Question

If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2600 rpm ?

Answer 1

Answer

Assuming

electric motor consumes  = 9.00 k J

electrical energy for time = 1.00 min

speed of engine = 2500 rpm

Power = $\tau_2\omega_2$

$E = \dfrac{2}{3}\times (P)$

$E = \dfrac{2}{3}\times (9)$

E = 6 k J

E = P t

$P = \dfrac{E}{t}$

$P = \dfrac{6000 J}{60\ s}$

P = 100 W

$\omega_2 = 2600 rpm = 2600 \times \dfrac{2\pi}{60}$

$\omega_2 = 272\ rad/s$

$\tau_2 = \dfrac{P}{\omega_2}$

$\tau_2 = \dfrac{100}{272}$

τ₂ = 0.368 N.m