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Torque developed by a motor while running at 1000 rpm is 106 N-m and the shaft torque available is 100 N-m. then the iron and mechanical losses in wat
Question

Torque developed by a motor while running at 1000 rpm is 106 N-m and the shaft torque available is 100 N-m. then the iron and mechanical losses in watts are:

A.

200π

B.

50π

C.

300π

D.

100π

Correct option is A

The difference between the developed torque and the shaft torque accounts for the iron and mechanical losses.

​Given:∙ Developed torque Td=106 N⋅m∙ Shaft torque Tsh=100 N⋅m∙ Speed N=1000 rpmTorque corresponding to losses:Tloss=Td−Tsh=106−100=6 N⋅mAngular speed:ω=2πN60=2π×100060=100π3 rad/sIron and mechanical losses (power):Ploss=Tloss×ω=6×100π3=200π \\[10pt]\textbf{Given:} \\[6pt]\bullet \ \text{Developed torque } T_d = 106 \, \text{N}\cdot\text{m} \\[4pt]\bullet \ \text{Shaft torque } T_{sh} = 100 \, \text{N}\cdot\text{m} \\[4pt]\bullet \ \text{Speed } N = 1000 \, \text{rpm} \\[14pt]\textbf{Torque corresponding to losses:} \\[8pt]T_{\text{loss}} = T_d - T_{sh} = 106 - 100 = 6 \, \text{N}\cdot\text{m} \\[14pt]\textbf{Angular speed:} \\[8pt]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1000}{60} = \frac{100\pi}{3} \, \text{rad/s} \\[14pt]\textbf{Iron and mechanical losses (power):} \\[8pt]P_{\text{loss}} = T_{\text{loss}} \times \omega = 6 \times \frac{100\pi}{3} = 200\pi \, \text{}​​

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