Question 12.2: Electric Motor Geared to Drive a Machine A motor at about n=......

a. From Equations (12.1) through (12.5), we have

p_n=p \cos \psi, \quad p_a=p \cot \psi=\frac{p_n}{\sin \psi}     (12.1)

P p=\pi, \quad P_n p_n=\pi, \quad P_n=\frac{P}{\cos \psi}, \quad P=\frac{N}{d}      (12.2)

m=\frac{1}{P} \quad m_n=\frac{1}{P_n}     (12.2′)

\tan \phi_n=\tan \phi \cos \psi       (12.3)

d=\frac{N p}{\pi}=\frac{N p_n}{\pi \cos \psi}=\frac{N}{P_n \cos \psi}      (12.4)

c=\frac{d_1+d_2}{2}=\frac{p}{2 \pi}\left(N_1+N_2\right)=\frac{N_1+N_2}{2 P_n \cos \psi}        (12.5)

P=\frac{1}{2 c}\left(N_1+N_2\right)=\frac{72}{18}

d_1=\frac{N_1}{P}=\frac{30}{72}(18)=7.5  in .

d_2=\frac{N_2}{P}=\frac{42}{72}(18)=10.5  in.

\cos \psi_1=\frac{N_1}{P_n d_1}=\frac{30}{5(7.5)}=0.8 \quad \text { or } \quad \psi_1=\psi_2=36.9^{\circ}

b. The virtual number of teeth, using Equation (12.7b), is

N^{\prime}=\frac{N}{\cos ^3 \psi}     (12.7b)

N^{\prime}=\frac{N}{\cos ^3 \psi}=\frac{30}{(0.8)^3}=58.6

Hence, interpolating in Table 11.2, Y =0.419. By Table 11.3, \sigma _o =32 ksi. Applying the Lewis equation (Equation (11.33), modified) with K_f =1,

F_b=\frac{\sigma_o b}{K_f} \frac{Y}{P}         (11.33)

F_b=\sigma_o b \frac{Y}{P_n}=32(2) \frac{0.419}{5}=5.363  kips

By Table 11.9, K =79 ksi. From Equation (11.40),

Q=\frac{2 N_g}{N_p+N_g}      (11.40)

Q=\frac{2 N_g}{N_p+N_g}=\frac{2(42)}{72}=\frac{7}{6}

The Buckingham formula, Equation ((11.38), modified), yields

F_w=d_p b Q K      (11.38)

F_w=\frac{d_1 b Q K}{\cos ^2 \psi}=\frac{7.2(2)(7)(79)}{6(0.8)^2}=2.16  kips

c. The horsepower capacity is based on F_w since it is smaller than F_b . The pitchline velocity equals

V=\frac{\pi d_1 n_1}{12}=\frac{\pi(7.5)(2400)}{12}=4712  fpm

The dynamic load, using Equation ((11.24c), modified), is

F_d=\frac{78+\sqrt{V}}{78} F_t \quad(\text { for } V>4000  fpm )        (11.24c)

F_d=\frac{78+\sqrt{4712}}{78} F_t=1.88 F_t

Equation (11.41), F_w \geq F_d, results in

2.16=1.88 F_t \quad \text { or } \quad F_t=1.15  kips

The corresponding gear power is therefore

hp =\frac{F_t V}{33,000}=\frac{1150(4712)}{33,000}=164

Comment: Observe that the dynamic load is about twice the transmitted load, as expected for reliable operation.