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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Question 4 The coefficient of performance (COP) for a vapor compression cycle chiller is 7.2. If the chiller can produce 54 tons of cooling. If the chiller operates for a total of 3509 hours annually, and the cost of electricity is $0.2 per kWh, how much does it cost to operate the chiller? Fill in the operating cost of the chiller in one year in $. Solution: Very similar to Q3, this question also examines your understanding of COP. Given: COP = 7.2 Cooling Output = 54 tons Time = 3509 hours Repeat what you have done in Question 3: Power Input = Cooling output / COP = 54 / 7.2 = 7.5 tons = 7.5 tons * 12000 Btuh = 90000 Btuh = 90000 Btuh / 3412 kw/Btuh = 26.38 kW Power Consumption = Power Input * Time = 26.38 kw * 3509 hours = 92559 kWh Cost = Power Consumption * Price Rate = 92559 kwh * 0.2 $/kwh = 18511 dollars

Question 5 An electric resistance heater is used to provide space heating at an efficiency level of 100%. The electricity is sold at $0.09 per kWh. Question: How much money is needed if 1010.5 MBH is demanded? Hint: 1 MBH equals 1,000 btuh Solution: Efficiency level = 100% Heating Output = 1010.5 MBH = 1010.5 [MBH] * 1000 [btu/MBH] = 1010500 Btu Power Input = Heting Output / Efficiency = 1010500 / 100% = 1010500 Btu Convert Btu to kWh Power Input = 1010500 Btu / 3412 Btu/kWh = 296.16 kWh Power Cost = Power Input * Cost = 296.16 kWh * 0.09 $/kWh = 26.65 $

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Question 6 A gas furnace is used to provide space heating at an efficiency level of 82%. The gas is sold at $1.35 per therm. Question: How much money is needed if 1125.3 MBH is demanded? Hint: 1 MBH equals 1,000 btuh Solution: Heating demand = 1125.3 MBH = 1125300 Btu 1 therm equals 100,000 Btu. Heating demand = 1125300 Btu / 100000 Btu/therm = 11.253 therm Heating input = Heating output / Efficiency = 11.253 / 82% = 13.723 therm Heating cost = Heating input * price = 13.723 therms * 1.35 $/therm = $18.53

Question 7 An air-cooled window DX unit is used for space cooling. The unit has a capacity of 4.8 tons and has a SEER of 14. If the unit operates for 2424 hours during the summer, how many kWh of electricity will be used? Hint: In the U.S., the SEER is the ratio of cooling in British thermal units (BTUs) to the electrical energy consumed in watt-hours. You may need to convert the unit in this process. Solution: Cooling capacity = 4.8 tons = 4.8 tons * 12000 Btuh/ton = 57,600 Btuh (See Examples above for the conversion) SEER = 14 = 14 Btuh/W Power Input = Cooling Capacity / SEER = 57600 [Btuh] / 14 [Btuh/W] = 4.1143 kW Time = 2424 hours Power consumption = Power input * time = 4.1143 kW * 2424 hours = 9973 kWh

Question 8 A heating plant for a large medical campus requires 6479 MBH (1MBH \= 1,000,000 btuh) of heating capacity. Which of the following pipe size and velocity in feet per second (fps) for the main would be needed if the plant is a 52 psig medium-pressure steam system? (Hint: Table 7.2) a.8" pipe at 130 fps b.4" pipe at 150 fps c.2" pipe at 100 fps d.4" pipe at 130 fps e. 6479 MBH; 52 psig Solution: Known: Heating demand = 6479 MBH, Pressure = 52 psig Check Table 7.2, the heating medium is 52 psig medium-pressure steam, so we are looking at the second to last row. The heating demand is 6479 MBH. 6479 is larger than 6000 (4 ’’ ), but way smaller than 25000 (8 ’’ ), so we are using 4 ’’ in this case. The answer is b, 4 ’’ pipe at 150 fps.

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