Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 8)
8.
Two shafts 'A' and 'B' transmit the same power. The speed of shaft 'A' is 250 r.p.m. and that of shaft 'B' is 300 r.p.m. The shaft 'B' has the greater diameter.
Discussion:
52 comments Page 1 of 6.
Vikash kumar said:
1 decade ago
The shaft having more speed will transmit less torque at same power level and for transmitting less torque less diameter shaft is required.
Abin baby said:
1 decade ago
Speed of the shaft have no relation to its diameter.
Barath said:
1 decade ago
The power transmitted by the shaft is directly proportional to speed where as the torque is inversely proportional so torque reduces with speed. While designing we take smaller dia for the shaft which has to transmit lesser torque. Remember the formula P = 2*3.14*N*T/60000 KW.
Suresh said:
1 decade ago
If size of shaft increases then speed decreases.
Chetan said:
1 decade ago
According to the equation of power (p) = (pi*d*n)/60, The speed of the shaft is directly proportional to the shaft diameter. Then how it can be possible?
Prasenjit said:
1 decade ago
Power transmission does not depend on diameter of shaft. Its depend on the torque. Torque is equal to force*radius of shaft. We don't know the force. So we can't say which dia is more than other.
VISHNUVARDAN CHADA said:
1 decade ago
Power = 2*pi*N*T/60. as power is same in both the cases (N*T)1=(N*T)2. But T=F*R.
=> (N*F*R)1 = (N*F*R)2.
For same force N*R = CONSTANT.
Hence 250d1 = 350d2.
=> d2 = 250d1/350.
So d2 is smaller than d1.
=> (N*F*R)1 = (N*F*R)2.
For same force N*R = CONSTANT.
Hence 250d1 = 350d2.
=> d2 = 250d1/350.
So d2 is smaller than d1.
Vrushang said:
1 decade ago
P = F*V .
P = F*omega*r.
So if both shaft transmit same power then omega*r=constant so,
for same power omega increase radius decreases.
P = F*omega*r.
So if both shaft transmit same power then omega*r=constant so,
for same power omega increase radius decreases.
ANUP said:
1 decade ago
P = 2(PI)NT/60.
HERE P1 = P2.
SO, 2(PI)N1T1/60 = 2(PI)N2T2/60
SINCE N1 < N2.
T1 > T2 TO MAINTAIN SAME POWER.
T = SHEAR STRESS * (PI) * D3 / 16.
SO, D1 > D2
HERE P1 = P2.
SO, 2(PI)N1T1/60 = 2(PI)N2T2/60
SINCE N1 < N2.
T1 > T2 TO MAINTAIN SAME POWER.
T = SHEAR STRESS * (PI) * D3 / 16.
SO, D1 > D2
Pari.siva said:
1 decade ago
Power is produced same,
We imagine,
Both pulleys are connected to same shaft.
And speed is inversely proposition to diameter in case greater speed pulley is the smaller diameter.
We imagine,
Both pulleys are connected to same shaft.
And speed is inversely proposition to diameter in case greater speed pulley is the smaller diameter.
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