Question No 1
A pulley system uses a flat belt of cross section area 1000 mm square and density 1150 kg/m^3. The angle of the lap on the smaller wheel is 130, the coefficient of friction is 0.32 and the maximum force allowed in the belt is 550N and velocity is 9 m/sec. Calculate
- The maximum power when centrifugal force is not included
- The maximum power when centrifugal force is included
- The initial belt tensions
Solution
- a. Maximum Power without centrifugal force
Power= P=(T_1- T_2 )v
T_1/e^uθ = T_2
550/e^(0.32×3.14) = T_2
T_2=201.36 N
P=(550- 201.36)9
P= 3137.71 WAns
- b. Maximum Power with centrifugal force
Power= P=(T- T_c ) v ×C
C=1- 1/e^uθ
C=1- 1/e^(0.32*3.14)
C=0.63
T_c=mv^2
m=area ×length ×density
m=0.001×1×1150=1.15 kg
T_c= 1.15×9^2=93.15 N
P=(550- 93.15) 9 ×0.63=2590.33 WAns
- c. initial belt tensions
Without Centrifugal force
T_o=((T_1+ T_2 ))/2
T_o=(550+201.36)/2=375.68 N
With centrifugal Force
T_o=((T_1+ T_2+2T_c ))/2
T_o=(550+201.36+2(93.15))/2 =468.83 N Ans
Question 2
The following data is for a conical clutch
Inside diameter 35 mm
Outside diameter 150 mm
Coefficient of friction 0.35
Included angle 120
Speed 2500 rev/min
Calculate the axial forced needed to allow the transmission of 1000 watts without slipping using
- The uniform pressure theory
- The uniform wear theory
- If the clutch is replaced by a multi-plate clutch with 3 contact surface for the same axial force what power can be transmitted by
- The uniform wear theory
- Uniform pressure theory
Solution
Power= P= τω
P/ω= τ
τ=1000/((2 ×3.14 ×2500)/60)=3.82 Nm Ans
Uniform pressure theory
T_r=2 π u p_n cosec α ((〖ro〗^3- 〖ri〗^3)/3)
3.82=2 ×3.14×0.35 × p_n× cosec 60 ×((〖0.075〗^3- 〖0.0175〗^3)/3)^
p_n=3.82/0.000152=25035 Pa
Axial Force = W = 2 π p_n r dr
W= 2 π ×25035×((0.075+ 0.0175)/2)×(0.075-0.0175)
W = 418.10 N Ans
Uniform Wear theory
T_r=2 π u C cosec α ((〖ro〗^2- 〖ri〗^2)/2)
3.82=2 ×3.14×0.35 × C× cosec 60 ×((〖0.075〗^2- 〖0.0175〗^2)/2)^
C = 3.82/0.002922 = 1307.03
Axial Force = W =2 π C dr
W = 2 π ×1307.03× (0.075-0.0175)
W = 471.96 N Ans
Multi disc clutch
Uniform pressure
Torque= T=uW/3 ×(Do^3-Di^3)/(Do-Di) ×n
T= (0.35×418)/3 ×(〖0.15〗^3-〖0.035〗^3)/(0.150-0.035) ×3
T=48 ×0.029×3=4.17 Nm
Power=P=((2×3.14×N))/60×T
P=1091.15 WAns
Uniform wear
Torque= T=uW/3 ×(Do+Di) ×n
T=(0.35×471.96 )/4 ×(0.15+0.035)×3
T=22.91 N
Power=P=((2×3.14×N))/60×T
P=5994.78 WAns
Question 3
3) The following gear train has an efficiency of 75% and the drum rotates at 75 rpm. If the torque input from the motor is 8Nm.Calculate
- The number of teeth on the gear X
- The speed at which the load is moving up as the rope on the drum
- The power input by the motor
- The power output by the gear train
- The load on the drum
Solution
a. Number of Teeth
Gear ratio= 50/x × 100/30 × 80/25=75/2000
50/x ×3.34 ×3.2=0.0375
50/x=0.0375/10.688
50/x=0.003508
x=50/0.00350=14222.22=14223 Teeths Ans
b. Speed
v=r × ω
v=0.2/2 × (2 ×3.14 ×75)/60
v= 0.785 m/sec Ans
c. Power of motor
P= τω
P=8 ×(2 ×3.14 ×2000)/60
P = 1674.66 W Ans
d. Power output
As efficiency of the train is 75% so
P_out=0.75 × P_motor
P_out=0.75 ×1674.66
P_out= 1256 W Ans
e. load torque
A power at the drum is the power of gear train so
P_out= τω
τ=P_out/ω
τ=1256/((2 ×3.14 ×75)/60)
τ=160 Nm Ans
Question 4
4) The diagram below shows two masses on two rotors in
planes B and C. Determine the masses to be added to the rotor in plane A and D
at a radius of 50 mm which will produce static and dynamic balance.
Solution
M mass kg
|
R distance mm
|
M*r
|
X distance from A
|
M*r*x
|
|
A
|
Ma
|
50
|
50Ma
|
0
|
0
|
B
|
5
|
20
|
100
|
80
|
8000
|
C
|
3
|
50
|
150
|
150
|
22500
|
D
|
Md
|
50
|
50Md
|
250
|
12500Md
|
From the polygon
for M*r*x we have
12500Md = 25000
Md = 2 Kg Ans
Mr of point D = 50*2
= 100 at an angle of 58 degree Ans
From the polygon
for M*r we have
50 Ma = 105
Ma = 2.1 Kg at
an angle of 35 degree Ans
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