**Q.1 2.5 kw power transmitted by a open belt drive linear velocity is 2.5 m/s.The angle of lap on small pulley 165 degree.
coefficient friction 0.3,Determine effect of power transmission in following case,
1)Initial tension belt increase by 8%
2)Angle of lap increased by 8%, use of an idler pulley for same speed and tension in tight side
3)coefficient friction increased by 8% by suitable decrease to friction surface of belt**

**Solution**
P=2.5kw, $\mu=0.3$, $\theta=165$, $\gamma=2.5$,

$p=(T_1-T_2)2.5$, $T_1-T_2=1000 N$.....................(1)

$\frac{T_1}{T_2}=e^{\mu\theta}=e^0.3$, $165\frac{T_2}{180}=2.37$....................(2)

$T_1=1729.9 N$ ,$T_2=729.9 N$

Initial tension $T_0=\frac{T_1+T_2}{2}=\frac{1729.9+729.9}{2}=1229.9 N$

**(1)** When $T_0$ used by 8%, $T_0'=1.08 \quad T_0=1328.3 N$ ,$\frac{T_1+T_2}{2}=1328.3 \quad so,T_1+T_2=2656.6$

$\mu$,$\theta$ remain unchange $\frac{T_1}{T_2}$ is same, so now $T_2=788.3N$ and $T_1=1868.3N$

$p=(T_1+T_2)2.5=2.7 kw$ so, % increased in power=$\frac{2.7-2.5}{2.5}=0.08=8$%

**(2)** $\theta=165(1.08)\frac{\pi}{180}$ ,Here $T_1$ remains the same ,so $\frac{1729.9}{T_2}=e^{\mu\theta}=2.54$

$T_2=680.5N,\quad p=(1729.9-680.5)\gamma=2.624kw$, so % unused in power $=\frac{p-2.5}{2.5}=0.0496=4.96$%

**(3)**$\frac{T_1}{T_2}=e^{0.3*1.08\theta}=2.54$ , $T_1=2.54T_2$.......(1)

$T_1+T_2=1229.9(2)=2459.8.............(2)$

$T_1=1764.9 \quad T_2=694.9N$

$p=(1764.9-694.9)2.5=2.675kw$ , % unused power=$\frac{p-2.5}{2.5}=0.07=7$%

**Q.2 The center to center distance of a two sprockets of chain drive are 600 mm ,chain drive to reduce the speed from 180 rpm to 90 rpm.The driving sprocket has 18 teeth and a pitch circle diameter of 480 mm define,
(1)No of teeth on driven sprocket
(2)Pitch and length of chain**

**Solution:**

$\frac{N_2}{N_1}=\frac{T_2}{T_1}$ where $T_2=36$ ,$k=\frac{c}{p}=14.342$

$p=2rsin(\frac{\phi}{2})=2rsin\frac{1}{2}(\frac{360}{T})$

$p=2(0.24)sin(\frac{180}{36})=41.88mm.$

T=no of teeth in sprocket

$\psi$=Angle subtended by chord link at center of sprocket $L=\pi(R+r)+\frac{R-r}{c}+2c..........(1)$

$r=\frac{p}{2}cosec\frac{180}{t} ; R=\frac{p}{2}cosec\frac{180}{T}$, but we know $\pi(R+r)=\frac{pT+pt}{2}$

put in equation (1)...$L=\frac{pT+pt}{2}+\frac{(\frac{p}{2}cosec\frac{180}{T}-\frac{p}{2}cosec\frac{180}{t})^2}{Kp}+2Kp$

$L=p[\frac{T+t}{2}+\frac{(cosec\frac{180}{T}-cosec\frac{180}{t})^2}{4K}+2K]$

$L=0.0418[\frac{36+18}{2}+\frac{(cosec\frac{180}{36}-cosec\frac{180}{18})^2}{4*14.342}+2*(14.342)]$

$L=2.351 m$

**Q.3 A chain drive is used in reduction of speed from 240 rpm. to 120 rpm. Find no of teeth on driven sprocket if pitch circle diameter of driven sprocket is 600mm.& center to center distance between the two sprocket is 800mm. determine pitch and lengthof the chain**

**Solution** $T_2$ =no of speed of driven sprocket,$N_1T_1=N_2T_2 \quad T_2=\frac{N_1T_1}{N_2}=\frac{240(20)}{120}=40$

Pitch of chain $0.3=\frac{p}{2}cosec\frac{180}{T_2}=6.37p$, $p=\frac{0.3}{6.37}=4.71 mm.$

Length of the chain,We know pitch circle radi of driving sprocket $r_1=\frac{p}{2}cosec\frac{180}{T_1}=\frac{47.1}{2}cosec\frac{180}{20}=150.5mm$.

$x=mp; m=\frac{c}{p}=\frac{800}{47.1}$ ;

We know m.f,$K=[\frac{T_1+T_2}{2}+\frac{(cosec\frac{180}{T_1}-cosec\frac{180}{T_2})^2}{4m}+2m]$

$=[\frac{20+40}{2}+\frac{(cosec\frac{180}{20}-cosec\frac{180}{40})^2}{4*(16.985)}+2(16.985)]$

$=64.5 \approx 65$

Length of the chain $pk=47.1(65)=3.0615m.$

**Q.4 The power is transmitted form a pulley 1 m diameter running at 200 rpm to pulley 2.25m diameter by means of a belt. Find the speed lost by the driven pulley as result of creep, if the stress on the tight and slack side of the belt is 1.4 MPs and 0.5 MPa respectively. The Young's Modulus for the material of the belt is 100 MPa.**

**Q.5 Two pulley, one 450 mm diameter and the other 200 mm diameter are on parallel shafts 1.95 m apart and are connected by a crossed belt. Find the
length of the belt required and the angle of contact between the belt and each pulley.
What power can be transmitted by the belt when the larger pulley rotates at
200 rpm, if the maximum permissible tension in the belt is 1 kN and the
co-efficient of friction between the belt and pulley is 0.25 ?**