Answer:
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' =
- ![w_{shaft](https://tex.z-dn.net/?f=w_%7Bshaft)
= 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![e^{(-0.05t)](https://tex.z-dn.net/?f=e%5E%7B%28-0.05t%29)
Now, the change in total energy, increment E, as a function of time;
ΔE = ![\int\limits^t_0}\frac{dE}{dt} . dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Et_0%7D%5Cfrac%7BdE%7D%7Bdt%7D%20%20.%20dt)
ΔE = ![\int\limits^t_0} 0.2e^{(-0.05t)} dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Et_0%7D%200.2e%5E%7B%28-0.05t%29%7D%20dt)
ΔE = ![\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0](https://tex.z-dn.net/?f=%5Cint%5Climits%5Et_0%7D%20%5Cfrac%7B0.2%7D%7B-0.05%7D%20%5Be%5E%7B%28-0.05t%29%7D%5D%5Et_0)
= 4[ 1 -
]
Therefore, the required expression is
= 4[ 1 -
]