Question

It is common to specify battery capacity, the amount of stored energy, with units of ampere-hours (Ah). A particular 12 V lead acid battery has a capacity of 80 Ah. That is, the battery stores enough energy to supply 1 A for 80 hours, or 80 A for 1 hour, or 20 A for 4 hours, and so on. Note that amperes times hours give units of charge, not energy, so specifying energy in this way requires knowledge of the nominal voltage of the battery in order to be meaningful.

The fully-charged, 12 V, 80 Ah battery is used to drive a motorized hoist to lift a 150 lb person into the air. The electromechanical energy conversion process, accounting for battery, motor, and mechanical losses, has an efficiency of 85%.

Required:
a. If the battery is fully-discharged during the lifting process, how high can the person be lifted? Assume constant battery voltage and no height limitation on the hoist.
b. The maximum allowable battery current is 15 A. How long will it take to lift the person to the height determined in part a?
c. What is the battery's output power during the lift? You may assume that the battery voltage remains constant.

Answer 1

a) 4401.7 m

b) 5.3 hours

c) 180 watts

Step-by-step explanation:

a) Determine how high a person can be lifted

since energy = voltage * amperes

= 12v * 80Ah = 960 watt h

960 watt h ≈ 960 * 3600 J

= 3456 KJ

next we will calculate the gravitational potential energy

Gpe = mass of person * g * h

= (150*0.454)kg * 9.8 * h

Gpe = 667.38 * h ------ ( 1 )

since efficiency is = 85% = 0.85

E (out ) from machine = energy * efficiency ------ ( 2 )

= 3456 * 0.85 = 2937.6 KJ

To determine (h) equate equation 1 and equation 2

667.38 * h = 2937.6 KJ

h = 2937.6 / 667.38

= 4401.7 m

b) Time taken to reach the height in part A

Eout = Gpe

Eout = ( 12v * 15A * x * 3600J * 0.85 )

Gpe = 667.38 * h = 667.38 * 4401.7 = 2937606.55

during the lift : 550800 x = 2937606.55

time taken x = 5.3 hours

c) Determine battery output power during lift

15A * 12v

= 180 watts