Question

A single-phase load takes 55 kw at 70% pf lagging from a 240V, 50 HZ supply. If the supply is made 60 Hz, with the voltage remaining the same what will be the KW load at 60 Hz?

Answer 1

Explanation & Calculations

• Since there is a power factor given ($\small \cos\theta =0.7\,\text{lagging}$) it can be understood that the load is not purely resistive, rather some inductance & capacitance is present.
• Therefore, the behavior of the load is frequency dependant.
• To calculate the new power the new current is needed but there is not enough information to estimate the reactance involved in the load.
• All we can do is generating information from the basic information given: AC voltage, lagging current.
• If we write for the instantaneous voltage, current & power,

$\qquad\qquad \begin{aligned} \small V&=V_m\sin\omega t\qquad i=i_m\sin(\omega t-\theta)\\ \small P&= Vi\\ &= \small V_m\sin\omega t\cdot i_m\sin(\omega t-\theta)\\ &= \small \frac{V_mi_m}{2}\cdot\big[cos\theta-\cos(2\omega t-\theta)\big]\\ &= \small V_{rms}i_{rms}\cos\theta-V_{rms}i_{rms}\cos(2\omega t-\theta) \end{aligned}$

• As it is seen the instantaneous power consists of two components: one is frequency-independent & the other is frequency-dependent.
• Since it is the average power that is considered long term wise, the behavior of $\small \cos(2\omega t-\theta)$ counts to zero yielding the power of the circuit to be

$\qquad\qquad \begin{aligned} \small P&= \small Vi\cos \theta \end{aligned}$

which remains constant regardless of the frequency.

• It is the active power on the other hand when the average is considered.

• What is asked in the question is the active power (as it is given in Watts).
• Therefore, the average\active power does not change even when the frequency is changed to 60hz.
• It is 55 kW.