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LECTURE 1- 1) Question 1 – Determine the July electric utility bill for a facility that used 112000 KWh during the month, and had a maximum power usage of 500 KW, during the peak period of time in that month. The utility has a fixed meter charge of $75 / month, and charges a flat rate of 5 cents/ KWh for energy and $12/KW for a maximum power usage during peak period in July Ans – Monthly Bill => Fixed Monthly + Rate of energy* Energy produced + Rate of max power * Max Power = 75+0.05*112000+12*500 = $11,675 2) Question 2 – Determine the rate at which heat must be added in Btu/hr to a 300 cfm airstream passing through a heating coil to change its temperature from 70 F to 120 F. Assume an inlet air at a specific volume of 13.5 ft 3 /lbm and a specific heat of 0.24 Btu/lbm.F The heat being added is sensible, as its contributing to the temp change of the airstream. Ans – q s = (Q*Cp)/V*(T e -T i ) = (300 cfm * 60 m/hr *0.24)/13.5*(120-70) = 16000 Btu/hr 3) Question 3 – Air at 1 atm and 76 F is flowing at the rate of 5000 cfm. At what rate must energy be removed, in Btu/hr, to change the temperature to 58 F assuming no dehumidification occurs? @ 76 F, density = 0.074 lbm/ft 3 C p = 0.34 Btu/lbmF Ans – q = mc p (T 2 -T 1 ) => m = density * Q = 0.074*5000*60 = 22200 lbm/hr q = 22200*0.24(58-76) = -95904 Btu/hr 4) Question 4 – Determine the operative temp. for a workstation in a room near a large window where the dry bulb temp. (temp of air) and globe temp. are measured to be 75 F and 81 F respectively. The air velocity is estimated to be 30 ft/min at the station. Ans – Operative temp depends on the T mrt (Mean radiant temperature) T mrt = (T g 4 + CV 1/2 * (T g – T a ) ) 1/4 = ((81+460) 4 + 0.103x10 9 * (30) 1/2 (81-75)) 1/4 = 546 R = 86 F Operating temp = (T mrt + T a ) /2 = (86 + 75)/2 = 80.5 F

5) Question 5 – A person breathes out CO 2 at a rate of 0.3 L/min, the concentration of CO 2 in the incoming ventilation air is 300 ppm (Parts per minute). It is desired to hold the concentration in the room below 1000 ppm. Assume that the air in the room is perfectly mixed. What is the minimum rate of air flow required to maintain the desired level. Ans - Q t C e + N = Q t C s => N = Q t C s - Q t C e Q t = N/(C s -C e ) = 0.3/(0.001-0.0003) = 7.1 L/s 6) Question 6 – Carbon Dioxide is being generated in an occupied space at the rate 0.25 cfm (0.1181 m 3 /s) and outdoor air with CO 2 concentration of 200 ppm is being supplied to the space at the rate of 900 cfm (0.425 m 3 /s). What will the steady-state concertation of CO 2 be in ppm if complete mixing is assumed? Ans - Q t C e + N = Q t C s => N = Q t C s - Q t C e C s = (Q t C e + N)/Q t = (900*200+0.25)/900 = 478 ppm ---------------------------------------------------------EOL------------------------------------------------------------ LECTURE 2 - 1) Question 1 - Find the heat transfer rate required to warm 1500 cfm (ft 3 /min) of air at 60 F and 90% relative humidity to 110 F without the addition of moisture. Ans – Q = 1500 cfm, State 1 - T1 = 60 F,  = 90% State 2 - T2 = 110 F c p = 0.244 From psychometric chart : - Specific volume = 13.3 ft 3 /lbm - Enthalpy of state 1 - i 1 or h 1 = 25.1 Btu/lbm - Enthalpy of State 2 - i 1 or h 1 = 37.4 Btu/lbm m a = (1500*60)/13.3 = 6752 Q = m a (i 1 -i 2 ) = m a c p (t 1 -t 2 ) = 6752*0.244(110-60) = 82.374 Btu/hr

2) Question 2 - 2000 cfm of air at 100 F db and 75 F wb are mixed with 1000 cfm of air at 60 F db and 50 F wb. The process is adiabatic, at a steady flow rate and at standard sea-level pressure. Find the condition of the mixed stream using the lever rule. State 1- DB = 60 F, WB = 50 F State 2- DB = 100 F, WB = 75 F From Chart: - Specific volume of state 1= 13.21 ft 3 /lbm - Specific volume of state 2= 14.4 ft 3 /lbm - Humidity rate of state 1  1 = 0.0054 lbm/lb - Humidity rate of state 2  2 = 0.013 lbm/lb m a1 = 1000*60/13.21 = 4511.28 m a2 = 2000*60/14.40 = 8333.33  3 =  1 + m a1 / m a2 = 0.0054 + 4511.28/8333.33 = 0.54 ---------------------------------------------------------EOL------------------------------------------------------------ LECTURE 3- 1) Question 1 – Q.5-4 What is the unit thermal resistance for an inside partition made up of 3/8 in. gypsum board on each side of 8 in. lightweight aggregate blocks with vermiculite-filled cores? Ans – R gypsum = 1/C = 1/3.1 = 0.32 (C found from table 5-1a in pg. 11 of lecture 3 slides) R air = 0.68 (Table 5-2a pg. 14 lecture 3 slides) R vermiculite = 1/C = 1/0.33 (C from table 5-1a pg.12 lecture 13 slides) R = 0.68+0.32+0.30+0.32+0.68 = 5.03 2) Question 2 - Q.5-5 Compute the thermal resistance per unit length for a 4 in. schedule 40 steel pipe with 1½ in. of insulation. The insulation has a thermal conductivity of 0.2 Btu-in./(hr-ft 2 -F). Ans - R = R steel + R ins = ln ( r 2 r 1 ) 2 π ∗ Lk + ln ( r 3 r 2 ) 2 π ∗ Lk  d 2 = Outer diameter = 4.5 in, r 2 = Outer Radius = 2.25 in (Table  d 1 = Inner diameter = 4.026 in, r 1 = Inner Radius = 2.013 in  r 3 = Outer Radius = r 2 + 1.5 = 3.75 in  k steel = 314 Btu-in./(hr-ft 2 -F). Vermiculite

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R = ln ( 2.25 2.013 ) 2 π ∗ 314 + ln ( 3.75 2.25 ) 2 π ∗ 0.2 = 0.4065 °F·ft²·hr/Btu. 3) Question 3 - Q.5-9 Compute the overall thermal resistance of a wall made up of 100 mm brick and 200 mm normal weight concrete block with a 20 mm air gap in between. There is 13 mm of gypsum plaster on the inside. Assume a 7 m/s wind velocity on the outside and still air inside. Ans – R mov-air = 0.029 m 2 C/W (table 5.2) R brick =  x/k brick = 0.1m/0.895 (table 5.1) = 0.112 m 2 C/W R air-gap = 0.18 m 2 C/W (Table 5.3) R conc = 1/C conc = 1/5.45 (table 5.1) = 0.183 m 2 C/W R gyp = 1/C = 1/12.6 = 0.0793 m 2 C/W R still-air = 0.12 m 2 C/W (table 5.2) R Total = R mov-air + R brick + R air-gap + R conc + R gyp + R still-air = 0.7033 m 2 C/W 4) Question 4 - Q.5-8 The pipe of question 2 has water flowing inside with a heat- transfer coefficient of 650 Btu/(hr-ft 2 -F) and is exposed to air on the outside with a film coefficient of 1.5 Btu/(hr-ft 2 -F). Compute the overall heat-transfer coefficient based on the outer area. Ans – From Question 4 = R steel + R ins = 0.4065 °F·ft²·hr/Btu. R total_new = 1 2 π ∗ r 1 ∗ Lh water + R steel + R ins + 1 2 π ∗ r 3 ∗ Lh air = 1 2 π ∗ 2.013 ∗ 1 ∗ 650 + 0.4065 + 1 2 π ∗ 3.75 ∗ 1 ∗ 1.5 = 0.4349 °F·ft²·hr/Btu. U = A 3 R total = 2 π ∗ r 3 R total = 2 π ( 3.75 ) 0.4349 = 54.17 Btu . / ° F·ft ² ·hr

5) Question 5 - Q.5-16 A wall is 20 ft wide and 8 ft high and has an overall heat-transfer coefficient of 0.07 Btu/(hr-ft 2 -F). It contains a solid urethane foam core steel door, 80 x 32 x 1¾ in., and a double glass window,120 x 30 in. The window is metal sash with no thermal break [U = 0.81 Btu/(hr-ft 2 -F)]. Assuming parallel heat-flow paths for the wall, door, and window, find the overall thermal resistance and overall heat-transfer coefficient for the combination with respect to the wall area. Assume winter conditions. Ans – A door = 80 x 32 12 2 = 17.7 ft 2 A window = 120 x 30 12 2 = 25 ft 2 A wall = 20 * 8 – 17.7 – 25 = 117.22 ft 2 U door = 0.4 Btu/(hr-ft 2 -F) (table 5-8) U = U wall A wall + U door A door + U window A window width ∗ height = 0.07 ∗ 117.22 + 17.77 ∗ 0.4 + 25 ∗ 0.81 20 ∗ 8 ¿ 0.222 Btu /( hr − ft 2 F ) R = 1/U = 4.5 °F·ft²·hr/Btu. ---------------------------------------------------------EOL------------------------------------------------------------ LECTURE 4 – 1) Question 1 - Q.6-13 An exposed wall in a building in Memphis, TN, has dimensions of 10 × 40 ft (3 × 12 m) with six 3 × 3 ft (0.9 × 0.9 m) windows of regular double glass, ½ in. air space in an aluminium frame without a thermal break. The wall is made of 4 in. (10 cm) lightweight concrete block and face brick. The block is painted on the inside. There is a ¾ in. (2 cm) air space between the block and brick. Estimate the heat loss for the wall and glass combination. Ans – ASHRAE – Memephis temp => t i = 70 F, t o = 21 F q = UA(t i – t o ) Windows : Area of single Window = 3x3 = 9 ft 2 Area of 6 windows = 6x9 = 54 ft 2 U = 0.81 Btu /( hr − ft 2 F ) (Table 5.5a) q window = 0.81 x 54 x (71 – 20) = 2143.26 Btu/hr Wall : Area of Wall = 10x40 = 400 ft 2 Area of wall removing windows = 400-54 = 346 ft 2 Resistance diagram:

R 1 = R INSIDE = 0.92 °F·ft²·hr/Btu. (Table 5.2a, horizontal orientation) R 2 = R PAINT = Negligible R 3 = R BLOCK = 1/0.565 = 1.77 °F·ft²·hr/Btu. R 4 = R AIR-GAP = 0.99 °F·ft²·hr/Btu. (Table 5-3a,  = 0.5, horizontal orientation, mean temp 50, temp diff 30 R 5 = R BRICK = 1.55 R 6 = R OUTSIDE = 0.17 °F·ft²·hr/Btu. R TOTAL = 0.92+1.77+0.99+1.55+0.17 = 5.4 °F·ft²·hr/Btu. U=1/R = 1/5.4 = 0.185 Btu/(hr-ft 2 -F) q wall = 0.185 x 346 x(70-21) = 3139.88 Btu/hr q total = q wall + q window = 3139.88 + 2143.26 = 5283.13 Btu/hr 2) Question 2 – Q.6-3 A large single-story business office is fitted with nine loose-fitting, double-hung wood sash windows 3 ft wide by 5 ft high. If the outside wind is 15 mph at a temperature of 0 F, what is the percent reduction in sensible heat loss if the windows are weather stripped? Assume an inside temperature of 70 F. Base your solution on a quartering wind. Ans – ∆ P w C p = ρ 2 ∗ g c ∗ V w 2 Cp = 0.35 (figure 6.2, 45 ° angle) ρ = 0.086 lbm ft 3 g c = 32.17 lbm − ft / lbf − s ∆ P w = 0.35 ∗ 0.086 2 ∗ 32.17 ∗ ( 15 mph ∗ 1.467 ft s mph 2 ) ∗ ( 0.1924 ∈ .of H 2 O 1 lbf ft 2 ) ∆ P w = 0.038 ∈ .of H 2 O Length of Window = (3x2+5x2)x9 windows = 144 ft K = 2.0 (Table 6.1, weather-stripped, loose-fitted window) Q/L = 0.23 (Figure 6.1, Pw = 0.038 in. of H 2 O, K=2.0) Q 1 = 0.23*144/2 = 16.56 cfm (divided by 2 due to double hang) K = 6.0 (Table 6.1, nonweather-stripped, loose-fitted window)

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Q/L = 0.75 (Figure 6.1, Pw = 0.038 in. of H 2 O, K=6.0) Q 1 = 0.75*144/2 = 54 cfm (divided by 2 due to double hang) Percentagereduction = Q 2 − Q 1 Q 2 ∗ 100 = 54 − 16.56 54 ∗ 100 = 69.3 % ---------------------------------------------------------EOL------------------------------------------------------------ LECTURE 4 – 1) Question 1 – Q.7-15 Determine magnitudes of direct, diffuse, and reflected clear-day solar radiation incident on a small vertical surface facing south on March 21 at solar noon for a location at 56 deg N latitude having a clearness number of 0.95. The reflecting surface is snow-covered ground of infinite extent with a diffuse reflectance of 0.7. Ans – Givens- Date: March 21 Latitude: l = 56 deg N Equation of time: -7.5 min (Table 7.2) Declination:  = 0 (Table 7.2) Solar Parameters: A= 1164 W/m 2 , B = 0.149, C = 0.109 (Table 7.2) Local Solar Time: LST = 12:00 pm Surface tilt:  = 90 deg (Vertical surface facing north) Surface Azimuth:  = 180 deg (Vertical surface facing north) Clearance Number: C N = 0.95 Diffuse Reflectance:  = 0.7 Hour Angle: h = 0 deg (noon) Solar Altitude Angle: sin(  ) = cos(l)*cos(h)*cos(  )+sin(l)*sin(  )  = sin -1 (cos(54)*cos(0)*cos(0)+sin(54)*sin(0)) = 34 deg Solar Azimuth :cos ( ϕ ) = sin ()∗ cos ( l ) − cos ()∗ sin ( l ) ∗ cos ( h ) cos () = ϕ = cos − 1 ( sin ( 0 ) ∗ cos ( 54 ) − cos ( 0 ) ∗ sin ( 54 ) ∗ cos ( 0 ) cos ( 34 ) ) = 180 Surface Solar Azimuth : γ = | ϕ − ψ | = | 180 − 180 | = 0 deg Angle of incident: cos(  ) = cos(  )*cos( γ )*sin(  )+sin(  )*cos(  )  = cos -1 (cos(34)*cos( 0 )*sin(90)+sin(34)*cos(90)) = 34 deg Normal Direct Irradiation: G ND = A e ( B sin () ) ∗ C N = 1164 e ( 0.149 sin ( 34 ) ) ∗ 0.95 = 847.142 W / m 2 Direct Radiation: G D = G ND *cos(  ) = 847.142*cos(34) = 702.31 W/m 2

Ratio of diffuse sky radiation on a vertical surface to that incident on a horizontal surface: G dV G dH = 0.55 + 0.437cos ()+ 0.313cos 2 ()= 0.55 + 0.437cos ( 34 ) + 0.313cos 2 ( 34 ) = 1.12 Diffuse Sky Radiation: G d  = G dV G dH ∗ C ∗ G ND = 1.12 ∗ 0.109 ∗ 847.142 = ¿ ¿ 103.42 W / m 2 Angle Factor from Wall to Ground: F wg = 1 − cos ( α ) 2 = 1 − cos ( 90 ) 2 = 0.5 Reflected Radiation: G R =  * F wg *(sin( ¿ + C)*G ND = 0.7*0.5*(sin(34)+0.109)*847.142 = 198.12 W/m 2