# The inertia constant of two groups of machines, which do not swing together are H1 and H2. The inertia constant of the system is

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The inertia constant of two groups of machines, which do not swing together are H1 and H2. The inertia constant of the system is

1.  $\frac{{{H_1}{H_2}}}{{{H_1} + {H_2}}}$
2. H1 – H2; H1 > H2
3.

$\sqrt {{H_1}{H_2}}$

4. $\frac{{{H_1} + \;{H_2}}}{{{H_1}{H_2}}}$

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Correct Answer - Option 1 :  $\frac{{{H_1}{H_2}}}{{{H_1} + {H_2}}}$

Two machine systems:

• In a two-machine system, the synchronous generator is connected to a synchronous motor by a loss-less network.
• Initially, both the machines are in stable condition. When the load on the motor increases suddenly, the electrical input cannot change simultaneously so that the motor will get oscillations.
• When the oscillations are very high when compared to the natural frequency of oscillations the motor has become unstable.
• When the motor has become unstable, the load on the synchronous generator is zero and the generator will make oscillations.
• Then the above two machine system becomes a single machine system.

Two machines are swinging together:

Two alternators are connected to a common bus, any change of load will change the rotor position of two alternators is called swinging of two machines.

Let's consider M1 and M2 are angular momentum of two generators.

δ1 = δ2 = δ and Paeq = Pa1 + Pa2

⇒ $M_{eq}\frac{d^2\delta}{dt^2}=M_1\frac{d^2\delta_1}{dt^2}+M_2\frac{d^2\delta_2}{dt^2}$

As δ1 = δ2 = δ

⇒ Meq = M1 + M2

And, the equivalent inertia constant (Heq) of two machines that are swinging together and having inertia constants H1 and H2 is given by Heq = H1 + H2

Two machines not swinging together:

When two alternators are not connected to the common bus, only a change of load will change the position of one alternator only, then the machines are not swinging together.

Let's consider M1 and M2 are angular momentum of two generators,

δ = δ2 - δ1

Where

δ2 is the load angle of generator 2, δ1 is the load angle of generator 1, δ is the resultant load angle.

apply derivation with respect to time on both sides we get,

$\frac{{{d^2}δ }}{{d{t^2}}} = \frac{{{d^2}{δ _2}}}{{d{t^2}}} - \frac{{{d^2}{δ _1}}}{{d{t^2}}}$

By applying the swing equation we can modify the above equation as

$\frac{{{d^2}δ }}{{d{t^2}}} = \frac{{{P_{{s_2}}} - {P_{{e_2}}}}}{{{M_2}\;\;}} - \frac{{{P_{{s_1}}} - {P_{{e_1}}}}}{{{M_2}\;\;}}$

Multiplying both sides by $\frac{{{M_1}{M_2}}}{{{M_1} + {M_2}}}$ we get,

$\frac{{{M_1}{M_2}}}{{{M_1} + {M_2}}}\frac{{{d^2}δ }}{{d{t^2}}} = \frac{{{M_1}{P_{{S_2}}} - {M_2}{P_{{S_1}}}}}{{{M_1} + {M_2}}} - \frac{{{M_1}{P_{{e_2}}} - {M_2}{P_{{e_1}}}}}{{{M_1} + {M_2}}}$

${P_{aeq}} = {P_{seq}} - {P_{eeq}}$

Where,

Paeq = equivalent accelerating power

Pseq = equivalent mechanical power

Peeq = equivalent electrical power

${P_{aeq}} = {M_{eq}}\frac{{{d^2}δ }}{{d{t^2}}}$ = $\frac{{{M_1}{M_2}}}{{{M_1} + {M_2}}}\frac{{{d^2}δ }}{{d{t^2}}}$

Hence the equivalent angular momentum of the system is given by,

${M_{eq}} = \frac{{{M_1}{M_2}}}{{{M_1} + {M_2}}}$

Similarly, if machines do not swing together ${H_{eq}} = \frac{{{H_1}{H_2}}}{{{H_1} + {H_2}}}$