Question

Question:Generator manufacturers often use the term, which is the weight in pounds of all the rotating parts of a generating unit (including the prime mover) multiplied by the square of the radius of gyration in feet. WR²/32.2 Is then the total moment of inertia of the rotating parts in . (a) Determine a formula for the stored kinetic energy in ft-lb of a generating unit in terms of and rotor angular velocity . (b) Show that

$\mathbf{H}=\frac{\mathbf{2}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{-\mathbf{4}}{\mathbf{WR}}^{\mathbf{2}}{\left(\mathbf{rpm}\right)}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{rated}}}\mathbf{per}\mathbf{unit}-\mathbf{seconds}$

where ${\mathbf{S}}_{\mathbf{rated}}$is the volt-ampere rating of the generator, and is the synchronous speed in r/min. Note that $\mathbf{1}\mathbf{}\mathbf{ft}\mathbf{-}\mathbf{1}\mathbf{b}\mathbf{=}\mathbf{746}\mathbf{/}\mathbf{550}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{356}\mathbf{}\mathbf{joules}$, (c) Evaluate for a three-phase generating unit rated 750 MVA, 3600 r/min , with ${\mathbf{WR}}^{\mathbf{2}}=\mathbf{4},\mathbf{000},\mathbf{000}\mathbf{lb}-{\mathbf{ft}}^{\mathbf{2}}$.

Short answer

Answer

(a)

The formula for the kinetic energy KE is$\frac{W{R}^2{\omega }_m^2}{64.4}\text{ft}-\text{lb}$ .

(b)

The expression of the inertia constant H is

$\frac{\left(2.31\times {10}^{-\mathbf{4}}\right){\text{WR}}^{\text{2}}{\left(\text{rpm}\right)}^{\text{2}}}{S_{\text{rated}}}\text{per}\text{unit}-\text{seconds}$, .

(c)

The value of inertia H is $15.967\text{per}\text{unit}-\text{seconds}$.

Step-by-step solution

Write the given data from the question

A three-phase generating unit rated 750 MVA.

The speed of three-phase generating unit rated at3600 r/min.

The value of given ${\text{WR}}^{\text{2}}=\text{4,000,000 lb}-{\text{ft}}^{\text{2}}$${\text{WR}}^{\text{2}}=\text{4,000,000 lb}-{\text{ft}}^{\text{2}}$.

Determine the formula of  ωsynand ωmsyn

Write the expression for the kinetic energy stored in the rotating bodies.

$\mathbf{KE}=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{I}{\omega }_{\mathbf{m}}^{\mathbf{2}}$ …… (1)

Here, I represent moment of inertia, w_m represent angular velocity.

Write the formula for the expression inertia constant.

$\mathbf{H}=\frac{\mathbf{KE}}{{\mathbf{S}}_{\mathbf{rated}}}$ …… (2)

Here, KE is kinetic energy and S_rated is rated volt-ampere rating.

Write the formula of for the inertia constant.

$\mathbf{H}=\frac{\left(\mathbf{2}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{-\mathbf{4}}\right){\mathbf{WR}}^{\mathbf{2}}{\left(\mathbf{rpm}\right)}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{rated}}}$ …… (3)

Here, WR² is square of the radius of gyration, rpm is rotation-per unit and _Srated is rated volt-ampere rating.

(a) Determine the expression for the kinetic energy stored in the rotating bodies.

Determine the value of kinetic energy stored in the rotating bodies.

Substitute $\frac{{\text{WR}}^{\text{2}}}{\text{32.2}}$ for into equation (1).

$$\begin{gathered} \text{KE}=\frac{1}{2}\left(\frac{{\text{WR}}^{\text{2}}}{32.2}\right){\omega }_{\text{m}}^{\text{2}} \\ =\frac{{\text{WR}}^{\text{2}}{\omega }_m^2}{64.4}\text{ft}-\text{lb} \end{gathered}$$

Therefore, the formula for the kinetic energy KE is$\frac{{\text{WR}}^{\text{2}}{\omega }_m^2}{64.4}\text{ft}-\text{lb}$ .

(b) Determine the expression for the inertia constant .

Determine the value of the inertia constant.

Substitute $\frac{{\text{WR}}^{\text{2}}{\omega }_m^2}{64.4}\text{ft}-\text{lb}$for KE into equation (2).

$\text{H}=\frac{1}{S_{\text{rated}}}\left(\frac{{\text{WR}}^{\text{2}}}{64.4}\right)\text{ft}-\text{lb}$

Substitute 1.356 Joules for 1 ft- Ib and $\frac{2\pi }{60}\text{rpm}$ for ${\omega }_{\text{m}}$ into above equation.

$$\begin{gathered} \text{H}=\frac{1}{S_{\text{rated}}}\left(\frac{W{R}^2}{64.4}\right){\left(\frac{2\pi }{60}\text{rpm}\right)}^2\left(1.356\right) \\ =\frac{1}{S_{\text{rated}}}\left(\frac{W{R}^2{\left(\text{rpm}\right)}^2}{64.4}\right){\left(0.1047\right)}^2\left(1.356\right) \\ =\frac{W{R}^2{\left(\text{rpm}\right)}^2}{S_{rated}}\left(0.01096\right)\left(0.02106\right) \\ =\frac{\left(2.31\times {10}^{-4}\right){\text{WR}}^{\text{2}}{\left(\text{rpm}\right)}^2}{S_{\text{rated}}}\text{per}\text{unit}-\text{seconds} \end{gathered}$$

Therefore, the expression of the inertia constant is .

$\frac{\left(2.31\times {10}^{-\mathbf{4}}\right){\text{WR}}^{\text{2}}{\left(\text{rpm}\right)}^{\text{2}}}{S_{\text{rated}}}\text{per}\text{unit}-\text{seconds}$

(c) Determine the expression for the inertia constant . 

Determine the value of inertia constant.

Substitute $4000000\text{lb}-{\text{ft}}^{\text{2}}$ for${\text{WR}}^{\text{2}}$, 3600 for rpm and 750 MVA for ${\text{S}}_{\text{rated}}$ into equation (3).

$$\begin{gathered} \text{H}=\frac{\left(2.31\times {10}^{-4}\right)\left(4000000\right)\left(3600\right)}{750\text{M}} \\ =\frac{\left(924\right)\left(12.96\text{M}\right)}{750\text{M}} \\ =\frac{11975.04}{750} \\ =15.967\text{per}\text{unit}-\text{seconds} \end{gathered}$$

Therefore, the value of the inertia constant H is $15.967\text{per}\text{unit}-\text{seconds}$.