MODULE 2 Introduction to Mechanical and Electrical Systems:
Energy, Sustainability, and Economics
The calculation examples are shown in an accordion-style display. Some questions are accompanied by equations and/or tables for quick reference. Solve the problem on your own and then click on the question to view the solution. This will help you make sense of the work as you read through the problem-solving process.
1.
How much heat is required to warm 1,500 gallons of water from 70°F to 140°F?
Specific heat of water is 1.0 Btu per deg. F per lb. One thousand and five hundred gallons of water weighs 12,525 at 8,350 pounds per gallon. Temperature rise is 70°F. Heat requirement will be 12,525 x 70, or 877,000 Btu.
2.
A heating system load is 300,000 Btuh. How much heating water flow is required to satisfy the load if the system is designed for a 25°F temperature drop? How much flow for a 35°F drop?
Use equation 2-6. GPM = 300,000 / (500 x 25), or 24 GPM for 20 deg. drop. GPM = 300,000 / (500 x 30), or 17 GPM for 35 deg. drop. This question demonstrates that using higher design temperature differences will result in lower flows, which can mean smaller piping and less pumping power.
3.
Approximately how much steam flow would be required for a 350,000 Btuh heating load?
Use equation 2-8. Steam flow = 350,000/1000, or 350 pounds per hour. Most practitioners assume a value of 1000 Btu per pound of low-pressure steam.
4.
Your client desires an interior temperature of 74°F for the design of his HVAC system in St. Louis (close to the airport weather station). Assuming that a 97.5% design will be satisfactory, what is the design temperature difference?
From Table 2.2A, 97.5% design temperature is 6°F. Temperature difference is 74 minus 6, or 68°F for heating load estimate.
5.
What will be the U-factor for a wall constructed as follows?
- 6” thick concrete wall.
- The wall is finished on the interior with 1/2” drywall, adhesively applied to the surface.
- The drywall is applied over 1.5” furring.
- The furring space is filled with insulation at R-4 per inch.
Use Tables 2-3A, 2-4, and 2-5 for this exercise.
U-factor, 6" concrete wall:
Add 1/2" drywall @ r = .45. Total R increases to 1.76; U = 0.57.
Add furred air space @ 1.12. Total R increases to 2.88; U = 0.35. (Note that the short circuit through the furring
strip is ignored for simplicity.)
Fill air space with R4 per inch insulation (delete air space, add insulation). Total R increases to 7.76; U = 0.13.
6.
A proposed building design has 15,000 sq. ft. of glass with shading coefficient 0.6, equally distributed north, south, east and west. What is the total solar load in tons July 21 at 4:00 pm?
Glass areas for the four orientations will be 15,000/4 sq. ft., or 3,750 sq. ft. in each direction. The following calculations
are based on the equation Qsolar = A x SC x SHGF. SC is 0.6, SHGF are found in Table 2-15A.
North Qsolar = 3,750 x 0.6 x 28 = 63,000 Btuh
South Qsolar = 3,750 x 0.6 x 29 = 65,300 Btuh
East Qsolar = 3,750 x 0.6 x 26 = 58,500 Btuh
East Qsolar = 3,750 x 0.6x 216 = 486,000 Btuh
Total solar load = (63,000 + 65,300 + 58,500 + 486,000) = 673,000 Btuh, which is 673,000 Btuh / (12,000 Btuh/ton),
or 56 tons.