Module 2 Calculated Problems

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Louisiana State University *

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3200

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Mechanical Engineering

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Dec 6, 2023

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The numbers may vary as they were generated randomly. Please focus on the equations applied to solve the problem. Quiz 2 Question 13 Find the expected infiltration CFM for a reception area that measures 124ft by 66ft with 10ft high walls. The windows are weatherized. There are exterior windows and doors on two sides of the reception area. Use the table below to assist with your answer. Solution: Construction dimension: 124 * 66 * 10 Construction volume = 124 * 66 * 10 = 81840 ft3 According to Table 2.8 (Provided in the question), if the construction is weatherized (second column), and there are exterior windows and doors on both sides (third row), ACH = 1. Total outdoor air in one air = Total Volume * ACH = 81840 ft3/h Convert to CFM: 81840 [ft3/h] / 60 [min/h] = 1364 ft3/min = 1364 CFM
Question 14 An air source heat pump is designed to have a HSPF of 8.9 btu/Wh. If the heat pump is to utilize 3085 kWh of electricity during the winter season, how many BTUs does the heat pump need to provide? Hint: Note that the capacity of the heat pump is in KW, not W. Solution: HSPF = 8.9 Btu/Wh Electricity consumption = 3085 kwh = 3,085,000 Wh Total Heating = Electricity consumption * HSPF = 3085000 Wh * 8.9 Btu/Wh = 274565000
Assignment 2 Question 3 The coefficient of performance (COP) for a vapor compression cycle chiller is 7.2. If the chiller can produce 45 tons of cooling, how much electric power (in kW) does the chiller consume? Fill in the electric power of the chiller in kW. Solution: COP = 7.2 Cooling output = 45 tons By its definition, COP measures the ratio between power input and cooling output (in the same unit). Power input = Cooling output / COP = 45 tons / 7.2 = 6.25 tons Then, convert tons to kW per the requirement. 6.25 tons = 6.25 tons * 12000 Btuh/ton = 75000 Btuh = 75000 Btuh / 3412 Btuh/ k W (See Section 1, screenshot attached below) = 21.98 kW
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