A}{1000(1-\eta)} [/latex] (20.40).

[latex] kW =\frac{k\left(t-t_{o} ight) A}{1000(1-\eta)}=\frac{15(50)(1.5)}{1000(1-0.7945)}=5.47 [/latex]

Question 20.7: A worm gear box with an effective surface area of 1.5 m^2 is...

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