University:
Conestoga College
Course:
ERGY8070 | Energy, Storage and Management
Academic year:

2021
Views:

155

Pages:

2
Author:

S T Cherukuri

Week 1 ERGY8070 Energy Storage Terminology & Foundational Concepts Practice Problems 1. As of 2020 in the United States, the majority of grid energy storage projects were pumped hydro: a) True b) False 2. Renewable energy sources present a challenge to the grid because: a) They sometimes produce power when it’s not needed. b) They sometimes do not produce enough power when it’s needed. c) Both a) and c) 3. Convert the following to the indicated unit: a) 120mA = ________A b) 1.9 kWh c) 12 MJ = ________Wh = ________J 4. A 24V battery has an energy capacity of 12 kWh. What is the storage capacity in Ah? 5. A battery has a storage capacity of 75 kWh. What is its maximum possible usable storage capacity? 6. A 420 Ah battery is able to run for 12 hours before being fully depleted. What was its current drain? 7. What is an example of an end-user to the electrical grid? 8. What is the storage capacity in kWh of a 24V DC battery specified at 500 Ah? 9. What system of current flow will we be using for ERGY8070? Week 1 ERGY8070 ANSWERS 1. The answer is FALSE. On Slide 19 (in the Week 1 PowerPoint file), we can see that pumped hydro represents only 7% of total grid energy storage projects, despite representing 94% of total grid storage capacity. 2. The answer is c). Renewable energy sources (like wind & solar) are non-dispatchable and can produce either too much or not enough power, depending on the grid’s current requirements. This is in contrast to fossil-fuel based generation technologies (coal & natural gas), which can be rapidly deployed or scaled-back to meet the grid’s current needs. Energy storage is therefore very important to the widespread adoption of renewable energy generation, 3. Recall that there are common prefixes used in the SI system (i.e. the metric system), as seen on Slide 38. Understanding these prefixes, we can convert the quantities as follows: a) 120mA = 120 mA x 1A/1000mA = 0.12A b) 1.9 kWh = 1.9 kWh x 1000Wh/1kWh = 1,900 Wh c) 12 MJ = 12 MJ x 1,000,000 J / 1MJ = 12,000,000 J 4. Recall that Wh = V x Ah (see Slide 52). Therefore: Ah = Wh / V Ah = (12 kWh x 1000 Wh/1kWh) / 24V = 500 Ah 5. Recall that 𝒖𝒔𝒂𝒃𝒍𝒆 𝒔𝒕𝒐𝒓𝒂𝒈𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 ≤ 𝒔𝒕𝒐𝒓𝒂𝒈𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 (see Slide 56). Therefore the maximum possible usable storage capacity of a 75 kWh battery is 75 kWh. 6. From Slide 63, recall that Ah = Amps x Hours. Therefore: Amps = Ah / Hours = 420 Ah / 12 h = 35 A 7. An end-user to the electrical grid could be a building (house, factory, office, etc.) or a device (street light, billboard, etc.) or even an electric vehicle charger 8. From Slide 62, recall that Volts x Ah = Wh for DC batteries. Therefore: kWh Capacity = 24V x 500 Ah x 1 kWh/1000Wh = 12 kWh 9. We will use conventional current flow, since this is what most specifications use