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Addition & Subtraction of Errors

Given a singl...

Question

Given a single area with two generating units.

Unit Rating Speed drop R (per unit on unit base)

1 400 0.04

2 800 0.05

The units share a load of $P_{1} = 200 M W, P_{2} = 500 M W$ . The units are operating in parallel to share a load of 700 MW at 50 Hz frequency. The load is increased by 130 MW with B = 0, The value of new frequency caused by steady state frequency deviation is _______ Hz.
49.75

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Solution

The correct option is A 49.75
$Δ f = \frac{- Δ P}{\frac{1}{R_{1}} + \frac{1}{R_{2}} + B}$

On a common base of 1000 MVA

$R_{1} = (0.04) (\frac{1000}{400}) = 0.1$

$R_{2} = (0.05) (\frac{1000}{800}) = 0.0625$

$Δ f = \frac{- 130 / 1000}{\frac{1}{0.1} + \frac{1}{0.0625}} = - 5 \times 10^{- 3} p . u$

$Δ f_{(a c t u a l)} = 50 \times Δ f p . u = 50 \times - 5 \times 10^{- 3}$

$f = f_{o} + Δ f_{(a c t u a l)} = 50 \times - (5 \times 10^{- 3}) (50)$

$= 49.75 H z$

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