Balancing of a Water and Air System (PDF

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14 The basic airflow equation for any free area is {airflow (cfm) = area (ft²) X velocity (fpm)} 2000 cfm = 2ft² X 1000 fpm {area (ft²) = airflow (cfm) / velocity (fpm)} 2ft²= 2000 cfm / 1000 fpm {velocity (fpm) = airflow (cfm) / area (ft²)} 1000 fpm = 2000 cfm / 2ft² {Velocity fpm = 4005 X √ Velocity pressure} 997.2 fpm = 4005 X √ 0.062 in. w.g. Free area defined as the total minimum area of openings in an air outlet or inlet device through which air can pass. Free area of return air or supply air grilles may be as low as 50% of the duct connection size. The free cross-sectional area of a duct normally is 100%. If other data are not available, it may be assumed that all similar return air or supply air grilles would have similar free areas when measured with the same instrument. Airflow equation for a return air or supply air grilles {Airflow (cfm) = area (ft²) X velocity (fpm) X % of free area} 2250 cfm = 3 ft² X 1000 fpm X 75% free area {% of free area = airflow (cfm) / area (ft²) X velocity (fpm)} 75% = 2250 cfm / 3 ft² X 1000 fpm Example 1: find the cfm of a duct of 24”X12” with a velocity of 1000 fpm. Find the cfm of the duct. Solution 24”X12”/144= 2 ft² 2 ft² X 1000 fpm = 2000 cfm Example 2: A 48”X36” return air grille has a measured average velocity of 370 fpm. A Pitot tube traverse of the connecting duct indicates airflow of 2975 cfm. Find the return air grille free area percentage. Solution Free area is equal to; duct airflow in cfm divided by grille velocity in fpm, Divided by the area of the grill times 100 2975 cfm / 370 fpm = 8.04 48”X36”/144= 12 ft² grill 8.04 / 12 ft² = 0.67 0.67 X 100 = 67% free area 14
15 15 Example 3: Find the nearest standard size round duct to handle 4600 cfm at a velocity of 1000 fpm. Solution Area of the duct in ft² is equal to: Airflow cfm / Velocity fpm Food square to inch square is equal to: ft² X 144 = in² In² to duct radius is equal to: √ square root of; area in² / ∏ (3.14) Radius to diameter is equal to: radius X 2 = diameter 4600 cfm / 1000 fpm = 4.6 ft² 4.6 ft² X 144 = 662.4 in² 662.4 in² / 3.14 ∏ = 210.95 Duct Flow √ of 210.95 = 14.52 14.52 X 2 = 29.05 29.05” diameter of a duct for 4600 cfm at 1000 fpm The preferred method of measuring duct volumetric flow is the pitot-tube traverse average. The maximum straight run should be obtained before and after the traverse station. To obtain the best duct velocity profile, measuring points should be located as shown in Chapter 14 of the 2001 ASHRAE Handbook—Fundamentals and ASHRAE Standard 111. When using factory-fabricated volume-measuring stations, the measurements should be checked against a pitot-tube traverse. Power input to a fan’s driver should be used as only a guide to indicate its delivery; it may also be used to verify performance determined by a reliable method (e.g., pitot-tube traverse of system’s main) that considers possible system effects. For some fans, the flow rate is not proportional to the power needed to drive them. In some cases, as with forward-curved-blade fans, the same power is required for two or more flow rates. The backward-curved-blade centrifugal fan is the only type with a flow rate that varies directly with the power input. Pitot Tube Traverses Procedures • To accomplish repeatable traverse measurements, take the measurements in a specific, measured pattern, as indicated in 3.2. below. • Duct size must not change in a traversed section. • Face the Pitot tube into the air stream, parallel to the ductwork at each measurement point, and measure the velocity. • Convert velocity pressure to fpm velocity before averaging if the traverse is taken at other than standard conditions. • Take traverse measurements at actual conditions and actual cubic feet per minute [Actual CFM]. Correct Actual CFM to standard CFM [Standard CFM] when specified by 15
Page 1 and 2: Balancing of a water and air system
Page 3 and 4: 3 8 TBA (Testing, Balancing and Adj
Page 5 and 6: 5 Air Balancing Multiple Hood syste
Page 7 and 8: 7 87 Series Loop 88 One pipe main 9
Page 9 and 10: 9 Testing and Balancing HVAC A syst
Page 11 and 12: 11 13. Filter frame properly instal
Page 13: 13 13. Measure and record all requi
Page 17 and 18: 17 For traverses taken in round duc
Page 19 and 20: 19 Flat Oval Duct Traverses For fie
Page 21 and 22: 21 Balancing Devices Volume Dampers
Page 23 and 24: 23 Turning Vanes Turning vanes (Fig
Page 25 and 26: 25 Balancing the VAV System The gen
Page 27 and 28: 27 maximum out-side air damper for
Page 29 and 30: 29 determines how much energy is in
Page 31 and 32: 31 the water removing capacity of a
Page 33 and 34: 33 Table 8A Dry Bulb on left, Wet B
Page 35 and 36: 35 60 58 56 54 52 50 48 46 44 42 40
Page 37 and 38: 37 Foot³ per 1lb Altitude 0 Feet A
Page 39 and 40: 39 Table 9 C Air Density at 29.29
Page 41 and 42: 41 41 Wet-Bulb (F) .0 .1 .2 .3 .4 .
Page 43 and 44: 43 Table 11A Enthalpy at saturation
Page 45 and 46: 45 Table 11 C Enthalpy at saturatio
Page 47 and 48: 47 Wet bulb temperature____________
Page 49 and 50: 49 and replacement should be shut o
Page 51 and 52: 51 duct may be great; the balancing
Page 53 and 54: 53 Design below the 250°F limit (A
Page 55 and 56: 55 Differential Pressure Gauges The
Page 57 and 58: 57 Venturi The venturi is a device
Page 59 and 60: 59 Pressure Total pressure Velocity
Page 61 and 62: 61 Table 1 shows that if the coil i
Page 63 and 64: 63 adjustment of these devices is n
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65 Grouting Motor and lubrication N
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67 For every outside temperature ot
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69 4. Identify the riser with the h
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71 2. Turn the pump off 3. Close th
Page 73 and 74:
73 Net positive suction head availa
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75 All automatic control valves mus
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77 77
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79 4. Read and record the actual pu
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81 13. Record the coils design and
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83 Air Control Air is always presen
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85 • allow draining of all or par
Page 87 and 88:
87 Valves have linear throttling ch
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89 to all the terminals down the li
Page 91 and 92:
91 Pump Circuits Primary-Secondary
Page 93 and 94:
93 • Evaporator: • the design a
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95 95
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97 Condenser • the design and act
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99 • Each reading of liquid tempe
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101 • Design and actual entering
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103 differs from the average by mor
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105 • Orifice plate • Turbine m
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107 BTUH = GPM X 60 X 8.337 X ∆T
Page 109 and 110:
109 Lam = lbs. of air per minute th
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111 When dry bulb temperature measu
Page 113 and 114:
113 GPM = FPM X Gallons per foot FP
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