A 24V, 48 AH battery in fully charged state is supplying a load of 2.4 kW. Calcu

Question

A 24V, 48 AH battery in fully charged state is supplying a load of 2.4 kW. Calculate: a) Current supplied by the battery. b) The percentage of charge is left in the battery after supplying the load for 15 minutes. c) For how long can the battery supply the load before its charge gets to 25% of its capacity ? State your answer in hour, minutes and seconds (eg 1 hour. 12 minutes and 25 seconds). d) If the battery is charged from its 25 % charge state with a charging current of 25 A, what is the charge state of the battery as percentage of its full charge at the end of 30 minutes and the end of after 60 minutes of charging e) The energy (in joules) stored in the battery at the end of 60 minutes of charging. f) Time taken before the battery is charged to 100% of its capacity. State your answer in hour, minutes and seconds (eg 1 hour, 12 minutes and 25 seconds).

Explanation / Answer

part (a)

Power = 2400 W

VI=2400

V=24 hence I=100 A.

part(b)

The rating is 48 Ah. Hence 100 A can be supplied for (48/100) h. = 0.48 h = 28 minutes 48 seconds. till full discharging.

Since the discharging curve is assumed to be linear (as in the cases of Li-ion battery), in 15 minutes, the charge remaining is,

(15 x 100) / (28.8) = 52.08 % charge remaining.

part (c)

The battery supplies 75% charge and 25% remains. Time taken for 75% charge supply is,

75 x 28.8 /100 = 21.6 minutes = 21 minutes 36 seconds.

part (d)

Charging curve like the discharging curve and is linear. Assuming 100% efficiency,we know,

Time taken for a full charging through charging current of 25 A,

= 48 Ah/ 25 A = 1.92 h

In 30 minutes, the percentage of battery charged,

0.5 x 100 / 1.92 =26.04 %. As the battery already had 25 % charge, the total charge now becomes

(26.04+25) % = 51.04%

In 60 minutes, the percentage of battery charged,

1 x 100 / 1.92 = 52.08%.

As the battery already had 25 % charge, the total charge now becomes

(52.08+25) % = 77.08%

part (e)

We know the battery was supplying 2.4 kW for 0.48 hours. So the total energy stored is

E= VIt

V=24 V

I=100 A

t=0.48 hours = 1728 seconds

E = 4147200 J

So, this is the energy at full charge. Energy after 60 mins of charging, i.e at 77.08 % of charge,

(77.08 x 4147200)/100 = 3196661.76 J

part (f)

Time taken for a full charging through charging current of 25 A,

= 48 Ah/ 25 A = 1.92 h

= 1 hours 55 minutes 12 seconds

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.