# Patent

## Title
Water-evacuator for air conditioner

## Abstract

A method for retrofitting an aircraft air conditioning unit drainage system comprises the step of removing drain pipe from the unit housing. A hole from which the drain pipe extends is plugged. A first location within a unit housing is determined where during operation of the unit air pressure is lower than at a second location. A second location below the first location is also determined where pooling occurs. A first and a second conduit with first and second internal diameters are respectively located at the first and the second locations. The first conduit and the second conduit are connected to a third conduit which extends generally vertically and which has a third internal diameter that is generally equal to the first internal diameter. A fourth conduit with a fourth internal diameter that is less than the second internal diameter is connected to the third conduit below the second location.

## Background
CROSS-REFERENCE TO RELATED APPLICATIONS
This Application is a continuation of U.S. patent application Ser. No. 13/664,026, filed Oct. 30, 2012, now allowed, which claims priority to U.S. Provisional Application No. 61/651,908 filed May 25, 2012. The disclosure of each is incorporated herein by reference in its entirety.


BACKGROUND OF THE INVENTION
1. Field of the Invention
The invention relates generally to the field of removing a liquid using pressure differentials and gravitational forces. More specifically, the invention relates to the field of removing condensation from confined air conditioning modules, e.g. aircraft air conditioning modules.
2. Description of the Related Art
Air conditioning units are well known in the art, and various methods have heretofore been utilized to drain and dispose the condensate that is generated by air conditioners during cooling cycles. For example, condensate generated by air conditioning units in commercial buildings may be drained using pumps and disposed into rain gutters. Condensate generated by air conditioning units in automobiles may be drained using gravity and disposed on the ground underneath the automobiles via drain tubes.
Air conditioning units used in aircraft may include vapor cycle evaporator drain systems to effectuate drainage of the condensate. FIG. 1 shows an exemplary aircraft air conditioning unit. As can be seen from this figure, the air conditioning unit includes a housing having an angled face, and a drain tube extending from this angled face is used to drain the condensate generated by the unit. If the blower motor is located close to the drain tube, effectiveness of the drain tube during operation of the air conditioning unit is significantly hampered. Specifically, air that is sucked into the unit's housing because of fan suction causes the pressure inside the housing to exceed the ambient pressure, which in-turn prevents the condensate from draining out the drain tube. The condensate thus forms a pool at the floor of the housing, and is often undesirably sprayed out the exhaust, causing damage to the aircraft's interior and inconveniencing the passengers. A common prior art solution to this problem is to locate the blower at an appreciable distance from the evaporator box. However, due to the limited space available in aircraft, this solution is not always feasible.
U.S. Pat. No. 7,543,458 to Wurth is directed to a portable vapor cycling air conditioning unit for use with small aircraft. Wurth ('458) discloses the use of a Venturi drainage system to drain the accumulated condensate. Condensate is conducted from the evaporator via a drain pipe to one or more extraction Venturi tubes that are positioned in a low pressure area in a blower. Forced air from the blower passes over the Venturi tubes and draws the moisture out of the unit. A ducting system is provided to draw air from outside the aircraft and to dispose of condensate and waste heat outside the aircraft.
SUMMARY
Systems and methods for draining condensate from air conditioning units are disclosed herein. According to an embodiment, a method for removing condensate from an air conditioning unit comprises the step of determining a first location within the unit housing where during operation of the air conditioning unit air pressure is lower than it is at a second location. The air conditioning unit has a housing, a blower, coils, an air intake duct, and an exhaust. The method comprises the step of determining the second location within the housing where pooling of condensate occurs. The second location is below the first location. The method includes the step of locating a first conduit with a first internal diameter at the first location and the step of locating a second conduit with a second internal diameter at the second location. The method comprises the step of connecting the first conduit and the second conduit to a third conduit with a third internal diameter that is generally equal to the first internal diameter. The third conduit extends generally vertically below the second location. The method further includes the step of connecting a fourth conduit with a fourth internal diameter to the third conduit below the second location. The fourth internal diameter is less than the first internal diameter and the second internal diameter. The fourth conduit has an orifice for allowing condensate to exit the fourth conduit.
According to an another embodiment, a method for retrofitting a drainage system in an aircraft air conditioning unit comprises the step of removing a drain pipe from the unit housing. The air conditioning unit has a housing, a blower, coils, an air intake duct, and an exhaust. The method includes the step of plugging a hole from which the drain pipe extended. A first location within the housing is determined where air pressure is lower than at a second location during operation of the unit. A second location within the housing where pooling of condensate occurs is also determined. The second location is below the first location. The method includes the step of locating a first conduit with a first internal diameter at the first location, and the step of locating a second conduit with a second internal diameter at the second location. The first conduit and the second conduit are connected to a third conduit with a third internal diameter that is generally equal to the first internal diameter. The third conduit extends generally vertically below the second location. The method comprises the step of connecting a fourth conduit with a fourth internal diameter to the third conduit below the second location. The fourth internal diameter is less than the first internal diameter and the second internal diameter. The fourth conduit has an orifice for allowing condensate to exit the fourth conduit.
According to yet another embodiment, a method for removing condensate from an air conditioning unit in an aircraft includes the step of determining a first location within a unit's housing where during operation of the air conditioning unit air pressure is lower than at a second location. The air conditioning unit, in addition to a housing, has a blower, coils, an air intake duct, and an exhaust. The method comprises the step of determining the second location within the housing where pooling of condensate occurs. The second location is below the first location. The method further comprises the step of providing a conduit. The conduit includes an outlet portion having a first segment, a second segment, and a first port. The first and the second segments each lie in a first horizontal plane that is above a second horizontal plane that passes through an internal region of the blower. The conduit also has an inlet portion having a first section, a second section, and a second port. The first and the second sections lie in a third horizontal plane that is below the first horizontal plane and the second horizontal plane. The method includes the step of operably coupling the first port to the first location and the second port to the second location. The method comprises the step of operably coupling a gravitational separator portion having an end to the outlet portion and the inlet portion such that the gravitational portion extends generally vertically. The method also includes the step of operably coupling a draining portion having an orifice to a gravitational separator portion above the end.



BRIEF DESCRIPTION OF THE SEVERAL VIEWS OF THE DRAWINGS
Illustrative embodiments of the present invention are described in detail below with reference to the attached drawing figures, which are incorporated by reference herein and wherein:
FIG. 1 is a perspective view of an aircraft air conditioning system having a prior art drainage system.
FIG. 2 is a perspective view of the aircraft air conditioning system of FIG. 1 having a water evacuator system in accordance with the teachings of the current invention.
FIG. 3 is a perspective view of the aircraft air conditioning system of FIG. 2 showing air intake and exhaust duct arrangements.
FIG. 4 is a perspective view of the aircraft air conditioning system of FIG. 2 with a part of a housing and coils of an internal chamber removed to reveal locations of ports of the water evacuator system.
FIG. 5 is an assembly drawing showing certain equipment used to install the water evacuator system.
FIG. 6 is a perspective view of the water evacuator system.
FIG. 7 shows the outside of the housing of the aircraft air conditioning system of FIG. 2 illustrating the location of installation of the water evacuator system.
FIG. 8 shows inner diameters of conduits and a drain of the water evacuator system, and includes arrows depicting the flow of air and condensation.
FIG. 9 shows a height of a drop between an induction tube and the drain of the water evacuator system.
FIG. 10 shows the inner diameters of the water evacuator system's conduits and drain, as well as the corresponding pressures.
FIG. 11 shows the inner diameters of the water evacuator system's conduits and drain, as well as the corresponding mass flow rates.
FIG. 12 shows the pressure loss per unit length.
FIGS. 13A-C show possible design variations of the water evacuator system in accordance with the teachings of the current invention.



DETAILED DESCRIPTION
Embodiments of the present invention provide systems and a method for removing condensate from an air conditioning unit in which size and/or weight are restricted.
A prior art version of an aircraft air conditioning drainage system is shown in FIG. 1. Referring to the figure, an air conditioning unit 10 has both a cooling portion 12 and an air-handling portion 14. The cooling portion 12 receives unconditioned air from an intake 16 into a chamber inside a housing 18 which includes cooling coils (coils 134 shown in FIG. 3). The air-handling portion 14 includes a motor with housing 20 used to operate a blower 22 in a known manner. Air received into intake 16 is drawn through the heat-exchanging coils in the housing 18 and then out an exhaust 24. The coils receive a circulation of a heat-transfer medium (e.g., refrigerant R-12, R134) from conduits, e.g., hoses or pipes, through connections made into couplers 26 and 28 in a known fashion.
In operation, the heat-exchanging coils in the chamber in the housing 18 develop condensate which drips to form a pool at the floor of the housing 18. This pooling of condensate has been conventionally handled using a drain tube 30 which extends out from an angled side face 32 of the prior art air conditioning unit 10 shown in FIG. 1. The effectiveness of the drain tube 30 is limited. With the blower motor on, the drain 30 does not eliminate pooling within the chamber in the housing 18, and therefore, a significant amount of condensate remains at the floor of the housing 18 and on the heat-exchanger coils. Specifically, the ambient barometric pressure at the far end of drain tube 30 is greater than pressure inside the housing chamber due to fan suction, so air is sucked into housing 18 and water does not drain properly. Because of this, sudden blasts of water are emitted out of the exhaust 24 when the amount of water collected is high. The sprayed water can cause water damage to the aircraft interior, and also disturb passengers. Water will only drain properly from the air conditioning unit 10 when the motor is off and the pressure is equalized.
FIGS. 2-13 illustrate an improvement over the arrangement shown in FIG. 1. Referring first to FIG. 2, it can be seen that much of the system is just like the prior art version shown in FIG. 1. As shown, air conditioning unit 110 of FIG. 2 has both cooling 112 and air-handling 114 portions, an air intake 116, a housing 118 for cooling coils 134 (see FIG. 3), a motor with housing 120, a blower 122, an exhaust 124, couplers 126 and 128, and an angled side face 132, all like in the conventional air conditioning unit 10 shown in FIG. 1.
Looking at the angled side face 132 in FIG. 2 reveals, however, that the drain tube 30 seen in FIG. 1 has been removed, and the hole where the stem existed has been eliminated. Where a retrofit is made to an existing air conditioning unit in the field, the tube 30 may be removed, and the hole from which drain tube 30 emanates may be plugged or otherwise sealed. Where the air conditioning unit 110 is a new fabrication, it can be designed with a housing 118 that does not include a drain stem (stem 30 is removed as shown in FIG. 2).
Referring to FIG. 3, it can be seen that a condensation removal device 130 (a system of conduits) has been added to the overall air conditioning unit 110 on the side of the housing 118 where the blower 122 is attached. Known through analysis of experimentation, device 130 accesses a low pressure location in the housing 118 and uses the resulting suction to draw condensate out of the unit 110. The condensate is then taken out of the flowing air using a gravitational drain. The front of the housing 118 and the internal heat exchanger coils have been cut away in FIG. 4 to reveal where an induction port 140 draws condensate from the floor 143 of the housing 118 where most of the condensate accumulation occurs. Another port 148 is connected into the chamber at a location where pressure is lower than at the induction port 140. The force from the weight of the water further reduces the pressure at the inlet 140, thus providing more potential energy, or pressure differential, so the water will actively seek the lower pressure inlet 148. It has been determined that the location of port 148 as shown in FIG. 4 is optimal because it is (i) immediately outside of the mouth 121 of the blower; (ii) above a horizontal plane which passes through a center axis of the blower; and, (iii) on the side opposite the blower exhaust 124, and as such, is exposed to especially low pressure. This creates the suction necessary to remove the condensate through port 140.
The specifics regarding the conduit system 130 which performs the condensation removal can be seen in FIGS. 5 and 6. Beginning with the induction port 140, a straight portion 141 may extend to an elbow 142, then into another straight portion 144 where it may be connected into an elbow 156 which is a part of the drainage portion of the device 130.
The low pressure/suction port 148 may be incorporated into the conduit system 130 via straight section 149, which may extend to an elbow 150. A straight section 152 may extend from the elbow 150 to an elbow 154. A vertical gravitational separation section 146 may extend from the elbow 154 to an elbow 155, which may connect to a reverse elbow 156. A straight drainage section 158 may extend beneath the reverse elbow 156. The straight drainage section 158 may be capped from below by a flat bottom 157. A metered drainage stem 160 may extend from section 158 and may have an orifice 164. Stem 160 may extends horizontally and backwards relative to the unit 110. A rib 162 may be used to receive a hose or tubing (not shown) and secure it over and onto the drainage stem 160. This hose (or tubing) may drain the condensate to: (i) outside the aircraft, (ii) a temporary storage vessel, or (iii) some other vessel or conduit network so that the condensate may be recycled and used for some purpose inside the aircraft.
FIG. 5 also shows the devices used to secure the condensate removal device 130 onto the housing 118. First, a bracket 165, which may comprise a large flange 167 and a short flange 166, may be fastened onto a shelf 117 made into the housing 118 as can be seen in FIG. 7. A pair of screws 176 and 178 may be installed into nuts (only one nut 186 is visible in FIG. 7) from inside the housing 118.
A large collar 168 and a small collar 170 may be used to secure ends of the low pressure/suction port 148 and the induction port 140, respectively, to the large flange 167 as shown in FIG. 5 using fasteners 172, 174 and receiving nuts 180, 182. The end of the low pressure port 148 may pass through a rectangular aperture 169 formed in the bracket 165. FIG. 7 shows where and how the ends of the low pressure and induction ports (148 and 140, respectively) are connected to the outer wall of the housing 118.
FIG. 8 shows the internals of conduit network 130, and the movement of both air (A) and condensate (C) using arrows so labeled. As can be seen in the figure, low pressures at port 148 may cause air (A) to be drawn into induction port 140, up vertical gravitational separation section 146, and then out the port 148. When this occurs, condensate (C) is pulled into port 140 along with the air (A). Once the vertical section 146 is reached, however, the condensate (C) may separate from the air (A) as shown. This is due to gravity, in that the relatively lower pressures existing at port 148 result in the air (A) to be drawn up and out port 148, while the condensate (C), which is too heavy, separates from the airflow. Once separated, the condensate (C) drops down into section 158 and exits from the drain stem 160 (i.e., orifice 164), where it may be recycled or discarded (e.g., by using drain hose or tubing (not shown)).
Referring now to FIG. 9, a value for height hc is determined which will result in the water condensate C to drain out tube 144 into tube 146 and be emitted out of the metered drainage system 160. The pressures P1, P2, and P0 (FIG. 10) at the orifices of device 130 drive the water evacuation process. Pressure seen at orifice 148 (P1) is less than the pressure seen at orifice 140 (P2) which is less than the ambient pressure (P0) seen at orifice 164 of the metered drainage system 160. That is:

P1<P2<P0 

Variable hc, the height necessary in tube 158 for proper draining, is measured from the lower part of the connection of tube 144 and 158 to the upper part of the connection of tube 164 and 158 (see FIGS. 9 and 10). hc can be determined according to the following relationship:

P1=λ·hc 

Resulting in:





  
    h
    c
  
  =
  
    abs
    ⁢
    
      
      
        
          P
          1
        
        λ
      
      
    
  




where:

    
    
        λ=Specific weight of the pooled fluid
    
    






  
    abs
    ⁢
    
      
      
        
      
      
    
  
  =
  
    Absolute
    ⁢
    
      
    
    ⁢
    value
    ⁢
    
      
        
      
      ⁢
      
        
      
    
    ⁢
    of
    ⁢
    
      
    
    ⁢
    
      
        P
        1
      
      λ
    
  




    
    
        P1=Gauge Pressure at port 148.

Variable hc, as determined, is the minimum height necessary. It is permissible (and recommended) to make the height, hc some percentage greater than the minimum (e.g., 10%-25% higher). This design factor of safety will account for any pressure or voltage transients.

    
    


The inner diameters (D1, D2, D3) of each of conduit sections 152, 144, and 160 play a significant role in draining the fluid properly. Inner diameter D1, in the preferred embodiment, exists not only at orifice 148, but also through portion 149 (see FIG. 6), around bend 150, through the length of portion 152, down through bend 154 and the length of vertical gravitational separator portion 146 all the way to floor 157. A second inside diameter D2 exists not only at orifice 140, but also the length of portion 144 until it feeds into the vertical separator portion 146. A third diameter D3 exists not only at orifice 164 but also along the entire length of drain stem 160.
These diameters (D1, D2, D3) have been proportioned in a manner that creates a desired result in the vertical separator portion 146. This effect is symbolized, and can be best understood by following the paths of the arrows representing airflow (A) and condensate flow (C) in FIG. 8. As the suction is administered at orifice 140 as the result of the relatively low pressure existing at orifice 148, both air (A) and any condensate in the proximity of orifice 140 are drawn into and along portion 144. Upon reaching the vertical separator portion 146, the air (A) will continue to rise, travel around bend 154 and exit the system at orifice 148 into the low pressure area in the chamber near the mouth 121 of the blower (see FIG. 4). The condensate (C), however, because it has a higher specific weight than the air (A), becomes separated upon reaching the vertical separator portion 146, drops to floor 157, collects and then passes through stem 160, and out orifice 164 into the drain tubing/hose (not shown).
Quantification of the relationship between inner diameters D1, D2 and D3 is critical in optimizing drainage capability. Referring to FIG. 11, {dot over (m)}1, {dot over (m)}2, {dot over (m)}3 are the mass flow rates at orifices 148, 140, and 164, and each instance, is determined according to the formula:

{dot over (m)}=ρ·V·A 

where: ρ=Fluid Density

    
    
        V=Fluid Velocity
        A=Tube Cross-Sectional Area
    
    


In order to optimize water drainage into port 140, and subsequently out of orifice 164, it is necessary to optimize D1, D2 and D3 such that {dot over (m)}2 is greatest.
The first mode of operation analyzed is when there is no water in the system.

{dot over (m)}1−{dot over (m)}2−{dot over (m)}3=0

ρ1·V1·A1−ρ2·V2·A2−ρ3·V3·A3=0

where: ρ1=ρ2=ρ3 
Knowing:



  
    
      A
      1
    
    =
    
      π
      ·
      
        
          (
          
            
              D
              1
            
            2
          
          )
        
        2
      
    
  




  
    
      A
      2
    
    =
    
      π
      ·
      
        
          (
          
            
              D
              2
            
            2
          
          )
        
        2
      
    
  




  
    
      A
      3
    
    =
    
      π
      ·
      
        
          (
          
            
              D
              3
            
            2
          
          )
        
        2
      
    
  



where:

    
    
        D1=Tube Diameter at orifice 148
        D2=Tube Diameter at orifice 140
        D3=Tube Diameter at orifice 164
    
    


Simplified:

V1·A1V2·A2−V3·A3=0

Q1−Q2−Q3=0

Where:

    
    
        Q1=Volumetric Flow Rate at orifice 148
        Q2=Volumetric Flow Rate at orifice 140
        Q3=Volumetric Flow Rate at orifice 164
    
    


Ideally Q3, or the volumetric flow rate through orifice 164, should be no more than 5% that of Q1. This means that there should be no more than a 5% pressure loss from Q1 as a result of D3, which in-turn means that D3 will be smaller than D1 or D2.
However, D3 should not be so small that it restricts the collected condensate from draining properly. The minimum drain diameter D3 for allowing collected condensate to properly drain may be computed as follows:

Q3=VJet·A3 

where: VJet=Fluid Exit Velocity Due to Gravity

    
    
        A3=Tube Cross-Sectional Area at D3 
    
    


and:

VJet=√{square root over (2·g·hc)}

where: g=Acceleration Due to Gravity

    
    
        hc=Height in tube 158
    
    


In small systems, typical values of Q3 will be between 0.5 and 1 in3/sec. The goal is to find a specific value or optimum range for D3. By using the equations above to solve for D3:




  
    D
    3
  
  =
  
    
      
        
          2
          ·
          
            Q
            3
          
          ·
          π
          ·
          g
          ·
          
            h
            c
          
        
      
      ·
      
        
          2
          ·
          g
          ·
          
            h
            c
          
        
        4
      
    
    
      π
      ·
      g
      ·
      
        h
        c
      
    
  




This is the minimum value of D3 such that the pooled condensate will adequately flow out of the system. People of skill in the art will appreciate that the equation above may be used to approximate or calculate the minimum value of D3, and that a conduit section of the next standard size above this approximated or calculated value may be utilized in the device 130.
As stated above Q3 should be no more than 5% of Q1. This sets the maximum value for Q3.
So:



  
    
      Q
      3
    
    =
    
      
        1
        20
      
      ·
      
        Q
        1
      
    
  




  
    
      
        π
        ·
        
          D
          3
          2
        
        ·
        
          V
          3
        
      
      4
    
    =
    
      
        (
        
          1
          20
        
        )
      
      ·
      
        
          π
          ·
          
            D
            1
            2
          
          ·
          
            V
            1
          
        
        4
      
    
  



Reduces to:





  
    
      D
      3
      2
    
    ·
    
      V
      3
    
  
  =
  
    
      (
      
        1
        20
      
      )
    
    ·
    
      D
      1
      2
    
    ·
    
      V
      1
    
  




For simplicity, if V3=V1 then,

D3=0.224·D1 

Or:




  
    
      
        
          2
          ·
          
            Q
            3
          
          ·
          π
          ·
          g
          ·
          
            h
            c
          
        
      
      ·
      
        
          2
          ·
          g
          ·
          
            h
            c
          
        
        4
      
    
    
      π
      ·
      g
      ·
      
        h
        c
      
    
  
  <
  
    D
    3
  
  <
  
    .224
    ·
    
      D
      1
    
  




Another way to ensure that pressure P3 is less than pressures P1 and P2 is to create a pressure drop.
The pressure drop caused by tube shear stress (see FIG. 12) is:




  
    Δ
    ⁢
    
      
    
    ⁢
    P
  
  =
  
    4
    ·
    
      τ
      w
    
    ·
    
      (
      
        L
        D
      
      )
    
  




where:

    
    
        ΔP=Pressure loss in tube
        L=Tube length
        D=Tube diameter
        τw=Tube wall shear stress

Referencing the formula above, in order to make ΔP→∞, either τw needs to be very large or

    
    






  (
  
    L
    D
  
  )




needs to be very large. L, or the length of tube 160, can be made to be long enough to cause a great enough pressure drop within tube 160.

Continuing from above for the equal density mode:



  
    
      
        Q
        1
      
      -
      
        Q
        2
      
      -
      
        Q
        3
      
    
    =
    0
  




  
    
      
        Q
        1
      
      -
      
        Q
        2
      
      -
      
        
          1
          20
        
        ⁢
        
          Q
          1
        
      
    
    =
    0
  




  
    
      
        
          19
          20
        
        ⁢
        
          Q
          1
        
      
      -
      
        Q
        2
      
    
    =
    0
  




  
    
      Q
      2
    
    =
    
      
        19
        20
      
      ⁢
      
        Q
        1
      
    
  



Where:

Q1=V1·A1 

Q2=V2·A2 

Expanded:




  
    
      V
      2
    
    ·
    
      A
      2
    
  
  =
  
    
      19
      20
    
    ⁢
    
      
        V
        1
      
      ·
      
        A
        1
      
    
  




Where:



  
    
      V
      1
    
    =
    
      
        (
        
          
            
              -
              2
            
            ·
            
              P
              1
            
          
          
            ρ
            1
          
        
        )
      
      
        1
        /
        2
      
    
  




  
    
      V
      2
    
    =
    
      
        (
        
          
            2
            ·
            
              (
              
                
                  P
                  2
                
                -
                
                  P
                  1
                
              
              )
            
          
          
            ρ
            2
          
        
        )
      
      
        1
        /
        2
      
    
  



    
    
        P1=Gauge Pressure at Orifice 148
        P2=Gauge Pressure at Orifice 140
        ρ1=Fluid density at Orifice 148
        ρ2=Fluid density at Orifice 140
    
    


Resulting:




  
    [
    
      
        D
        2
      
      
        D
        1
      
    
    ]
  
  =
  
    
      
        [
        
          
            -
            
              P
              1
            
          
          
            [
            
              
                P
                2
              
              -
              
                P
                1
              
            
            ]
          
        
        ]
      
      
        1
        /
        4
      
    
    ·
    
      
        [
        
          19
          20
        
        ]
      
      
        1
        /
        2
      
    
  




The above equation is determined for the system without pooled condensate, thus there is nothing to drain. Generally the higher the pressure difference between P1 and P2 the smaller the change there needs to be between D1 and D2. Under normal conditions P1<P2 and D2 will need to be greater than D1. Generally,

D2≈1.15·D1 

Again, this is for a system that has similar fluid densities in all tubes.
Now for the analysis of the system with fluid condensate in tube D2 

{dot over (m)}1−{dot over (m)}2−{dot over (m)}3=0

ρ1·V1·A1−ρ2V2·A2−ρ3·V3·A3=0

Where:

ρ1≠ρ2 

ρ3=ρ1 

V1=V3 

D3=0.224·D1 

Resulting:

ρ1·V1·A1−ρ2·V2·A2ρ1V3·A3=0

Simplifying:

0.95·ρ1·V1·D12=ρ2·V2·D22 

Resulting




  
    [
    
      
        D
        2
      
      
        D
        1
      
    
    ]
  
  =
  
    
      
        [
        
          
            
              -
              
                ρ
                1
              
            
            ·
            
              P
              1
            
          
          
            
              ρ
              2
            
            ·
            
              [
              
                
                  P
                  2
                
                -
                
                  P
                  1
                
              
              ]
            
          
        
        ]
      
      
        1
        /
        4
      
    
    ·
    
      
        [
        
          19
          20
        
        ]
      
      
        1
        /
        2
      
    
  




The goal is to get some value for D1 and D2 that will work efficiently when there are similar and dissimilar fluid densities. Taking the root mean square obtains an average value of the two varying quantities.




  
    
      [
      
        
          D
          2
        
        
          D
          1
        
      
      ]
    
    Total
    2
  
  =
  
    
      
        1
        2
      
      ·
      
        
          
            [
            
              
                
                  [
                  
                    
                      D
                      2
                    
                    
                      D
                      1
                    
                  
                  ]
                
                
                  
                    ρ
                    1
                  
                  =
                  
                    ρ
                    2
                  
                
                2
              
              +
              
                
                  [
                  
                    
                      D
                      2
                    
                    
                      D
                      1
                    
                  
                  ]
                
                
                  
                    ρ
                    1
                  
                  ≠
                  
                    ρ
                    2
                  
                
                2
              
            
            ]
          
          ⁢
          
            

          
          [
          
            
              D
              2
            
            
              D
              1
            
          
          ]
        
        Total
        2
      
    
    =
    
      
        [
        
          19
          40
        
        ]
      
      ·
      
        
          [
          
            
              
                -
                
                  P
                  1
                
              
              
                [
                
                  
                    P
                    2
                  
                  -
                  
                    P
                    1
                  
                
                ]
              
            
            +
            
              [
              
                
                  
                    -
                    
                      ρ
                      1
                    
                  
                  ·
                  
                    P
                    1
                  
                
                
                  
                    ρ
                    2
                  
                  ·
                  
                    [
                    
                      
                        P
                        2
                      
                      -
                      
                        P
                        1
                      
                    
                    ]
                  
                
              
              ]
            
          
          ]
        
        
          1
          /
          2
        
      
    
  




The equation above will be better used as an inequality. In this case:




  
    D
    3
  
  <
  
    
      [
      
        
          D
          2
        
        
          D
          1
        
      
      ]
    
    Total
    2
  
  ≤
  
    
      [
      
        19
        40
      
      ]
    
    ·
    
      
        [
        
          
            
              -
              
                P
                1
              
            
            
              [
              
                
                  P
                  2
                
                -
                
                  P
                  1
                
              
              ]
            
          
          +
          
            [
            
              
                
                  -
                  
                    ρ
                    1
                  
                
                ·
                
                  P
                  1
                
              
              
                
                  ρ
                  2
                
                ·
                
                  [
                  
                    
                      P
                      2
                    
                    -
                    
                      P
                      1
                    
                  
                  ]
                
              
            
            ]
          
        
        ]
      
      
        1
        /
        2
      
    
  




Or simply stated:




  
    D
    3
  
  <
  
    D
    2
  
  ≤
  
    
      D
      1
    
    ·
    
      
        [
        
          
            [
            
              19
              40
            
            ]
          
          ·
          
            
              [
              
                
                  
                    -
                    
                      P
                      1
                    
                  
                  
                    [
                    
                      
                        P
                        2
                      
                      -
                      
                        P
                        1
                      
                    
                    ]
                  
                
                +
                
                  [
                  
                    
                      
                        -
                        
                          ρ
                          1
                        
                      
                      ·
                      
                        P
                        1
                      
                    
                    
                      
                        ρ
                        2
                      
                      ·
                      
                        [
                        
                          
                            P
                            2
                          
                          -
                          
                            P
                            1
                          
                        
                        ]
                      
                    
                  
                  ]
                
              
              ]
            
            
              1
              /
              2
            
          
        
        ]
      
      
        1
        /
        2
      
    
  




Thus, as can be appreciated from the equation above, D2 needs to be greater than D3 but less than some quantity times D1.
Generally speaking, D1 will be about 1.5 to 2.5 times the value of D2. Using a ratio that is considerably smaller, for instance,

D1≈0.5·D2 

will choke down the flow rate of the denser fluid, thus causing the flow rate to be limited through orifice 140. The effect will be a slower drain time out of orifice 140. Conversely, if the ratio is larger,

D1≈3·D2 

the flow rate out of orifice 140 will be above maximum but excess pressure from P1 might be lost through orifice 140. The final diameters of D1 and D2 should be determined through experimentation in line with manufacturing limitations.

One thing to note is that for the water to properly flow, the pressure seen at orifice 148 needs to be less than the pressure seen at orifice 140. If P1>P2 but less than P0 (ambient pressure) the water seen at port 140 will not flow out, thus causing spray. So the ratio should be:




  
    (
    
      
        
          P
          1
        
        -
        
          P
          0
        
      
      
        
          P
          2
        
        -
        
          P
          0
        
      
    
    )
  
  >
  1




Also, for more efficient drainage, (P1−P0) needs to be greater than (P2−P0). Drain potential increases with the greater (P1−P0) is as compared to (P2−P0).

In summary, the design process begins with determining the negative pressure within air conditioning console. Specifically, the location within the console where air pressure is the lowest is determined, and the fluid induction conduit inlet (e.g., orifice 1301a in FIG. 13A; orifice 1301b in FIG. 13B) is located as near as possible to this low pressure area. Next the location in the console where condensate fluid has the propensity to collect is determined, and the fluid induction tube is located there; for example, orifice 1302a in FIG. 13A and orifice 1302b in FIG. 13B are each located at locations where condensed water will collect. Pressure values at location P1 and P2 are then determined.
Next, the height drop between induction tube 1301c (see FIG. 13c) and drain 1302c is determined. This is done using the relation:




  
    h
    c
  
  =
  
    abs
    ⁢
    
      
      
        
          P
          1
        
        λ
      
      
    
  




The drop height typically ranges from 0.5 inches to 3 inches depending on how much negative pressure is within the unit. Experimentation will determine the negative pressure within the unit.
Next, a value for D3 is determined. D3 needs to be large enough to allow the fluid to drain out depending on the rate of pooling, but small enough to not allow pressure loss from P1 or P2.




  
    
      
        
          2
          ·
          
            Q
            3
          
          ·
          π
          ·
          g
          ·
          
            h
            c
          
        
      
      ·
      
        
          2
          ·
          g
          ·
          
            h
            c
          
        
        4
      
    
    
      π
      ·
      g
      ·
      
        h
        c
      
    
  
  <
  
    D
    3
  
  <
  
    .224
    ·
    
      D
      1
    
  




Based off the information above, an adequate value for diameter D1 is computed.

D1≈4.5·D3 

The final criterion is to determine drain diameter D2, which can be gleaned by substantially satisfying the relation:




  
    D
    3
  
  <
  
    D
    2
  
  ≤
  
    
      D
      1
    
    ·
    
      
        [
        
          
            [
            
              19
              40
            
            ]
          
          ·
          
            
              [
              
                
                  
                    -
                    
                      P
                      1
                    
                  
                  
                    [
                    
                      
                        P
                        2
                      
                      -
                      
                        P
                        1
                      
                    
                    ]
                  
                
                +
                
                  [
                  
                    
                      
                        -
                        
                          ρ
                          1
                        
                      
                      ·
                      
                        P
                        1
                      
                    
                    
                      
                        ρ
                        2
                      
                      ·
                      
                        [
                        
                          
                            P
                            2
                          
                          -
                          
                            P
                            1
                          
                        
                        ]
                      
                    
                  
                  ]
                
              
              ]
            
            
              1
              /
              2
            
          
        
        ]
      
      
        1
        /
        2
      
    
  




The following example illustrates these computations.
Example



  
    
      ρ
      1
    
    =
    
      
        air
        ⁡
        
          (
          
            .0024
            ⁢
            
              slugs
              
                ft
                3
              
            
          
          )
        
      
      =
      
        (
        
          0.0000013889
          ⁢
          
            slugs
            
              in
              3
            
          
        
        )
      
    
  




  
    
      ρ
      2
    
    =
    
      
        water
        ⁡
        
          (
          
            1.94
            ⁢
            
              slugs
              
                ft
                3
              
            
          
          )
        
      
      =
      
        (
        
          0.0011227
          ⁢
          
            slugs
            
              in
              3
            
          
        
        )
      
    
  




  
    
      λ
      water
    
    =
    
      
        62.30
        ⁢
        
          lb
          
            ft
            3
          
        
      
      =
      
        .0361
        ⁢
        
          lb
          
            in
            3
          
        
      
    
  




  
    
      P
      1
    
    =
    
      
        -
        .060
      
      ⁢
      
        
      
      ⁢
      
        psi
        ⁡
        
          (
          gauge
          )
        
      
    
  




  
    
      P
      2
    
    =
    
      
        -
        .030
      
      ⁢
      
        
      
      ⁢
      
        psi
        ⁡
        
          (
          gauge
          )
        
      
    
  




  
    
      Q
      3
    
    =
    
      .75
      ⁢
      
        
          in
          3
        
        sec
      
    
  




  
    
      h
      c
    
    =
    
      
        abs
        ⁢
        
          
          
            
              P
              1
            
            λ
          
          
        
      
      =
      
        
          abs
          ⁢
          
            
            
              
                
                  -
                  .060
                
                ⁢
                
                  lb
                  
                    in
                    2
                  
                
              
              
                .0361
                ⁢
                
                  lb
                  
                    in
                    3
                  
                
              
            
            
          
        
        =
        
          1.66
          ⁢
          
            
          
          ⁢
          inches
        
      
    
  



Add 10% to account for transients
1.66·1.1=1.83 inches
Now determine diameter D3.



  
    
      
        
          
            2
            ·
            
              Q
              3
            
            ·
            π
            ·
            g
            ·
            
              h
              c
            
          
        
        ·
        
          
            2
            ·
            g
            ·
            
              h
              c
            
          
          4
        
      
      
        π
        ·
        g
        ·
        
          h
          c
        
      
    
    <
    
      D
      3
    
  




  
    
      
        
          
            
              2
              ·
              .75
              ·
              π
              ·
              386.1
              ·
              1.66
            
          
          ·
          
            
              2
              ·
              386.1
              ·
              1.66
            
            4
          
        
        
          π
          ·
          386.1
          ·
          1.66
        
      
      <
      
        D
        3
      
    
    ⁢
    
      

    
    →
    
      
        
          
            
              3020.3
            
            ·
            
              1281.9
              4
            
          
          2013.5
        
        <
        
          D
          3
        
      
      ⁢
      
        

      
      →
      
        .1633
        <
        
          D
          3
        
      
    
  



So D3 needs to be larger than 0.1633 inches.
For Manufacturing D3=0.17
Determine diameter D1.

D1≈4.5·D3 

D1≈4.5·0.17

D1·0.765 inches

    
    
        Rounding up:
        D1 will be 0.80 inches in diameter
    
    


Determine diameter D2.



  
    
      
    
    ⁢
    
      
        D
        2
      
      ≤
      
        
          D
          1
        
        ·
        
          
            [
            
              
                [
                
                  19
                  40
                
                ]
              
              ·
              
                
                  [
                  
                    
                      
                        -
                        
                          P
                          1
                        
                      
                      
                        [
                        
                          
                            P
                            2
                          
                          -
                          
                            P
                            1
                          
                        
                        ]
                      
                    
                    +
                    
                      [
                      
                        
                          
                            -
                            
                              ρ
                              1
                            
                          
                          ·
                          
                            P
                            1
                          
                        
                        
                          
                            ρ
                            2
                          
                          ·
                          
                            [
                            
                              
                                P
                                2
                              
                              -
                              
                                P
                                1
                              
                            
                            ]
                          
                        
                      
                      ]
                    
                  
                  ]
                
                
                  1
                  /
                  2
                
              
            
            ]
          
          
            1
            /
            2
          
        
      
    
  




  
    
      D
      2
    
    ≤
    
      .8
      ·
      
        
          [
          
            
              [
              
                19
                40
              
              ]
            
            ·
            
              
                [
                
                  
                    
                      -
                      
                        (
                        
                          -
                          .06
                        
                        )
                      
                    
                    
                      [
                      
                        
                          (
                          
                            -
                            .03
                          
                          )
                        
                        -
                        
                          (
                          
                            -
                            .06
                          
                          )
                        
                      
                      ]
                    
                  
                  +
                  
                    [
                    
                      
                        
                          -
                          
                            (
                            0.0000013889
                            )
                          
                        
                        ·
                        
                          (
                          
                            -
                            .06
                          
                          )
                        
                      
                      
                        
                          (
                          0.0011227
                          )
                        
                        ·
                        
                          [
                          
                            
                              (
                              
                                -
                                .03
                              
                              )
                            
                            -
                            
                              (
                              
                                -
                                .06
                              
                              )
                            
                          
                          ]
                        
                      
                    
                    ]
                  
                
                ]
              
              
                1
                /
                2
              
            
          
          ]
        
        
          1
          /
          2
        
      
    
  




  
    
      
    
    ⁢
    
      
        D
        2
      
      ≤
      
        .8
        ·
        
          
            [
            
              
                [
                
                  19
                  40
                
                ]
              
              ·
              
                
                  [
                  
                    2
                    +
                    .002474
                  
                  ]
                
                
                  1
                  /
                  2
                
              
            
            ]
          
          
            1
            /
            2
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        D
        2
      
      ≤
      
        .8
        ·
        
          
            [
            .6722
            ]
          
          
            1
            /
            2
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        D
        3
      
      ≤
      
        D
        2
      
      ≤
      
        .656
        ⁢
        
          
        
        ⁢
        inches
      
    
  




  
    
      
    
    ⁢
    
      
        D
        2
      
      ≈
      
        .63
        ⁢
        
          
        
        ⁢
        inches
      
    
  




  
    
      
    
    ⁢
    where
  




  
    
      
    
    ⁢
    
      
        D
        3
      
      =
      
        .17
        ⁢
        
          
        
        ⁢
        inches
      
    
  



It should be understood that instead of providing conduits having internal diameters which are constant lengthwise, the internal diameters could be varied and the desired fluid mechanics obtained by restricting flow through the minimum internal diameters at locations (e.g., at the orifices). Thus, it is possible that this sort of variation could be made and the desired drainage arrangement maintained.
Many different arrangements of the various components depicted, as well as components not shown, are possible without departing from the spirit and scope of the present invention. Embodiments of the present invention have been described with the intent to be illustrative rather than restrictive. Alternative embodiments will become apparent to those skilled in the art that do not depart from its scope. A skilled artisan may develop alternative means of implementing the aforementioned improvements without departing from the scope of the present invention.
It will be understood that certain features and subcombinations are of utility and may be employed without reference to other features and subcombinations and are contemplated within the scope of the claims. Not all steps listed in the various figures need be carried out in the specific order described.

## Claims

1. A method for removing condensate from an air conditioning unit, the air conditioning unit having a housing, a blower, coils, an air intake duct, and an exhaust, the method comprising:
determining a first location within the housing where during operation of the air conditioning unit air pressure is lower than at a second location, wherein the first location is adjacent a mouth of the blower and above a horizontal plane that passes through a center of the blower;
determining the second location within the housing where pooling of condensate occurs, the second location being below the first location;
locating a first conduit with a first internal diameter at the first location;
locating a second conduit with a second internal diameter at the second location;
connecting the first conduit and the second conduit to a third conduit with a third internal diameter that is substantially equal to the first internal diameter, the third conduit extending substantially vertically below the second location; and
connecting a fourth conduit with a fourth internal diameter to the third conduit below the second location;
wherein the fourth internal diameter is less than each of the first internal diameter and the second internal diameter;
wherein the fourth conduit has an orifice for allowing condensate to exit the fourth conduit.

2. The method of claim 1 wherein:
the first internal diameter is about 0.80 inches;
the second internal diameter is about 0.63 inches; and
the fourth internal diameter is about 0.17 inches.

3. A method for removing condensate from an air conditioning unit, the air conditioning unit having a housing, a blower, coils, an air intake duct, and an exhaust, the method comprising:
determining a first location within the housing where during operation of the air conditioning unit air pressure is lower than at a second location;
determining the second location within the housing where pooling of condensate occurs, the second location being below the first location;
locating a first conduit with a first internal diameter at the first location;
locating a second conduit with a second internal diameter at the second location;
connecting the first conduit and the second conduit to a third conduit with a third internal diameter that is substantially equal to the first internal diameter, the third conduit extending substantially vertically below the second location; and
connecting a fourth conduit with a fourth internal diameter to the third conduit below the second location;
wherein the fourth internal diameter is less than each of the first internal diameter and the second internal diameter;
wherein the fourth conduit has an orifice for allowing condensate to exit the fourth conduit;
wherein the first location is:
adjacent a mouth of the blower; and
above a horizontal plane that passes through a center of the blower;

the second conduit has a second conduit top surface and a second conduit bottom surface;
the fourth conduit has a fourth conduit top surface and a fourth conduit bottom surface; and
a height drop in the third conduit between the second conduit bottom surface and the fourth conduit top surface is determined by dividing an absolute value of a pressure at the first location by a fluid density of water.

4. The method of claim 3 wherein the fourth internal diameter is proportional to the height drop and a volumetric flow rate of condensate in the fourth conduit.

5. The method of claim 3 wherein a minimum value for the fourth internal diameter is approximated using the relationship:




  
    D
    4
  
  =
  
    
      
        
          2
          ·
          
            Q
            4
          
          ·
          π
          ·
          g
          ·
          
            h
            c
          
        
      
      ·
      
        
          2
          ·
          g
          ·
          
            h
            c
          
        
        4
      
    
    
      π
      ·
      g
      ·
      
        h
        c
      
    
  



wherein:
D4 is the fourth internal diameter;
Q4 is a volumetric flow rate of condensate in the fourth conduit;
g is the acceleration due to gravity; and
he is the height drop.

6. The method of claim 5 wherein the first internal diameter, the second internal diameter, and the fourth internal diameters are chosen to substantially satisfy the following relationship:




  
    D
    4
  
  <
  
    D
    2
  
  ≤
  
    
      D
      1
    
    ·
    
      
        [
        
          
            [
            
              19
              40
            
            ]
          
          ·
          
            
              [
              
                
                  
                    -
                    
                      P
                      1
                    
                  
                  
                    [
                    
                      
                        P
                        2
                      
                      -
                      
                        P
                        1
                      
                    
                    ]
                  
                
                +
                
                  [
                  
                    
                      
                        -
                        
                          ρ
                          1
                        
                      
                      ·
                      
                        P
                        1
                      
                    
                    
                      
                        ρ
                        2
                      
                      ·
                      
                        [
                        
                          
                            P
                            2
                          
                          -
                          
                            P
                            1
                          
                        
                        ]
                      
                    
                  
                  ]
                
              
              ]
            
            
              1
              /
              2
            
          
        
        ]
      
      
        1
        /
        2
      
    
  



wherein:
D2 is the second internal diameter;
D1 is the first internal diameter;
P1 is the pressure at the first location;
P2 is the pressure at the second location;
ρ1 is the fluid density of air; and
ρ2 is the fluid density of water.

7. The method of claim 6 further comprising a step of connecting the fourth conduit to a tube for routing outside the air conditioning unit for recycling condensate exiting the orifice.

8. A method for removing condensate from an air conditioning unit in an aircraft, the air conditioning unit having a housing, a blower, coils, an air intake duct, and an exhaust, the method comprising:
determining a first location within the housing where during operation of the air conditioning unit air pressure is lower than at a second location;
determining the second location within the housing where pooling of condensate occurs, the second location being below the first location;
providing a conduit, the conduit comprising:
an outlet portion having a first segment, a second segment, and a first port; the first and the second segments each lying in a first horizontal plane that is above a second horizontal plane that passes through an internal region of the blower;
an inlet portion having a first section, a second section, and a second port; the first and the second sections lying in a third horizontal plane that is below the first horizontal plane and the second horizontal plane;

operably coupling the first port to the first location and the second port to the second location;
operably coupling a gravitational separator portion to both the outlet portion and the inlet portion such that the gravitational separator portion extends substantially vertically; and
operably coupling a draining portion having an orifice to the gravitational separator portion.

9. The method of claim 8 further comprising a step of closing the end with a cap.

10. The method of claim 8 wherein the first and the second segments each lie in different vertical planes.

11. The method of claim 10 wherein the first and the second sections each lie in different vertical planes.

12. The method of claim 8 further comprising a step of coupling tubing to the orifice to route water exiting the orifice away from the air conditioning unit for recycling.

13. The method of claim 8 wherein the outlet portion has a first diameter that is greater than a second diameter of the inlet portion.