solucionario de stocker, Ejercicios de Ingeniería

¡Descarga solucionario de stocker y más Ejercicios en PDF de Ingeniería solo en Docsity! CHAPTER 2 - THERMAL PRINCIPLES Page 1 of 5 2-1. Water at 120 C and a pressure of 250 kPa passes through a pressure-reducing valve and then flows to a separating tank at standard atmospheric pressure of 101.3 kPa, as shown in Fig. 2-14. (a) What is the state of the water entering the valve (subcooled liquid, saturated liquid, or vapor)? (b) For each kilogram that enters the pressure-reducing valve, how much leaves the separating tank as vapor? Solution: (a) From Fig. 2-2, a temperature of 120 C and pressure of 250 kPa water lies in the sub-cooled regiom. so it is a sub- cooled liquid. (b) At 120 C, h 1 = 503.72 kJ/kg from Table A-1 For Pressuring Reducing Valve Dh = 0 h 2 = h 1 At 101.3 kPa, Table A-1, h f = 419.06 kJ/kg h g = 2676 kJ/kg Let x be the amount of vapor leaving the separating tank. h = h f + x(h g - h f ) 419.062676 419.06503.72 hh hh x fg f − − = − − = x = 0.0375 kg/kg - - - Ans. 2-2. Air flowing at a rate of 2.5 kg/s is heated in a heat exchanger from -10 to 30 C. What is the rate of heat transfer? Solution: q = mc p (t 2 - t 1 ) m = 2.5 kg/s c p = 1.0 kJ/kg.K t 2 = 30 C t 1 = -10 C Chapter2 1 CHAPTER 2 - THERMAL PRINCIPLES Page 2 of 5 Then, q = (2.5)(1.0)(30 + 10) q = 100 kw - - - Ans. 2-3. One instrument for measuring the rate of airflow is a venturi, as shown in Fig. 2-15, where the cross-sectional area is reduced and the pressure difference between position A and B measured. The flow rate of air having a density of 1.15 kg/m 3 is to be measured in a venturi where the area of position A is 0.5 m 2 and the area at b is 0.4 m 2 . The deflection of water (density = 1000 kg/m3) in a manometer is 20 mm. The flow between A and B can be considered to be frictionless so that Bernoulli’s equation applies. (a) What is the pressure difference between position A and B? (b) What is the airflow rate? Solution: (a) Bernoulli equation for manometer B B A A gz p gz p + ρ =+ ρ p A - p B = ρg(z B -z A ) z B - z A = 20 mm = 0.020 m g = 9.81 m/s 2 ρ = 1000 kg/m 3 p A - p B = (1000 kg/m 3 )(9.81 m/s 2 )(0.020 m) p A - p B = 196.2 Pa - - - Ans. (b) Bernoulli Equation for Venturi constant 2 Vp 2 =+ ρ Chapter2 2 CHAPTER 2 - THERMAL PRINCIPLES Page 5 of 5 Human Body: A = 1.5 to 2.5 m 2 use 1.5 m 2 t s = 31 to 33 C use 31 C C = (5.8762 W/m 2 .K)(1.5 m 2 )(31 C - 24 C) C = 61.7 W Order of Magnitude ~ 60 W - - - Ans. 2-7 What is the order of magnitude of radiant heat transfer from a human body in a comfort air-conditioning situation? Solution: Eq. 2-10. ( )4 2 4 1A21 TTFAFq −εσ=− Surface area of human body = 1.5 to 2.5 m 2 use 1.5 m 2 AF ε F A = (1.0)(0.70)(1.5 m 2 ) - 1.05 m 2 s = 5.669x10 -8 W/m 2 .K 4 T 1 =31 C + 273 = 304 K T 2 = 24 C + 273 = 297 K q 1-2 = (5.669x10 -8 )(1.05)(304 4 - 297 4 ) q 1-2 = 45 W Order of Magnitude ~ 40 W - - - Ans. 2-8. What is the approximate rate of heat loss due to insensible evaporation if the skin temperature is 32 C, the vapor pressure is 4750 Pa, and the vapor pressure of air is 1700 Pa? The latent heat of water is 2.43 MJ/kg; C diff = 1.2x10 -9 kg/Pa.s.m 2 . Solution: Equation 2-19. q ins = h fg AC dif f( p s - p a ) Where: A = 2.0 m 2 average for human body area h fg = 2.43 MJ/kg = 2,430,000 J/kg p s = 4750 Pa p a = 1700 Pa C diff = 1.2x10 -9 kg/Pa.s.m 2 q ins = (2,430,000)(2.0)(1.2x10 -9 )(4750 - 1700) q ins = 18 W - - - Ans. - 0 0 0 - Chapter2 5 CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER Page 1 of 9 3-1 Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air when the following conditions prevail: t = 30 C, W = 0.015 kg/kg, and p t = 90 kPa. Solution: Equation 3-4. st a a a pp TR p TR − ==ν T = 30 C + 273 = 303 K R a = 287 J/kg.K P t = 90 kPa = 90,000 Pa Equation 3-2 st s pp 0.622p W − = s s p90 0.622p 0.015 − = 1.35 - 0.15p s = 0.622p s p s = 2.1193 kPa ( )( ) 2119.390000 303287 pp TR st a − = − =ν ν = 0.99 m 3 /kg - - - Ans. 3-2. A sample of air has a dry-bulb temperature of 30 C and a wet-bulb temperature of 25 C. The barometric pressure is 101 kPa. Using steam tables and Eqs. (3-2), (303), and (3-5), calculate (a) the humidity ration if this air is adiabatically saturated, (b) the enthalpy of air if it is adiabatically saturated, (c) the humidity ratio of the sample using Eq. (3-5), (d) the partial pressure of water vapor in the sample, and (e) the relative humidity. Solution: Eq. 3-2. st s pp 0.622p W − = Eq. 3-3. h = c p t + Wh g Eq. 3-5 h 1 = h 2 - (W 2 - W 1 )h f h 1 = c p t 1 + Wh g1 h g1 at 30 C = 2556.4 kJ/kg t1 = 30 C cp = 1.0 kJ/kg.K h 1 = (1)(30) + 2556.4W 1 h 1 = 30 + 2556.4W 1 Chapter3 6 CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER Page 2 of 9 h 2 = c p t 2 + Wh g2 hg 2 at 25 C = 2547.3 kJ/kg t 2 = 25 C c p = 1.0 kJ/kg.K h 2 = (1)(25) + 2547.3W 2 h 2 = 25 + 2547.3W 2 h f at 25 C = 125.66 kJ/kg Then: h 1 = h 2 - (W 2 - W 1 )h f 30 + 2556.4W 1 = 25 + 2547.3W 2 - (W 2 - W 1 )(125.66) 5 = 2421.64W 2 - 2430.74W 1 But, st s 2 pp 0.622p W − = ps at 25 C = 3.171 kPa ( ) 3.171-101 3.1710.622 W2 = W 2 = 0.0201 kg/kg 5 = 2421.64(0.0201) - 2430.74W 1 W 1 = 0.018 kg/kg (a) Humidity Ratio W 2 = 0.0201 kg/kg - - - Ans. (b) h 2 = c p t 2 + W 2 hg 2 h 2 = (1)(25) + (0.0201)(2547.3) h 2 = 76.2 kJ/kg - - - Ans. (c) Humidity Ratio W 1 = 0.018 kg/kg - - - Ans. (d) p s1 st s 1 pp 0.622p W − = s s p101 0.622p 0 − =018. p s1 = 2.84 kPa p s1 = 2840 kPa - - - Ans. Chapter3 7 CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER Page 5 of 9 standard atmospheric pressure enters the unit. The leaving condition of the air is 13 C dry-bulb temperature and 90 percent relative humdity. Using properties from the psychrometric chart, (a) calculate the refrigerating capacity inkilowatts and (b) determine the rate of water removal from the air. Solution: At 27 C dry-buld, 5 Percent Relative Humidity h 1 = 55.311 kJ/kg, ν 1 = 0.86527 m 3 /kg W 1 = 0.0112 kg/kg At 13 C Dry-Bulb, 90 Percent Relative Humidity h 2 = 33.956 kJ/kg W 2 = 0.0084 kg/kg m = (3.5 m 3 /s)/(0.86526 m 3 /kg) = 4.04498 kg/s (a) Refrigerating Capacity = m(h 1 - h 2 ) = (4.04498)(55.311 - 33.956) = 86.38 kW - - - Ans. (b) Rate of Water Removal = m(W 1 - W 2 ) = (4.04498)(0.0112 - 0.0084) = 0.0113 kg/s - - - Ans. 3-7. A stream of outdoor air is mixed with a stream of return air in an air-conditioning system that operates at 101 kPa pressure. The flow rate of outdoor air is 2 kg/s, and its condition is 35 C dry-bulb temperature and 25 C wet-bulb temperature. The flow rate of return air is 3 kg/s, and its condition is 24 C and 50 percent relative humidity. Determine (a) the enthalpy of the mixture, (b) the humidity ratio of the mixture, (c) the dry-bulb temperature of the mixture from the properties determined in parts (a) and (b) and (d) the dry-bulb temperature by weighted average of the dry-bulb temperatures of the entering streams. Solutions: Use Fig. 3-1, Psychrometric Chart At 35 C Dry-Bulb, 24 C Wet-Bulb h 1 = 75.666 kJ/kg, m 1 = 2 kg/s W 1 = 0.0159 kg/kg At 24 C Dry-Bulb, 50 Percent Relative Humidity h 2 = 47.518 kJ/kg, m 2 = 3 kg/s W 2 = 0.0093 kg/kg (a) ( )( ) ( )( ) 32 47.518375.6662 hm + + = h m = 58.777 kJ/kg - - - Ans. Chapter3 10 CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER Page 6 of 9 (b) ( )( ) ( )( ) 32 0.009330.01592 Wm + + = W m = 0.1194 kg/kg - - - Ans. (c) At 58.777 kJ/kg and 0.01194 kg/kg. From Psychrometric Chart, Fig. 3-1. Dry-Bulb Temperature = 28.6 C - - - Ans. (d) ( )( ) ( )( ) 32 243352 tm + + = t m = 28.4 C - - - Ans. 3-8. The air conditions at the intake of an air compressor are 28 C, 50 percent relative humidity, and 101 kPa. The air is compressed to 400 kPa, then sent to an intercooler. If condensation of water vapor from the compressed air is to be prevented, what is the minimum temperature to which the air can be cooled in the intercooler? Solution: At 28 C, p s = 3.778 kPa At 50 percent relative humidity, p s = (0.5)(3.778 kPa) = 1.889 kPa st s pp 0.622p W − = Moisture ratio is constant at 101 kPa ( ) 1.889101 1.8890.622 W − = W = 0.011855 kg/kg at 400 kPa, determine p s s s p 0.622p 0.011855 − = 400 p s = 7.4812 kPa From Table A-1. Dew-Point = 40.3 C - - - Ans. Chapter3 11 CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER Page 7 of 9 3-9. A winter air-conditioning system adds for humidification 0.0025 kg/s of saturated steam at 101 kPa pressure to an airflow of 0.36 kg/s. The air is initially at a temperature of 15 C with a relative humidity of 20 percent. What are the dry- and wet-bulb temperatures of the air leaving the humidifier? Solution: At 15 C Dry-Bulb, 20 Percent Relative Humidity h 1 = 20.021 kJ/kg W 1 = 0.0021 kg/kg At 101 kPa steam, h fg = 2675.85 kJ/kg m s = 0.0025 kg/s m = 0.36 kg/s m s = m(W 2 - W 1 ) 0.0025 = 0.36(W 2 - 0.002) W 2 = 0.00894 kg/kg m(h 2 - h 1 ) = m s h g (0.36)(h2 - 20.021) = (0.0025)(2675.85) h 2 = 38.6 kJ/kg Fig. 3-1, Psychrometric Chart W 2 = 0.00894 kg/kg h 2 = 38.6 kJ/kg Dru-Bulb Temperature = 16.25 C Wet-Bulb Temperature = 13.89 C 3-10. Determine for the three cases listed below the magnitude in watts and the direction of transfer of sensible heat [ using Eq. (3-8)], latent heat [ using Eq. (3-9)], and total heat [ using Eq. (3-14)]. the area is 0.15 m 2 and h c = 30 W/m 2 .K. Air at 30 C and 50 percent relative humidity is in contact with water that is at a temperature of (a) 13 C, (b) 20 C, and (c) 28 C. Solution: Equation 3-8. dq s = h c dA(t i - t a ) Equation 3-9. dq L = h D dA(W i - W a )h fg Equarion 3-14. )h(h c dAh dq ai pm c t −= At 30 C, 50% Relative Humidity h a = 63.965 kJ/kg = 63,965 J/kg W a = 0.0134 kg/kg (a) 13 C dq s = h c dA(t i - t a ) dq s = (30)(0.15)(13 - 30) Chapter3 12 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 1 of 8 4-1 The exterior wall of a single-story office building near Chicago is 3 m high and 15 m long. The wall consists of 100-mm facebrick, 40-mm polystyrene insulating board. 150-mm lightweight concrete block, and an interior 16-mm gypsum board. The wall contains three single-glass windows 1.5 n high by 2 m long. Calculate the heat loss through the wall at design conditions if the inside temperature is 20 C. Solution: Table 4-3, Design Outdoor is -18 C for Chicago. For the wall: Area, A = (3 m)(15 m) - (3)(1.5 m)(2 m) = 36 m 2 . Resistance: Table 4-4. Outside Air Film 0.029 Facebrick, 100 mm 0.076 Polystyrene Insulating Board, 40 mm 1.108 Lightweight Concrete Block, 150 mm 0.291 Gypsum Board, 16 mm 0.100 Inside Air Film 0.120 ==== R tot = 1.724 m 2 .K/ W Wall: ( )2018 1.724 36 t Rtot A q −−=∆= q = -794 Watts For the glass: Area A = (3)(1.5 m0(2 m) = 9 m 2 Table 4-4, U = 6.2 W/m 2 .K q = UA∆t = (6.2)(9)(-18 - 20) q = -2,120 Watts Total Heat Loss Thru the Wall = -794 W -2,120 W = -2,194 Watts - - - Ans . 4-2. For the wall and conditions stated in Prob. 4-1 determine the percent reduction in heat loss through the wall if (a) the 40 mm of polystyrene insulation is replaced with 55 mm of cellular polyurethane, (b) the single-glazed windows are replaced with double-glazed windows with a 6-mm air space. (c) If you were to choose between modification (a) or (b) to upgrade the thermal resistance of the wall, which would you choose and why? Solution (a) Resistance: Table 4-4 Outside Air Film 0.029 Facebrick, 100 mm 0.076 Cellular Polyurethane, 55 mm 2.409 Lightweight Concrete Block, 150 mm 0.291 Gypsum Board, 16 mm 0.100 Inside Air Film 0.120 ===== R tot = 3.025 m 2 .K/W Chapter4 15 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 2 of 8 Wall: ( )2018 3.025 36 t Rtot A q −−=∆= q = - 452 Watts New Total Heat Loss Thru Wall q = - 452 W - 2,120 W q = - 2,572 W ( ) ( ) ( )%100 W2,914 W2,572W2,914 %Reduction − −−− = % Reduction = 11.74 %- - - Ans. (b) For the glass: (Double-Glazed) Table 4-4, U = 3.3 W/m 2 .K q = UA∆t = (3.3)(9)(-18 - 20) q = -1,129 Watts New Total Heat Loss Thru Wall q = - 794 W - 1,129 W ( ) ( ) ( )%100 W2,914 W,923W2,914 %Reduction − −−− = 1 % Reduction = 34 %- - - Ans. (c) Choose letter b --- Ans. 4-3 An office in Houston, Texas, is maintained at 25 C and 55 percent relative humidity. The average occupancy is five people, and there will be some smoking. Calculate the cooling load imposed by ventilation requirements at summer design conditions with supply air conditions set at 15 C and 95 percent relative humidity if (a) the recommended rate of outside ventilation air is used and (b) if a filtration device of E = 70 precent is used. Solution: Table 4-3, Houston Texas Summer Deisgn Conditions Dry-Bulb = 34 C Wet-Bulb = 25 C At 34 C Dry-Bulb, 24 C Wet-Bulb h o = 76 kJ/kg. W o = 0.0163 kg/kg At 15 C Dry-Bulb, 95 percent relative humidity h s = 40.5 kJ/kg, W s = 0.010 kg/kg At 25 C, 55 percent relative humidity h i = 53.2 kJ/kg, W i = 0.011 kg/kg (a) V = Vo Table 4-1, 10 L/s per person V = (10 L/s)(5) = 50 L/s q s = 1.23V(t o - t s ) q s = 1.23(50)(34- 15) Chapter4 16 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 3 of 8 q s = 1,168.5 W q L = 3000V(W o - W i ) q L = 3000(50)(0.0163 - 0.010) q L = 945 W q t = q s + q L q t = 1,168.5 W + 945 W q t = 2,113.5 W q t = 2.1 kw - - - Ans. (a) V1 = Vm Table 4-1, 2.5 L/s per person V1 = (2.5 L/s)(5) = 12.5 L/s E VV VV mo r2 − == 0.7 12.550 V2 − = V 2 = 53.5714 L/s q s = 1.23V 1 (t o - t s ) + 1.23V 2 (t i - t s ) q s = 1.23(12.5)(34 - 15) + 1.23(53.5714)(25 - 15) q s = 951 W q L = 3000V 1 (W o - W s ) + 3000V 2 (W i - W s ) q L = 3000(12.5)(0.0163 - 0.010) + 3000(53.5714)(0.011 - 0.010) q L = 397 W q t = q s + q L q t = 951 W + 397 W q t = 1,348 W q t = 1.35 kw - - - Ans. 4-4 A computer room located on the second floor of a five-story office building is 10 by 7 m. The exterior wall is 3.5 m high and 10 m long; it is a metal curtain wall (steel backed with 10 mm of insulating board), 75 mm of glass-fiber insulation, and 16 mm of gypsum board. Single-glazed windows make up 30 percent of the exterior wall. The computer and lights in the room operate 24 h/d and have a combined heat release to the space of 2 kw. The indoor temperature is 20 C. (a) If the building is located in Columbus, Ohio, determine the heating load at winter design conditions. (b) What would be the load if the windows were double-glazed? Solution: (a) Table 4-3, Columbus, Ohio, Winter Design Temperature = -15 C. Thermal Transmission: Wall: Chapter4 17 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 6 of 8 Transmission = UA(to - t1) = (3.5)(3)(37 - 25) = 126 W Solar: q sg = (SHGF max )(SC)(CLF)A 1 q sg = (580)(0.45)(0.79)(1.46) = 301 W Heat Gain = 126 W + 301 W = 427 W - - - Ans. 4-8. Compute the total heat gain for the south windows of an office building that has no external shading. The windows are double-glazed with a 6-mm air space and with regular plate glass inside and out. Draperies with a shading coefficient of 0.7 are fully closed. Make Calculation for 12 noon in (a) August and (b) December at 32 o North Latitude. The total window area is 40 m 2 . Assume that the indoor temperatures are 25 and 20 C and that the outdoor temperatures are 37 and 4 C. Solution: Tabkle 4-7 Double-glazed, 6-mm air space, U-value Summer - 3.5 W/m 2 .K Winter - 3.3 W/m 2 .K A = 40 m 2 (a) August, SUmmer, Indoor = 25 C, Outdoor = 37 C Thermal Transmission: q 1 = UA(t o -t i ) q 1 = (3.5)(40)(37 - 25) q 1 = 1,680 W Solar: q sg = (SHGFmax)(SC)(CLF)A Table 4-10, 32 o North Latitude, Facing South SHGF = 355 W/m 2 Table 4-12, Facing South at 12 Noon. CLF = 0.83 and SC = 0.7 q sg = (355)(0.7)(0.83)(40) q sg = 8,250 W q t = q 1 + q sg q t = 1,680 W + 8,250 W q t = 9,930 W - - - Ans. (b) December, Winter, Indoor = 20 C, Outdoor = 4 C Chapter4 20 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 7 of 8 Thermal Transmission: q 1 = UA(t o -t i ) q 1 = (3.3)(40)(4 - 20) q 1 =-2,112 W Solar: q sg = (SHGFmax)(SC)(CLF)A Table 4-10, 32 o North Latitude, Facing South, December SHGF = 795 W/m 2 Table 4-12, Facing South at 12 Noon. CLF = 0.83 and SC = 0.7 q sg = (795)(0.7)(0.83)(40) q sg = 18,476 W q t = q 1 + q sg q t = -2,112 W + 18,476 W q t = 16,364 W - - - Ans. 4-9. Compute the instantaneous heat gain for the south wall of a building at 32 o north latitude on July 21. The time is 4 p.m. sun time. The wall is brick veneer and frame with an overall heat-transfer coefficient of 0.35 W/m 2 .K. The wall is 2.5 by 6 m with a 1- x 2-m window. Solution: Wall: A = (2.5 m)(5 m ) - (1 m)(2 m) = 10.5 m 2 U = 0.35 W/m 2 .K q w = UA(CLTD) Table 4-11, South, Type F, 4 P.M. CLTD = 22 q w = (0.35)(10.5)(22) q w = 80.85 Watts. - - - Ans 4-10. Compute the peak instantaneous heat gain per square meter of area for a brick west wall similar to that in Example 4-3. Assume that the wall is located at 40 o north latitude. The date is July. What time of the day does the peak occur? The outdoor daily average temperature of 30 C and indoor design temperature is 25 C. Solution: Ex. 4-3, U = 0.356 W/m 2 .K Table 4-15, Type F, West Wall CLTD max = 33 at 1900 h or 7 P.M. Chapter4 21 CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION Page 8 of 8 CLTD adj = CLTD + (25 - T i ) + (T av - 29) CLTD adj = 33 + (30 - 29) = 34 C q max / A = U(CLTD) q max / A = (0.356)(34) q max / A = 12.1 W/m 2 at 7 P.M. - - - - Ans. - 0 0 0 - Chapter4 22 CHAPTER 5 - AIR CONDITIONING SYSTEMS Page 3 of 4 (b) ( ) ( )( )13241.0 42 tt1.0 q m 21 S − = − = m = 3.82 kg/s - - - - Ans. 5-4. In discussing outdoor-air control Sec. 5-3 explained that with outdoor conditions in the X and Y regions on the psychrometric chart in Fig. 5-5 enthalpy control is more energy-efficient. We now explore some limitations of that statement with respect to the Y region. Suppose that the temperature setting of the outlet air form the cooling coil is 10 C and that the outlet air is essentially saturated when dehumidification occurs in the coil. If the condition of return air is 24 C and 40 percent relative humidity and the outddor conditions are 26 C and 30 percent relative humidity, would return air or outside air be the preferred choice? Explain why. Solution: See Fig. 5-5 and Sec. 5-3. Outside Air: At 26 C, 30 percent relativw humidity h o = 42 kJ/kg Coil outlet = 10 C saturated q = 42 kJ/kg - 29.348 kJ/kg q = 12.652 kJ/kg Recirculated air: At 24 C, 40 percent relative humidity h i = 43 kJ/kg With 10% outdoor air. h m = (0.10)(42) + (0.90)(43) = 42.9 kJ/kg q = 42.9 kJ/kg - 29.348 kJ/kg q = 13.552 kJ/kg > 12.652 kJ/kg. Ans. Outside air is preferred due to lower cooling required. 5-5. A terminal reheat system (Fig. 5-9) has a flow rate of supply air of 18 kg/s and currently is operating with 3 kg/s of outside air at 28 C and 30 percent relative humidity. The combined sensible load in the spaces is 140 kw, and the latent load is negligible. The temperature of the supply air is constant at 13 C. An accountant of the firm occupying the building was shocked by the utility bill and ordered all space thermostat be set up from 24 to 25 C. What is the rate of heat removal in the cooling coil before and after the change and (b) the rate of heat supplied at the reheat coils before and after change? Assume that the space sensible load remains at 140 kw? Solution: See Fig. 5-9. Outside air at 28 C and 30 percent relative humidity h o = 46 kJ/kg At 24 C Set-Up. Coil entering temperature, t m t m = [(3)(28) + (18 - 3)(24)] / 18 = 24.667 C Coil supply temperature = 13 C constant Cooling rate = (18)(24.667 - 13) = 210 kw Chapter5 25 CHAPTER 5 - AIR CONDITIONING SYSTEMS Page 4 of 4 Space sensible load = 140 kw constant Reheat supply temperature, t s . t s = 24 - 140 / 18 = 16.222 C Heating Rate = (18)(16.222 - 13) Heating Rate = 58 kw At 25 C Set-Up. Coil entering temperature, t m t m = [(3)(28) + (18 - 3)(25)] / 18 = 25.5 C Coil supply temperature = 13 C constant Cooling rate = (18)(25.5 - 13) = 225 kw Space sensible load = 140 kw constant Reheat supply temperature, t s . t s = 25 - 140 / 18 = 17.222 C Heating Rate = (18)(17.222 - 13) Heating Rate = 76 kw Answer: (a) Before = 210 kw After = 225 kw 15 kw increase in cooling rate. (b) Before = 58 kw After = 76 kw 18 kw increase in heating rate - 0 0 0 - Chapter5 26 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 1 of 13 6-1. Compute the pressure drop of 30 C air flowing with a mean velocity of 8 m/s in a circular sheet-metal duct 300 mm in diameter and 15 m long using (a) Eqs. (6-1) and (6-2) and (b) Fig. 6-2. Solution: Equation 6-1. ρ=∆ 2 V D L fp 2 Equation 6-2. ( ) 2 f D Re 9.3D 2log14 1 f                           ε +− ε + = 1log2.1 D = 300 mm = 0.3 m V = 8 m/s At 30 C, Table 6-2. µ = 18.648 mPa.s = 1.8648 x10 -5 Pa.s ρ = 1.1644 kg/m 3 Table 6-1, ε = 0.00015 m ε/D = 0.00015 m / 0.3 m = 0.0005 Reynolds Number µ ρ = VD Re ( )( )( ) 149,860 101.8648 1.16440.38 Re 5 = × = − (a) Equation 6-2. ( )( ) 2 f0.0005149860 9.3 12log 0.0005 1 2log1.14 1 f                     +−      + = 2 f 0.124116 12log7.74206 1 f                     +− = By trial and error; f = 0.01935 Equation 6-1 Chapter6 27 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 4 of 13 Equation 6-11: Pa A A 1 2 V p 2 2 1 2 1 loss       − ρ = ( ) loss 2 2 2 1 12 p 2 VV pp − ρ− =− Table 6-2. At 20 C, ρ = 1.2041 kg/m3 A 1 V 1 = A 2 V 2 D 1 = 0.2 m D 2 = 0.4 m D 1 2 V 1 = D 2 2 V 2 (0.2) 2 V 1 = (0.4) 2 V2 V 2 = 0.25V 1 0.25 D D A A 2 2 2 1 2 1 == p 2 - p 1 = 200 Pa - 150 Pa = 50 Pa ( ) Pa A A 1 2 V 2 VV pp 2 2 1 2 1 2 2 2 1 12       − ρ − ρ− =− ( )[ ]( ) ( ) ( )2 2 1 2 1 2 1 0.251 2 1.2041V 2 1.20410.25VV 50 −− − = V 1 = 14.88171 m/s Q = A1V1 1 2 14 VDQ π= ( ) ( )14.881710.2Q 2 4 π= Q = 0.4675 m 3 /s - - - Ans. 6-5. A duct 0.4 m high and 0.8 m wide, suspended from a ceiling in a corridor, makes a right-angle turn in horizontal plane. The inner radius is 0.2 m, and the outer radius is 1.0 m, measured from the same center. The velocity of air in the duct is 10 m/s. To how many meters of straight duct is the pressure loss in this elbow equivalent? Solution: Inner radius = 0.2 m Outer radius = 1.0 m W = 0.8 m H = 0.4 m Figure 6-8: W / H = 0.8 / 0.4 = 2.0 Ratio of inner to outer radius = 0.2 / 1.0 = 0.2 Chapter6 30 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 5 of 13 Then: 0.35 2 V p 2 loss = ρ ( )( ) m0.533 0.40.8 0.40.82 ba 2ab Deq = + = + = Friction loss for the D eq = 1.95 Pa/m Then: ρ= 2 V 0.35p 2 loss ρ = 1.2041 kg/m 3 ( ) ( ) Pa211.2041 2 10 0.35p 2 loss == Equivalent Length = 21 Pa / (1.95 Pa/m) Equivalent lengtn = 10.8 m - - - - Ans. 6-6. An 0.3- by 0.4-m branch duct leaves an 0.3- by 0.6-m main duct at an angle of 60 o . The air temperature is 20 C. The dimensions of the main duct remain constant following the branch. The flow rate upstream is 2.7 m 3 /s, and the pressure is 250 Pa. The branch flow rate is 1.3 m 3 /s. What is the pressure (a) downstream in the main duct and (b) in the branch duct? Solution: p1 = 250 Pa, See Fig. 6-10. β = 60 o ( )( ) m/s15 0.60.3 2.7 Vu == ( )( ) m/s7.78 0.60.3 1.3-2.7 Vd == ( )( ) m/s10.83 0.40.3 1.3 Vb == at 20 C, ρ = 1.2041 kg/m 3 . (a) Eq. 6-16. ( ) Pa V V 10.4 2 V p 2 u d 2 d loss       − ρ = Chapter6 31 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 6 of 13 ( ) ( ) ( ) Pa 15 7.78 10.4 2 1.20417.78 p 22 loss       −= p loss = 3.377 Pa Bernoulli Equation 6-10         ρ −−+ ρ ρ= loss 2 2 2 11 2 p 2 V 2 Vp p ( )       −−+= 1.2041 3.377 2 7.78 2 15 1.2041 250 1.2041p 22 2 p 2 = 346 Pa - - - Ans. (b) Fig. 6-11 0.722 15 10.83 V V u b == β = 60 o 583.1 2 V p 2 loss = ρ ( ) ( ) Pa111.81.2041 2 10.83 1.583p 2 loss ==         ρ −−+ ρ ρ= loss 2 2 2 11 2 p 2 V 2 Vp p ( )       −−+= 1.2041 111.8 2 10.83 2 15 1.2041 250 1.2041p 22 2 p 2 = 203 Pa - - - Ans. 6-7. In a branch entry, an airflow rate of 0.8 m 3 /s joins the main stream to give a combined flow rate of 2.4 m 3 /s. The air temperature is 25 C. The branch enters with an angle of β = 30 o (see Fig. 6-12). The area of the branch duct is 0.1 m 2 , and the area of the main duct is 0.2 m 2 both upstram and downstream. What is the reduction in pressure between points u and d in the main duct? Solution: At 25 C, Table 6-2, ρ = 1.18425 kg/m3 Equation 6-17. ( ) ddub 2 bd 2 ud 2 d AppcosAVAVAV −=βρ−ρ−ρ β = 30 o Chapter6 32 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 9 of 13 ( )( )( ) ( )22 -4 3 16 1.20410.02101.283422 C π × = C 3 = 8.122739 x 10 -6 6 1 1 3 3 opt C HQ5C D       = ( )( )( ) 6 1 36- opt 361.3 0.810000108.1227395 D       × = D opt = 0.289 m - - - - Ans. 6-10. Measurements made on a newly installed air-handling system were: 20 r/s fan speed, 2.4 m 3 /s airflow rate, 340 Pa fan discharge pressure, and 1.8 kw supplied to the motor. These measurements were made with an air temperature of 20 C, and the system is eventually to operate with air at a temperature of 40 C. If the fan speed remains at 20 r/s, what will be the operating values of (a) airflow be the operating values of (a) airflow rate, (b) static pressure, and (c) power? Solution: At 20 C, ω 1 = 20 r/s Q 1 = 2.4 m 3 /s SP 1 = 340 Pa P 1 = 1.8 kw ρ 1 = 1.2041 kg/m 3 At 40 C ω 2 = 20 r/s ρ 2 = 1.1272 kg/m 3 (a) Since ω is constant also Q is constant, Q 2 = 2.4 m 3 /s (b) Law 2, Q = constant SP ~ ρ 1 1 2 2 SPSP       ρ ρ = ( )Pa 340 1.2041 1.1272 SP2       = SP 2 = 318 Pa - - - Ans. (c) Law 2, Q = constant P ~ ρ 1 1 2 2 PP       ρ ρ = Chapter6 35 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 10 of 13 ( )kw1.8 1.2041 1.1272 P2       = P 2 = 1.685 kw - - - Ans. 6-11. A fan-duct system is designed so that when the air temperature is 20 C, the mass flow rate is 5.2 kg/s when the fan speed is 18 r/s and the fan motor requires 4.1 kw. A new set of requirement is imposed on the system. The operating air temperature is changed to 50 C, and the fan speed is increased so that the same mass flow of air prevails. What are the revised fan speed and power requirement? Solution: At 20 C, Table 6-2 ρ 1 = 1.2041 kg/m 3 m 1 = 5.2 kg/s ω 1 = 18 r/s P 1 = 4.1 kw At 50 C, Table 6-2 ρ 2 = 1.0924 kg/s m2 = 5.2 kg/s /sm4.3186 1.2041 5.2m Q 3 1 1 1 == ρ = /sm4.7602 1.0924 5.2m Q 3 2 2 2 == ρ = Revised fan speed, Equation 6-29 Q α ω or ω α 1/ρ 2 1 12 ρ ρ ω=ω ( ) ( ) ( )0924.1 2041.1 18=ω2 ω 2 = 19.84 r/s - - - Ans. Revised power requirement, Equation 6-31. ( ) 2 PQV SPQP 2 += Equation 6-30 2 V P 2ρ α Then 2 QV P 2ρ α 2QαP Chapter6 36 CHAPTER 6 - FAN AND DUCT SYSTEMS Page 11 of 13 2 P ρ α 1 ( ) 2 2 2 2 11 2 1.0924 1.2041 4.1 P P       = ρ ρ = P 2 = 4.98 kw - - - Ans. 6-12. An airflow rate of 0.05 m 3 /s issues from a circular opening in a wall. The centerline velocity of the jet is to be reduced to 0.75 m/s at a point 3 m from the wall. What should be the outlet velocity u o of this jet? Solution: Equation 6-32. ( )[ ]22 2 oo x r57.5x A7.41u u + = 1 u = 0.75 m/s x = 3 m r = 0 x A7.41u u oo = o o u Q A = Q = 0.05 m 3 /s x Qu7.41 u o = ( ) 3 0.05u7.41 0.75u o == u o = 1.84 m/s - - - Ans. 6-13. Section 6-19 points out that jets entrains air as they move away from their inlet into the room. The entrainment ratio is defined as the ration of the air in motion at a given distance x from the inlet to the airflow rate at the inlet Q x /Q o . Use the expression for the velocity in a circular jet, Eq. (6-32), multiplied by the area of an annular ring 2πrdr and integrate r from 0 to h to find the expression for Q x /Q o . Solution: Equarion 6-32. ( )[ ]22 2 oo x r57.5x A7.41u u + = 1 ( )∫ ∞ π= 0 x rdr2uQ Chapter6 37 CHAPTER 7 - PIPING SYSTEMS Page 1 of 6 7-1. A convector whose performance characteristics are shown in Fig. 7-4 is supplied with a flow rate of 0.04 kg/s of water at 90 C. The length of the convector is 4 m, and the room-air temperature is 18 C. What is the rate of heat transfer from the convector to the room air? Solution: See Fig. 7-4 m = 0.04 kg/s t 1 = 90 C L = 4 m t i = 18 C p 12 mc q tt −= c p = 4.19 kJ/kg.K ( ) ( )( )41900.04 q 90t 2 −= q0059666.090t 2 −= Mean Water Temp. ( )212 1 m ttt += ( )0.0059666q-9090t 2 1 m += 0.0029833q90tm −= Equation for Fig. 7-4. W/m56016t L q m −= ( )( )56016t4q m −= 224064tq m −= Substituting: t m = 90 - 0.0029833 (64t m - 2240) t m = 81.182 C q = 64t m - 2,240 Watts q = 64(81.182) - 2,240 Watts q = 2,956 Watts q = 2.956 kW ---- Ans. 7-2. Compute the pressure drop in pascals per meter length when a flowrate of 8 L/s of 60 C water flows through a Schedule 40 steel pipe of nominal diameter 75 mm (a) using Eq. (7-1) and (b) using Figs. 7-6 and 7-7. Solution: (a) Eq. 7-1. ρ=∆ 2 V D L fp 2 From Table 7-3 at 60 C. ρ = 983.19 kg/m 3 µ = 0.476 mPa.s = 0.000476 Pa.s 75-mm Schedule 40 Steel Pipe, Table 7-1, ID = 77.92 mm Chapter7 40 CHAPTER 7 - PIPING SYSTEMS Page 2 of 6 ( ) m/s1.678 4m0.07792 /sm0.008 V 2 3 = π = Table 6-1, ε = 0.000046 commercial steel. 0.00059 0.07792 0.000046 D == ε ( )( )( ) 0.000476 983.190.077921.678DV Re = µ π = Re = 270,067 From the Moody Chart, Fig. 6-1. Re = 270,067, 0.00059 D = ε f = 0.019 ρ=∆ 2 V D L fp 2 ( ) ( ) ( )983.19 2 1.678 0.07792 1 0.019 L p 2       = ∆ ∆p/L = 338 Pa/m ---- Ans. 7-3. In the piping system shown schematically in Fig. 7-14 the common pipe has a nominal 75 mm diameter, the lower branch 35 mm, and the upper branch 50 mm. The pressure of water at the entrance is 50 kPa above atmospheric pressure, and both branches discharged to atmospheric pressure. The water temperature is 20 C. What is the water flow rate in liters per second in each branch? Solution: ∆p = 50 kPa - 0 - 50 kPa = 5000 Pa Use Fig. 7-6, water temperature of 20 C Table 7-4. For 75-mm pipe Elbow = 4 x 3 m = 12 m Straight Pipe = 8 m + 4 m + 5 m + 7 m + 15 m = 39 m L 1 = 12 m + 39 m = 51 m Chapter7 41 CHAPTER 7 - PIPING SYSTEMS Page 3 of 6 For 50-mm pipe Straight Branch = 0.9 m Straight Pipe = 30 m L 2 = 0.9 m + 30 m = 30.9 m For 35-mm pipe Side Branch = 4.6 m Straight Pipe = 6 m + 18 m = 24 m Elbow = 1 x 1.2 m = 1.2 m L 3 = 4.6 m + 24 m + 1.2 m = 29.8 m Q 1 = Q 2 + Q 3 pL L p L L p 2 2 2 1 1 1 ∆=      ∆ +      ∆ pL L p L L p 3 3 3 1 1 1 ∆=        ∆ +      ∆ 32 2 2 L L p L L p         ∆ =      ∆ 3 3 ( ) ( )29.8 L p 30.9 L p 2 2         ∆ =      ∆ 3 3       ∆ =        ∆ 2 2 L p 1.036913 L p 3 3 Assume f = 0.02 r = 998.21 kg/m3. For 75-mm pipe, ID = 77.92 mm = 0.07792 m For 50-mm pipe, ID = 52.51 mm = 0.05251 m For 35-mm pipe, ID = 35.04 mm = 0.03504 m ρ      = ∆ 2 V D 1 f L p 2 1 VD 4 1 Q 2 π= 2D 4Q V π = ρ         π       = ∆ 4 1 2 2 1 1 1 D 8Q D 1 f L p ρ         π = ∆ 5 1 2 2 1 1 1 D 8Q f L p ( ) ( ) 2 152 2 1 1 1 5,633,748Q 0.07792 8Q L p =         π = ∆ 21.99802.0 ( ) ( ) 2 252 2 2 2 2 Q 0.05251 8Q L p 176,535,4021.99802.0 =         π = ∆ Chapter7 42 CHAPTER 7 - PIPING SYSTEMS Page 6 of 6 - 0 0 0 - Chapter7 45 CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS Page 1 of 6 8-1. A cooling and dehumidifying coil is supplied with 2.4 m 3 /s of air at 29 C dry-bulb and 24 C wet-bulb temperatures, and its cooling capacity is 52 kW. The face velocity is 2.5 m/s. and the coil is of the direct- expansion type provided with refrigerant evaporating at 7 C. The coil has an air-side heat-transfer area of 15 m 2 per square meter of face area per row of tubes. The ratio of the air-side to refrigerant-side area is 14. The values of h r and h c are 2050 and 65 W/m 2 .K, respectively. Calculate (a) the face area, (b) the enthalpy of outlet air, (c) the wetted-surface temperatures at the air inlet, air outlet, and at the point where the enthalpy of air is midway between its entering and leaving conditions, (d) the total surface area, (e) the number of rows of tubes, and (f) the outlet dry-bulb temperature of the air. Solution: At 29 C dry-bulb and 24 C wet-bulb h a,1 =72.5 kJ/kg g a,1 = 0.88 m 3 /kg (a) Face Area = (2.4 m 3 /s) / (2.5 m/s) Face Area = 0.96 m 2 (b) Enthalpy of outlet air, h a,2 m = (2.4 m 3 /s) / (0.88 m 3 /kg) = 2.7273 kg/s m q hh t a,1a,2 −= kg/s2.7273 kW 52 kJ/kg72.5ha,2 −= h a,2 = 53.4 kJ/kg (c) Wetted Surface Temperature Eq. 8-1. ( )ia pm c hh c dAh dq −= Eq. 8-2. ( )riir ttdAhdq −= Eq. 8-3. irpm c ia ri Ahc Ah hh tt R = − − = t r = 7 C A/A i = 14 h r = 2050 W/m 2 .K h c = 65 W/m 2 .K c pm = 1.02 kJ/kg.K ( )( ) ( )( ) 0.4352 20501.02 1465 hh tt R ia ri == − − = h a and h i in kJ/kg Eq. 8-4. h i = 9.3625+1.7861t i +0.01135t i 2 +0.00098855t i 3 Eq. 8-5. Chapter8 46 CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS Page 2 of 6 0t0.000988550.01135t1.7861t9.3625h R t R t 3 i 2 iia ri =++++−− At the air inlet, h a,1 = 72.5 kJ/kg 0t0.000988550.01135t1.7861t9.362572.5 0.4352 7 04352 t 3 i 2 ii i =++++−− By trial and error: t i = 17.31 C and enthalpy h i = 48.8 kJ/kg at air inlet. At the air outlet, h a,3 = 53.4 kJ/kg 0t0.000988550.01135t1.7861t9.3625 0.4352 7 04352 t 3 i 2 ii i =++++−− 4.53 By trial and error: t i = 13.6 C and enthalpy h i = 38.23 kJ/kg at air outlet. At the midway enthalpy, h a,2 =(1/2)(72.5 kJ/kg + 53.4 kJ/kg) = 62.95 kJ/kg 0t0.000988550.01135t1.7861t9.3625 0.4352 7 04352 t 3 i 2 ii i =++++−− 95.62 By trial and error: t i = 15.5 C and enthalpy h i = 43.46 kJ/kg at midway enthalpy. Answer - - - 17.31 C, 15.5 C, and 13.6 C. (d) Total surface area. Between 1 and 2. ( ) ( )differenceenthalpymean c Ah hhmq pm 21c 2121 −=−= − − c pm = 1020 J/kg.K ( )( ) ( )             +       + =− − 2 43.4648.8 - 2 62.9572.5 1020 A65 62.9572.52.7273 21 A 1-2 = 18.93 m 2 Between 2 and 3. ( ) ( )differenceenthalpymean c Ah hhmq pm 2c 322 −=−= − − 3 3 c pm = 1020 J/kg.K ( )( ) ( )             +       + =− − 2 38.2343.46 - 2 62.95 1020 A65 62.952.7273 2 4.53 4.53 3 A 2-3 = 23.59 m 2 Surface Area of Coil = 18.93 m 2 + 23.59 m 2 Chapter8 47 CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS Page 5 of 6 ( ) ( ) A Ah tthtt ir ri,2ci,22 −=− h c = 54 W/m2.K h r = 2000 W/m2.K t r = 9 C A/A i = 15 ( )( ) ( )( ) 15 20009-13.8 5413.8t 2 =− t 2 = 25.7 C - - - - Ans. (b) ( )       − + + =− − r 21 21 ric 21pm t 2 tt A hA A h 1 1 ttGc c pm = 1020 J/kg.K G = 0.53 kg/s ( )( )( )       − + + =− − 9 2 25.735 A 2000 15 54 1 1 25.73210200.53 21 A 1-2 = 4.47 m2 - - - Ans. 8-4. For a coil whose performance and conditions of entering air are shown in Table 8-1, when the face velocity is 2 m/s and the refrigerant temperature is 4.4 C, calculate (a) the ratio of moisture removal to reduction in dry- bulb temperature in the first two rows of tubes in the direction of air flow in the last two rows and (b) the average cooling capacity of the first two and the last two rows in kilowatts per square meter of face area. Solution: Use Table 8-1. Face velocity = 2 m/s Refrigerant Temperature = 4.4 C. (a) First 2-rows: At 30 C dry-bulb, 21.7 C wet-bulb temperature h 1 = 63 kJ/kg W 1 = 0.013 kg/kg γ 1 = 0.08735 m 3 /kg Final DBT = 18.2 C Final WBT = 17.1 C h 2 = 48.5 kJ/kg W 2 = 0.0119 kg/kg Ratio for the first two rows = (W 1 - W 2 ) / (t 1 - t 2 ) = (0.013 - 0.0119) / (30 - 18.2) = 0.0000932 kg/kg.K - - - Ans. For the last two rows. Rows of tube = 4 in Table 8-1. Final DBT = 14.3 C Final WBT = 13.8 C Chapter8 50 CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS Page 6 of 6 h 3 = 38.5 kJ.kg W 3 = 0.0095 kg/kg Ratio for the last two rows = (W 2 - W 3 ) / (t 2 - t 3 ) = (0.0119 - 0.0095) / (18.2 - 13.8) = 0.00055 kg/kg.K - - - Ans. (b) First two rows. kW per sq m of face area = [(2 m/s)/(0.8735 m 3 /kg)](h 1 - h 2 ) = (2 / 0.8735)(63 - 48.5) = 33.2 kw - - - Ans. For the last two rows. kW per sq m of face area = [(2 m/s)/(0.8735 m 3 /kg)](h 2 - h 3 ) = (2 / 0.8735)(48.5 - 38.5) = 22.9 kw - - - Ans. 8-5. An airflow rate of 0.4 kg/s enters a cooling and dehumidifying coil, which for purpose of analysis is divided into two equal areas, A 1-2 and A 2-3 . The temperatures of the wetted coil surfaces are t i,1 = 12.8 C, t i,2 = 10.8 C , and t i,3 = 9.2 C. The enthalpy of entering air h a,1 = 81.0 and ha,2 = 64.5 kJ/kg. Determine ha,3 . Soution: G = 0.4 kg/s Then equation:                 + −        + =                 + −        + −− 2 hh 2 hh q 2 hh 2 hh q i,3i,2a,3a,2 21 i,2i,1a,2a,1 21 Eq. 8-4. 3 i 2 iii t0.000988550.01135t1.786t9.3625h +++= At t i,1 = 12.8 C h i,1 = 36.16 kJ/kg At t i,2 = 10.8 C h i,2 = 31.22 kJ/kg At t i,3 = 9.2 C h i,3 = 27.52 kJ/kg Then:                 + −        + =                 + −        + 2 hh 2 hh h- h 2 hh 2 hh h-h i,3i,2a,3a,2 a,3a,2 i,2i,1a,2a,1 a,2a,1               + −        + =             + −      + 2 27.5231.22 2 h64.5 h- 64.5 2 31.2236.16 2 64.581 64.5-81 a,3 a,3 0.422427(0.5h a,3 + 2.88) = 64.5 - h a,3 1.211214h a,3 = 63.28341 h a,3 = 52.25 kJ/kg - - - Ans. - 0 0 0 - Chapter8 51 CHAPTER 9 - AIR-CONDITIONING CONTROLS Page 1 of 4 9-1. A space thermostat regulates the damper in the cool-air supply duct and thus provides a variable air flow rate. Specify whether the damper should be normally open or normally closed and whether the thermostat is direct- or reverse-acting. Answer: Use normally closed damper and reverse-acting thermostat since as the space temperature increases the volume rate of air will increase the pressure will reduce. 9-2. On the outdoor-air control system of Example 9-4, add the necessary features to close the outdoor-air damper to the minimum position when the outdoor temperature rises above 24 C. Answer: Add a diverting relay. Pressure will divert to 68 kPa (20 %) minimum position when the outdoor temperature rises above 24 C. 9-3. The temperature transmitter in an air-temperature controller has a range of 8 to 30 C through which range the pressure output change from 20 to 100 kPa. If the gain of the receiver-controller is set at 2 to 1 and the spring range of the cooling-water valve the controller regulates is 28 to 55 kPa, what is the throttling range of this control? Solution: Output of temperature transmitter = (100 -20 kPa) / (30 - 8 C) = 3.6364 kPa/K Throttling Range = (55 kPa - 28 kPa) / [(2)(3.6364 kPa/K)] Throttling Range = 3.7 K . . . Ans. 9-4 The air supply for a laboratory (Fig. 9-29) consists of a preheat coil, humifidier, cooling coil, and heating coil. The space is to be maintained at 24 C, 50 percent relative humidity the year round, while the outdoor supply air may vary in relative humidity between 10 and 60 percent and the temperature from -10 to 35 C. The spring ranges available for the valves are 28 to 55 and 62 to 90 kPa. Draw the control diagram, adding any additional components needed, specify the action of the thermostat(s) and humidistat, the spring ranges of the valves, and whethet they are normally open or n ormally closed. Answer: Spring range = 28 to 55 kPa = 62 to 90 kPa (a) Limitiations: Chapter9 52 CHAPTER 9 - AIR-CONDITIONING CONTROLS Page 4 of 4 2. Damper is closing when the temperature is less than 24 C and opening when temperature is greater than 24 C. 9-6. Section 9-18 described the flow characteristics of a coil regulated by a valve with linear characteristics. The equation of the flow-steam position for another type of valve mentioned in Sec. 9-18, the equal-percentage valve, is 1 100 strokestemofpercent xwhereA pC Q x v −== ∆ If such a valve with an A value of 20 and a Cv of 1.2 is applied to controlling the coil in Fig. 9-25 with Dcoil = 2.5Q2 and the total pressure drop across the valve and coil of 80 kPa, what is the flowrate when the valve stem stroke is at the halfway position? (Compare with a linear-characteristic valve in Fig. 9-27.) Solution: 22.5QkPa80p −=∆ C v = 1.2 A = 20 at percent of stem stroke = 50 %. 0.51 100 50 x −=−= x v A pC Q = ∆ 0.5- 2 20 2.5Q801.2 Q = − 22.5Q800.26833Q −= Q = 2.21 L/s - - - Ans. Comparing to linear-characteristic valve in Fig. 9-27. pC 100 strokepercent Q v ∆= ( ) 22.5Q801.2 100 50 Q −= Q = 3.893 L/s > 2.21 L/s - 0 0 0 - Chapter9 55 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 1 of 10 10-1. A Carnot refrigeration cycle absorbs heat at -12 C and rejects it at 40 C. (a) Calculate the coefficient of performance of this refrigeration cycle. (b) If the cycle is absorbing 15 kW at the -12 C temperature, how much power is required? (c) If a Carnot heat pump operates between the same temperatures as the above refrigeration cycle, what is the performance factor? (d) What is the rate of heat rejection at the 40 C temperature if the heat pump absorbs 15 kW at the - 12 C temperature? Solution: (a) Coefficient of performance = T 1 / (T 2 - T 1 ) T 1 = -12 C + 273 = 261 K T 2 = 40 C + 273 = 313 K Coefficient of performance = 261 / (261 + 313) Coefficient of performance = 5.02 - - - Ans. (b) Coefficient of performance = useful refrigeration / net work 5.02 = 15 kw / net work net work = 2.988 kW - - - Ans. (c) Performance factor = coefficient of performance + 1 Performance factor = 6.01 - - - Ans. (d) Performance factor=heat rejected from cycle/work required. 15kwrejectedheat rejectedheat factorePerformanc − = 15kwrejectedheat rejectedheat 6.02 − = Heat rejected = 17.988 kw - - - Ans. 10-2. If in a standard vapor-compression cycle using refrigerant 22 the evaporating temperature is -5 C and the condensing temperature is 30 C, sketch the cycle on pressure-enthalpy coordinates and calculate (a) the work of compression, (b) the refrigerating effect, and (c) the heat rejected in the condenser, all in kilojoules per kilograms , and (d) the coefficient of performance. Solution. At pont 1, Table A-6, -5 C, h 1 = 403.496 kJ/kg s 1 = 1.75928 kJ/kg.K At point 2, 30 C condensing temperature, constant entropy, Table A-7. Chapter10 56 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 2 of 10 h 2 = 429.438 kJ/kg At point 3, Table A-6, 30 C h 3 = 236.664 kJ/kg h 4 = h 3 = 236.664 kJ/kg (a) Work of compression = h 2 - h 1 = 429.438 - 403.496 = 25.942 kJ/kg - - - Ans. (b) Refrigerating effect = h 1 - h 4 = 403.496 - 236.664 = 166.832 kJ/kg - - - Ans. (c) Heat rejected = h 2 - h 3 = 429.438 - 236.664 = 192.774 kJ/kg - - - Ans. (d) Coefficient of performance 12 41 hh hh eperformancoftCoefficien − − = 496.403438.429 664.236496.403 − − =eperformancoftCoefficien Coefficient of performance = 6.43 - - - Ans. 10-3. A refrigeration system using refrigerant 22 is to have a refrigerating capacity of 80 kw. The cycle is a standard vapor-compression cycle in which the evaporating temperature is -8 C and the condensing temperature is 42 C. (a) Determine the volume flow of refrigerant measured in cubic meter per second at the inlet to the compressor. (b) Calculate the power required by the compressor. (c) At the entrance to the evaporator what is the fraction of vapor in the mixture expressed both on a mass basis and a volume basis? Solution: At 1, Table A-6, -8 C. h 1 = h g1 = 402.341 kJ/kg h f1 = 190.718 kJ/kg Chapter10 57 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 5 of 10 At 2, 30 C condensing temperature, saturated, Table A-3. h 2 = 1486.14 kJ/kg s 2 = 5.2624 kJ/kg.K At 1, s 1 = s 2 . 0.878 0.654365.9025 0.654365.2624 ss ss x fg f1 = − − = − − = h 1 = h f + x (h g - h f ) h 1 = 108.599 + (0.878)(1437.23 - 108.599) h 1 = 1275.14 kJ/kg h 3 = 341.769 kJ/kg h 4 = 319.586 kJ/kg 1275.14-1486.14 319.5861275.14 hh hh eperformancoftCoefficien 12 41 − = − − = Coefficient of performance = 4.53 Ans. 4.53 wet versus 4.19 dry. 10-5. In the vapor-compression cycle a throttling device is used almost universally to reduce the pressure of the liquid refrigerant. (a) Determine the percent saving in net work of the cycle per kilograms of refrigerant if an expansion engine would be used to expand saturated liquid refrigerant 22 isentropically from 35 C to the evaporator temperature of 0 C. Assume that compression is isentropic from saturated vapor at 0 C to a condenser pressure corresponding yo 35 C. (b) Calculate the increase in refrigerating effect in kilojoules per kilograms resulting from use of expansion engine. Solution: Vapor-Compression Cycle: Chapter10 60 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 6 of 10 At 1, 0 C, Table A-6. h 1 = 405.361 kJ/kg s 1 = s g1 = 1.75279 kJ/kg.K At 2, 35 C, constant entropy, Table A-7. h 2 = 430.504 kJ/kg At 3, Table A-6 h 3 = 243.114 kJ/kg h 4 = h 3 = 243.114 kJ/kg Net Work = h 2 - h 1 = 430.504 - 405.361 = 25.143 kJ/kg Refrigerating Effect = h 1 - h 4 = 405.361 - 243.114 = 162.247 kJ/kg For expansion engine: At a, 0 C, Table A-6. h a = h ga = 405.361 kJ.kg h fa = 200 kJ/kg s a = s ga = 1.75279 kJ/kg.K s fa = 1.00000 kJ/kg.k At b, constant entropy, Table A-2 Chapter10 61 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 7 of 10 h b = 430.504 kJ/kg At c, Table A-6. h c = 243.114 kJ/kg s c = 1.14594 kJ/kg At d, constant entropy. 0.193866 1.000001.75279 1.000001.14594 ss ss x faga fad = − − = − − = h d = h fa + x(h ga - h fa ) h d = 200 + (0.193866)(405.361 - 200) h d = 239.813 kJ/kg Net Work = (h b - h a ) - (h c - h d ) Net Work = (430.5 - 405.361) - (243.114 - 239.813) Net Work = 21.838 kJ/kg Refrigerating Effect = h a - h d = 405.361 - 239.813 = 165.548 kJ/kg (a) Percent Saving ( )100% 25.143 21.83825.143 − = = 13.1 % - - - Ans. (b) Increase in refrigerating effect. = 165.548 kJ/kg - 162.247 kJ/kg = 3.301 kJ/kg - - - Ans. 10-6. Since a refrigeration system operates more efficiently when the condensing temperature is low, evaluate the possibility of cooling the condenser cooling water of the refrigeration system in question with another refrigeration system. Will the compressor performance of the two systems be better, the same, or worse than one individual system? Explain why. Solution: Chapter10 62 CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE Page 10 of 10 (c) Refrigerating capacity without heat exchanger At 1, ν = 65.3399 L/kg Refrigerating Capacity ( )51 hh L/kg65.3399 L/s12.0 −      = ( )664.236−      = 401.555 L/kg65.3399 L/s12.0 = 30.3 kW - - - - Ans. (d) Refrigerating capacity with heat exchanger At 1, ν = 70.2751 L/kg Refrigerating Capacity ( )51 hh L/kg70.2751 L/s12.0 −      = ( )664.236−      = 411.845 L/kg70.2751 L/s12.0 = 29.9 kW - - - - Ans. - 0 0 0 - Chapter10 65 CHAPTER 11 - COMPRESSORS Page 1 of 6 11-1. An ammonia compressor has a 5 percent clearance volume and a displacement rate of 80 L/s and pumps against a condensing temperature of 40 C. For the two different evaporating temperatures of -10 and 10 C, compute the refrigerant flow rate assuming that the clearance volumetric efficiency applies. Solution: Equation 11-7. suc vc 100ratentdisplacemew ν η ×= (a) At -10 C, Table A-3. s 1 = 5.7550 kJ/kg ν suc = 417.477 L/kg At 40 C, constant entropy, Fig. A-1 ν dis = 112.5 L/kg m = 5 % Equation 11-4 and Equation 11-5.         − ν ν −= 1m100η dis suc vc 86.4451 112.5 417.477 5100ηvc =      −−= suc vc 100ratentdisplacemew ν η ×= ( ) ( ) 417.477 100 86.445 L/s80w ×= w = 0.166 kg/s at -10 C - - - Ans. (b) At 10 C, Table A-3 s1 = 5.4924 kJ/kg.K ν suc = 205.22 L/kg At 40 C, constant entropy, Fig. A-1 ν dis = 95 L/kg m = 5 % Equation 11-4 and Equation 11-5.         − ν ν −= 1m100η dis suc vc 94.1991 95 205.22 5100ηvc =      −−= Chapter11 66 CHAPTER 11 - COMPRESSORS Page 2 of 6 suc vc 100ratentdisplacemew ν η ×= ( ) ( ) 205.22 100 94.199 L/s80w ×= w = 0.367 kg/s at 10 C - - - Ans. 11-2. A refrigerant 22 compressor with a displacement rate of 60 L/s operates in a refrigeration system that maintains a constant condensing temperature of 30 C. Compute and plot the power requirement of this compressor at evaporating temperatures of -20, -10, 0, 10 and 20 C. Use the actual volumetric efficiencies from Fig. 11-12 and the following isentropic works of compression for the five evaporating temperatures, respectively, 39.9, 30.2, 21.5, 13.7, and 6.5 kJ/kg. Solution: (a) At -20 C evaporating temperature, Table A-6. ν suc = 92.8432 L/kg p suc = 244.83 kPa Table A-7, 30 C p dis = 1191.9 kPa Ratio = p dis / p suc = 1191.9 kPa / 244.82 kPa = 4.87 Figure 11-12 η va = volumetric efficiency = 67.5 % suc va 100ratentdisplacemew ν η ×= ( ) ( ) 92.8432 100 67.5 L/s60w ×= w = 0.4362 kg/s at -20 C P = w∆h i ∆h i = 39.9 kJ/kg P = (0.4362)(39.9) P = 17.4 kw at -20 C (b) At -10 C evaporating temperature, Table A-6. ν suc = 65.3399 L/kg p suc = 354.3 kPa Table A-7, 30 C p dis = 1191.9 kPa Ratio = p dis / p suc = 1191.9 kPa / 354.30 kPa = 3.364 Figure 11-12 η va = volumetric efficiency = 77.5 % suc va 100ratentdisplacemew ν η ×= Chapter11 67 CHAPTER 11 - COMPRESSORS Page 5 of 6 At 3, 40 C condensing temperature, Table A-6, 8 C Subcooling t = 40 -8 = 32 C h 3 = 239.23 kJ/kg h 4 = h 3 = 239.23 kJ/kg (a) For actual volumetric efficiency Displacement rate = (4 cyl)(29 r/s)(0.087 2 π / 4 m 3 /cyl.r)(0.070 m) = 0.04827 m 3 /kg = 48.27 L/kg Actual rate of refrigerant flow = 115 kw / (403.876 - 239.23 kJ/kg) = 0.6985 kg/s Actual volumetric flow rate at the compressor suction = (0.6985 kg/s)(53.5682 L/kg) = 37.42 L/s 100 /smn,compressioofratentdisplaceme /sm,compressorenteringrateflowvolume 3 3 va ×=η η va = (37.42 L/s)(100) /(48.27 L/s) = 77.5 % - - - Ans. (b) For compression efficiency. Actual work of compression = 0.9 (34.5 kW) / (0.6985 kg/s) = 44.45 kJ/kg 100 kJ/kgn,compressioofworkactual kJ/kgn,compressioof workisentropic c ×=η 100 kJ/kg44.45 kJ/kg403.876-435.391 c ×=η η c = 70.9 % - - - Ans. 11-4. An automobile air conditioner using refrigerant 12 experiences a complete blockage of the airflow over the condenser, so that the condenser pressure rises until the volumetric efficiency drops to zero. Extrapolate the actual volumetric-efficiency curve of Fig. 11-12 to zero and estimate the maximum discharge pressure, assuming an evaporating temperature of 0 C. Solution: Figure 11-12. At actual volumetric efficiency = - ( ) ( ) ( ) 17.1857 6756 670 5ratioPressure =− − − += Table A-5, 0 C, p suc = 308.61 kPa p dis = (17.18)(308.61 kPa) p dis = 5302 kPa - - - Ans. 11-5. Compute the maximum displacement rate of a two-vane compressor having a cylinder diameter of 190 mm and a rotor 80 mm long with a diameter of 170 mm. The compressor operates at 29 r/s. Chapter11 70 CHAPTER 11 - COMPRESSORS Page 6 of 6 Solution: Use Fig. 11-20 (a) θ = 3.3525 radians Crosshatched area = (1/2)(3.3525)(0.095) 2 + (1/2)(0.094472)(0.010)(2) - (π/2)(0.085) 2 Crosshatched area = 0.004724 m 2 . Displacement rate for two=vane compressor D = 2(Crosshatched area)(L)(rotative speed) D = (2)(0.004724)(0.080)(29) D = 0.0219 m 3 /s D = 21.9 L/s - - - Ans. 11-6. A two-stage centrifugal compressor operating at 60 r/s is to compress refrigerant 11 from an evaporating temperature of 4 C to a condensing temperature of 35 C. If both wheels are to be of the same diameter, what is this diameter? Solution: At 4 C evaporating temperature, Table A-4. h 1 = 390.93 kJ/kg s 1 = 1.68888 kJ/kg.K At 35 C condensing temperature, Fig. A-2, constant entropy, h 2 = 410 kJ/kg w = 60 r/s Equation 11-16, V 2t 2 = 1000∆h i V 2t 2 = 1000(410 - 390.93)/2 V 2t = 97.65 m/s per stage Section 11-25. Refrigerant 11. 113.1 m/s tip speed, wheel diameter = 0.60 m then at 97.65 m/s tip speed. wheel diameter = (97.65 / 113.1)(0.6 m) wheel diameter = 0.52 m - - - Ans. - 0 0 0 - Chapter11 71 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 1 of 11 12-1. An air-cooled condenser is to reject 70 kw of heat from a condensing refrigerant to air. The condenser has an air-side area of 210 m 2 and a U value based on this area is 0.037 kW/m 2 .K; it is supplied with 6.6 m3/s of air, which has a density of 1.15 kg/m3. If the condensing temperature is to be limited to 55 C, what is the maximum allowable temperature of inlet air? Solution: A o = 210 m 2 U o = 0.037 kW/m 2 .K q = 70 kw ρ = 1.15 kg/m 3 Condensing Temperature = 55 C w = (6.6 m 3 /s) / (1.15 kg/m 3 ) = 5.739 kg/s c p =1.0 kJ/kg.K ( ) ( ) ( ) ( )    − − −−− = oc ic ocic tt tt ln tttt LMTD q = U o A o LMTD ( )( ) K9.009 2100.037 70 AU q LMTD oo === But q = wc p (t o - t i ) ( )( ) K12.197 115.739 70 wc q tt p io ===− ( ) ( ) ( ) ( )    − − −−− = oc ic ocic tt tt ln tttt LMTD ( ) ( )    − − = o i t55 t55 ln 12.197 9.009 3.8724 t55 t55 o i = − − 55 - t i = 3.8724(55 - 12.197 - t i ) t i = 38.6 C - - - Ans. 12-2. An air-cooled condenser has an expected U value of 30 W/m 2 .K based on the air-side area. The condenser is to transfer 60 kW with an airflow rate of 15 kg/s entering at 35 C. If the condenser temperature is to be 48 C, what is the required air-side area? Solution: q = U o A o LMTD q = wc p (t o - t i ) w = 15 kg/s c p = 1.0 kJ/kg.K p io wc q tt += ( )( )115 60 35t o += Chapter12 72 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 4 of 11 2 1 2 12 w w pp       ∆=∆ ( )( )22 250p =∆ ∆p 2 = 200 kPa - - - Ans. 12-5. (a) Compute the fin effectiveness of a bar fin made of aluminum that is 0.12 mm thick and 20 mm long when h f = 28 W/m 2 .K, the base temperature is 4 C, and the air temperature is 20 C. (b) If you are permitted to use twice as much metal for the fin as originally specified in part (a) and you can either double the thickness or double the length, which choice would be preferable in order to transfer the highest rate of heat flow. Why? Solution: (a) Aluminum fins k = 202 W/m.K 2y = 0.12 mm = 0.00012 m y = 0.00006 m L = 20 mm = 0.020 m ky h M f= ( )( )0.00006202 28 M = M = 48.1 m -1 ML tanhML =η ML = (48.1 m -1 )(0.020 m) = 0.962 ( ) 0.962 0.962tanh =η η = 0.7746 - - - - Ans. (b) If the fin thickness is doubled. 2y = 0.24 m = 0.00024 m y = 0.00012 m ( )( )0.00012202 28 M = M = 33.99 m -1 ML tanhML =η ML = (33.99 m -1 )(0.020 m) = 0.6798 ( ) 0.6798 0.6798tanh =η η = 0.87 > 0.7746 If the length L is doubled L = 40 mm = 0.040 m Chapter12 75 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 5 of 11 ( )( )0.00006202 28 M = M = 48.1 m -1 ML tanhML =η ML = (48.1 m -1 )(0.040 m) = 1.924 ( ) 1.924 1.924tanh =η η = 0.498 < 0.7746 Ans. Therefore double the fin thickness to improve rate of heat flow with an efficiency of 87 % compared to 77.46 %. 12-6. Compute the fin effectiveness of an aluminum rectangular plate fin of a finned air-cooling evaporator if the fins are 0.18 mm thick and mounted on a 16-mm-OD tubes. The tube spacing is 40 mm in the direction of air flow and 45 mm vertically. The air-side coefficient is 55 W/m 2 .K. Solution: h f = 55 W/m 2 .K Alumimum Fins, k = 202 W/m.K 2y = 0.00018 mm y = 0.00009 mm ky h M f= ( )( )0.00009202 55 M = M = 55 m -1 . Equivalent external radius. ( ) ( )( ) 22 2 e 2 16 4540 2 16 r       π−=               −π r e = 23.94 mm = 0.02394 m r i = 8 mm = 0.008 m (r e - r i )M = (0.02394 - 0.008)(55) - 0.88 r e /r i = 23.94 mm / 8 mm = 3 From Fig. 12-8/ Fin Effectiveness = 0.68 - - - Ans. 12-7. What is the UA value of a direct-expansion finned coil evaporator having the following areas: refrigerant side, 15 m 2 ; air-side prime, 13.5 m 2 , and air-side extended, 144 m 2 ? The refrigerant-side heat-transfer coefficient is 1300 W/m 2 .K, and the air-side coefficient is 48 W/m 2 .K. The fin effectiveness is 0.64. Solution: η = 0.64 A i = 15 m 2 h i = 1300 W/m 2 .K Chapter12 76 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 6 of 11 h f = 48 W/m 2 .K A p = 13.5 m 2 A e = 144 m 2 Eq. 12-20 neglect tube resistance. ( ) iiepfoo Ah 1 AAh 1 AU 1 + η+ = ( ) ( )( ) ( )( )151300 11 AU 1 oo + + = 14464.05.1348 U o A o = 4,025 W/K - - - Ans. 12-8. A refrigerant 22 system having a refrigerating capacity of 55 kW operates with an evaporating temperature of 5 C and rejects heat to a water-cooled condenser. The compressor is hermetically sealed. The condenser has a U value of 450 W/m 2 .K and a heat-transfer area of 18 m 2 and receives a flow rate of cooling water of 3.2 kg/s at a temperature of 30 C. What is the condensing temperature? Solution: Eq. 12-26. ( ) ( ) ( ) ( )    − − −−− = oc ic ocic tt tt ln tttt LMTD Heat Rejection: q = UALMTD = wc p (t o - t i ) c p = 4190 J/kg.K ( )( ) ( ) ( ) ( ) ( )( )( )30t41903.2 tt t ln t 18450q o oc c o −=                 − − − = 30 30 ( ) ( ) 0.60412 tt t ln oc c =    − − 30 t c - 30 = 1.82964 (t c - t o ) t o = 16.397 + 0.45345 t c - - - Eq. No. 1 Figure 12-12. At Heat-rejection ratio = 1.2 Condensing Temperature = 36 C At Heat-rejection ratio = 1.3 Condensing Temperature = 49 C Heat-rejection ratio = 0.92308 + 0.0076923 t c ( )( ) ( )iopc ttwc550000.0076923t0.92308q −=+= ( )( ) ( )( )( )30t41903.2550000.0076923t0.92308 oc −=+ coc 0.45345t16.397t0.031554t33.7865 +==+ t c = 41.22 C - - - Ans. 12-9. Calculate the mean condensing heat-transfer coefficient when refrigerant 12 condenses on the outside of the horizontal tubes in a shell-and-tube condenser. The outside diameter of the tubes is 19 mm, and in the Chapter12 77 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 9 of 11 Uo, W/sq m.K 2300 2070 1930 1760 1570 1360 1130 865 V, m/s 1.22 0.975 0.853 0.731 0.61 0.488 0.366 0.244 Water flowed inside the tubes, and the tubes were 51 mm OD and 46 mm ID and had a conductivity of 60 W/m.K. Using a Wilson plot, determine the condensing coefficient. Solution: Wilson plot 0.8 2 1 o V C C U 1 += Tabulation: 1/U o 1/V 0.8 0.000434783 0.852928 0.000483092 1.020461 0.000518135 1.13564 0.000568182 1.28489 0.000636943 1.485033 0.000735294 1.775269 0.000884956 2.234679 0.001156069 3.090923 By linear regression: C 1 = 0.000153033 C 2 = 0.000325563 But: m o o 1 kA xA h 1 C += ( ) 1.05155 24651 51 A A m o = + = x = (1/2)(51 - 46) = 2.5 mm = 0.0025 m k = 60 W/m.K ( )( ) 60 1.051550.0025 h 1 30.00015303 o += h o = 9,156 W/m 2 .K - - - Ans. 12-14. Develop Eq. (12-23) from Eq. (12-22). Solution: Eq. 12-22. 4 1 3 fg 2 cv tk4 xhg k xh         ∆µ ρ = Chapter12 80 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 10 of 11 L dxh h L 0 cv cv ∫ = L dx tk4 xhg x k h L 0 4 1 3 fg 2 cv ∫         ∆µ ρ = ∫        ∆µ ρ       = L 0 4 1- 4 1 fg 2 cv dxx tk4 hg L k h L 0 4 3 4 1 fg 2 cv x 3 4 tk4 hg L k h                   ∆µ ρ       = 4 3 4 1 fg 2 cv L 3 4 tk4 hg L k h               ∆µ ρ       = 4 1 3 fg 2 4 1 cv tLk khg 3 4 4 1 h         ∆µ ρ             = Ans. Eq. 12-23. 4 1 3 fg 2 cv tLk khg 0.943h         ∆µ ρ = 12-15. From Fig. 12-21, determine C and b in the equation h = C∆T b applicable to values in the middle of the typical range. Solution: Use Fig. 12-21 Tabulation: Heat-transfer Coefficient Heat flux ∆t W/m 2 .K, h W/m 2 K 400 710 1.775 600 1550 2.583 800 2820 3.525 1000 4170 4.170 1500 9000 6.000 h = C∆t b By Curve-Fitting: C = 212.8 b = 1.08 h = 212.8∆t 1.08 - - - Ans. 13-16. Section 12-18 makes the statement that on a graph of the performance of a water chilling evaporator with the coordinates of Fig. 12-23, a curve for a given entering water temperature is a straight line if the heat-transfer Chapter12 81 CHAPTER 12 - CONDENSERS AND EVAPORATORS Page 11 of 11 coefficients are constant. prove this statement. Solution: Use Fig. 12-23. t e = evaporating temperature t a = entering-water temperature (constant) U = heat-transfer coefficient (constant) ( )       − + =−= e ba abp t 2 tt UAttwcq ebaapbp UAt0.5UAt0.5UAttwctwc −+=− ( ) ( )0.5UAwc UAtt0.5UAwc t p eap b − −+ = ( ) ( )         − − −+ = a p eap p t 0.5UAwc UAtt0.5UAwc wcq ( ) ( ) ( )       − −−−+ = 0.5UAwcp t0.5UAwcUAtt0.5UAwc wcq apeap p ( )        − − = 0.5UAwc UAtUAt wcq p ea p ( )ea p p tt 0.5UAwc UAwc q − − = If U is constant. q = (constant)(t a - t e ) At constant t 0 , this is a straight line. - - - Ans. - 0 0 0 - Chapter12 82 CHAPTER 13 - EXPANSION DEVICES Page 3 of 14 ( ) ( )296 4 37108.86937101.7150.0002367 −− ×+×−=µ 0.00018544 =µ ( ) ( ) ( ) 43,1094792.2 0.0001812 0.00163 Re == Eq. 13-9. 0.25Re 0.33 f = ( ) 0.02303 42,132 0.33 f 0.254 == ( ) 0.02290 43,109 0.33 f 0.255 == 0.022965 2 0.022900.02303 fm = + = m/s6.2085 2 6.5225.895 Vm = + = Eq. 13-4 ( ) ( )45 2 54 VVwA 2 V D L f-p-p −=         ν ∆ Eq. 13-7 A w 2 V D L f 2 V D L f 2 ∆ = ν ∆ ( ) ( )4554 VV A w A w 2 V D L f-p-p −= ∆ ( ) ( )5.8956.5224792.2 A w 2 V D L f-1390.3-1425.81000 −= ∆ 32,495.3 A w 2 V D L f = ∆ ( ) ( ) ( ) ( ) 32,495.34792.2 2 6.2085 0.00163 L 0.022965 = ∆ ∆L 4-5 = 0.155 m - - - Ans. 13-2. A capillary tube is to be selected to throttle 0.011 kg/s of refrigerant 12 from a condensing pressure of 960 kPa and a temperature of 35 C to an evaporator operating at -20 C. (a) Using Figs. 13-7 and 13-8, select the bore and length of a capillary tube for this assignment. (b) If the evaporating temperature had been 5 C rather than -20 C, would the selection of part (a) be suitable? Discuss assumptions that have been made. Solution: Table A-5, p = 960 kPa, t sat = 40 C, Subcooling = 40 C - 35 C = 5 C (a) Use bore diameter D = 1.63 mm Fig. 13-7, 960 kPa inlet pressure, saturated. Flow rate = 0.0089 kg/s Fig. 13-8. Flow correction factor = (0.011 kg/s)/(0.0089 kg/s) Flow correction factor = 1.24 Then Length = 1,230 mm = 1.23 m L - - - Ans. (b) Use positions from 35 C to -20 C at 5 C increment. Table A-5, 35 C, sat. p = 847.72 kPa. Chapter13 85 CHAPTER 13 - EXPANSION DEVICES Page 4 of 14 At position 1, h 1 = 233.50 kJ/kg ν 1 = 0.78556 L/kg = 0.000786 m 3 /kg Table 15-5, µ 1 = 0.000202 Pa.s p 1 = 960 kPa ( ) 2 2 kg/s.m5271.4 40.00163 0.011 A w = π = ( )( )0.0007865271.4 A w V 11 =ν= V 1 = 4.143 m/s       µ       = νµ = 111 1 1 D A wDV Re ( )( ) 42,537 0.000202 0.001635271.4 Re1 == ( ) 0.02298 42537 0.33 Re 0.33 f 0.250.25 1 1 === At position 2, 30 C p 2 = 744.90 kPa h f2 = 228.54 kJ/kg h g2 = 363.57 kJ/kg ν f2 = 0.77386 L/kg = 0.000774 m 3 /kg ν g2 = 23.5082 L/kg = 0.02351 m 3 /kg µ f2 = 0.0002095 Pa.s µ g2 = 0.00001305 Pa.s 2a 4acbb x 2 −±− = ( ) 2 1 A w a 2 2 f2g2       ν−ν= ( ) ( ) 1.7182=−= 2 1 5271.40.0007740.02351a 22 ( ) ( ) 2 f2g2f2f2g2 A w hh1000b       ν−νν+−= ( ) ( )( )2 5271.40.0007740.023510.000774228.54363.571000b −+−= b = 135,519 ( ) 2 V 2 1 A w hh1000c 2 1 f2 2 1f2 −ν      +−= 2 ( ) ( ) ( ) ( ) 2 4.143 0.000774 2 1 5271.4233.50228.541000c 2 22 −+−= c = -4,960.3 2a 4acbb x 2 −±− = Chapter13 86 CHAPTER 13 - EXPANSION DEVICES Page 5 of 14 ( )( ) ( ) 0.0365 7182.12 4960.3-7182.14135,519135,519 x 2 = −±− = Then: h 2 = h f2 + x(h g2 - h f2 ) h 2 = 233.47 kJ/kg ν 2 = ν f2 + x(ν g2 - ν f2 ) ν 2 = 0.001604 m 3 /kg ( )f2g2f22 x µ−µ+µ=µ µ 2 = 0.0002023 Pa.s ( )( )0.0016045271.4 A w V 22 =ν= V 2 = 8.455 m/s       µ = µν = A wDVD Re 2 2 ( )( ) 42,474 0.0002023 0.001635271.4 Re 2 == 0.252 Re 0.33 f = ( ) 0.02299 42,474 0.33 f 0.252 == 0.022985 2 0.022990.02298 fm = + = m/s 2 4.142 Vm 299.6 455.8 = + = ( ) ( )1221 VV A w A w 2 V D L f-p-p −= ∆ ( ) ( ) ( ) ( ) ( ) ( )4.1438.4555271.45271.4 2 6.299 0.00163 L 0.022985-744.9-9601000 −= ∆ ∆L 1-2 = 0.8217 m At position 3, 25 C p 2 = 651.62 kPa h f2 = 223.65 kJ/kg h g2 = 361.68 kJ/kg ν f2 = 0.76286 L/kg = 0.000763 m 3 /kg ν g2 = 26.8542 L/kg = 0.026854 m 3 /kg µ f2 = 0.000217 Pa.s µ g2 = 0.0000128 Pa.s 2a 4acbb x 2 −±− = ( ) 2 1 A w a 2 2 f3g3       ν−ν= Chapter13 87 CHAPTER 13 - EXPANSION DEVICES Page 8 of 14 0.254 Re 0.33 f = ( ) 0.0230 42,285 0.33 f 0.254 == 0.0230 2 0.02300.0230 fm = + = m/s 2 13.748 Vm 993.16 237.20 = + = ( ) ( )3434 VV A w A w 2 V D L f-p-p −= ∆ ( ) ( ) ( ) ( ) ( ) ( )13.74820.2375271.45271.4 2 16.993 0.00163 L 0.0230-567.29-651.621000 −= ∆ ∆L 3-4 = 0.0793 m At position 5, 15 C p 5 = 491.37 kPa h f5 = 214.05 kJ/kg h g5 = 357.73 kJ/kg ν f5 = 0.74262 L/kg = 0.00074262 m 3 /kg ν g5 = 35.4133 L/kg = 0.0354133 m 3 /kg µ f5 = 0.0002355 Pa.s µ g5 = 0.0000124 Pa.s 2a 4acbb x 2 −±− = ( ) 2 1 A w a 2 2 f5g5       ν−ν= ( ) ( ) 16,701 2 1 5271.40.000742620.0354133a 22 =−= ( ) ( ) 2 f5g5f5f5g5 A w hh1000b       ν−νν+−= ( ) ( )( )2 5271.40.000742620.03541330.00074262214.05357.731000b −+−= b = 144,396 ( ) 2 V 2 1 A w hh1000c 2 4 f5 2 4f5 −ν      +−= 2 ( ) ( ) ( ) ( ) 2 20.237 0.00074262 2 1 5271.4233.31214.051000c 2 22 −+−= c = -19,457 2a 4acbb x 2 −±− = ( )( ) ( ) 0.1327 16,7012 19,457-16,7014144,396144,396 x 2 = −±− = Then: h 5 = h f5 + x(h g5 - h f5 ) Chapter13 90 CHAPTER 13 - EXPANSION DEVICES Page 9 of 14 h 5 = 233.12 kJ/kg ν 5 = ν f5 + x(ν g5 - ν f5 ) ν 5 = 0.005343 m 3 /kg ( )f5g5f55 x µ−µ+µ=µ µ 5 = 0.0002059 Pa.s ( )( )0.0053435271.4 A w V 55 =ν= V 5 = 28.165 m/s       µ = µν = A wDVD Re 5 5 ( )( ) 41,731 0.0002059 0.001635271.4 Re 5 == 0.255 Re 0.33 f = ( ) 0.02309 41,731 0.33 f 0.255 == 0.02305 2 0.023090.0230 fm = + = m/s 2 20.237 Vm 201.24 165.28 = + = ( ) ( )4554 VV A w A w 2 V D L f-p-p −= ∆ ( ) ( ) ( ) ( ) ( ) ( )20.23728.1655271.45271.4 2 24.201 0.00163 L 0.02305-491.37-567.291000 −= ∆ ∆L 4-5 = 0.0378 m At position 6, 10 C p 6 = 423.30 kPa h f6 = 209.32 kJ/kg h g6 = 355.69 kJ/kg ν f6 = 0.73326 L/kg = 0.00073326 m 3 /kg ν g6 = 40.9137 L/kg = 0.0409137 m 3 /kg µ f6 = 0.000246 Pa.s µ g6 = 0.0000122 Pa.s 2a 4acbb x 2 −±− = ( ) 2 1 A w a 2 2 f6g6       ν−ν= ( ) ( ) ,431 2 1 5271.40.000733260.0409137a 22 22=−= ( ) ( ) 2 f6g6f6f6g6 A w hh1000b       ν−νν+−= Chapter13 91 CHAPTER 13 - EXPANSION DEVICES Page 10 of 14 ( ) ( )( )2 5271.40.000733260.04091370.00073326209.32355.691000b −+−= b = 147,189 ( ) 2 V 2 1 A w hh1000c 2 5 f6 2 5f6 −ν      +−= 2 ( ) ( ) ( ) ( ) 2 28.165 0.00073326 2 1 5271.4233.12209.321000c 2 22 −+−= c = -24,189 2a 4acbb x 2 −±− = ( )( ) ( ) 0.1604 22,4312 24,189-22,4314147,189147,189 x 2 = −±− = Then: h 6 = h f6 + x(h g6 - h f6 ) h 6 = 232.80 kJ/kg ν 6 = ν f6 + x(ν g6 - ν f6 ) ν 6 = 0.007178 m 3 /kg ( )f6g6f66 x µ−µ+µ=µ µ 6 = 0.0002085 Pa.s ( )( )0.0071785271.4 A w V 66 =ν= V 6 = 37.838 m/s       µ = µν = A wDVD Re 6 6 ( )( ) 41,211 0.0002085 0.001635271.4 Re 6 == 0.256 Re 0.33 f = ( ) 0.02316 41,211 0.33 f 0.256 == 0.02313 2 0.023160.02309 fm = + = m/s 2 28.165 Vm 33 838.37 = + = ( ) ( )5665 VV A w A w 2 V D L f-p-p −= ∆ ( ) ( ) ( ) ( ) ( ) ( )28.16537.8385271.45271.4 2 33.0 0.00163 L 0.02313-423.30-491.371000 −= ∆ ∆L 5-6 = 0.0138 m At position 7, 5 C p 7 = 363.55 kPa h f7 = 204.64 kJ/kg h g7 = 353.60 kJ/kg Chapter13 92 CHAPTER 13 - EXPANSION DEVICES Page 13 of 14 13-4. Refrigerant 22 at a pressure of 1500 kPa leaves the condenser and rises vertically 10 m to the expansion valve. The pressure drop due to friction in the liquid line is 20 kPa. In order to have no vapor in the refrigerant entering the expansion valve, what is the maximum allowable temperature at that point? Solution: say ν = ν 1 p 2 = p 1 - gH/ν 1 - ∆p H = 10 m g = 9.81 m/s ∆p = 20 kPa p 1 = 1500 kPa Table A-6. ν 1 =0.8808 L/kg = 0.0008808 m 3 /kg p 2 = (1500)(1000) - (9.81)(10) / (0.0008808) - (20)(1000) p 2 = 1,368.62 kPa Table A-6. t 2 = 35.4 C - - - Ans. 13-5. A superheat-controlled expansion valve in a refrigerant 22 system is not equipped with an external equalizer. The valve supplies refrigerant to an evaporator coil and comes from the factory with a setting that requires 5K superheat in order to open the valve at an evaporator temperature of 0 C. (a) What difference in pressure on opposite sides of the diaphragm is required to open the valve? (b) When the pressure at the entrance of the evaporator is 600 kPa, how much superheat is required to open the valve if the pressure drop of the refrigerant through the coil is 55 kPa? Solution: Using Fig. 13-15 and deriving equation by assuming parabolic curve. Let y - pressure, kPa and x = temperature , C. y 2 - y 1 = A (x 2 2 - x 1 2 ) + B(x 2 - x 1 ) At 5 C evaporator temperature, 5 K superheat 100 kPa pressure differential x 1 = 5 C, x 2 = 5 C + 5 = 10 C y 2 - y 1 = 100 kPa 100 = A (10 2 - 5 2 ) + B (10 - 5) 100 = 75A + 5B - - Eq. 1 At -30 C evaporator temperature 12 C superheat 100 kPa pressure differential x 1 = -30 C, x 2 = -30 C + 12 = -18 C y 2 - y 1 = 100 kPa 100 = A ((-18) 2 -(-30) 2 ) + B ((-18) -(-30)) 100 = -576A + 12 B - - Eq. 2 But 5B = 100 - 75A Then 100 = -576A + 12 (20 - 15A) A = 0.185185 B = 17.222222 Chapter13 95 CHAPTER 13 - EXPANSION DEVICES Page 14 of 14 Therefore: y 2 - y 1 = 0.185185 (x 2 2 - x 1 2 ) + 17.222222(x 2 - x 1 ) (a) At 0 C evaporator temperature, 5 K superheat x 1 = 0 C x 2 = 0 C + 5 = 5 C y 2 - y 1 = 0.185185 (5 2 -0 2 ) + 17.222222(5 -0) y 2 - y 1 = 90.74 kPa - - - Ans. (b) At 0 C evaporator temperature, p = 497.59 kPa ∆p = 600 kPa + 55 kPa - 497.59 kPa = 157.41 kPa. x 1 = 0 C Then 157.41 = 0.185185 (x 2 2 -0 2 ) + 17.222222(x 2 -0) x 2 = 8.4 C x 2 - x 1 = 8.4 K - - - Ans. 13-6. The catalog of an expansion valve manufacturer specifies a refrigerating capacity of 45 kW for a certain valve when the pressure difference across the valve is 500 kPa. The catalog ratings apply when vapor-free liquid at 37.8 C enters the expansion valve and the evaporator temperature is 4.4 C. What is the expected rating of the valve when the pressure difference across it is 1200 kPa? Solution: Eq. 13-22 ( ) m/sfferencepressuredi2CVelocity = With all other data as constant except for pressure difference and refrigerating capacity. ( ) m/sfferencepressuredi2Capacity ingRefrigerat α Then: New Refrigerating Capacity ( ) kPa500 kPa1200 kW45= = 69.7 kW - - - - Ans. - 0 0 0 - Chapter13 96 CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS Page 1 of 11 14-1. Either graphically or by using the computer, for an ambient temperature of 30 C develop the performance characteristics of a condensing unit (of the form of Fig. 14-6 or Table 14-3) if the compressor has performance shown by Fig. 14-1 [ or Eq. (14-1) and (14-2)] and the condenser has characteristics shown by Fig. 14-3 [ or Eq. (14-4)]. Solution: Use mathematical computation: Use Fig. 14-3 or Eq. 14-4 q c = (9.39 kW/K)(t c - t amb ) at 30 C q c = (9.39 kW/K)(t c -30) Range of Evaporator Temperature, Fig. 14-1. -10 C, -5 C, 0 C, 5 C, and 10 C. Eq. (14-1), constant at Table 14-1, Fig. 14-1. 2 c 2 e9 2 ce8c 2 e7ce6 2 c5c4 2 e3e21e ttcttcttcttctctctctccq ++++++++= Eq. (14-2) constant at Table 14-1, Fig. 14-1. 2 c 2 e9 2 ce8c 2 e7ce6 2 c5c4 2 e3e21 ttdttdttdttdtdtdtdtddP ++++++++= Eq. (14-3) q c = q e + P Solving for t c at t e = -10 C ( ) ( ) 2 cc 2 e 0.001525t1.118157t-100.061652104.60437137.402q −−+−+= ( ) ( ) ( ) 2 cc 2 c t100.00026682t100.00040148t100.0109119 −−−−−− ( ) 2 c 2 t1030.00000387 −+ 2 cce 0.0015305t1.049186t97.5235q +−= ( ) ( ) 2 cc 2 0.0063397t-0.870024t100.01426100.8932221.00618P +−−−−= ( ) ( ) ( ) 2 cc 2 c t100.00014746t100.00023875t10-0.033889 −−−−+ ( ) 2 c 2 t10620.00000679 −+ 2 cc t0.004185480.507259t8.5124P −+= q c = q e + P 2 ccc t0.002654980.541927t-106.0359q −= Then, q c = (9.39 kW/K)(t c -30) 2 ccc t0.002654980.541927t-106.0359281.7-9.39t −= 0387.73599.931927tt0.00265498 c 2 c =−+ t c = 38.64 C ( ) ( )2e 38.640.001530538.641.04918697.5235q +−= Chapter14 97