Module 1 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 1 Question 10 Suppose PFT is considering implementing a new energy efficiency measure (EEM). The EEM involves sensors that cost $4406 in total to install, and are expected to last 15 years. Savings are identical at $1176 per year in utilities. The new system requires $248 per year for maintenance and repairs. These costs are assumed for first year of operation. Estimate the Simple Payback Period. Solution: Total Cost = $4406 Eneregy_Savings_Per_Year = $1176 Maintanance_Per_Year = $248 According to Section 1.4.4: Net_Savings_Per_Year = Eneregy_Savings_Per_Year - Maintanance_Per_Year = 1176-248 = 928 Payback = Total Cost / Net_Savings_Per_Year = 4406 / (928 / year) = 4.75 Years Section 1.4.4
Question 11 An electric motor running a large copier draws 1.5 kW. The motor runs 6 hours in a day. How much heat is produced in the space PER DAY as a result of the copier’s operation? Convert the result to Btu. Solution: Power = 1.5 kW Time = 6 Hours Work = Power * Time = 1.5 kW * 6 Hours = 9 kWh According to Table 2.2, 1kWh = 3412 Btu Work_btuh = Work_kWh * 3412Btuh/Kwh = 9 * 3412 = 30708 Btu You can also refer to Example 2.3 for work calculation and 2.4 for unit conversion.
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Question 12 A heating system requires an 79% efficient boiler to generate 746 thousand Btuh to counteract a heating load. If [Propane Gas] is the fuel source utilized by the boiler, what would be the input rate measured in gallons per hour? (Refer to Table 2.3 for the heating value) Solutions: Eff = 0.79 Heat_load = 746,000 Btuh From Table 2.3, Heat_Value_Propane = 93000 Btu Total_Heat_Needed = Heat_Load / Eff = 746000/0.79 = 944304 Btuh Fuel_Need = Total_Heat_Needed / Heat_Value_Propane = 944304 Btuh / 93000 Btu = 10.15 hr
Assignment 1 Question 3 Find the summer heat gain for this wall due to conduction using the TETD method. Hint: Table 3.7 & Example 3.2 Dimension: 17' by 49' Orientation: South Construction: Medium-weight Time: 4:00 PM R-value: 7 Solution: Analysis: This is a wall. According to Equation 3.6, Q = U*A*TETD U = 1 / R, by Eq. 3.1 and 3.2. -> U = 1 / 7 = 0.143Btuh/(F * ft2) A = 17 * 49 = 833 ft2 TETD = 34 F Hence, Q = 0.143Btuh/(F * ft2) * 833 ft2 * 34 F = 4050 Btuh Calculate U from R: Get TETD
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Example
Question 4 A heating system that uses hot air has a total heating load of 500 thousand btuh. How much hot air flow in CFM is required to satisfy the total load if the temperature difference is 20? Hint: Section 2.5.1. Solution: Heating_Load = 500,000 Btuh Dealta_T = 20 F This is Hot Air, CFM = Q / (1.1 * Delta T) by Eq. 2.12 CFM = 500000 / 1.1 / 20 = 22727
Question 5 A heating system that uses heated water has a total heating load of 514 thousand btuh. How much heated water flow in gallons per minute (GPM) is required to satisfy the total load if the system is designed for 34 °F temperature change? Hint: Section 2.5.2. Solution: Heat_Load = 514,000 Btuh Delta_T = 34 F This is Hot Water, GPM = Q / (500 * Delta T) by Eq. 2.11 GPM = 514000 / 500 / 34 = 30.24
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Question 6 A heating system that uses hot steam has a total heating load of 554 thousand btuh. How much heated Steam flow in pounds per hour (lb/hr) is required to satisfy the total load? Hint: Section 2.5.3. Solution: Heat_Load = 554,000 Btuh This is Heated Steam, SFR = Q/1000 by Eq. 2.13 GPM = 554000/1000 = 554 lb/hr
Question 7 An electric motor running a large copier draws 2.8 kW. The motor runs 6 hours in a day. How much heat is produced in the space PER DAY as a result of the copier’s operation? Convert the result to Btu. Solution: Load = 2.8 kW Time = 6 hours Work = Load * Time = 2.8 * 6 = 16.8 kWh Convert kWh to Btu by Table 2.2, or using Eq. 3.10 Work = 16.8 kWh * 3412 Btu/kWh = 57322 Btu You can also refer to Example 2.3 for work calculation and 2.4 for unit conversion.
Question 8 Calculate the total heat gains of the window below. Hint: Example 3.2. Dimension: 9' by 13' Materials: 1" insulating glass, 1/4" lites, clear (see Table 3.5) Note: SC = 0.23 due to blinds Orientation: South Time: When SHGC peaks for the South orientation on July 21, 40 N Temperature difference between design and outdoor: 31 F Solution: Analysis: This is a window. Window s heat gain has two parts conduction and radiation (Eq. 3.5). First, calculate the conduction part by Q = U * A *TETD (Eq. 3.6) A = 9 * 13 = 117 ft2 U = 0.47 by Table 3.5 TETD = 31 F (Given) Q_TETD = 117 * 0.47 * 31 = 1705 Second, calculate the radiation part by A * SC * SHGF (Eq. 3.6) SHGF = 109 (Table 3.6, South, Peak) SC = 0.23 (Given) Q_SHGF = 117 * 109 * 0.23 = 2933 Finally, combine the two portions. Q = Q_TETD + Q_SHGF = 1705 + 2933 = 4638
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