HE7

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California State University, Fullerton *

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301

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Mechanical Engineering

Date

Apr 3, 2024

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pdf

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Uploaded by GrandAntPerson261

HOMEWORK 4: Chapter 3 page 90 16) How many pounds of coal have the same heating value as 20 gal of gasoline? 1 gallon of gasoline = heating value of 125,000 BTU 20 gallons = 2.5 million BTU. 90 lbs of coal =heating value of 1 million BTU 2.5 million BTU = 2.5 x 90 = 225 pounds of coal 20 gallons of gasoline x ( 90 lbs coal/ 8 gal gas) = 225 lbs of coal. 17)A household furnace has an output of 100,000 Btu/hr. What size electrical heating unit (in kW) would be needed to replace this? 1 BTU/hr= 0.0002930711 kw 100,000 BTU/ hr x 0.00029307107 kW/Btu = 29.307 kW = ~ 29.3 kW 100,000 BTU/ h x(1 kWh 3413 BTU) = 29.3 kW Chapter 5 page 147 - 149 7) To keep an iced drink cold for the longest period of time, what color cup would you use, and why? To keep an iced drink cold for the longest period of time, the color cup I would use is silver, because the reflective surface will preserve temperatures at a consistent degree and the ice will melt at a slower rate, allowing it to stay colder for a longer time. A Silver cup will both absorb and radiate the least radiant energy, thus minimizing heat transfer by radiation. 15) A wall is made up of four elements, as follows ½” wood siding (lapped) ½” plywood sheathing 3 ½ fiberglass ½” sheetrock Using the R- values of Table 5.2. how many Btu per hour per square foot will be lost through the wall when the outside temperature is 50*F colder than the inside? R = 0.81 + 0.62 + 10.9 + 0.45 = 12.78 Qc/t a = (Delta)T/ R = 50*/ 12.78 = 3.91 BTU/ hft^2 Find the total R- value. R = 0.81 + 0.62 + 10.9 + 0.45 = 12.78 Qc/ (ta) = 1/R (delta)t = 50*F/ 12.78 = 3.91 BTU/ (hft^2) 18) The following situation (As dealt with by one of the authors) can illustrate the economics of insulating the floor above a vented crawl space.

a) If the composite structure of the floor is made up of carpet (with an R-value of ½ in. ft^2-h-*F/Btu), subfloor (R= 1.0), and air space between the joists (R=0.8), then find the total R - value. Add the R values. R = 0.5 + 1 + 0.8 = 2.3 b) If 6 in. of fiberglass (R= 19) is added between the joists, then find the percentage reduction in the heat loss. R value.. R= 2.3 + 19 = 21.3 The heat loss depends on 1/R, so the % reduction is given by (1/2.3 – 1/21.3)/(1/2.3) x 100% = 89.2% c) If there are 6500 degree – days in this area and the price of fuel is $10 per million Btu, then find the heating cost per square food per heating season for the uninsulated floor. Qc/A = 1/R (Delta T) t = (1/2.3)(6500 degree-day)(24 h/day) = 67830 Btu/ft^2 each season 67830 Btu/ft^2 x $10/ (1,000,000 Btu) = $0.678/ft^2 each season D) If 6 -in. fiberglass costs $0.40 per square foot, what will be the payback time (As a result of energy savings) on this installation? The insulation reduces heat loss by 89.2% so the new cost is $0.678/ ft^2 x(1 – 0.892) = $0.073/ ft^2 Payback will occur after $0.40/ $0.605 = 66% of a heating season has passed.

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