Answer:
0.458 N/m
Explanation:
Power = Energy/time
From the question,
Note: two-third of the energy into the motor,
Therefore,
P = 2/3(E)/t................. Equation 1
Where P = power of the motor, E = Electrical Energy, t = time
Given: E = 10.8 kJ = 10800 J, t = 1.00 min = 1/60 s
Substitute into equation 1
P = 10800(1/60)(2/3)
P = 120 W.
But,
The power of a rotating motor is given as,
P = Tω................ Equation 2
Where T = Torque of the engine, ω = angular velocity of the engine
Make T the subject of the equation
T = P/ω............... Equation 3
Given: P = 120 W, ω = 2500 rpm = (2500×0.1047) rad/s = 261.75 rad/s
Substitute into equation 3
T = 120/261.75
T = 0.458 N/m
Then the torque develop in the engine = 0.458 N/m
The amount of torque this engine will develop if you run it at 2500 rpm is 0.46 Newton.
Given the following data:
Energy = 10.8 kJ = 10800 Joules
Time = 1 minutes = 60 seconds
Angular velocity = 2500 rpm
Conversion:
2500 rpm to rad/s = 261.8 rad/s
To find how much torque this engine will develop if you run it at 2500 rpm:
Power output = [tex]1 - \frac{1}{3} =\frac{2}{3}E[/tex] ...equation 1.
Mathematically, power is given by the formula:
[tex]Power = \frac{Energy}{Time}[/tex] ...equation 2.
Substituting eqn. 1 into eqn. 1, we have:
[tex]Power = \frac{\frac{2}{3} E}{Time} \\\\Power = \frac{\frac{2}{3} \times 10800}{60}\\\\Power = \frac{7200}{60} \\\\Power = 120[/tex]
Power = 120 Watts.
In a rotating motor, power is given by the formula:
[tex]Power = T\omega\\\\120 = T \times261.8 \\\\T= \frac{120}{261.8}[/tex]
Torque, T = 0.46 Newton.
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