Chapter 17, Problem 20P

(a)

To determine

Design a Gates Rubber V-belt drive to transmit power from the motor to the compressor flywheel.

Explanation of Solution

Consider a $C-270$ belt.

Write the expression for center to center distance between pulleys.

$C=0.25\left\{\left[L_p-\frac{\pi }{2}\left(D+d\right)\right]+\sqrt{{\left[L_P-\frac{\pi }{2}\left(D+d\right)\right]}^2-2{\left(D-d\right)}^2}\right\}$ (I)

Here, length of the belt is $L_P$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$.

Write the expression for contact angle for smaller pulley.

${\theta }_d=180°-2{\sin }^{-1}\left(\frac{D-d}{2C}\right)$ (II)

Here, the center distance between pulley is $C$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$.

Write the expression for contact angle for bigger pulley.

${\theta }_d=180°+2{\sin }^{-1}\left(\frac{D-d}{2C}\right)$ (III)

Here, the center distance between pulley is $C$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$.

Write the expression for contact correction factor

$K_1=0.143543+0.007468{\theta }_d-0.000015052{\left({\theta }_d\right)}^2$ (III)

Here, the contact angle for smaller pulley is ${\theta }_d$.

Write the expression for velocity of the smaller pulley.

$V=\frac{\pi dn}{12}$ (IV)

Here, the rotational speed of the smaller pulley is $n$.

Write the expression for allowable power per belt

$H_a=K_1{K}_2{H}_{tab}$ (V)

Here, the angle of wrap correction factor is $K_1$, belt correction factor is $K_2$, and table horse power ratings of standards v-belts is $H_{tab}$.

Write the expression for design power.

$H_d=K_s{H}_{nom}{n}_d$ (VI)

Here, nominal power is $H_{nom}$, the service factor is $K_s$ and the design factor is $n_d$.

Write the expression for number of belts.

$N_b=\frac{H_d}{H_a}$ (VII)

Write the expression for torque

$T=\frac{63025H_d}{n}$ (VIII)

Here, the design power is $H_d$, and the speed of the rotation is $n$.

Write the expression for difference in tension in tight side an slack side of the belt.

$\Delta F=\frac{2T}{d}$ (IX)

Here, the torque is $T$ and diameter of the smaller pulley is $d$.

Write the expression for centrifugal tension

$F_c=K_c{\left(\frac{V}{1000}\right)}^2$ (X)

Here, the speed of the belt  is $V$ and constant term is $K_c$.

Write the expression for initial tension in belt.

$F_i=\left(\frac{T}{d}\right)\left[\frac{e^{f{\theta }_d}+1}{e^{f{\theta }_d}-1}\right]$ (XI)

Here, the coefficient of the friction is $f$ and angle of the contact for the smaller pulley is ${\theta }_d$, the torque developed by the belt is $T$ and the diameter of the smaller of the pulley is $d$.

Write the expression for tension in tight side of the belt.

$F_1=F_c+F_i\left(\frac{2e^{f{\theta }_d}}{e^{f{\theta }_d}+1}\right)$ (XII)

Here, the centrifugal tension in the belt is $F_c$ and initial tension in the belt is $F_i$.

Write the expression for tension in slack side of the belt.

$F_2=F_1-\Delta F$ (XIII)

Here, the difference in tension in tight side an slack side of the belt is $\Delta F$ and tension in tight side of the belt is $F_1$.

Write the expression for factor of the safety.

$n_{fs}=\frac{H_a{N}_b}{H_d}$ (XIV)

Here, allowable power per belt is $H_a$ and design power is $H_d$ and the number of the belt is $N_b$.

Write the expression for number of the belt passes.

$N_P={\left[{\left(\frac{K}{F_1+\frac{K_b}{d}}\right)}^{-b}+{\left(\frac{K}{F_2+\frac{K_b}{D}}\right)}^{-b}\right]}^{-1}$ (XV)

Here, belt parameter is $K_b$ and durability parameter for the V-belt is $K$ and $b$.

Write the expression for the belt life.

$t=\frac{N_p{L}_p}{720V}$ (XVI)

Conclusion:

Substitute.$270\text{in}+2.9\text{in}=272.9\text{in}$ for $L_P$ $60\text{in}$ for $D$ and $11\text{in}$ for $d$ in Equation (I).

$\begin{array}{c}C=0.25\left\{\begin{array}{l}\left[272.9\text{in}-\frac{\pi }{2}\left(60\text{in}+11\text{in}\right)\right]+\\ \sqrt{{\left[272.9\text{in}-\frac{\pi }{2}\left(60\text{in}+11\text{in}\right)\right]}^2-2{\left(60-11\right)}^2}\end{array}\right\}\\ =0.25\times 307.12\text{in}\\ =76.78\text{in}\end{array}$

Substitute $60\text{in}$ for $D$ and $11\text{in}$ for $d$ and $76.78\text{in}$ $C$ in Equation.(II).

$\begin{array}{c}{\theta }_d=180°-2{\sin }^{-1}\left(\frac{60\text{in}-11\text{in}}{2\times 76.78\text{in}}\right)\\ =180°-37.2°\\ =142.8°\times \frac{\pi }{180°}\text{rad}\\ =2.492\text{rad}\end{array}$

Substitute $60\text{in}$ for $D$ and $11\text{in}$ for $d$ and $76.78\text{in}$ $C$ in Equation.(III).

$\begin{array}{c}{\theta }_d=180°+2{\sin }^{-1}\left(\frac{60\text{in}-11\text{in}}{2\times 76.78\text{in}}\right)\\ =180°+37.2°\\ =217.2°\times \frac{\pi }{180°}\text{rad}\\ =3.79\text{rad}\end{array}$

Substitute $142.8°$ for ${\theta }_d$ in Equation.(IV).

$\begin{array}{c}{K}_1=0.143543+0.007468\left(142.8°\right)-0.000015052{\left(142.8°\right)}^2\\ =0.143543+0.759457\\ =0.903\end{array}$

Substitute $11\text{in}$ for $d$ and $875\text{rev}/\min$ for $n$ in Equation (V).

$\begin{array}{c}V=\frac{\pi \times 11\text{in}\times 875\text{rev}/\min }{12}\\ =\frac{9625\pi }{12}\text{ft}/\min \\ =2520\text{ft}/\min \end{array}$

Substitute $0.903$ for $K_1$, $1.15$ for $K_2$, and $7.83\text{hp}$ for $H_{tab}$ in Equation.(VI).

$\begin{array}{c}{H}_a=0.903\times 1.15\times 7.83\text{hp}\\ =0.903\times 9.0045\\ =8.13\text{hp}\end{array}$

Substitute $1.2$ for $K_s$, $1.1$ for $n_d$ and $50\text{hp}$ for $H_{nom}$ in Equation (VII).

$\begin{array}{c}{H}_d=1.2\times 50\text{hp}\times 1.1\\ =1.32\times 50\text{hp}\\ =66\text{hp}\end{array}$

Substitute $8.13\text{hp}$ for $H_a$ and $66\text{hp}$ for $H_d$ in Equation (VIII)

$\begin{array}{c}{N}_b=\frac{66\text{hp}}{8.13\text{hp}}\\ =8.118\\ \approx 9\end{array}$

Substitute $8.13\text{hp}$ per belt for $H_d$ and $875\text{rev}/\min$ for $n$ in Equation (IX).

$\begin{array}{c}T=\frac{63025\times 8.13\text{hp}}{875\text{rev}/\min }\\ =\frac{512393.25}{875}\text{lbf}\cdot \text{in}\\ =585.6\text{lbf}\cdot \text{in}\end{array}$

Substitute $585.6\text{lbf}\cdot \text{in}$ for $T$ and $11\text{in}$ for $d$ in Equation (X).

$\begin{array}{c}\Delta F=\frac{2\times 585.6\text{lbf}\cdot \text{in}}{11\text{in}}\\ =\frac{1171.2}{11}\text{lbf}\\ =106.5\text{lbf}\end{array}$

Substitute $1.716$ for $K_c$ and $2520\text{ft}/\min$ for $V$ in Equation (XI).

$\begin{array}{c}{F}_c=1.716\times {\left(\frac{2520\text{ft}/\min }{1000}\right)}^2\\ =1.716\times 6.3504\text{lbf}\\ =10.9\text{lbf}\end{array}$

Substitute $586.9\text{lbf}\cdot \text{in}$ for $T$, $11\text{in}$ for $d$, $0.5123$ for $f$ and $2.492\text{rad}$ for ${\theta }_d$ in Equation (XII).

$\begin{array}{c}{F}_i=\left(\frac{586.9\text{lbf}\cdot \text{in}}{11\text{in}}\right)\left[\frac{e^{0.5123\times 2.492\text{rad}}+1}{e^{0.5123\times 2.492\text{rad}}-1}\right]\\ =53.35\text{lbf}\times \left[\frac{3.5846+1}{3.5846-1}\right]\\ =94.6\text{lbf}\end{array}$

Substitute $10.9\text{lbf}$ for $F_c$, $94.6\text{lbf}$ for $F_i$, $0.5123$ for $f$ and $2.492\text{rad}$ for ${\theta }_d$ in Equation (XIII).

$\begin{array}{c}{F}_1=10.9\text{lbf}+\left(94.6\text{lbf}\right)\times \left(\frac{2e^{0.5123\times 2.492\text{rad}}}{e^{0.5123\times 2.492\text{rad}}+1}\right)\\ =10.9\text{lbf}+\left(94.6\text{lbf}\right)\times \frac{2\times 3.5846}{4.5846}\\ =158.8\text{lbf}\end{array}$

Substitute $158.8\text{lbf}$ for $F_1$, $106.5\text{lbf}$ for $\Delta F$, in Equation (XIII).

$\begin{array}{c}{F}_2=158.8\text{lbf}-106.7\text{lbf}\\ =52.1\text{lbf}\end{array}$

Substitute $84\text{hp}$ for $H_d$, $8.13\text{hp}$ for $H_a$ and $9$  for $N_b$ in Equation (XIV).

$\begin{array}{c}{n}_{fs}=\frac{8.13\text{hp}\times 9}{66\text{hp}}\\ =1.11\end{array}$

Thus, the factor of the safety is $1.11$.

Substitute $158.8\text{lbf}$ for $F_1$, $52.1\text{lbf}$ for $F_2$, $1600$ for $K_b$, $60\text{in}$ for $D$ and $11\text{in}$ for $d$, $2038$ for $K$ and $11.173$ for $b$ in Equation (XV).

$\begin{array}{c}{N}_P={\left[{\left(\frac{2038}{158.8\text{lbf}+\frac{1600}{11\text{in}}}\right)}^{-111.173}+{\left(\frac{2038}{52.1\text{lbf}+\frac{1600}{60\text{in}}}\right)}^{-11.173}\right]}^{-1}\\ ={\left[{\left(\frac{2038}{304.25}\right)}^{-111.173}+{\left(\frac{2038}{78.76}\right)}^{-11.173}\right]}^{-1}\\ =1.68\times {10}^9\end{array}$

Thus, the belt life in passes is $1.68\times {10}^9$.

Substitute ${10}^9$ for $N_p$ , $272.9\text{in}$ for $L_P$ and $2520\text{ft}/\text{min}$ in Equation (XVI).

$\begin{array}{c}t=\frac{{10}^9\times 272.9\text{in}}{720\times 2520\text{ft}/\text{min}}\\ =\frac{{10}^9\times 272.9}{1814400}\text{hr}\\ =150407.84\text{hr}\end{array}$

Thus, the tool life in hour is $150407.84\text{hr}$.

Thus, the selected belt is C-270.

(b)

To determine

Whether the cutting of the V-belt grooves in the flywheel is avoided by using a V-flat drive.

Answer to Problem 20P

The cutting of the V-belt grooves in the flywheel cannot be avoided by using a V-flat drive.

Explanation of Solution

Write the expression for the fully developed friction torque on the flywheel using the flats of the V-belts.

$T_{\text{flat}}=F_{i}D\left[\frac{e^{f{\theta }_D}-1}{e^{f{\theta }_D}+1}\right]$ (XVII)

Here the friction coefficient of flat on flywheel is $f$.

Write the expression for the flywheel torque.

$T_{\text{fly}}=m_{\text{G}}T$ (XVIII)

Here, the torque constant is $m_{\text{G}}$.

Conclusion:

Substitute $94.6\text{lbf}$ for $F_i$, $60\text{in}$ for $D$, $0.13$ for $f$ and $3.791\text{rad}$ for ${\theta }_D$ in Equation (XVII).

$\begin{array}{c}{T}_{\text{flat}}=94.6\text{lbf}\times 60\text{in}\times \left[\frac{e^{0.13\times 3.791}-1}{e^{0.3\times 3.791}+1}\right]\\ =5676\times 0.2415\text{lbf}\cdot \text{in}\\ =1371\text{lbf}\cdot \text{in}\end{array}$

Substitute $5.147$ for $m_{\text{G}}$, $586.9\text{lbf}\cdot \text{in}$ for $T$ , in Equation (XVIII).

$\begin{array}{c}{T}_{\text{fly}}=5.147\times 586.9\text{lbf}\cdot \text{in}\\ =3021\text{lbf}\cdot \text{in}\end{array}$

Since, the torque produced by V-belt is lower than the torque given by flywheel. Thus, he cutting of the V-belt grooves in the flywheel cannot be avoided by using a V-flat drive.