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
= 4[ 1 -
]

Explanation:

Given the data in the question;

Q = -0.2[ 1 -
]

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' =
-

= T × ω

we substitute

= 18 × 100

= 1800 W

= 1.8 kW

so

w' =
-

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 -
] + 0.2

dE/dt = -0.2 + 0.2
] + 0.2

dE/dt = 0.2

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

ΔE =

ΔE =

ΔE =

= 4[ 1 -
]

Therefore, the required expression is
= 4[ 1 -
]