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

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}(dE)/(dt) . dt`

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

ΔE =
`\int\limits^t_0} (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))` ]