PhEn-602 Pharmaceutical Facility Design Notes #7 J. Manfredi 1
HVAC Design and the Psychrometric Chart 2
HVAC Considerations Pharmaceutical Facility Design Process 1. SPACE REQUIREMENTS Temperature Humidity Air quality Pressurization 2. HEATING & COOLING REQUIREMENTS Air Conditioning loads Process equipment Loads Temperature 3. HVAC SYSTEM DESIGN Air systems Power plant (utility) Systems 4. TESTING/ EVALUATION Design stage Consturcion Post construction testing and commissioning Qualification/Validation 3
Design Process: Load Calculation Heat loss calculation is used to determine the heating requirements Calculate the heat lost from the building to the outside during a winter design condition. The common winter design condition is the coldest winter night in a particular climate that is surpassed only 2.5% of the time. 4
Design Process: Heating Load Types of Heat Loss: Conduction: (transmission) Heat transfer thru solid surfaces such as walls, floors, roofs. Convection (ventilation) Warm molecules move from one place to another typically by air to air exchange of energy Outside air entering (Doors, fans, etc.) Infiltration leakage through the building envelope is also a convective heat loss Radiation (typically electromagnetic) does not play a large role in calculating the heating load as it is generally a very small amount 5
Design Process: Heating Load Conduction Heat Loss: Conduction: Hc= UA (Δ T) U= overall heat transfer coefficient, (btu/hr x area x deg F) U= (1/R) R= resistance of the building material R = R1 + R2 + R3 A = surface area, in square feet Δ T= temperature difference 6
Design Process: Heating Load Convection Heat Loss: Convection: Hv=1.08x cfm x (ΔT) 1.08 = Air factor CFM = Rate air enters the building ΔT= Temperature difference Sources: Infiltration: outside air entering though cracks in the building envelope. Ventilation 7
Design Process: Heating Load Total Heat Loss: Ht = Hc + Hv 8
Design Process: Cooling Load Cooling load is the energy that must be removed to counteract the heat gain. Heat gain components: Conduction heat gain Convection heat gain Solar heat gain Internal heat gain (e.g. lights, motors, equipment) People Cooling load is measured in Btu/hr, or in tons 1 Ton = 12,000 Btu/hr 1 Btu = amount of energy required to raise the temperature of one pound of water 1 degree Fahrenheit. 9
Design Process: Cooling Load Conduction heat gain: Hc = UA(CLTD) U= overall heat transfer coefficient A = surface area, in square feet CLTD= cooling load temperature difference Calculated similar to conduction heat loss Recognizes that the exterior temperature of some building materials may exceed the outside temperature. (e.g. tar and gravel roof temperature may exceed 250 deg F during peak outside air conditions). 10
Design Process: Cooling Load Convection Heat Gain: (Two (2) parts) Sensible Heat Gain H sens = 1.08 x cfm x (Δ T) 1.08 is called the air factor. Air factor units: btu/hr/(cfm) deg F Latent heat gain (Moisture load) H latent = 4840 cfm (Δ W) W= heat gain from moisture generated in the space Measured in lb of water per lb of dry air 11
Design Process: Cooling Load Internal Heat Gain: From equipment, lights, motors, computers, copy machines Generally adds sensible heat On some occasion, can have equipment that generates a latent load (e.g. vial washing machine). 12
Design Process: Cooling Load Heat Gain from People (Two parts) Sensible Heat Gain Latent Heat Gain Hp = People Heat Gain (btu/hr) Hp= N x P s (CLF) + N x P L N= Number of occupants P s =Sensible Heat Gain per person CLF= Cooling load factor; usually 1.0 Dependent on number of hours in the space P L =Latent Heat Gain per person 13
People Heat Gain: Design Process: Cooling Load Function of activity performed Example; typical male factory worker, performing light bench work generates about 400 Btu/hr of Sensible Heat Gain and about 400 Btu/hr Latent Heat Gain For most activities, approximately 1,000 Btu/hr Heat Gain Combined 400 Sensible and 600 Latent Typical factory worker performing light work, generates approx. 900 Btu/hr (400 Btu/hr Sensible & 500 Btu/hr Latent) Dancing results in about 1,400 Btu/hr Gym activity 2,000 Btu/hr Females generate on average; 85% of male Heat Gain 14
Common Building Heat Gains ROOF Hc Lights Hv Hc Hi He Hp Hi Hc Hc Total heat Gain is sum of: Hp = People Heat Gain Hc = Conduction Heat gain=ua (CLTD) Hv = Convective Heat Gain = 1.1 cfm (T) Hi = Heat Gain from Lights He = Heat Gain from Equipt. 15
Design Process: Cooling Load In pharmaceutical facilities: Conduction Heat Gain may be significant Convection Heat Gain may be significant depending on process Ventilation load can be significant Solar Heat Gain is typically small Locale may have impact Internal Heat Gain (i.e.: lights, motors, equipment) will probably be significant Heat Gain from people can vary Generally small, but must be evaluated 16
Design Process: Cooling Load In pharmaceutical facilities the infiltration heat gain is typically negligible for two reasons: 1) generally tight construction 2) pressurization of rooms 17
Calculating the Cooling Load Across a Coil Q T = 4.5 x CFM x Δh Where Δh = h m h o (enthalpy of air entering the coil minus enthalpy of air leaving the coil in btu/lb) h m Enthalpy of mixed air entering the cooling coil in btu/lb h o Enthalpy of air leaving the cooling coil in btu/lb Q T in btu/hr 18
Total cooling load across coil Includes sensible and moisture load Q T = 4.5 x CFM x Δh Δh = h m h o h m = enthalpy of mixed air h o = enthalpy of air leaving coil hm Mixed air ho Leaving air HEPA Filter Outside air Return air from space Cooling Coil 95% Filter Supply air to the space 19
Total cooling load across coil Total load across the coil is the sum of the sensible load and the latent load. Q T =Q s + Q L 20
Sensible cooling load calculation Qs = AF x CFM x ΔT (AF = air factor) (Equal to 1.08) Qs= 1.08 (CFM) ΔT ΔT = Temperature difference of air side in degrees Fahrenheit Qs in Btu/hr 21
Latent cooling load calculation Q L = 4840 x CFM (ΔW) ΔW = moisture difference (moisture level entering coil in lb of water/lb of dry air, minus moisture level of air leaving the coil) in lb of water/lb of dry air Q L in Btu/hr 22
Latent cooling load calculation If moisture level is given in grains of water per lb of dry air: Latent cooling load calculation Q L = 0.69 CFM ΔW or Q L = 0.69 CFM ΔGrains ΔW = moisture difference (moisture level entering coil in grains of water/lb of dry air, minus moisture level of air leaving the coil) in grains of /lb of dry air Q L in Btu/hr Typically use grains of moisture per lb of dry air Note: 7,000 grains per lb (4,840/7,000 = 0.69) 23
Example: An air handling unit serves a Class 10,000 clean room. Flow through the unit is 5,000 CFM. The enthalpy of the air leaving the coil is 20 Btu/lb. The enthalpy of the air entering the coil is a mixture of return air and outside air, and has an enthalpy of 27 Btu/lb. What is the total cooling load across the coil? 24
Answer: Q T = 4.5 x CFM x Δh Q T = 4.5 x 5,000 x (27 20) Q T = 157,500 Btu/hr = 13.1 tons 25
Example (cont d): If the dry-bulb temperature exiting the coil was 52 F, and the mixed air temperature entering the coil was 70 F, what was the sensible load across the coil? Qs= 1.08 CFM x ΔT Qs = 1.08 x 5,000 x (70-52) Qs = 97,200 Btu/hr = 8.1 tons Based on the above, what was the latent load across the coil? Q T = sensible load + latent load 13.1 tons = 8.1 tons + latent load Latent load = 5 tons 26
Sensible Heat Ratio: SHR: ratio of sensible load divided by the internal total load. 27
Example: A clean room has a sensible heat gain of 90,000 BTU/HR, and a latent load of 30,000 BTU/HR. What is the sensible heat ratio? SHR = 90,000 / (90,000 +30,000) SHR = 90,000 / 120,000 = 0.75 28
Mixed air entering the coil Mixed-air airflow (CFM) is sum of outside-air (CFM) and return-air (CFM) Weighted average of the drybulb temperature of outsideair and return-air CFM ma x T ma = CFM oa x T oa + CFM ra x T ra 29
Example: Outside air at 500 CFM and 90 F dry-bulb. Return air at 3,000 CFM and 75 F dry-bulb. What is the temperature of the mixed air entering the cooling coil? CFM ma x T ma = CFM oa x T oa + CFM ra x T ra 3,500 x T ma = 500 x 90 + 3,000 x 75 3,500 x T ma = 270,000 T ma = 270,000/3,500 = 77.14 30
Psychrometric Chart Discussion Psychrometric chart Chart that displays the properties of air and water vapor Essential tool for designing temperature and humidity control systems 31
Psychrometric Chart Dry bulb temperature on the x-axis Humidity Ratio on vertical axis If we know two properties, we can obtain all other properties. Humidity Ratio also referred to as Specific Humidity. Units are grains of water per lb of dry air. 32
Reading the Psych Chart 33
Wet Bulb Temperature: Measured with wet wick reading Measures the temperature of the air as the moisture evaporates 34
Dew Point Temperature The temperature at which existing moisture will condense from the air (forming dew) 50 F 86 80 70, 50% rh = 50 Dew Point 70 60 50 40 30 20 50 35
Humidity Ratio, Specific Humidity Absolute measure of moisture The Ratio of the Weight of Water to the Weight of the Dry Air 36
Relative Humidity: The moisture content of air expressed as a percent of maximum moisture at the specified dry bulb temperature 37
Vapor Pressure: The pressure exerted by the water vapor molecules on the surrounding environment 0.37" 1.3 1.1 0.9 70, 50% rh = 0.37" Hg. 0.7 0.5 0.3 0.37" 0.1 38
Enthalpy Total energy in the air The higher the sensible temperature, the higher the enthalpy of the air. The higher the moisture content of the air, the higher the enthalpy. Measured in Btu/lb 39
Enthalpy Total energy in the air the higher the sensible temperature, the higher the enthalpy of the air. The higher the moisture content of the air, the higher, the enthalpy. 40
Air Conditioning Processes 41
HVAC Systems: Air properties Understand properties: Dry-bulb and wet-bulb temperature Humidity ratio Moisture content Enthalpy Dew point 42
HVAC Systems: Psychrometric chart Identify processes on the chart: Sensible heating and cooling Cooling and dehumidifying Desiccant dehumidifying chemical drying Humidifying Mixing 43
Psychrometric Chart Note: Actual process flow for cooling 200 Moisture (gr/lb) (Humidity Ratio) Temperature 0 44
Psychrometric Chart Actual process flow for cooling: 1) First sensible cooling occurs 2) Then dehumidification occurs a) ride along 100% saturation curve Cooling Process 200 Dehumidification 99% RH Sensible cooling Moisture (gr/lb) (Humidity Ratio) Temperature 0 45
Psychrometric Chart Cooling-based dehumidification Desiccant dehumidification 46
Sample problem: Solve using Psychrometric Chart Background Data: Assume we wish to design an air handling system to serve a sterile fill room. Room design conditions are 66 deg F DB, 45% RH. Assume a design outside air condition of 88 DB, 77 WB. The air handler serving the room supplies 22,000 CFM. Make-up air for pressurization and ventilation is 4,500 CFM. Assume that the exfiltration from the space is equal to the pressurization air of 4,500 CFM, with the balance returning to the air handler. Assume zero duct leakage. Due to the heat gain across the supply fan and supply ductwork, there is a 6 degree rise in supply air temperature. Due to the extensive amount of duct pressure losses, a return fan is installed to boost the pressure in the return duct, to ensure a consistent flow through the air handler. The return fan and return ductwork add 3.5 degrees to the air returning to the air handler. Assume the heat gain in the space is strictly sensible (negligible moisture gain). 47
Sample problem: Cooling coil Make-up Air 4,500 CFM Hin Hout AIRHANDLING UNIT Return air fan 22,000 CFM Return Airwall Class 100 Region Design T= 66 deg F RH=45% 48
Sample problem: Based on the information provided, answer the following: a) What is the Dry-bulb temperature and wet-bulb temperature of the air entering the cooling coil? b) What is the cooling load across the coil? 49