Question

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 1

Answer:

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

Explanation:

Given the data in the question;

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

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' = $w_{elect$ - $w_{shaft$

$w_{shaft$ = T × ω

we substitute

$w_{shaft$= 18 × 100

$w_{shaft$ = 1800 W

$w_{shaft$ = 1.8 kW  

so

w' = $w_{elect$ - $w_{shaft$

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 - $e^{(-0.05t)$ ] + 0.2

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

dE/dt = 0.2$e^{(-0.05t)$

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

ΔE = $\int\limits^t_0}\frac{dE}{dt} . dt$

ΔE = $\int\limits^t_0} 0.2e^{(-0.05t)} dt$

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

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

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