ERGY8070. Week 2: Practice Problems

Week 2 ERGY8070 Energy Storage Terminology & Foundational Concepts Practice Problems 1. A 70 Ah battery is half discharged in 15 minutes. a. At what c-rate has it been discharged? b. What is its discharge rate in mA? c. What is its depth of discharge? d. What is its state of charge. 2. As a battery is discharged, its terminal voltage will: a. Increase b. Stay the same c. Decrease 3. What does an e-rate of 1E mean? 4. In a single cycle, a 2 MWh storage system loses 50 kWh while charging and 100 kWh while discharging. a. What is the charging efficiency? b. What is the discharging efficiency? c. What is the roundtrip efficiency? 5. Your boss requires 2100 kWh of energy storage to fit within a 5 m3 area. You receive quotes from two suppliers. Which storage system is more suitable? Supplier A: Energy density of 500 kWh/ m3, specific energy of 90 Wh/kg Supplier B: Energy density of 350 kWh/ m3, specific energy of 120 Wh/kg 6. Elon Musk is looking to reduce the weight in the latest Tesla vehicle. Which of the following batteries would you recommend to him? Battery A: Specific energy of 150 Wh/kg Battery B: Specific energy of 0.001 MJ/g 7. How long would it take to charge a 75 kWh battery, with a 6 kW charger that is 98% efficient? 8. How long would it take to fully discharge a 50 kWh battery to a 1 kW discharge system that is 90% efficient? 9. True or False: roundtrip efficiency can exceed charging or discharging efficiency? 10. If a battery has a usable storage capacity of 90% and has discharged all of its usable storage capacity, what is its depth of discharge? 11. In class, we saw that cutting the C-rate in half (going from 1C to 0.5C) doubled the discharge time. What effect would doubling the C-rate from 1C to 2C have on discharge time? Week 2 ERGY8070 ANSWERS 1. A 70 Ah battery is half discharged in 15 minutes a. At what c-rate has it been discharged? Recall that time to fully discharge in hours (t) = 1/c-rate If the battery is 50% discharged in 15 minutes, it will be fully discharged in 30 minutes. Therefore, t=0.5 hours Rearranging the above formula for c-rate: c-rate = 1/t = 1/0.5 = 2C b. What is its discharge rate in mA? 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑟𝑎𝑡𝑒 = 𝑟𝑎𝑡𝑒𝑑 𝐴ℎ 70 𝐴ℎ 1000 𝑚𝐴 = = 140 𝐴 ∗ = 140,000 𝑚𝐴 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑡𝑖𝑚𝑒 (ℎ) 0.5ℎ 1𝐴 c. What is the depth of discharge? – As stated in the problem ,the battery is half discharged. Therefore, its depth of discharge (DoD) is 50%. d. What is the state of charge? – Recall that state of charge (SoC) + DoD = 100%. Since DoD = 50%, SoC must also be 50%. 2. The answer is c). Recall that a battery’s terminal voltage will gradually decrease as it is discharged. 3. Recall that e-rate is to power, as c-rate is to current. Since 1C refers to the discharge current that will discharge an entire battery in 1 -hour, 1E therefore refers to the discharge power that will discharge the entire battery in 1-hour. 4. In a single cycle, a 2 MWh storage system loses 50 kWh while charging and 100 kWh while discharging. a. What is the charging efficiency? 2 𝑀𝑊ℎ = 2𝑀𝑊ℎ ∗ 𝜂!" = B1 − 1000𝑘𝑊ℎ = 2,000 𝑘𝑊ℎ 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 1𝑀𝑊ℎ 𝐸#$%%!" 50 𝑘𝑊ℎ E ∗ 100% = B1 − E ∗ 100% = 97.5% ( 𝐸!" 2,000 + 50)𝑘𝑊ℎ b. What is the discharging efficiency? 𝐸#$%%$&' 100 𝑘𝑊 𝜂$&' = B1 − E ∗ 100% = B1 − E ∗ 100% = 95% 𝐸% 2,000 𝑘𝑊 c. What is the roundtrip efficiency? 𝜂(' = 0.975 ∗ 0.95 = 0.926 = 92.6% Week 2 ERGY8070 5. ),+,, ./0 Option A: 1,, ./0/3! = 4.2 𝑚4 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ),+,, ./0 Option B: 41, ./0/3! = 6 𝑚4 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 Option A would be suitable. 6. To reduce weight, we want the battery with the higher specific energy. However, we cannot compare two batteries with specific energy in different units. In order to compare these two batteries, we will need to convert them to the same units. Let’s convert Battery B to Battery A’s units. Recall from Week 1, Slide 50, that 1 Wh = 3600 J. Therefore. 𝐵𝑎𝑡𝑡𝑒𝑟𝑦 𝐵 = 0.001 𝑀𝐽 1,000,000 𝐽 1𝑊ℎ 1000𝑔 ∗ ∗ ∗ = 277.78 𝑊ℎ/𝑘𝑔 𝑔 1𝑀𝐽 3600𝐽 1𝑘𝑔 Therefore, Battery B is the one you should recommend. 7. Recall that Therefore, 𝑡5 = 𝑡5 = 7 8. Recall that 𝑡? = Therefore, 𝑡? = 6" 91 ./0 #$ 8#$ 7&'( 6" 8&'( 7&'( 6" 8&'( 6" 7#$ 8#$ = ,.;<∗> ./ = 12.76 ℎ𝑜𝑢𝑟𝑠 , = ,.;∗1, ./0 + ./ = 45 ℎ𝑜𝑢𝑟𝑠 9. Charging and discharging efficiency are both less than 100%. Since roundtrip efficiency is the product of charging efficiency and discharging efficiency and both values, when expressed as decimals, are below 1, roundtrip efficiency must therefore be less than either charging or discharging efficiency. Therefore this statement is false. 10. DoD = Energy Discharged / Storage Capacity Let storage capacity = x. Therefore, Energy Discharged =0.9x DoD = 0.9 x / x = 0.9 or 90% 11. From Week 2 slide 20, time to fully discharge in hours (t) = 1 / c-rate (C) For 1C, t = 1/C = 1 hours For 2C, t = 1/2C = 0.5 hours Therefore, doubling the C-rate cuts the discharge time in half (i.e. reduces discharge time by 50%)