Doug Jefferson, Business Development Engineer System Design & Optimization Problems Doug Jefferson, Business Development Engineer

System Design Tools and Definitions Conversion Factors 7,000 gr (‘grains’) per lb(water vapor) 12,000 BTU/hr per Ton of cooling Cooling water GPM = (Btu/hr) / (500 x ΔT(water-side)) Steam condensate load in lb/hr = (Btu/hr) / (latent heat of steam @ X psig). 100 psig steam latent heat = 880 Btu/lb

System Design Tools and Definitions Moisture Load Equation = 4.5 x SCFM x ΔM = gr/hr Comments: Used to turn any humidity ratio (moisture) depression into a rate/load 0.075 lb/ft³ = ‘standard’ density of air 4.5 constant = 0.075 lb/ft³ x 60 min/hr SCFM = ‘standardized’ volume of air in ft³/min ΔM = change in moisture = gr/lb2 – gr/lb1 Across a room = gr/lb(set point) – gr/lb(delivered) Across a component = gr/lb(inlet) – gr/lb(outlet)

System Design Tools and Definitions Moisture Load Equation = 4.5 x SCFM x ΔM = gr/hr SCFM = 5,000 Room set point = 70°F , 30% RH = 33 gr/lb Delivered air condition = 10 gr/lb Moisture Removal = ?? Moisture Removal = 4.5 x 5,000 x (33 - 10) Moisture Removal = 517,500 gr/hr Moisture Removal = 73.9 lbs/hr

System Design Tools and Definitions Sensible Energy Equation = 1.08 x SCFM x ΔT = Btu/hr Comments: Used to turn any temperature depression into a rate/load 0.075 lb/ft³ = ‘standard’ density of air 0.24 Btu/lb-ºF = thermo conductivity of air 1.08 constant = 0.075 lb/ft³ x 60 min/hr x 0.24 Btu/lb-ºF SCFM = ‘standardized’ volume of air in ft³/min ΔT = change in temperature = T2 – T1 Across a room = T(set point) – T(delivered) Across a cooling component = T(inlet) – T(outlet) Across a heating component = T(outlet) – T(inlet)

System Design Tools and Definitions Sensible Energy Equation = 1.08 x SCFM x ΔT = Btu/hr SCFM = 5,000 Room set point = 70°F , 30% RH = 33 gr/lb Delivered air condition = 55°F Sensible Energy Removal = ?? Sensible Energy = 1.08 x 5,000 x (70 - 55) Sensible Energy = 81,000 Btu/hr Sensible Energy = 6.7 Tons

System Design Tools and Definitions Total Energy Equation = 4.5 x SCFM x Δh = Btu/hr Comments: Used to turn any enthalpy depression into a rate/load Use whenever you have a change in moisture and temperature 0.075 lb/ft³ = ‘standard’ density of air 4.5 constant = 0.075 lb/ft³ x 60 min/hr SCFM = ‘standardized’ volume of air in ft³/min Δh = change in enthalpy = h2 – h1 Across a system/unit = h(inlet) –h(outlet) Across a ‘wet’ cooling coil = h(inlet) – h(outlet)

System Design Tools and Definitions Total Energy Equation = 4.5 x SCFM x Δh = Btu/hr SCFM = 5,000 Room set point = 70°F , 30% RH = 33 gr/lb Delivered air condition = 55°F Enthalpy @ 70°F , 33 gr/lb = 22 Enthalpy @ 55°F , 33 gr/lb = 18.5 Total Energy Removal = ?? Total Energy = 4.5 x 5,000 x (22 - 18.5) Total Energy = 78,750 Btu/hr Total Energy = 6.6 Tons

System Design Tools and Definitions Mixing Two Air Streams Together Example: Cooling Coil 6,000 scfm ?F ? gr/lb 2,000 scfm 50F 53 gr/lb 2,000 scfm 95F 120 gr/lb 4,000 scfm 70F 20 gr/lb Return Air From Space

System Design Tools and Definitions Mixing Two Air Streams Together Solution, Temperature: 1. (2,000 scfm) x (50F) + (4,000 scfm) x (70F) = (6,000 scfm) x (?F) 2. Solve for ?F by Dividing Both Side of Equation By 6,000 scfm: (2,000 scfm) x (50F) + (4,000 scfm) x (70F) = (6,000 scfm) x (?F) (6,000 scfm) (6,000 scfm) Answer = 63.3F

System Design Tools and Definitions Mixing Two Air Streams Together Solution, Moisture: 1. (2,000 scfm) x (53 gr/lb) + (4,000 scfm) x (20 gr/lb) = (6,000 scfm) x (?gr/lb) 2. Solve for ?gr/lb by Dividing Both Side of Equation By 6,000 scfm: (2,000 scfm) x (53 gr/lb) + (4,000 scfm) x (20 gr/lb) = (6,000 scfm) x (?gr/lb) (6,000 scfm) (6,000 scfm) Answer = 31 gr/lb

System Design Tools and Definitions Mixing Two Air Streams Together Final Solution: Cooling Coil 6,000 scfm 63.3F 31 gr/lb 2,000 scfm 50F 53 gr/lb 2,000 scfm 95F 120 gr/lb 4,000 scfm 70F 20 gr/lb Return Air From Space

System Design Method Follow these steps while working through the following problems: Translate given information into values that can be used with the energy equations Iterate with moisture and sensible energy formulas to determine the system airflow. Remember, there is only one air stream to handle both the internal sensible and moisture loads in the space Optimize a system to deliver your calculated values to the space

System Design Method HINT: Don’t worry about the system layout before you’ve completed the iteration process. Use the internal sensible and moisture loads to calculate the process outlet Airflow, Temperature & Moisture. Then, design a system which best achieves these values and meets other customer requirements (i.e. minimum outside air volume, etc.) The Black Box AHU Control Space Set Points: 70F & 30% RH

System Optimization Problem #1 Room constraints and loads Set Points = 70F & 30% RH Summer ASHRAE design condition = 95F & 120 gr/lb Available utilities = 100 psig steam & 45F chilled water Internal moisture load = 27 lb/hr Internal sensible load = 8.1 tons Application requires 100% outside air

System Optimization Problem #1 Translate Set Points = 70F & 30% RH = 33 gr/lb Summer ASHRAE design condition = 95F & 120 gr/lb = 41.74 Btu/lb Available utilities = 100 psig steam & 45F chilled water Internal moisture load = 27 lb/hr x 7,000 gr/lb = 189,000 gr/hr Internal sensible load = 8.1 tons x 12,000 Btu/hr/ton = 97,200 Btu/hr Application requires 100% outside air - no return air from space

System Optimization Problem #1 Iterate One of the internal loads (moisture or sensible) will dictate the process airflow. HINT: You can tell it’s the ‘other’ load when your calculations require a -20F leaving air temperature and you only have 45F chilled water available for cooling.

System Optimization Problem #1 Iterate Pick a load to start with. In this example, we’ll use the internal moisture load of 189,000 gr/hr first: Moisture Load Equation = 4.5 x SCFM x ΔM = gr/hr 189,000 gr/hr = 4.5 x SCFM x (33 gr/lb – 10 gr/lb) (10 gr/lb is an educated guess for the process outlet moisture. It is a relatively easy value to achieve and will be a good starting point for this example.) Solve for SCFM = 1,826 scfm

System Optimization Problem #1 Iterate Use this value to see if it makes sense for the internal sensible load: Sensible Energy Equation = 1.08 x SCFM x ΔT = Btu/hr 97,200 Btu/hr = 1.08 x 1,826 scfm x (70ºF – ?ºF) Solve for ?ºF = 20.7ºF There is only 45F chiller water available so it will not be possible to deliver -20F to the space (plus, delivering very cold air to a space is typically not desired). Therefore, the internal sensible load (not the internal moisture load) must be determining system airflow.

System Optimization Problem #1 Iterate Recalculate: Sensible Energy Equation = 1.08 x SCFM x ΔT = Btu/hr 97,200 Btu/hr = 1.08 x SCFM x (70ºF – 55ºF) (55F is a reasonable educated guess for the process outlet temperature. It is relatively easy to design a dry final coil with a 10-degree ‘approach’ temperature.) Solve for SCFM = 6,000 scfm

System Optimization Problem #1 Iterate Recalculate for process outlet moisture: Moisture Load Equation = 4.5 x SCFM x ΔM = gr/hr 189,000 gr/hr = 4.5 x 6,000 scfm x (33 gr/lb – ? gr/lb) Solve for gr/lb = 26 gr/lb

System Optimization Problem #1 Iteration Summary Delivering 6,000 scfm @ 55F & 26 gr/lb will overcome BOTH internal loads. The Black Box AHU 6,000 SCFM 55ºF 26 gr/lb Control Space Set Points: 70F & 30% RH

System Optimization Problem #1 Now that the process outlet volume and conditions are determined, the system design can begin. The project assumptions state that 100% outside air is required for the application, so we know there is no return air from the space. Filtration, fans and other system components can be added later – we will just focus on the basic temperature and moisture psychrometrics of the system. We’ll consider three optimization possibilities in this example. Optimization is in the eye of the beholder and many designs have more than one ‘correct’ answer. The most ‘correct’ solution depends on the customer’s requirements and utility availability.

System Optimization Problem #1 Solution #1 Put all the 100% outside air through the dehumidifier: Process outlet conditions = 152F & 59 gr/lb. The requirement for this design is 26 gr/lb, so this design will not meet the requirements of the project.

System Optimization Problem #1 Solution #2 Put all the 100% outside air through the dehumidifier but add a pre-cooling coil (i.e. a pre-dehumidifier): Process outlet conditions = 140F & 26 gr/lb. This meets the requirements of the project. A pre-cooling coil will cool/pre-dehumidify to 67F sat., just enough to achieve the required system performance. A final cooling coil will be needed to cool from 140F (plus fan heat) to 55F.

System Optimization Problem #1 Solution #2 System Flow Schematic

System Optimization Problem #1 Solution #2 Load summary and calculations: Reactivation steam heater = 546,000 Btu/hr (from wheel run) Reactivation steam load = 546,000 / 880 = 620.5 lb/hr Pre-cool coil: Outside air enthalpy @ 95F & 120 gr/lb = 41.74 Btu/lb Pre-cool set point enthalpy @ 67F sat. = 31.59 Btu/lb Pre-cool Load = 4.5 x 6,000 x (41.74 – 31.59) / 12,000 = 22.8 Tons Post-cool Load = 1.08 x 6,000 x (145 – 55) / 12,000 = 48.6 Tons

System Optimization Problem #1 Solution #3 Treat all the 100% outside air with the system but add a pre-cooling coil (i.e. a pre-dehumidifier), reduce the dehumidifier size and add a face and bypass section: Face and Bypass Blend Net process outlet conditions = 93F & 26 gr/lb. This meets the requirements of the project. A pre-cooling coil will cool/pre-dehumidify to 55F sat. A final cooling coil will be needed to cool from 93F (plus fan heat) to 55F.

System Optimization Problem #1 Here is what the system will look like:

System Optimization Problem #1 Solution #3 Load summary and calculations: Reactivation steam heater = 285,000 Btu/hr (from wheel run) Reactivation steam load = 285,000 / 880 = 324 lb/hr Pre-cool coil: Outside air enthalpy @ 95F & 120 gr/lb = 41.74 Btu/lb Pre-cool set point enthalpy @ 55F sat. = 23.21 Btu/lb Pre-cool Load = 4.5 x 6,000 x (41.74 – 23.21) / 12,000 = 41.7 Tons Post-cool Load = 1.08 x 6,000 x (98 -55) / 12,000 = 23.2 Tons

System Optimization Problem #1 Summary of Designs Solution 1 Solution 2 Solution 3 Dehumidifier Size HCD-9000 HCD-4500 Reactivation (Btu/hr) N/A 546,000 285,000 Pre-cool (tons) 22.8 41.7 Post-cool 48.6 23.2 Total Tons 71.4 64.9

System Optimization Problem #1 Comments on Problem #1 System Optimization Solution 1: Didn’t work. Many desiccant application require a mixed system with pre-cooling ‘wet’ coils upstream of a desiccant. This maximizes the effectiveness of the desiccant by delivering cool and dry air to the wheel. Solution 2: This design achieves all of the requirements of the control space, delivering 6,000 scfm @ 55F and 26 gr/lb. However, a larger desiccant section is required to overcome the additional moisture load not removed by the pre-cooling coil. This option would seem to be acceptable due to the decreased pre-cooling load (vs. Solution 3). However, referring to the load summary, this is not the case. While the pre-cool load is decreased, the post-cooling coil load is increased due to the latent-to-sensible heat conversion across the desiccant (more ΔM) plus the higher entering temperature. Solution 3: This design would be considered to be the ‘most’ optimized. The overall cooling tonnage is lower than Solution 2 (6.4 tons and 9% saved), there is a 261 MBH savings of reactivation energy (48%) plus first cost savings due to the smaller desiccant section With mixed systems, maximizing the outside air pre-cooling before the desiccant almost always saves first system costs and operating costs.

System Design & Optimization Problem #2

System Optimization Problem #2 Room constraints and loads Set Points = 68F & 20% RH Summer ASHRAE design condition = 95F & 120 gr/lb Available utilities = 100 psig steam & 45F chilled water Internal moisture load = 4.3 lb/hr Internal sensible load = 42,800 Btu/hr Application requires a minimum of 500 scfm for pressurization

System Optimization Problem #2 Translate Set Points = 68F & 20% RH = 20 gr/lb Summer ASHRAE design condition = 95F & 120 gr/lb = 41.74 Btu/lb Available utilities = 100 psig steam & 45F chilled water Internal moisture load = 4.3 lb/hr x 7,000 gr/lb = 30,100 gr/hr Internal sensible load = 42,800 Btu/hr Application requires 500 cfm for pressurization

System Optimization Problem #2 Iterate One of the internal loads (moisture or sensible) will dictate the process airflow. HINT: You can tell it’s the ‘other’ load when your calculations require a -20F leaving air temperature and you only have 45F chilled water available for cooling.

System Optimization Problem #2 Iterate Pick a load to start with. In this example, we’ll use the internal sensible load of 42,800 Btu/hr first: Sensible Energy Equation = 1.08 x SCFM x ΔT = Btu/hr Assume 50 ºF delivered air condition 42,800 Btu/hr = 1.08 x scfm x (68ºF – 50ºF) Solve for ? Scfm = 2,200 scfm

System Optimization Problem #2 Iterate Calculate for process outlet moisture: Moisture Load Equation = 4.5 x SCFM x ΔM = gr/hr 30,100 gr/hr = 4.5 x 2,200 scfm x (20 gr/lb - ? gr/lb) Solve for ? gr/lb = 17 gr/lb

System Optimization Problem #2 Iteration Summary Delivering 2,200 scfm @ 50F & 17 gr/lb will overcome BOTH internal loads. Control Space Set Points: 68F & 20% RH The Black Box AHU 2,200 SCFM 50ºF 17 gr/lb

System Optimization Problem #2 Now that the process outlet volume and conditions are determined, the system design can begin. The project assumptions state that 500 cfm of make up air is required for pressurization. Return air blended with this makeup air will be used to satisfy internal moisture and sensible loads. Filtration, fans and other system components can be added later – we will just focus on the basic temperature and moisture psychrometrics of the system.

Dehumidify Only the Make-up Air Pre-cool Make-up—Blend—Then DH

Blend Make-up And Return Air Before Cooling Blend Before DH - No Pre-cool

System Optimization Problem #2 Solution #1 Mix the return and make up air and put 100% of the air through the dehumidifier:

System Design Tools and Definitions Mixing Two Air Streams Together Example: Return Air From Space 500 scfm 95F 120 gr/lb 1,700 scfm 68F 20 gr/lb 2,200 scfm ?F ? gr/lb Make-up Outside

System Design Tools and Definitions Mixing Two Air Streams Together Example: 2,200 scfm 74F 43 gr/lb 500 scfm 95F 120 gr/lb Make-up Air From Outside 1,700 scfm 68F 20 gr/lb Return Air From Space

System Optimization Problem #2 Solution #1a Mix the return and make up air and put 100% of the air through the dehumidifier: Process outlet conditions = 109F & 12 gr/lb. The requirement for this design is 17 gr/lb, so this design is more than required

System Optimization Problem #2 Solution #1b Mix the return and make up air and put 100% of the air through the dehumidifier: Blend outlet conditions = 106.5F & 17 gr/lb. The requirement for this design is 17 gr/lb, so this design meets the requirement.

System Optimization Problem #2 Solution #1 Load Summary and calculations: Reactivation steam heater = 90,000 Btu/hr (from wheel run) Reactivation steam load = 90,000 / 880 = 102.3 lb/hr Cooling Coil: Post-cool load = 1.08 x 2,200 x (106.5-50) / 12,000 = 11.2 tons

System Optimization Problem #2 Solution #2 Treat Make-Up Air with a cooling coil before mixing with Return Air:

System Design Tools and Definitions Mixing Two Air Streams Together Example: Cooling Coil Return Air From Space 500 scfm 95F 120 gr/lb 50F 53 gr/lb 1,700 scfm 68F 20 gr/lb 2,200 scfm ?F ? gr/lb

System Design Tools and Definitions Mixing Two Air streams Together Example: Cooling Coil 2,200 scfm 64F 28 gr/lb 500 scfm 50F 53 gr/lb 500 scfm 95F 120 gr/lb 1,700 scfm 68F 20 gr/lb Return Air From Space

System Optimization Problem #2 Solution #2 Put all the 100% outside air through the dehumidifier and add a pre-cooling coil on the make up air: Meets design condition of 17 gr/lb

System Optimization Problem #2 Solution #2 Load summary and calculations: Reactivation steam heater = 42,000 Btu/hr (from wheel run) Reactivation steam load = 42,000 / 880 = 47.7 lb/hr Pre-cool coil: Outside air enthalpy @ 95F & 120 gr/lb = 41.74 Btu/lb Pre-cool set point enthalpy @ 50F sat. = 20.3 Btu/lb Pre-cool load = 4.5 x 500 x (41.74 - 20.3) / 12,000 = 4 Tons Post-cool load = 1.08 x 2,200 x (80 – 50) / 12,000 = 6 Tons

System Optimization Problem #2 Solution #3 Blend make up air and return air, precool air to 50F.

System Design Mixing Two Air streams Together Example: Cooling Coil 2,200 scfm ?F ? gr/lb 2,200 scfm 74F 43 gr/lb 500 scfm 95F 120 gr/lb Make-up Air From Outside 1,700 scfm 68F 20 gr/lb Return Air From Space

System Design Tools and Definitions Mixing Two Air Streams Together Example: Cooling Coil 2,200 scfm 50F 43 gr/lb 2,200 scfm 74F 43 gr/lb 500 scfm 95F 120 gr/lb Make-up Air From Outside 1,700 scfm 68F 20 gr/lb Return Air From Space

System Optimization Problem #2 Solution #3a Blend make up air and return air, precool air to 50F, process through dehumidifier Does not meet design condition of 17 gr/lb

System Optimization Problem #2 Solution #3b Blend make up air and return air, precool air to 50F, process air through dehumidifier Meets design condition of 17 gr/lb

System Optimization Problem #2 Solution #3 Load summary and calculations: Reactivation steam heater = 90,000 Btu/hr (from wheel run) Reactivation steam load = 90,000 / 880 = 102.3 lb/hr Cooling coils: Pre-cool load = 1.08 x 2,200 x (74 – 50) / 12,000 = 4.8 Tons Post-cool load = 1.08 x 2,200 x (82.5 – 50) / 12,000 = 6.5 Tons

System Optimization Problem #2 Summary of Designs Solution 1 Solution 2 Solution 3 Dehumidifier Size HCD-2250 HCD-1125 Reactivation (lbs/hr) 102.3 47.7 Pre-cool (tons) N/A 4 4.8 Post-cool 11.2 6 6.5 Total Tons 10 11.3

System Optimization Problem #2 Comments on Problem #2 System Optimization Solution 1: System is able to deliver required moisture content. Reactivation energy is high when compared to Solution 2. One advantage of Solution 1 is the use of only one cooling coil versus two cooling coils. This would save the end user on first cost and installed cost. Solution 2: This design is the most efficient for both steam and cooling requirements. A smaller drier is used which typically results in a lower first cost. Solution 3: This design has a relatively high requirement for steam and cooling. This unit would also use a larger drier than Solution 2 and a higher first cost with two cooling coils versus one. This design would be the worst of the three. With mixed systems, maximizing the outside air pre-cooling before the desiccant almost always saves first system costs and operating costs.

System Optimization Summary Determine sensible and moisture loads Determine system airflow to meet loads What are the objectives and limitations of project Lowest energy Lowest first cost Smallest footprint Future capacity Utilities available Determine system layout When to blend make up and return air Where to locate cooling coils What is the correct drier size