The daily load demand of a grid is given by
$$P=P_{max} \cdot cos(\pi t/24)$$
Here \$-12hrs ≤ t ≤ 12hrs\$. Maximum power occurs at (\$t=0\$) and minimum occurs at (\$t= ±12\$). A \$10 GW\$ hydro power plant with pump storage is available for base load operation. The turn-around efficiency of the storage system is \$0.7\$. Find the peak load.
I found the average power in terms of \$P_{max}\$, by integration of the above over -12 to 12 and then dividing by 24, which comes out to be: \$P_{avg}=2 P_{max}/\pi\$. but now I couldn't figure out how to proceed. Any help will be greatly appreciated.
Best Answer
Preamble
One of the benefits of the hydro power plant is that it can be throttled all the way down to zero. Hydro power plants are good at supplying peaky loads. If a standalone hydro power plant has spare capacity, it just throttles down and conserves the water, rather than pumping it back up into the reservoir. These considerations make your problem (in its original form) look less intuitive and realistic.
However, your problem will look more intuitive and realistic if you imagine that the nominal power Pnom = 10 GW is coming from an external supply other than the hydroelectric plant. This external supply can not throttle up or down. Also, there is a local hydroelectric facility, which can pump the reservoir when Pload < Pnom and use it to augment the power supply when Pload > Pnom. This scheme can support Pload,max > Pnom.
Approach
At τsw the hydroelectric system switches from pumping storage to using it. $$P_{nom}=P_{load,max} \cdot cos(\pi \tau_{sw}/24)$$
Energy is in [GW⋅h] units. Energy balance of the pumped storage. $$E_{storage,out}=0.7 E_{storage,in}$$
Substitute the appropriate integrals for the energies, and you get a system of two equations with two variables τsw and Pload,max.