1 kW power at 720 rpm is supplied to the worm shaft. The number of starts for threads of the worm is four with a 50 mm pitch–circle diameter. The worm wheel has 30 teeth with 5 mm module. The normal pressure angle is 20°. Calculate the efficiency of the worm gear drive and the power lost in

Question

1 kW power at 720 rpm is supplied to the worm shaft. The number of starts for threads of the worm is four with a 50 mm pitch–circle diameter. The worm wheel has 30 teeth with 5 mm module. The normal pressure angle is 20°. Calculate the efficiency of the worm gear drive and the power lost in friction.

Accepted Answer

[latex] ext { Given } k W=1 \quad n=720 rpm \quad \alpha=20^{\circ} [/latex].

[latex] d_{1}=50 mm \quad z_{1}=4 \quad z_{2}=30 ext { teeth } \quad m=5 mm [/latex].

Step I Effi ciency of worm gear drive
From Eq. (20.5),

[latex] l=\pi m z_{1} [/latex] (20.5).

[latex] l=\pi m z_{1}=\pi(5)(4)=(20 \pi) mm [/latex].

[latex] an \gamma=\frac{l}{\pi d_{1}}=\frac{20 \pi}{\pi(50)}=0.4 \quad ext { or } \quad \gamma=21.8^{\circ} [/latex].

From Eq. (20.33),

[latex] V_{s}=\frac{\pi d_{1} n_{1}}{60000 \cos \gamma} [/latex] (20.33).

[latex] V_{s}=\frac{\pi d_{1} n_{1}}{60000 \cos \gamma}=\frac{\pi(50)(720)}{60000 \cos (21.8)}=2.03 m / s [/latex].

From Fig. 20.9, the coefficient of friction is 0.035.
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)-0.035 an (21.8)]}{[\cos (20)+0.035 \cot (21.8)]} [/latex].

[latex] =0.9012=90.12 \% [/latex].

Step II Power lost in friction
Power lost in friction = (1 – η) kW = (1 – 0.9012) (1)
= 0.0988 kW or 98.8 W.

Question 20.3: 1 kW power at 720 rpm is supplied to the worm shaft. The num...

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