4. A hi-tech company is designing a semi-truck powered by Li-ion batteries. The

Question

4. A hi-tech company is designing a semi-truck powered by Li-ion batteries. The weight (mass) of the batteries, mbat, needed to drive a distance of x km can be estimated using the following equation mbar.(x)-( mvehicle +-Gdveh.) (expeba-r-1 The empty weight of the truck without batteries is 13000 kg, constants a-025 wh/km-kg·bs 0.1 wh/m2-km-kg, and Cd.Aveh = 5 n", and ebat is in wh/kg. The available Li-ion batteries are listed in the following table. NMC LiNi Mn,Co04 LMO LFP Acronym Chemistry LCO (LiC002) 3.8-4.4 LiFePO4 3.2-3.5 LiMn2Og 3.8-4.1 3.8-4.0 Voltage/V (Theor.) Sp. Energy (530) 190 (440) 150 (550) 160 /Whkg Vol. Energy/Whl- 560 Cost, 2015/16 USS (NOK) h (1.5) (1.3) (1.9) Corresp. Cycle Cost 2-182-15 2-22 /cents (ore) kWh- (13-134) (11-114) (17-167) . Singlecycle cont is bred on a. 90% SoC "indon, Nn. Soll lfetime, and 2 HO 2U 1", cycle, (590) 160 418 0,17 260 0.25 260 0.16 0.2 1-14 (11-107) For a driving distance of 800 km (a) Which type of battery listed in the table willadd the LEAST additional weight to the truck? What is the weight of the batteries? (10 pt (b) W hat is least expensive option? What is the cost of the batteries in US dollar? (10 pt) (d) If the maximum total weight of a truck is 50000 kg (per US highway regulation), what is the maximum load these Li-ion battery trucks can carry? (10 pt)

Explanation / Answer

A high tech company designing a semi- truck powered by lion batteries. The mass of the battery can be found by the given equation.
mbat(x)= (mvehicle+(b/a)*CdAveh)(exp(a/ebat)*x-1)-------------------(1)
empty vehicle mass without batteries is given 13000kg i.e mvehicle= 13000kg
a= 0.25Wh/km-kg, b=0.1 Wh/m2-km-kg, Cd*Aveh= 5m2
Given that ebat is in Wh/kg.
Given driving distance is x= 800km
(1)
The battery which adds least additional weight to truck is LiFePo4 Battery since it has more sp. Energy. i.e it has higher value of ebat.
to find weight of the battery we have:
mbat(x)= (mvehicle+(b/a)*CdAveh)(exp(a/ebat)*x-1)
= (13000+0.4*5)* (1.40-1)
= 5200.8 kg------------ for battery LiFePo4
mbat(x)= (mvehicle+(b/a)*CdAveh)(exp(a/ebat)*x-1)
= (13000+0.4*5)* (1.45-1)
= 5850.9kg---------------- for battery LiCoO2
mbat(x)= (mvehicle+(b/a)*CdAveh)(exp(a/ebat)*x-1)
= (13000+0.4*5)* (1.57-1)
=7411kg----------------------for battery LiMn2O4
mbat(x)= (mvehicle+(b/a)*CdAveh)(exp(a/ebat)*x-1)
= (13000+0.4*5)* (1.43-1)
= 5590 Kg---------------For battery LiNi1/2Mn1/2CoO4

(2) The least expensive option is LiFePo4 as it has cost 2015/16 as 0.16 which is cheaper compared to the other batteries . The cost in USD $ is 111.55 USD
(3)
given maximum total weight of the truck is 50000 kg, the empty truck weight is 13000kg .
Hence let us consider the truck has LiFePo4 batteries of quantity '7'
therefore, total weight of truck is 13000+(5200*7) = 49400 (approx 50000)
It can carry the maximum voltage of 3.5*7 = 24.5 V
and maximum load of 107Kw/h * 7 = 749 Kw/h

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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