Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
