TUT-7 (by Kuldeep)

Question-4) 2 kW of power is transmitted by an open belt drive. The linear velocity of the belt is 2 m/s.
The angle of lap on the smaller pulley is 160°. The coefficient of friction is 0.25. Determine the effect on
power transmission in the following cases : (i) Initial tension in the belt is increased by 8%. (ii) Initial
tension in the belt is decreased by 8%. (iii) Angle of lap is increased by 8% by the use of an idler pulley,
for the same speed and the tension on the tight side. (iv) Coefficient of friction is increased by 8% by
suitable dressing to the friction surface of the belt.

ANS-)

P= 2.5 kw, μ =0.3, θ =165∘, ν = 2.5 m/s

P=(T1 – T2) ν

2500 = (T1 – T2) ×2.5 or T1 – T2 =1000N.

T1T2=eμθ=e0.3×165π180=2.37

or T1 = 2.37T2 - T2 =1000 OR T2 =729.9N.

T2 =729.9 ×2.37 = 1729.9 N.

Initial Tension, T0 = (T1 + T2)/2 =(1729.9+729.9)/2 = 1229.9N.

(i)When initial tension increase by 8%

T′0 = 1229.9 × 1.08 = 1328.3N.

As μ and θ remain unchanged, eμθ or T1T2 is same,

2.37 T2 + T2 =2556.6

T2 =788.3N, T1 =1868.3N

P = (T1 –T2) ν = (1868.3-788.3)2.5=2700 W or 2.7 kw.

Increase power = (2.7-2.5)/2.5=0.08 or 8%

(ii) When initial tension is decreased by 8%

T′0 =1229.9 × (1-0.08) =1131.5

or T1 + T2 = 2663

3.37T2 = 2263

T2 = 671.5N, T1 = 1591.5N.

P =(1591.5-671.5)2.5=2300W or 2.3kw.
Decrease in power =(2.5-2.3)/2.5 = 0.08 or 8%

(iii) T1T2=eμθ

T1 is the same as before, whereas θ increases by 8%

1729.9/T2 =e0.3×((165×1.08π)/180) =2.54

T2 =680.5N.

P =(1729.9-680.5)2.5 =2624W or 2.624kw

Increase in power = (2.624-2.5)/2.5 = 0.0496 or 4.96%

(iv) T1T2=eμθ=e0.3×1.08×165π180=2.54

Or T1 = 2.54T2

T1 + T2 = 1229.9 × 2 =2459.8

T2 = 694.9 N.

T1 =694.9 × 2.54 =1764.9 N.

P = (1764.9-694.9)2.5 = 2675W or 2.675kW

Question 2)Two pulleys, one 450 mm dia and other 200 mm dia are on parallel shafts 2 m apart are
connected by a crossed belt. Find the length of belt required_ What power can be transmitted by the
belt when the larger pulley rotates at 200 rpm and maximum permissible tension in the belt is 100 kgf ?
Given the coefficient of friction between the belt and pulley is 0.25.

Given :- d1 = 450 mm= 0.45 m or

r1 =0.225 m; d2 = 200 mm= 0.2 m; or r2 =0.1 m; x =1.95 m, N1 =200 rpm; T1 = 1kN= 1000N ; μ =0.25

We know that speed of the belt,

v=πd1N160=π×0.45×20060 = 4.714 m/s.

Length of the belt

We know that length of the crossed belt,

L=π(r1+r2)+2x+(r1+r2)2x

=π(0.225+0.1)+2×1.95+(0.225+0.1)2/1.95

= 4.975 m.

Angle of contact between the belt and each pulley

Let θ= Angle of contact between the belt and each pulley.

sinα=(r1+r2)x = (0.225+0.1)/1.95 = 0.1667 or α = 9.6∘

θ = 180∘ +2α = 180∘+2 × 9.6∘ = 199.2∘

=199.2 × π/180 = 3.477 rad.

Power Transmitted,

Let T2 = Tension in the slack side of the belt,

We know that

logT1T2=μθ

= 0.25 × 3.477 =0.8692

logT1T2 =0.8692/2.3 =0.378 or T1T2 = 2.387 …(taking antilog of 0.378)

T2 = T1/2.387 = 1000/2.387 =419N

We know that power transmitted,

P=(T1–T2)ν= (1000-419)4.714 =2740W = 2.74kw.

Question 5) An open belt drive is required to transmit 10 kW from a motor running at 400 rpm. The belt
is 12 mm thick and has a mass density of 0.001 gm/mm3. Safe stress in the belt is not to exceed 2.5
N/mm2. Diameter of the driving pulley is 240 mm whereas the speed of the driven pulley is 200 rpm.
The two shafts arc 2 m apart. The coefficient of friction is 0.3. Determine the width of the belt.
ANS- values change kr lena ...

Question 6) A pulley used to transmit power by means of ropes has a diameter of 3.6 meters and

has 15 grooves of 45° angle. The angle of contact is 170 and the coefficient of friction between the

ropes and the groove sides is 0.28. The maximum possible tension in the ropes is 960 N and the

mass of the rope is 1.5 kg per meter length. What is the speed of pulley in rpm and the power

transmitted if the condition of maximum power prevail.

Given data:

Dia. Of pulley(D) = 3.6m

Number of groove(or ropes) = 15

Angle of groove(2a) = 45°, α = 22.50º

Angle of contact(θ) = 170° = 1700 X (Π/180°) = 2.97 rad

Coefficient of friction(µ) = 0.28

Max. Tension(Tmax) = 960N

For maximum power: Tc = 1/3Tm

= 1/3 X 960 = 320N ....(i)

Tm = T1 + TC

960 = T1 + 320 T1 = 640N ..(ii)

Now Tc = (1/3)Tm = mV^2

V = (Tm/3m)1/2 = [960/(3 X 1.5)]1/2 = 14.6m/sec ...(iiii)

Since V = πDN/60 = 14.6, N = 77.45R.P.M.

Now, Ratio of Tension in V-Belt::

T1/T2 = eµ.θ.cosec α 640/T2 = e(0.28).(2.97).cosec22.5

T2 = 73.08N ...(iv)

Maximum power transmitted(P) = (T1–T2).v/1000 Kw

P = [(640 – 73.08) X 14.6]/1000 KW

P = 8.277KW

Total maximum power transmitted = Power of one rope X No. of rope

P = 8.277 X 15 = 124.16KW

Question 7) The gearing of a machine tool is shown in figure. The motor shaft is connected to A and
rotates at 975 r.p.m. The gear wheels B, C, D and E arc fixed to parallel shafts rotating together. The final
gear F is fixed on the output shaft G. What is the speed of F ? The number of teeth on each wheel is as
given below :

Gear ABCDEF

No. of teeth 20 50 25 75 26 65

Solution. Given : NA = 975 r.p.m. ;

TA = 20 ; TB = 50 ; TC = 25 ; TD = 75 ; TE = 26 ;

TF = 65

Let NF = Speed of gear F

We know that

Speed of the first driver/Product of no. of teeth on drivens = Speed of the last driven/Product of no. of
teeth on drivers

Question 1) A flat belt running on a pulley 1 m in diameter is to transmit 10 H.P. at a speed of 200

rpm. Taking angle of lap as 170o and coefficient of friction as 0.25 find the necessary width of the

belt, if the pull is not to exceed 20 kg/cm width of belt. Neglect centrifugal tension.

ANS- isme value change krna h or jaha 8 ko divide kiya h waha apne ko 20 se divide krna h ...

not sure pr width 69.04 mm aani chahiye...