Calculate the HEAT LOSS of the given office building, Use the tables and the charts from...
Fantastic news! We've Found the answer you've been seeking!
Question:
Transcribed Image Text:
Calculate the HEAT LOSS of the given office building, Use the tables and the charts from the hand-outs on the referenced pages. DESIGN CONDITIONS Winter temperature is 10F (for location "O" in Figure 15.8). An indoor temperature of 70F is as- sumed; temperature difference (TD) is 60. PROJECT CONDITIONS Outdoor air: Estimate infiltration for the example office in Figure 15.11 at 0.75 air changes per hour in winter (0.5 air changes per hour in summer). Estimate ventilation rate at 15 CFM per person. Fresh air CFM due to infiltration is the largest value in winter (pp. 202-203). Glass: Double glass. U = (p. 200). Lighting: Do not take heat gain credit because lights will be off on cold nights. People: Estimate 160 occupants, but do not take heat gain credit because people are absent on cold nights. Ceiling-roof: Detail is shown in Figure 15.12. U = 1/R total = Wall: See detail in Figure 15.12. U 1/R total = (p. 200) QUANTITIES Refer again to Figures 15.11 and 15.12 for dimensions. Outdoor air: Building volume is air changes per hour, the expected infiltration is . Ventilation rate for 160 people at 15 CFM each is 2,400 CFM. Use the larger infiltration value for winter heat loss (pp. 202 and 203). Glass: Windows are 4 ft. high. Total window area is 2,400 sq. ft. Ceiling-roof: Area is half floor area (two floors), 10,800 sq. ft. Walls: Net wall is is Gross wall area . Add for soffits and sills: ft.; deduct for windows and doors Floor: Slab on grade area = 10,800 sq. ft. Slab edge: 494 linear feet. Doors: Two at 5'x7' = 70 sq. ft. (excluding fire exits they are counted as wall area). WINTER HEAT LOSS = BTUH Insert appropriate U values, constants, factors, and temperature difference; then complete calculations and total heat loss BTUH. Add 10% only if ducts are outside the conditioned space. Floor: Slab on grade. Factor is 2 BTUH per sq. ft. (p. 205). Slab edge: Not shown. No insulation; no heating ducts. U = (pp. 200 and 205). Doors: Single glass. U = (p. 200). Equipment: Do not take a heat gain credit for equipment because it does not operate on cold nights. Office Plan FIGURE 15.11 160 upants 150' 210' 2'-0" www www FL 11'-0" TYP. to. FLR. Roof 3" urethane on stl. deck Wall 4" brick, 2" air, space "gbd. sheathing, R-11 Insulation, 1/2" gbd. Windows 40" high, fixed dbl. glass Soffit & Sill Similar to walls Construction FIGURE 15.12 winter Fdb indoor Fdb outdoor TD Glass U value Fdb outdoor Lighting Fdb indoor People project name Design Conditions: summer location floor area sqft. calculated by date Project Conditions: winter Outdoor Air; calculate 1&2 below but use only the largest CFM value: 1. Infiltration based on air change rate_ 2. Ventilation based on CFM per person SC (Shading Coefficient) summer total watts operating at peak heat gain time _total occupants at peak heat gain time TD Ceiling-Roof U value color weight (lbs/sqft) Fwb outdoor Walls % RH indoor GD time of peak gain Floor Equipment Other U value U value, Door color weight (lbs/sqft) U value, Slab Edge U value _watts or hp, Appliances Item Quantities Winter Heat Loss = BTUH Summer Heat Gain = BTUH Outdoor Air winter CFM (1.1) (TD) = summer CFM Glass total sqft ( U ) ( TD) = (1.1) (TD) (0.68) GD): U) TD) = N sqft E sqft S sqft W sqft horiz sqft Lighting watts People ( SF) (___ SC) = ( SF) (_ SC) ( SF) (___ SC) SF)(_SC) _SF) (_SC) (3.4) = _sens) = latent) = = = _U) (_ TETD)= _U)(_TETD)= = Ceiling-Roof, sqft Walls sqft _U)____TD) _U)(__TD) = Floor bsmt sqft factor)= slab sqft factor) = crawl space sqft U) TD) = above grade sqft U)L TD)= Slab Edge linft U) TD) = Doors sqft _U)(__ TD) = Equipment watts hp (3.4)= (2500) Appliances Other Subtotals If ducts are outside the conditioned space add 10% TOTAL BTUH heat loss _U)(__TD) = = sensible) latent) = = heat gain Check; heat loss = 20-60 BTUH persqft. (south to north); heat gain = 15-60 BTUH per sqft. (north to south). Allow 4% of area served for fan rooms, and 2% of gross building area for central plant equipment. Calculate the HEAT GAIN of the given office building, Use the tables and the charts from the hand-outs on the referenced pages. DESIGN CONDITIONS Maps show a summer dry-bulb temperature of 95F and a wet-bulb temperature of 76F (Figures 15.17 and 15.18, location "O"). Desired indoor conditions are 75F and 50% relative humidity. Time of peak heat gain is estimated as 4 PM. Temperature difference (TD) 20. Grain difference (GD) = 40 (p. 216). See Figure 15.21 for the example office building plan. Refer to Figure 15.22 for construction details. PROJECT CONDITIONS Outdoor air: Select largest of infiltration at 0.5 air changes per hour or ventilation at 15 CFM per person (pp. 202 and 203). Glass: Double glass U = The SC (shading coefficient) for clear double glass is (p. 215). Lighting: Estimate high, 3 watts per sq. ft. of floor area for this office occupancy, and revise as necessary after completing lighting design. Office building lighting installations can range from 1 to 5 watts per sq. ft. depending on luminous intensity, lamp type, and fixture design (pp. 215-216). People: 160 occupants seated doing light work. Ceiling-roof: U = 0.05; weight is 7 pounds per sq. ft.; color is light; ETD = 36 (p. 215). Walls: U = weight is 45 pounds per sq. ft.; color is dark. The west wall is hot at 4 PM; its ETD is 40; the weighted average ETD for all walls at 4 PM is 23 (p. 215). Doors: U = (tempered single glass). Solar gain for doors is included in Glass. Equipment: Allow 1 watt per sq. ft. for electrical equipment operating at the time of peak heat gain. Also allow 10 hp for air handler fans. QUANTITIES Outdoor air: Use largest value. Infiltration at 0.5 air changes per hour = 1,980 CFM. (21,600) (11) (0.5) 60 = 1,980. Ventilation at 15 CFM per person = 2,400 CFM. (160) (15) = 2,400 (pp. 202 and 203) Glass: 1,030 sq. ft. face north (includes 70 sq. ft. of glass door area); 1,440 sq. ft. face south. Lighting: Totals 64.800 watts at 3 watts per sq. ft. People: Each occupant produces 250 BTUH sensible plus 150 BTUH latent (p. 218). Ceiling-roof: Area is 10,800 sq. ft. Wall: Net wall area is Doors: Area 70 sq. ft. Equipment: Allow 21,600 watts at 1 watt per sq. ft.; plus 10 hp for air handler fans. SUMMER HEAT GAIN = BTUH Insert appropriate factors and temperature differ- ence; then complete calculations and total BTUH. Add 10% if ducts are located outside the conditioned zone. Refer back to the heat loss example on page 000 for more detail. Office Plan FIGURE 15.11 160 occupants 150' 210' 60: N 2'-0" www www 11'-0" TYP. FLR to. FLR. Roof 3" urethane on stl. deck Wall 4" brick, 2" air space, "gbd. sheathing, R-11 Insulation, 1/2" gbd. Windows 40" high, fixed dbl. glass Soffit Sill Similar to walls Construction FIGURE 15.12 winter Fdb indoor Fdb outdoor TD Glass U value Fdb outdoor Lighting Fdb indoor People project name Design Conditions: summer location floor area sqft. calculated by date Project Conditions: Outdoor Air; calculate 1&2 below but use only the largest CFM value: 1. Infiltration based on air change rate_ 2. Ventilation based on CFM per person winter SC (Shading Coefficient) summer total watts operating at peak heat gain time _total occupants at peak heat gain time TD Ceiling-Roof color Fwb outdoor Walls color % RH indoor GD Floor Equipment U value U value U value, Door U value, Slab Edge U value _watts or hp, Appliances weight (lbs/sqft) weight (lbs/sqft) time of peak gain Other Item Quantities Winter Heat Loss = BTUH Summer Heat Gain = BTUH Outdoor Air winter CFM (1.1)TD) = summer CFM Glass total sqft ( U) (___ TD) = N sqft E sqft S sqft W sqft horiz sqft Lighting watts People (1.1) (TD) = (0.68) GD) = U) TD) SF)(SC) = = =30 ( SF) (_ SC) _SF)(__SC) SF) (SC) = _SF) (SC) (3.4) = sens) latent) = _U)(__ TETD)= _U)_TETD)= = = Ceiling-Roof. sqft Walls sqft Floor bsmt sqft slab sqft crawl space sqft above grade sqft Slab Edge linft L _U) (__ TD) _U)____TD) factor) _factor) = = U) TD) _U) (TD) = = U)L TD) = = Doors sqft _U) (TD) = _U)(__TD) Equipment watts hp (3.4) = (2500) = Appliances sensible) _latent) = Other Subtotals If ducts are outside the conditioned space add 10% TOTAL BTUH heat loss heat gain = Check; heat loss = 20-60 BTUH persqft. (south to north); heat gain = 15-60 BTUH per sqft. (north to south). Allow 4% of area served for fan rooms, and 2% of gross building area for central plant equipment. Calculate the HEAT LOSS of the given office building, Use the tables and the charts from the hand-outs on the referenced pages. DESIGN CONDITIONS Winter temperature is 10F (for location "O" in Figure 15.8). An indoor temperature of 70F is as- sumed; temperature difference (TD) is 60. PROJECT CONDITIONS Outdoor air: Estimate infiltration for the example office in Figure 15.11 at 0.75 air changes per hour in winter (0.5 air changes per hour in summer). Estimate ventilation rate at 15 CFM per person. Fresh air CFM due to infiltration is the largest value in winter (pp. 202-203). Glass: Double glass. U = (p. 200). Lighting: Do not take heat gain credit because lights will be off on cold nights. People: Estimate 160 occupants, but do not take heat gain credit because people are absent on cold nights. Ceiling-roof: Detail is shown in Figure 15.12. U = 1/R total = Wall: See detail in Figure 15.12. U 1/R total = (p. 200) QUANTITIES Refer again to Figures 15.11 and 15.12 for dimensions. Outdoor air: Building volume is air changes per hour, the expected infiltration is . Ventilation rate for 160 people at 15 CFM each is 2,400 CFM. Use the larger infiltration value for winter heat loss (pp. 202 and 203). Glass: Windows are 4 ft. high. Total window area is 2,400 sq. ft. Ceiling-roof: Area is half floor area (two floors), 10,800 sq. ft. Walls: Net wall is is Gross wall area . Add for soffits and sills: ft.; deduct for windows and doors Floor: Slab on grade area = 10,800 sq. ft. Slab edge: 494 linear feet. Doors: Two at 5'x7' = 70 sq. ft. (excluding fire exits they are counted as wall area). WINTER HEAT LOSS = BTUH Insert appropriate U values, constants, factors, and temperature difference; then complete calculations and total heat loss BTUH. Add 10% only if ducts are outside the conditioned space. Floor: Slab on grade. Factor is 2 BTUH per sq. ft. (p. 205). Slab edge: Not shown. No insulation; no heating ducts. U = (pp. 200 and 205). Doors: Single glass. U = (p. 200). Equipment: Do not take a heat gain credit for equipment because it does not operate on cold nights. Office Plan FIGURE 15.11 160 upants 150' 210' 2'-0" www www FL 11'-0" TYP. to. FLR. Roof 3" urethane on stl. deck Wall 4" brick, 2" air, space "gbd. sheathing, R-11 Insulation, 1/2" gbd. Windows 40" high, fixed dbl. glass Soffit & Sill Similar to walls Construction FIGURE 15.12 winter Fdb indoor Fdb outdoor TD Glass U value Fdb outdoor Lighting Fdb indoor People project name Design Conditions: summer location floor area sqft. calculated by date Project Conditions: winter Outdoor Air; calculate 1&2 below but use only the largest CFM value: 1. Infiltration based on air change rate_ 2. Ventilation based on CFM per person SC (Shading Coefficient) summer total watts operating at peak heat gain time _total occupants at peak heat gain time TD Ceiling-Roof U value color weight (lbs/sqft) Fwb outdoor Walls % RH indoor GD time of peak gain Floor Equipment Other U value U value, Door color weight (lbs/sqft) U value, Slab Edge U value _watts or hp, Appliances Item Quantities Winter Heat Loss = BTUH Summer Heat Gain = BTUH Outdoor Air winter CFM (1.1) (TD) = summer CFM Glass total sqft ( U ) ( TD) = (1.1) (TD) (0.68) GD): U) TD) = N sqft E sqft S sqft W sqft horiz sqft Lighting watts People ( SF) (___ SC) = ( SF) (_ SC) ( SF) (___ SC) SF)(_SC) _SF) (_SC) (3.4) = _sens) = latent) = = = _U) (_ TETD)= _U)(_TETD)= = Ceiling-Roof, sqft Walls sqft _U)____TD) _U)(__TD) = Floor bsmt sqft factor)= slab sqft factor) = crawl space sqft U) TD) = above grade sqft U)L TD)= Slab Edge linft U) TD) = Doors sqft _U)(__ TD) = Equipment watts hp (3.4)= (2500) Appliances Other Subtotals If ducts are outside the conditioned space add 10% TOTAL BTUH heat loss _U)(__TD) = = sensible) latent) = = heat gain Check; heat loss = 20-60 BTUH persqft. (south to north); heat gain = 15-60 BTUH per sqft. (north to south). Allow 4% of area served for fan rooms, and 2% of gross building area for central plant equipment. Calculate the HEAT GAIN of the given office building, Use the tables and the charts from the hand-outs on the referenced pages. DESIGN CONDITIONS Maps show a summer dry-bulb temperature of 95F and a wet-bulb temperature of 76F (Figures 15.17 and 15.18, location "O"). Desired indoor conditions are 75F and 50% relative humidity. Time of peak heat gain is estimated as 4 PM. Temperature difference (TD) 20. Grain difference (GD) = 40 (p. 216). See Figure 15.21 for the example office building plan. Refer to Figure 15.22 for construction details. PROJECT CONDITIONS Outdoor air: Select largest of infiltration at 0.5 air changes per hour or ventilation at 15 CFM per person (pp. 202 and 203). Glass: Double glass U = The SC (shading coefficient) for clear double glass is (p. 215). Lighting: Estimate high, 3 watts per sq. ft. of floor area for this office occupancy, and revise as necessary after completing lighting design. Office building lighting installations can range from 1 to 5 watts per sq. ft. depending on luminous intensity, lamp type, and fixture design (pp. 215-216). People: 160 occupants seated doing light work. Ceiling-roof: U = 0.05; weight is 7 pounds per sq. ft.; color is light; ETD = 36 (p. 215). Walls: U = weight is 45 pounds per sq. ft.; color is dark. The west wall is hot at 4 PM; its ETD is 40; the weighted average ETD for all walls at 4 PM is 23 (p. 215). Doors: U = (tempered single glass). Solar gain for doors is included in Glass. Equipment: Allow 1 watt per sq. ft. for electrical equipment operating at the time of peak heat gain. Also allow 10 hp for air handler fans. QUANTITIES Outdoor air: Use largest value. Infiltration at 0.5 air changes per hour = 1,980 CFM. (21,600) (11) (0.5) 60 = 1,980. Ventilation at 15 CFM per person = 2,400 CFM. (160) (15) = 2,400 (pp. 202 and 203) Glass: 1,030 sq. ft. face north (includes 70 sq. ft. of glass door area); 1,440 sq. ft. face south. Lighting: Totals 64.800 watts at 3 watts per sq. ft. People: Each occupant produces 250 BTUH sensible plus 150 BTUH latent (p. 218). Ceiling-roof: Area is 10,800 sq. ft. Wall: Net wall area is Doors: Area 70 sq. ft. Equipment: Allow 21,600 watts at 1 watt per sq. ft.; plus 10 hp for air handler fans. SUMMER HEAT GAIN = BTUH Insert appropriate factors and temperature differ- ence; then complete calculations and total BTUH. Add 10% if ducts are located outside the conditioned zone. Refer back to the heat loss example on page 000 for more detail. Office Plan FIGURE 15.11 160 occupants 150' 210' 60: N 2'-0" www www 11'-0" TYP. FLR to. FLR. Roof 3" urethane on stl. deck Wall 4" brick, 2" air space, "gbd. sheathing, R-11 Insulation, 1/2" gbd. Windows 40" high, fixed dbl. glass Soffit Sill Similar to walls Construction FIGURE 15.12 winter Fdb indoor Fdb outdoor TD Glass U value Fdb outdoor Lighting Fdb indoor People project name Design Conditions: summer location floor area sqft. calculated by date Project Conditions: Outdoor Air; calculate 1&2 below but use only the largest CFM value: 1. Infiltration based on air change rate_ 2. Ventilation based on CFM per person winter SC (Shading Coefficient) summer total watts operating at peak heat gain time _total occupants at peak heat gain time TD Ceiling-Roof color Fwb outdoor Walls color % RH indoor GD Floor Equipment U value U value U value, Door U value, Slab Edge U value _watts or hp, Appliances weight (lbs/sqft) weight (lbs/sqft) time of peak gain Other Item Quantities Winter Heat Loss = BTUH Summer Heat Gain = BTUH Outdoor Air winter CFM (1.1)TD) = summer CFM Glass total sqft ( U) (___ TD) = N sqft E sqft S sqft W sqft horiz sqft Lighting watts People (1.1) (TD) = (0.68) GD) = U) TD) SF)(SC) = = =30 ( SF) (_ SC) _SF)(__SC) SF) (SC) = _SF) (SC) (3.4) = sens) latent) = _U)(__ TETD)= _U)_TETD)= = = Ceiling-Roof. sqft Walls sqft Floor bsmt sqft slab sqft crawl space sqft above grade sqft Slab Edge linft L _U) (__ TD) _U)____TD) factor) _factor) = = U) TD) _U) (TD) = = U)L TD) = = Doors sqft _U) (TD) = _U)(__TD) Equipment watts hp (3.4) = (2500) = Appliances sensible) _latent) = Other Subtotals If ducts are outside the conditioned space add 10% TOTAL BTUH heat loss heat gain = Check; heat loss = 20-60 BTUH persqft. (south to north); heat gain = 15-60 BTUH per sqft. (north to south). Allow 4% of area served for fan rooms, and 2% of gross building area for central plant equipment.
Expert Answer:
Related Book For
Fundamentals of Heat and Mass Transfer
ISBN: 978-0471457282
6th Edition
Authors: Incropera, Dewitt, Bergman, Lavine
Posted Date:
Students also viewed these mechanical engineering questions
-
Managing Scope Changes Case Study Scope changes on a project can occur regardless of how well the project is planned or executed. Scope changes can be the result of something that was omitted during...
-
Shortly after Murray began working in the tax department of the public accounting firm of Dewey, Cheatham, and Howe, he was preparing a tax return and discovered an error in last year's work papers....
-
When an interrupt or a system call transfers control to the operating system, a kernel stack area separate from the stack of the interrupted process is generally used. Why?
-
Express the equation y = x + 4 as a function machine. Here, whatever value x is given, 4 is added to it to produce the y-value. Therefore x is the input and y the output. As a function machine this...
-
Consider the 2016 major league baseball data in Table B.22. While team ERA was useful in predicting the number of games that a team wins, there are some other measures of team performance, including...
-
Media outlets such as ESPN and Fox Sports often have Web sites that provide indepth coverage of news and events. Portions of these Web sites are restricted to members who pay a monthly subscription...
-
1. Let f (t) = t 4t - (t - 5t + 6) u (t) (t 6) u(t) a. (2 points) Write f(t) as a piecewise-defined function, that is, f(t) = b. (2 points) On the axes, sketch the graph of f(t) f(t) + 2 3 4 5 6 7 8...
-
Ron Carver created a nongrantor trust in 2016. During 2020, the trust earned $4,000 in ordinary dividends, $1,600 in interest income, and $2,400 in tax-exempt municipal bond interest. The only...
-
The electromagnetic wave travels 5 m and its magnitude is reduced by a factor of 10 dB. Calculate the attenuation coefficient of this wave.
-
An 85.0kg man is trying to push over a wall. The man pushes on the wall with a force of 1200N for 12s but the wall does not move. How much work does he do on the wall?
-
1. What was Dam Blue Stuff? 2. How did the area get the name Comstock Load? 3. Who invented square formwork? 4. Did the inventor ever get a patient? 5. Who was William Sutro?
-
An Olympian archer does 40 joules of work to pull her arrow back to be ready to shoot. What would be the ideal kinetic energy of the arrow when it is launched?
-
Implementation is often a complicated process. How do unclear objectives impact implementation? Discuss THREE obstacles that agencies face when implementing policy. Why are they problematic? Which...
-
Imagine you have been assigned the responsibility of preparing a paper for the governor's next economic conference. Prepare a paper addressing the following: Explain why equilibrium of supply and...
-
1A. If the researcher is concerned about the number of variables, the nature of the analysis, and completion rates, then, he/she is at which stage of the sampling design process (Figure 11.1 in the...
-
Using an economic balance sheet, which of the Laws current financial assets is most concerning from an asset allocation perspective? A. Equities B. Real estate C. Fixed income Raye uses a costbenefit...
-
Using the economic balance sheet approach, the Laws economic net worth is closest to: A. $925,000. B. $1,425,000. C. $1,675,000. Raye uses a costbenefit approach to rebalancing and recommends that...
-
Raye believes the previous advisers asset class specifications for equity and derivatives are inappropriate given that, for purposes of asset allocation, asset classes should be: A. diversifying. B....
Study smarter with the SolutionInn App