P.2) Using an allowable shear stress of 50 MPa, determine the power that can be transmitted at 2000 rpmthrough a shaft with a 30-mm diameter.Hint: (Power) = (Torque)(Angular velocity) [See Lec_02, or pg. 117 eq. (3-44)]:2πP = Tw=Tn= Tn/954960 × 1000Note: in the formula, P is in kW, T is in Nm, n is in rpm

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Answered: P.2) Using an allowable shear stress of…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Q.1 A rotary shaft carries distributed load of 250 w/m with a long 3m. The shaft transmits 35 KW at 290 rpm. A shaft is simply supported shaft. The mux. shear stresses 56 MPa and the allowable Compression stress 45 MPa, the safety factor is 1.5. Find the size of the shaft if it subjected to suddenly applied load.
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An electric motor's torque is a function of the speed at which it rotates. Many brushed AC motors have a linear toque curve as shown below. If the power output of the motor is equal to the torque (T) multiplied by the rotational speed in rad/s (w), what is the maximum power you can get out of the motor if the stall torque (Tstall) is 0.001 Nm and the no load speed (wno load) of 14,332 RPM? Give your answer in Watts. There are two ways to solve this: Use a stall torque and no load speed and plot the torque profile below and then plot the power versus rotational speed to find the maximum. The second method is to determine the mathematical relationship for power given the information at the start of the question and set the differential of power with respect to RPM equal to zero, and solve for the RPM to get the maximum. Tstall Wno load W
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