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The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals negative 0.2 open square brackets 1 minus e to the power of negative (0.05 t )end exponent close square brackets , where t is in seconds and Q with dot on top is in kW. The shaft of the motor rotates at a constant speed of omega equals 100 space r a d divided by s (about 955 revolutions per minute, or RPM) and applies a constant torque of tau equals 18 space N m to an external load. The motor draws a constant electric power input equal to 2.0 kW. For the motor, provide an expression to show the change in total energy, increment E, as a function of time.

Answer :

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Answer:

the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

[tex]w_{shaft[/tex] = T × ω

we substitute

[tex]w_{shaft[/tex]= 18 × 100

[tex]w_{shaft[/tex] = 1800 W

[tex]w_{shaft[/tex] = 1.8 kW  

so

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = 0.2[tex]e^{(-0.05t)[/tex]

Now, the change in total energy, increment E, as a function of time;

ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]

ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]

ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]

[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

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