Chapter 16, Problem 30P

To determine

The peak power in the gear train.

The power and shock conditions that the gear train must transmit.

Answer to Problem 30P

The peak power is $28.6\text{hp}$.

The gear train has to be sized for $28.6\text{hp}$ under the power and shock conditions.

Explanation of Solution

Write the expression for load torque.

    $T_L=nT$                                                                         (I)

Here, the torque on crank shaft is $T$ and the step down ratio is $n$.

Write the expression for rated motor torque

    $T_r=\frac{63025Pn}{N}$                                                              (II)

Here, the power of motor is $P$ and the motor revaluations is $N$.

Write the expression for torque of an induction motor has a linear characteristic.

    $T=a\omega +b$                                                                    (III)

Here, constant is $a$, constant is $b$, mean angular velocity is $\omega$.

Write the expression for the rated angular velocity.

    ${\omega }_r=\frac{2\pi N_r}{60n}$                                                                  (IV)

Here, the rated rpm of is $N_r$.

Write the expression for the synchronous angular velocity.

    ${\omega }_s=\frac{2\pi N_s}{60n}$                                                                     (V)

Here, the synchronous rpm of is $N_s$.

Write the expression for constant $a$.

    $a=\frac{-T_r{n}^2}{{\omega }_s-{\omega }_r}$                                                                    (VI)

Here, rated angular velocity is ${\omega }_r$ and synchronous angular velocity is ${\omega }_s$.

Write the expression for constant $b$.

    $b=\frac{n{T}_r{\omega }_s}{{\omega }_s-{\omega }_r}$                                                                 (VII)

Write the expression for the ratio between torques for different time interval.

    $\frac{T_2}{T_r}={\left(\frac{T_L-T_r}{T_r-T_2}\right)}^{\left(t_2-t_1\right)/t_1}$                                                 (VIII)

Here the rated torque is $Tr$ and the load torque is $T_L$.

Write the expression for inertia of the flywheel.

    $I=n^2\frac{a\left(t_2-t_1\right)}{\ln \left(T_2/T_r\right)}$                                                     (IX)

Here, the time interval is $t_2$ and $t_1$.

Write the expression for maximum angular velocity of the flywheel.

    ${\omega }_{\max }=\frac{T_2-b}{a}$                                                          (X)

Here the constants are $a$ and $b$.

Write the expression for minimum angular velocity of the flywheel.

    ${\omega }_{\min }={\omega }_r$                                                               (XI)

Here, the rated angular velocity is ${\omega }_r$.

Write the expression for the mean angular velocity of the flywheel.

    $\omega =\frac{{\omega }_{\max }+{\omega }_{\min }}{2}$                                                         (XII)

Here, the maximum angular velocity is ${\omega }_{\max }$ and minimum angular velocity is ${\omega }_{\min }$.

Write the expression for the coefficient of speed fluctuation.

    $C_s=\frac{{\omega }_{\max }-{\omega }_{\min }}{\omega }$                                                      (XIII)

Write the expression for the maximum fluctuation in energy.

    $\Delta E=I{\omega }^2{C}_s$                                                            (XIV)

Here, the coefficient of speed fluctuation is $C_s$.

Write the expression for the transmitted horse power.

    $H=\frac{\left(60/2\pi \right)\cdot T_L\cdot \omega }{63025}$                                                         (XIV)

Here, the load torque is $T_L$.

Write the expression for the outer diameter of the flywheel.

    $d_o=d+2t$                                                                     (XVI)

Here, the mean diameter of the flywheel is $d$ and thickness is $t$.

Write the expression for the inner diameter of the flywheel.

    $d_i=d-2t$                                                                     (XVII)

Here, the mean diameter of the flywheel is $d$ and thickness is $t$.

Write the expression for the mass moment of inertia of the hollow cylinder.

    $I=\frac{W}{8g}\left(d_o{}^2+d_i{}^2\right)$                                                         (XVIII)

Here, the weight of the flywheel is $W$, the outer diameter of the hollow cylinder is $d_o$ and inner diameter the hollow cylinder is $d_i$

Write the expression for the ring volume.

    $V=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\times b$                                                  (XIX)

Here, the width of the flywheel is $b$.

Write the expression for the ring volume.

    $V=\frac{W}{\gamma }$                                                                   (XX)

Here, the weight density of the flywheel material is $\gamma$.

Conclusion:

Substitute $\left(1300\times 12\right)\text{in}$ for $T$ and $10$ for $n$ in Equation (I).

    $\begin{array}{c}{T}_L=\left(1300\times 12\right)\text{lbf}\cdot \text{in}\times 10\\ =15600\text{lbf}\cdot \text{in}\end{array}$

Substitute $10$ for $n$, $3\text{bhp}$ for $P$ and $1125\text{rev}/\min$ for $N$ in Equation (II).

    $\begin{array}{c}{T}_r=\frac{10\times 63025\times 3\text{bhp}}{1125\text{rev}/\min }\\ =\frac{1890750}{1125}\text{lbf}\cdot \text{in}\\ \text{=1681}\text{lbf}\cdot \text{in}\end{array}$

Substitute $1125\text{rev}/\min$ for $N_r$, $10$ for $n$ in Equation (IV).

    $\begin{array}{c}{\omega }_r=\frac{2\pi \times 1125\text{rev}/\min }{60\times 10}\\ =\frac{2250\pi }{600}\text{rad}/\text{s}\\ =11.781\text{rad}/\text{s}\end{array}$

Substitute $1200\text{rev}/\min$ for $N_s$ , $10$ for $n$ in Equation (V).

    $\begin{array}{c}{\omega }_s=\frac{2\pi \times 1200\text{rev}/\min }{60\times 10}\\ =\frac{2400\pi }{600}\text{rad}/\text{s}\\ =12.566\text{rad}/\text{s}\end{array}$

Substitute $\text{1681}\text{lbf}\cdot \text{in}$ for $T_r$, $11.781\text{rad}/\text{s}$ for ${\omega }_r$ and $12.566\text{rad}/\text{s}$ for ${\omega }_s$ in Equation (VI).

    $\begin{array}{c}a=\frac{-\text{1681}\text{lbf}\cdot \text{in}}{12.566\text{rad}/\text{s}-11.781\text{rad}/\text{s}}\\ =\frac{-\text{1681}\text{lbf}\cdot \text{in}}{7.85\text{rad}/\text{s}}\\ =-2141\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\end{array}$

Substitute $\text{1681}\text{lbf}\cdot \text{in}$ for $T_r$, $11.781\text{rad}/\text{s}$ for ${\omega }_r$ and $12.566\text{rad}/\text{s}$ for ${\omega }_s$ in Equation (VII).

    $\begin{array}{c}b=\frac{\text{1681}\text{lbf}\cdot \text{in}\times 12.566\text{rad}/\text{s}}{12.566\text{rad}/\text{s}-11.781\text{rad}/\text{s}}\\ =\frac{\text{1681}\text{lbf}\cdot \text{in}\times 12.566\text{rad}/\text{s}}{7.85\text{rad}/\text{s}}\\ =26903.8\text{lbf}\cdot \text{in}\end{array}$

Substitute $-2141\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}$ for $a$ and $26903.8\text{lbf}\cdot \text{in}$ for $b$ in Equation (III).

    $T=-2141\omega +26903.8$

Substitute $0.5\text{s}$ for $t_1$ and $10\text{s}$ for $t_1$, $\text{1681}\text{lbf}\cdot \text{in}$ for $T_r$ and $15600\text{lbf}\cdot \text{in}$ for $T_L$ in Equation (VIII).

    $\begin{array}{l}\frac{T_2}{\text{1681}\text{lbf}\cdot \text{in}}={\left(\frac{15600\text{lbf}\cdot \text{in}-\text{1681}\text{lbf}\cdot \text{in}}{\text{1681}\text{lbf}\cdot \text{in}-T_2}\right)}^{\left(10\text{s}-0.5\text{s}\right)/0.5\text{s}}\\ T_2=\text{1681}\text{lbf}\cdot \text{in}\times {\left(\frac{15600\text{lbf}\cdot \text{in}-\text{1681}\text{lbf}\cdot \text{in}}{\text{1681}\text{lbf}\cdot \text{in}-T_2}\right)}^{19}\\ T_2=267.71\text{lbf}\cdot \text{in}\end{array}$

Substitute $10$ for $n$, $-21.41\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}$ for $a$, $0.5\text{s}$ for $t_1$ and $10\text{s}$ for $t_1$, $\text{168}\text{.1}\text{lbf}\cdot \text{in}$ for $T_r$ and $26.771\text{lbf}\cdot \text{in}$ for $T_2$ in Equation (IX).

    $\begin{array}{c}I=\frac{{10}^2\times -21.41\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\times \left(10\text{s}-0.5\text{s}\right)}{\ln \left(26.771\text{lbf}\cdot \text{in}/\text{168}\text{.1}\text{lbf}\cdot \text{in}\right)}\\ =\frac{100\times -21.41\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\times 9.5\text{s}}{\ln \left(26.771\text{lbf}\cdot \text{in}/\text{168}\text{.1}\text{lbf}\cdot \text{in}\right)}\\ =11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\end{array}$

Substitute $267.71\text{lbf}\cdot \text{in}$ for $T_2$, $-2141\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}$ for $a$ and $26903.8\text{lbf}\cdot \text{in}$ for $b$ in Equation (X).

    $\begin{array}{c}{\omega }_{\max }=\frac{267.71\text{lbf}\cdot \text{in}-26903.8\text{lbf}\cdot \text{in}}{-2141\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}}\\ =\frac{-26636.1\text{lbf}\cdot \text{in}}{-2141\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}}\\ =12.44\text{rad}/\text{s}\end{array}$

Substitute $11.781\text{rad}/\text{s}$ for ${\omega }_r$ in Equation (XI).

    ${\omega }_{\min }=11.781\text{rad}/\text{s}$

Substitute $11.781\text{rad}/\text{s}$ for ${\omega }_{\min }$ and $12.44\text{rad}/\text{s}$ for ${\omega }_{\max }$ in Equation (XII).

    $\begin{array}{c}\omega =\frac{12.44\text{rad}/\text{s}+11.781\text{rad}/\text{s}}{2}\\ =\frac{24.221\text{rad}/\text{s}}{2}\\ =12.1105\text{rad}/\text{s}\end{array}$

Substitute $121.105\text{rad}/\text{s}$ for $\omega$, $117.81\text{rad}/\text{s}$ for ${\omega }_{\min }$ and $124.4\text{rad}/\text{s}$ for ${\omega }_{\max }$ in Equation (XIII).

    $\begin{array}{c}{C}_s=\frac{124.4\text{rad}/\text{s}-117.81\text{rad}/\text{s}}{2}\\ =\frac{242.21\text{rad}/\text{s}}{121.11\text{rad}/\text{s}}\\ =\text{0}\text{.0544}\end{array}$

Substitute $11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}$ for $I$, $\text{0}\text{.0544}$ for $C_s$ and $12.1105\text{rad}/\text{s}$ for $\omega$ in Equation (XIV).

    $\begin{array}{c}\Delta E=\left(11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\right)\times {\left(12.1105\text{rad}/\text{s}\right)}^2\times \text{0}\text{.0544}\\ =0.0544\times 1626985\text{lbf}\cdot \text{in}\\ =88508\text{lbf}\cdot \text{in}\end{array}$

Substitute $12.1105\text{rad}/\text{s}$ for $\omega$, $15600\text{lbf}\cdot \text{in}$ for $T_L$ in Equation (XV).

    $\begin{array}{c}H=\frac{\left(60/2\pi \right)\times 15600\text{lbf}\cdot \text{in}\times 12.1105\text{rad}/\text{s}}{63025}\\ =\frac{1803270}{63025}\text{hp}\\ =28.6\text{hp}\end{array}$

Thus, the peak power is $28.6\text{hp}$.

Substitute $30\text{in}$ for $d$ and $1\text{in}$ for $t$ in Equation (XVI).

    $\begin{array}{c}{d}_o=30\text{in}+2\times 1\text{in}\\ =30\text{in}+2\text{in}\\ =32\text{in}\end{array}$

So, the outer diameter of the flywheel is $32\text{in}$.

Substitute $30\text{in}$ for $d$ and $1\text{in}$ for $t$ in Equation (XVII).

    $\begin{array}{c}{d}_i=30\text{in}-2\times 1\text{in}\\ =30\text{in}-2\text{in}\\ =28\text{in}\end{array}$

So, the inner diameter of the flywheel is $28\text{in}$.

The gear train has to be sized for $28.6\text{hp}$ under power and shock conditions since the flywheel on the motor shaft.

Substitute $11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}$ for $I$, $32.2\times 12\text{in}/{\text{s}}^2$ for $g$ , $32\text{in}$ for $d_o$ and $28\text{in}$ for $d_i$ in Equation (XVIII).

    $\begin{array}{l}11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}=\frac{W}{8\times 32.2\times 12\text{in}/{\text{s}}^2}\times \left({\left(32\text{in}\right)}^2+{\left(28\text{in}\right)}^2\right)\\ W=\frac{11072\text{lbf}\cdot \text{in}\cdot \text{s}/\text{rad}\times 8\times 32.2\times 12\text{in}/{\text{s}}^2}{\left({\left(32\text{in}\right)}^2+{\left(28\text{in}\right)}^2\right)}\\ W=1891\text{lbf}\end{array}$

Substitute $1891\text{lbf}$ for $W$, $0.260\text{lbf}/{\text{in}}^3$ for $\gamma$ in Equation (XX).

    $\begin{array}{c}V=\frac{1891\text{lbf}}{0.260\text{lbf}/{\text{in}}^3}\\ =7273{\text{in}}^3\end{array}$

Substitute $7273{\text{in}}^3$ for $V$, $32\text{in}$ for $d_o$ and $28\text{in}$ for $d_i$ in Equation (XIX).

    $\begin{array}{l}7273{\text{in}}^3=\frac{\pi }{4}\left(32{\text{in}}^2-28{\text{in}}^2\right)\times b\\ b=\frac{7273{\text{in}}^3}{60\pi }\\ b=38.6\text{in}\end{array}$

Thus, the gear train has to be sized for $28.6\text{hp}$ under power and shock conditions since the flywheel on the motor shaft.