Question 1 |
The torque provided by an engine is given by T(\theta ) = 12000 + 2500 sin(2\theta ) N.m, where \theta is the angle turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a machine that provides a constant resisting torque. If variation of the speed from the mean speed is not to exceed \pm 0.5%, the minimum mass moment of inertia of the flywheel should be _______ kg.m^2 (round off to the nearest integer).
245 | |
360 | |
570 | |
640 |
Question 1 Explanation:
\begin{aligned} \omega &=\frac{\pi \times 200}{30}=20.9439 \mathrm{rad} / \mathrm{s} \\ \Delta E &=\int_{0}^{\frac{\pi}{2}}\left(T-T_{\text {mean }}\right) d \theta=2500 \int_{0}^{\frac{\pi}{2}} \sin 2 \theta d \theta \\ &=2500 \times 1=2500 \mathrm{~J} \\ \Delta E &=I \omega^{2} C_{s} \\ 2500 &=I \times 20.9439^{2} \times 0.01 \\ I &=569.934 \mathrm{kgm}^{2} \simeq 570 \mathrm{~kg} . \mathrm{m}^{2} \end{aligned}
Question 2 |
The controlling force curves P, Q and R for a spring controlled governor are shown in the figure, where r_1 and r_2 are any two radii of rotation.
The characteristics shown by the curves are
The characteristics shown by the curves are
P - Unstable; Q - Stable; R - Isochronous | |
P - Unstable; Q - Isochronous; R - Stable | |
P- Stable; Q - Isochronous; R - Unstable | |
P - Stable; Q - Unstable; R - Isochronous |
Question 2 Explanation:
F(r)|_P=ar+b \quad \rightarrow \text{Unstable}
F(r)|_Q=ar+b \quad \rightarrow \text{Isochronous}
F(r)|_R=ar-b \quad \rightarrow \text{Stable}
F(r)|_Q=ar+b \quad \rightarrow \text{Isochronous}
F(r)|_R=ar-b \quad \rightarrow \text{Stable}
Question 3 |
The turning moment diagram of a flywheel fitted to a fictitious engine is shown in the
figure.
The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within\pm 2 \% of the average speed, the mass moment of inertia of the flywheel is _________ kgm^2.
The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within\pm 2 \% of the average speed, the mass moment of inertia of the flywheel is _________ kgm^2.
8.25 | |
5.56 | |
2.14 | |
3.58 |
Question 3 Explanation:
N = 1000 rpm
\begin{aligned} \omega &=\frac{2 \pi \times 1000}{60}=104.7195 \mathrm{rad} / \mathrm{s} \\ \Delta E_{\max } &=\left(\pi-\frac{\pi}{2}\right) \cdot(3000-2000)=\left(\frac{\pi}{2} \times 1000\right) \mathrm{J} \\ C_{s} &=\pm 2 \%=4 \%=0.04 \\ \Delta E_{\max } &=I \cdot \omega^{2} \cdot C_{s} \\ \frac{\pi}{2} \times 1000 &=I \times(104.7195)^{2} \times 0.04 \\ I &=\frac{1570.795}{(104.7195)^{2} \times 0.04}=3.58 \mathrm{kg}-\mathrm{m}^{2} \end{aligned}
\begin{aligned} \omega &=\frac{2 \pi \times 1000}{60}=104.7195 \mathrm{rad} / \mathrm{s} \\ \Delta E_{\max } &=\left(\pi-\frac{\pi}{2}\right) \cdot(3000-2000)=\left(\frac{\pi}{2} \times 1000\right) \mathrm{J} \\ C_{s} &=\pm 2 \%=4 \%=0.04 \\ \Delta E_{\max } &=I \cdot \omega^{2} \cdot C_{s} \\ \frac{\pi}{2} \times 1000 &=I \times(104.7195)^{2} \times 0.04 \\ I &=\frac{1570.795}{(104.7195)^{2} \times 0.04}=3.58 \mathrm{kg}-\mathrm{m}^{2} \end{aligned}
Question 4 |
A flywheel is attached to an engine to keep its rotational speed between 100 rad/s and
110 rad/s. If the energy fluctuation in the flywheel between these two speeds is 1.05 kJ
then the moment of inertia of the flywheel is _________________kg.m^2 (round off to 2
decimal places).
1 | |
2 | |
2.5 | |
4 |
Question 4 Explanation:
\begin{aligned} \Delta E_{\max } &=\frac{1}{2} I\left(\omega_{\max }^{2}-\omega_{\min }^{2}\right) \\ \therefore \qquad 1.05 \times 10^{3} \times 2 &=I\left(110^{2}-100^{2}\right) \\ I &=1 \mathrm{kg} . \mathrm{m}^{2} \end{aligned}
Question 5 |
A thin uniform rigid bar of length L and mass M is hinged at point O, located at a distance of \frac{L}{3} from one of its ends. The bar is further supported using springs, each of stiffness k, located at the two ends. A particle of mass m=\frac{M}{4} is fixed at one end of the bar, as shown in the figure. For small rotations of the bar about O, the natural frequency of the systems is
\sqrt{\frac{5k}{M}} | |
\sqrt{\frac{5k}{2M}} | |
\sqrt{\frac{3k}{2M}} | |
\sqrt{\frac{3k}{M}} |
Question 5 Explanation:
Taking Mass moment of inertia about point 0,
\begin{aligned} I &=\frac{M l^{2}}{12}+M\left(\frac{l}{2}-\frac{l}{3}\right)^{2}+m \times\left(\frac{2 l}{3}\right)^{2} \\ &=\frac{M l^{2}}{12}+\frac{M l^{2}}{36}+\frac{4 M l^{2}}{9} \\ &=\frac{M l^{2}}{9}+\frac{4 M l^{2}}{4 \times 9}=\frac{2 M l^{2}}{9} \end{aligned}
Now we will balance torque about 0
\begin{aligned} I \alpha &=k \times \frac{2 L}{3} \times\left(\frac{2 L}{3} \theta\right)+k \times \frac{L}{3} \times\left(\frac{L}{3} \theta\right) \\ \frac{2 M l^{2}}{9} \frac{d^{2} \theta}{d t^{2}} &=\frac{5 k}{2 M}=\omega_{n}^{2} \theta \\ \therefore \quad \omega_{n}&=\sqrt{\frac{5 k}{2 M}} \end{aligned}
\begin{aligned} I &=\frac{M l^{2}}{12}+M\left(\frac{l}{2}-\frac{l}{3}\right)^{2}+m \times\left(\frac{2 l}{3}\right)^{2} \\ &=\frac{M l^{2}}{12}+\frac{M l^{2}}{36}+\frac{4 M l^{2}}{9} \\ &=\frac{M l^{2}}{9}+\frac{4 M l^{2}}{4 \times 9}=\frac{2 M l^{2}}{9} \end{aligned}
Now we will balance torque about 0
\begin{aligned} I \alpha &=k \times \frac{2 L}{3} \times\left(\frac{2 L}{3} \theta\right)+k \times \frac{L}{3} \times\left(\frac{L}{3} \theta\right) \\ \frac{2 M l^{2}}{9} \frac{d^{2} \theta}{d t^{2}} &=\frac{5 k}{2 M}=\omega_{n}^{2} \theta \\ \therefore \quad \omega_{n}&=\sqrt{\frac{5 k}{2 M}} \end{aligned}
Question 6 |
The torque (in N-m) exerted on the crank shaft of a two stroke engine can be described as T=10000+1000\sin 2\theta -1200\cos 2\theta, where \theta is the crank angle as measured from inner dead center position. Assuming the resisting torque to be constant, the power (in kW) developed by the engine at 100 rpm is ________
104.719kM | |
125.325kM | |
865.265kM | |
456.235kM |
Question 6 Explanation:
\begin{array}{l} T=10000+1000 \sin 2 \theta-1200 \cos 2 \theta \\ N=100 \mathrm{rpm} \\ W=\frac{2 \pi \times 100}{60}=10.472 \mathrm{rad} / \mathrm{s} \\ T_{\text {man }}=\frac{1}{2 \pi} \int_{0}^{2 \pi}[10,000+1000 \sin 2 \theta- \\ =10,000[2 \pi-0]+\frac{1000}{2} \times[\cos 2 \pi]_{0}^{2 \pi} \\ -\frac{1200}{2}[\sin2\theta]_{0}^{2\pi} \\ =\frac{1}{2 \pi}[10,000 \cdot 2 \pi+500 \cdot[\cos 4 \pi-\cos 0] \\ \quad -600[\sin 4 \pi-\sin 0] \\ (10000\times2\pi)+500[1-1]-600[0-0]\\ =\frac{10000 \times 2 \pi}{2 \pi}=10000\\ \text{Power} = 10000\times\frac{2\pi\times100}{60}=104.719\mathrm{kM} \end{array}
Question 7 |
Torque and angular speed data over one cycle for a shaft carrying a flywheel are shown in the figures. The moment of inertia (in kg.m^{2}) of the flywheel is _______
98.65 | |
25.65 | |
31.26 | |
78.98 |
Question 7 Explanation:
\begin{aligned} I &=? \\ E &=3000 \times\left[\pi-\frac{\pi}{2}\right]-1500 \times[2 \pi-\pi] \\ &=3000 \times \frac{\pi}{2}-1500 \pi=0 \\ T_{\text {mean }} \times & 2 \pi=E \\ T_{\text {mean }} &=\frac{E}{2 \pi}=0 \\ \omega_{\min } &=10 \mathrm{rad} / \mathrm{s} \mathrm{at} 0=\pi / 2 \\ \omega_{\max } &=20 \mathrm{rad} / \mathrm{s} \mathrm{at} 0=\pi \\ C_{s} &=\frac{\omega_{\max }-\omega_{\min }}{\frac{\omega_{\max }+\omega_{\min }}{2}} \\ &=\frac{2\left\{\omega_{\max }-\omega_{\min }\right\}}{\omega_{\max }+\omega_{\min }} \\ &=\frac{2 \times\{20-10\}}{20+10}=0.67 \\ \Delta E &=3000 \times \frac{\pi}{2}=1500 \pi \\ \Delta E &=I \omega^{2} C_{s}\\ 1500 \pi &=I \times \omega_{\text {mean }}^{2} \times 0.67 \\ 1500 \pi &=I \times\left\{\frac{20+10}{2}\right\}^{2} \times 0.67 \\ I &=\frac{1500 \pi}{15^{2} \times 0.67}=31.26 \mathrm{kg} / \mathrm{m}^{2} \end{aligned}
Question 8 |
Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg.m^2) of a flywheel to keep the speed fluctuation within \pm 0.5% of the average speed is _______
592.73 | |
986.48 | |
986.44 | |
125.69 |
Question 8 Explanation:
\begin{aligned} \Delta E &=2600 \mathrm{J} \\ N_{\text {average }} &=200 \mathrm{rpm} \\ I &=? \\ C_{s} &=\pm 0.5 \% \\ &=2 \times \frac{0.5}{100}=0.01 \\ \Delta E &=I \omega^{2} C_{s} \\ 2600 &=I \times\left\{\frac{2 \pi}{60} \times 200\right\}^{2} \times 0.01 \\ 2600 &=I \times 438.6491 \times 0.01 \\ I &=\frac{2600}{4.3865}=592.73 \mathrm{kgm}^{2} \end{aligned}
Question 9 |
A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 rad/s. If the total fluctuation of speed is not to exceed \pm 2%, the mass moment of inertia of the flywheel in kg-m^{2}
is
25 | |
50 | |
100 | |
125 |
Question 9 Explanation:
\begin{aligned} \Delta E &=400 \mathrm{Nm} \\ \omega_{\text {mean }} &=20 \mathrm{rad} / \mathrm{s} \\ C_{s} &=\pm 2 \% \\ &=2 \times \frac{2}{100}=0.04 \\ \Delta E &=I \omega_{\text {mean }}^{2} C_{s} \\ 400 &=I \times\{20\}^{2} \times\{0.04\} \\ \text{or}\quad I &=25 \mathrm{kgm}^{2} \end{aligned}
Question 10 |
The speed of an engine varies from 210 rad/s to 190 rad/s. During a cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kgm^{2}
is
0.1 | |
0.2 | |
0.3 | |
0.4 |
Question 10 Explanation:
\begin{aligned} \omega_{\min } &=190 \mathrm{rad} / \mathrm{s} \\ \omega_{\max } &=210 \mathrm{rad} / \mathrm{s} \\ \Delta E &=400 \mathrm{Nm} \\ C_{s} &=\frac{\omega_{\max }-\omega_{\min }}{\frac{\left(\omega_{\max }+\omega_{\min }\right)}{2}} \\ C_{s} &=\frac{210-190}{\frac{(210+190)}{2}}=\frac{20}{200}=0.1 \\ \omega_{\text {mean }} &=\frac{\omega_{\min }+\omega_{\max }}{2}=\frac{190+210}{2} \\ &=200 \mathrm{rad} / \mathrm{s} \\ \Delta E &=I \omega_{\text {mean }}^{2} C_{s} \\ 400 &=I \times(200)^{2} \times 0.1 \\ I &=\frac{400}{200^{2} \times 0.1}=0.1 \mathrm{kgm}^{2} \end{aligned}
There are 10 questions to complete.
Sir,2nd and 5th questions not from the flywheel portion
Sir, question 8, either Cs in question is wrong or Cs in solution is wrong..plz correct it