# Electrical power with kW and kWh - Pwr avail?

1. Sep 15, 2014

### jmc

1. The problem statement, all variables and given/known data
Req'd to calculate power avail for propulsion.
Given 2x 475kW fuel cells, 6400 kWh lead-acid battery.
Ships machinery power requirement is 1100 kW at any given time, and battery max discharge rate is 1.2 kW/kWh.

2. Relevant equations
I'm not sure I fully understand the how all the units fit together to determine what power if left over for propulsion given the battery discharge rate, and the consumption of 1100 kW. It obviously would have to be a positive value. The kW and kWh units is what seems to be confusing me.

3. The attempt at a solution
475kW x2 = 950 kW avail from fuel cells, subtract the 1100 kW consumption load = -150 kW. ?
How does the battery add to that?

2. Sep 15, 2014

### AlephZero

kW are units of power (energy per unit time).
kWh are units of energy = (kW) x (time in hours).

"battery max discharge rate is 1.2 kW/kWh." means the maximum power you can get from the battery depends on how energy remains in it. If there is 1000 kWh of energy (charge) left it can produce a maximum of 1200 kW (1.2 x 1000) of power. If there is 100 kWh of energy left it can only produce 120 kW of power.

You calculated the battery needs to produce 150kW of power to make up the total consumption of 1100 kW.

So the question is, how much energy can you take from the battery until 150kW is its maximum power output, and how long does it take to reach that point.