Question

A turbine – generator set a regulation constant of 6% on the generator rating of 120 MVA, 50 Hz. The generator frequency decreases by 0.01 Hz. The increase in turbine output for steady state operation is:

Answer

0.8 MW

0.4 MW

0.2 MW

40 MW​

Answer 1

Explanation:

A turbine – generator set a regulation constant of 6% on the generator rating of 120 MVA, 50 Hz. The generator frequency decreases by 0.01 Hz. The increase in turbine output for steady state operation is: Answer. 0.8 MW.

Answer 2

Answer:

The increase in turbine output for steady-state operation is approximately -3.33 MW or -3330 kW.

Explanation:

The regulation constant of a turbine-generator set is 6% based on the generator rating of 120 MVA and 50 Hz.

If the generator frequency decreases by 0.01 Hz, the increase in turbine output for steady-state operation can be calculated using the formula Δp_mp.u = -Δf_p.u / R, where Δf_p.u is the steady-state frequency deviation in per unit and R is the regulation constant.

To calculate Δf_p.u, we need to convert the given frequency deviation into per unit.

The steady-state frequency deviation (∆ f) for an Area Control system can be calculated using the formula

∆ f = (ΔP x Rated power) / (% Change in frequency x Rated frequency) in MW/Hz[4].

Here, ΔP is the change in power output, which we need to calculate using the given information.

Since R = 6%, we can convert it into per unit as R = 0.06 p.u. Using these values, we get:

Δf_p.u = (0.01/50) = 2 x 10^-4 p.u.

Now we can calculate Δp_mp.u:

Δp_mp.u = -Δf_p.u / R

         = -(2 x 10^-4) / 0.06

        ≈ -3.33 x 10^-3 p.u.

Therefore, the increase in turbine output for steady-state operation is approximately -3.33 MW or -3330 kW[3].

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