Abstract:
A rolling diaphragm pump includes a housing, a rolling seal diaphragm disposed in the housing, a piston for driving the diaphragm, and a valve for regulating the flow of working fluid in a portion of the housing. A constant differential pressure is maintained across the diaphragm independent of discharge pressure of the pump. A method of pumping a viscous medium includes pumping the viscous medium by maintaining a constant differential pressure across a rolling seal diaphragm independent of discharge pressure of the viscous medium, with the diaphragm disposed between the viscous medium and a working medium and being driven by a piston.

Description:
CROSS-REFERENCE TO RELATED APPLICATION 
       [0001]    The benefits of U.S. Provisional Application No. 60/886,919 filed Jan. 26, 2007 and entitled “Rolling Diaphragm Pump” are claimed under 35 U.S.C. § 119(e), and the entire contents of this provisional application are expressly incorporated herein by reference thereto. 
     
    
     FIELD OF THE INVENTION 
       [0002]    The invention relates to a pump and product delivery. More particularly, the invention relates to rolling diaphragm type pumps. 
       BACKGROUND OF THE INVENTION 
       [0003]    In a rolling diaphragm type pump, a small amount of positive differential pressure is needed to keep the diaphragm in the correct orientation (convoluted orientation). However, if the differential pressure is too high, the diaphragm will wear out at a faster rate or even burst in extreme cases. 
         [0004]    Prior art rolling diaphragm pump designs create a positive differential pressure by sizing the driving cylinders to match the diaphragm diameter. This limits the diaphragm size to one that will match with commercial cylinders. Additionally, the greater the mismatch, the more the differential pressure will vary with the pump&#39;s output pressure. 
         [0005]    In a prior art rolling diaphragm pump, the differential pressure across the diaphragm is determined by the pump dimensions and the working pressure. For a given pump size, the differential pressure increases as the working pressure increases. This not only reduces the lifetime of the diaphragms but also limits the maximum pressure of the pump. 
         [0006]    A dual-unit pump, e.g., a rolling diaphragm piston pump, is disclosed in U.S. Pat. No. 4,543,044, the entire contents of which are incorporated herein by reference thereto. The pump is suitable for pumping an abrasive high-viscosity slurry, and is adapted to operate at a constant flow rate by means for detecting and correcting a pressure differential in the two units before the units switch from the pumping cycle to the filling cycle and vice versa. The flow of liquids is controlled by valves of the type which switch the flow to and from the units with essentially no volume change in the liquid inlet and outlet lines. 
         [0007]    Turning to  FIG. 1 , a rolling diaphragm pump  10  of the prior art is shown. Piston  12 , which for example may be formed of nylon, is disposed within a cylindrical housing  13  and seated with respect to top-hat shaped rubber diaphragm  14 . A working fluid  16  such as oil and a discharge fluid  18  (the fluid that is being pumped) are shown. A standard hydraulic cylinder  19  (such as a double-rodded cylinder with a vented top) includes a fluid region  20  such as having oil therein, rods  22   a  and  22   b , and a vented region  24 . Piston  12  is used to maintain the shape of diaphragm  14 . Diaphragm  14  is coupled along its circumference to housing  13  at regions  27  along axis  25   b  which is normal to axis  25   a  (along which rods  22   a ,  22   b  for example are axially disposed). 
         [0008]    In  FIG. 1 , P 1  is the discharge pressure of the medium that is being pumped (e.g., to a packaging machine so that the medium may be used to fill a chub), P 2  is the pressure of the hydraulic fluid under piston  12  (e.g., the working fluid pressure), P 3  is vented to atmosphere and assumed as zero pressure with respect to atmosphere, and P 4  is connected to P 2  and thus is the same as the pressure of the hydraulic pressure P 2 . In addition, for the purposes of this analysis, A 1  is the effective area that pressure P 1  acts upon to produce force in a direction parallel to axis  25   a , A 2  is the effective area that pressure P 2  acts upon to produce force in a direction parallel to axis  25   a , and A 3  is the internal area of the hydraulic cylinder  19  about a plane normal to axis  25   a . Product is discharged from pump  10  in direction E. Preferably, P 1 &gt;P 2  in  FIG. 1 . 
         [0009]    According to the design of pump  10  in  FIG. 1 , the downward force is determined by the following Equation 1 below: 
         [0000]        F   down   =P   1   ·A   1   +P   2   ·A   4   +P   3 ·( A   3   −A   4 )  (Eq. 1) 
         [0000]    The upward force is determined by Equation 2 below: 
         [0000]        F   up =( P   2   ·A   2 )+[ P   4 ·( A   3   −A   4 )]  (Eq. 2) 
         [0000]    Area A 1  is the same as area A 2 , pressure P 2  is the same as pressure P 4 , and pressure P 3  is zero pressure with respect to atmosphere. Thus, the upward force must balance the downward force as in Equation 3 below: 
         [0000]      ( P   1   ·A   1 )+( P   2   ·A   4 )=( P   2   −A   1 )+[ P   2 ·( A   3   −A   4 )]  (Eq. 3) 
         [0010]    This balance can be simplified as shown in Equations 4-6 below: 
         [0000]    
       
         
           
             
               
                 
                   
                     [ 
                     
                       
                         A 
                         1 
                       
                       · 
                       
                         ( 
                         
                           
                             P 
                             1 
                           
                           - 
                           
                             P 
                             2 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   = 
                   
                     
                       [ 
                       
                         
                           P 
                           2 
                         
                         · 
                         
                           ( 
                           
                             
                               A 
                               3 
                             
                             - 
                             
                               A 
                               4 
                             
                           
                           ) 
                         
                       
                       ] 
                     
                     - 
                     
                       ( 
                       
                         
                           P 
                           2 
                         
                         · 
                         
                           A 
                           4 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     4 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     [ 
                     
                       
                         A 
                         1 
                       
                       · 
                       
                         ( 
                         
                           
                             P 
                             1 
                           
                           - 
                           
                             P 
                             2 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   = 
                   
                     
                       P 
                       2 
                     
                     · 
                     
                       [ 
                       
                         
                           A 
                           3 
                         
                         - 
                         
                           ( 
                           
                             2 
                             · 
                             
                               A 
                               4 
                             
                           
                           ) 
                         
                       
                       ] 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     5 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     Δ 
                      
                     
                         
                     
                      
                     P 
                   
                   = 
                   
                     
                       
                         P 
                         1 
                       
                       - 
                       
                         P 
                         2 
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             A 
                             3 
                           
                           - 
                           
                             2 
                             · 
                             
                               A 
                               4 
                             
                           
                         
                         ) 
                       
                       
                         A 
                         1 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     6 
                   
                   ) 
                 
               
             
           
         
       
     
         [0011]    Thus, as shown in Equation 6, ΔP is dependent on working pressure. 
         [0012]    Pumping high viscosity liquids and slurries at high pressure and/or at constant pressure or constant flow rate is particularly difficult. High viscosities slurries, for example, may be between 10,000 and 5,000,000 centipoise, and may be abrasive and include large particulates such as rocks ⅛ inch in general size. However, pumping pressures in prior art rolling diaphragm pumps are limited by the pressure that the diaphragm can withstand. The differential pressure ΔP varies with discharge pressure P 1  and therefore if P 1  becomes too high, ΔP can become so high that the diaphragm integrity is lost and the diaphragm breaks. 
         [0013]    It is desired that the differential pressure ΔP is in the range of 10 psi to 20 psi so that the diaphragm is maintained in the correct shape and position, while not being overstressed. The prior art rolling diaphragm pump permits this but only for a fixed range of discharge pressure P 1  as will be further described herein. 
       SUMMARY OF THE INVENTION 
       [0014]    A rolling diaphragm pump may include a housing, a rolling seal diaphragm disposed in the housing, a piston for driving the diaphragm, and a valve for regulating the flow of working fluid in a portion of the housing. A constant differential pressure may be maintained across the diaphragm independent of discharge pressure of the pump. In some embodiments, the rolling seal diaphragm is top hat shaped. Also, the constant differential pressure may be between 1 psi and 100 psi, between 10 psi and 50 psi, or between 10 psi and 20 psi. The discharge pressure may be greater than 1000 psi or greater than 500 psi. 
         [0015]    A method of pumping a viscous medium may include: pumping the viscous medium by maintaining a constant differential pressure across a rolling seal diaphragm independent of discharge pressure of the viscous medium, with the diaphragm disposed between the viscous medium and a working medium and being driven by a piston. The viscous medium may be discharged at a constant flow rate or discharged at a constant pressure. The method may further include: regulating the flow of the working medium. In the method, the rolling seal diaphragm may be top hat shaped. Also in the method, the constant differential pressure may be between 1 psi and 100 psi, between 10 psi and 50 psi, or between 10 psi and 20 psi. In the method, the discharge pressure may be greater than 1000 psi or greater than 500 psi. Further, in the method, the viscous medium may be a slurry with a viscosity between 10,000 and 5,000,000 centipoise. Also, the viscous medium may include aggregate with a maximum lateral dimension of ¼ inch or aggregate with a maximum lateral dimension of ⅛ inch. 
         [0016]    In one embodiment of the invention, a rolling diaphragm pump creates a positive differential pressure through the use of an adjustable check valve that regulates the flow of a working fluid, such as oil, between the driving cylinder and the bottom of the diaphragm by opening when a threshold pressure is met. This provides control of the differential pressure (ΔP) while being independent of the discharge pressure (P 1 ). Advantageously, an increased pressure range is realized in which the rolling diaphragm pump can operate, and variable control of diaphragm stress is permitted. 
         [0017]    In some embodiments, a rolling diaphragm allows continuously variable control of the differential pressure across the diaphragm, which is independent of the discharge pressure. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0018]    Preferred features of embodiments are disclosed in the accompanying drawings, wherein: 
           [0019]      FIG. 1  shows a prior art rolling diaphragm pump; and 
           [0020]      FIG. 2  shows an exemplary embodiment of an inventive rolling diaphragm pump. 
       
    
    
     DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS 
       [0021]    Turning to  FIG. 2 , an exemplary embodiment of an inventive rolling diaphragm pump  100  is shown. Pump  100  is suitable, for example, for use in pumping mine roof bolt anchoring compositions, water-bearing explosives, food products, concrete, fraccing fluids for oil and gas wells, coal/water slurries, nuclear waste slurries, asphalt, paint, and filled epoxy resins. However, this list is non-exhaustive and a variety of high viscosity liquids and slurries are amendable to pumping in accordance with the exemplary embodiment. 
         [0022]    Inventive rolling diaphragm pump  100  includes a piston  112 , which for example may be formed of nylon, is disposed within a cylindrical housing  113 , and is seated with respect to a rolling seal diaphragm  114  such as a top-hat shaped rubber diaphragm. A working medium  116  such as oil fluid and a discharge medium  118  (the medium that is being pumped) are shown. A standard hydraulic cylinder  119  (such as a single-rodded cylinder) includes a fluid region  120  such as having oil therein, and a rod  122 . Portion  121  is in communication with housing  113 . Piston  112  is used to maintain the shape of diaphragm  114 . Diaphragm  114  is coupled along its circumference to housing  113  at regions  127  along axis  125   b  which is normal to axis  125   a  (along which rod  122  for example is axially disposed). 
         [0023]    In  FIG. 2 , Pt is the discharge pressure of the medium that is being pumped (e.g., to a packaging machine so that the medium may be used to fill a chub), P 2  is the pressure of the hydraulic fluid under piston  112  (e.g., the working fluid pressure), and P 4  is connected to P 2  and is the pressure in fluid region  120  and is greater than P 2  by the setting of check valve  126 . In addition, for the purposes of this analysis, A 1  is the effective area that pressure P 1  acts upon to produce force in a direction parallel to axis  125   a , A 2  is the effective area that pressure P 2  acts upon to produce force in a direction parallel to axis  125   a , and A 3  is the internal area of the hydraulic cylinder  119  about a plane normal to axis  125   a . Product is discharged from pump  100  in direction E. Preferably, P 1 &gt;P 2  in  FIG. 2 , so that diaphragm  114  does not invert (resulting in accelerated wear of the diaphragm). Moreover, P 4 &gt;P 2 . 
         [0024]    Because rod  122  is threadably associated with piston  112 , oil flows around the threads and on top of rod  122  so that A 4  does not effect A 2 . Therefore, A 1  is the same as A 2 . The pressure of check valve  126 , P check , is fully adjustable to suit a given need, the check valve being designed to open when a threshold differential pressure is met. Thus, a constant pressure can be created across diaphragm  114  regardless of the pumping pressure. In other words, regardless of whether the pumping pressure is 500 psi, 1000 psi, or 10,000 psi, the differential pressure ΔP, calculated as P 1 −P 2 , is always the same. In contrast, the prior art pump  10  would not function properly at wide ranges of pressures (e.g., 1,000 psi as compared to 10,000 psi) because the differential pressure ΔP would increase as P1 increases and become so great as to compromise the diaphragm. Pump  100  provides constant flow rate or constant pressure performance. Unlike prior art pump  10 , inventive pump  100  advantageously permits pumping of viscous mediums with large aggregates (1) at high pressure and/or (2) at constant pressure or constant flow rate over wide pressure ranges. In addition, inventive pump  100  advantageously may permit longer life of operation in high pressure usage than rotating or progressive-type pumps which suffer from substantial wear when pumping media having large aggregates. 
         [0025]    The theory of operation of exemplary inventive pump  100  now will be explained. In pump  100 , the downward force is determined by the following Equation 7: 
         [0000]        F   down =( P   1   ·A   1 )+( P   2   −A   3 )  (Eq. 7) 
         [0000]    The upward force is determined by Equation 8 below: 
         [0000]        F   up =( P   2   ·A   2 )+( P   4   ·A   3 )  (Eq. 8) 
         [0000]    Area A 2  is the same as area A 1 , and the check valve pressure P check  is P 4 −P 2 . The upward force must balance the downward force as in Equation 9 below: 
         [0000]      ( P   1   ·A   1 )+( P   2   ·A   3 )=( P   2   ·A   1 )+( P   4   ·A   3 )  (Eq. 9) 
         [0000]    This balance can be simplified as shown in Equations 10-12 below: 
         [0000]    
       
         
           
             
               
                 
                   
                     [ 
                     
                       
                         A 
                         1 
                       
                       · 
                       
                         ( 
                         
                           
                             P 
                             1 
                           
                           - 
                           
                             P 
                             2 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   = 
                   
                     
                       ( 
                       
                         
                           P 
                           4 
                         
                         · 
                         
                           A 
                           3 
                         
                       
                       ) 
                     
                     - 
                     
                       ( 
                       
                         
                           P 
                           2 
                         
                         · 
                         
                           A 
                           3 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     [ 
                     
                       
                         A 
                         1 
                       
                       · 
                       
                         ( 
                         
                           
                             P 
                             1 
                           
                           - 
                           
                             P 
                             2 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   = 
                   
                     
                       A 
                       3 
                     
                     · 
                     
                       ( 
                       
                         
                           P 
                           4 
                         
                         - 
                         
                           P 
                           2 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     11 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     Δ 
                      
                     
                         
                     
                      
                     P 
                   
                   = 
                   
                     
                       
                         P 
                         1 
                       
                       - 
                       
                         P 
                         2 
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               P 
                               4 
                             
                             - 
                             
                               P 
                               2 
                             
                           
                           ) 
                         
                         · 
                         
                           ( 
                           
                             
                               A 
                               3 
                             
                             
                               A 
                               1 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         
                           P 
                           check 
                         
                         · 
                         
                           ( 
                           
                             
                               A 
                               3 
                             
                             
                               A 
                               1 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                      
                     12 
                   
                   ) 
                 
               
             
           
         
       
     
         [0000]    Thus, as shown in Equation 12, ΔP is independent of the working pressure. 
         [0026]    A theoretical performance comparison, based on the above Equations 1-12, is presented below for an exemplary resin pump assuming the following: diaphragm area A 1  of 101.6234 in. 2 , cylinder area A 3  of 8.295768 in. 2 , rod area A 4  of 1.484893 in. 2 , check pressure of 90 psi, diaphragm diameter 11.75 in., piston diameter 11 in., cylinder diameter 3.25 in., and rod diameter 1.375 in. Table 1 shows the theoretical performance of the prior art rolling diaphragm pump while Table 2 shows the performance according to the inventive design, with P 1  being the discharge pressure of the medium that is being pumped, P 2  being the pressure of the hydraulic fluid under the piston, and ΔP being P 1 −P 2 . 
         [0000]    
       
         
               
               
               
             
               
               
               
             
           
               
                 TABLE 1 
               
               
                   
               
               
                 P 1  (psi) 
                 P 2  (psi) 
                 ΔP (psi) 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 100 
                 95.02009 
                 4.979909 
               
               
                 200 
                 190.0402 
                 9.959817 
               
               
                 300 
                 285.0603 
                 14.93973 
               
               
                 400 
                 380.0804 
                 19.91963 
               
               
                 500 
                 475.1005 
                 24.89954 
               
               
                 600 
                 570.1205 
                 29.87945 
               
               
                 700 
                 665.1406 
                 34.85936 
               
               
                 800 
                 760.1607 
                 39.83927 
               
               
                 900 
                 855.1808 
                 44.81918 
               
               
                 1000 
                 950.2009 
                 49.79909 
               
               
                 1100 
                 1045.221 
                 54.77899 
               
               
                 1200 
                 1140.241 
                 59.75890 
               
               
                 1300 
                 1235.261 
                 64.73881 
               
               
                 1400 
                 1330.281 
                 69.71872 
               
               
                 1500 
                 1425.301 
                 74.69863 
               
               
                 1600 
                 1520.321 
                 79.67854 
               
               
                 1700 
                 1615.342 
                 84.65845 
               
               
                 1800 
                 1710.362 
                 89.63836 
               
               
                 1900 
                 1805.382 
                 94.61826 
               
               
                 2000 
                 1900.402 
                 99.59817 
               
               
                   
               
             
          
         
       
     
         [0000]    
       
         
               
               
               
             
               
               
               
             
           
               
                 TABLE 2 
               
               
                   
               
               
                 P 1  (psi) 
                 P 2  (psi) 
                 ΔP (psi) 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 100 
                 92 
                 8.0 
               
               
                 200 
                 192 
                 8.0 
               
               
                 300 
                 292 
                 8.0 
               
               
                 400 
                 392 
                 8.0 
               
               
                 500 
                 492 
                 8.0 
               
               
                 600 
                 592 
                 8.0 
               
               
                 700 
                 692 
                 8.0 
               
               
                 800 
                 792 
                 8.0 
               
               
                 900 
                 892 
                 8.0 
               
               
                 1000 
                 992 
                 8.0 
               
               
                 1100 
                 1092 
                 8.0 
               
               
                 1200 
                 1192 
                 8.0 
               
               
                 1300 
                 1292 
                 8.0 
               
               
                 1400 
                 1392 
                 8.0 
               
               
                 1500 
                 1492 
                 8.0 
               
               
                 1600 
                 1592 
                 8.0 
               
               
                 1700 
                 1692 
                 8.0 
               
               
                 1800 
                 1792 
                 8.0 
               
               
                 1900 
                 1892 
                 8.0 
               
               
                 2000 
                 1992 
                 8.0 
               
               
                   
               
             
          
         
       
     
         [0027]    As evident from Table 1, in the prior art design the ΔP is dependent on the working pressure, while in the exemplary inventive design ΔP is independent of the working pressure. 
         [0028]    A theoretical performance comparison, based on the above Equations 1-12, also is presented below for an exemplary catalyst pump assuming the following: diaphragm area A 1  of 44.17875 in. 2 , cylinder area A 3  of 8.295768 in., rod area A 4  of 1.484893 in. 2 , check pressure of 35 psi, diaphragm diameter 7.75 in., piston diameter 7.25 in., cylinder diameter 3.25 in., and rod diameter 1.375 in. Table 3 shows the theoretical performance of the prior art rolling diaphragm pump while Table 4 shows the performance according to the inventive design, with P 1  being the discharge pressure of the medium that is being pumped, P 2  being the pressure of the hydraulic fluid under the piston, and ΔP being P 1 −P 2  as in the examples above. 
         [0000]    
       
         
               
               
               
             
               
               
               
             
           
               
                 TABLE 3 
               
               
                   
               
               
                 P 1  (psi) 
                 P 2  (psi) 
                 ΔP (psi) 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 100 
                 89.24147 
                 10.75853 
               
               
                 200 
                 178.4829 
                 21.51706 
               
               
                 300 
                 267.7244 
                 32.27559 
               
               
                 400 
                 356.9659 
                 43.03412 
               
               
                 500 
                 446.2074 
                 53.79265 
               
               
                 600 
                 535.4488 
                 64.55118 
               
               
                 700 
                 624.6903 
                 75.30971 
               
               
                 800 
                 713.9318 
                 86.06824 
               
               
                 900 
                 803.1732 
                 96.82677 
               
               
                 1000 
                 892.4147 
                 107.5853 
               
               
                 1100 
                 981.6562 
                 118.3438 
               
               
                 1200 
                 1070.898 
                 129.1024 
               
               
                 1300 
                 1160.139 
                 139.8609 
               
               
                 1400 
                 1249.381 
                 150.6194 
               
               
                 1500 
                 1338.622 
                 161.3779 
               
               
                 1600 
                 1427.864 
                 172.1365 
               
               
                 1700 
                 1517.105 
                 182.895 
               
               
                 1800 
                 1606.346 
                 193.6535 
               
               
                 1900 
                 1695.588 
                 204.4121 
               
               
                 2000 
                 1784.829 
                 215.1706 
               
               
                   
               
             
          
         
       
     
         [0000]    
       
         
               
               
               
             
               
               
               
             
           
               
                 TABLE 4 
               
               
                   
               
               
                 P 1  (psi) 
                 P 2  (psi) 
                 ΔP (psi) 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 100 
                 91.90837 
                 8.091632 
               
               
                 200 
                 191.9084 
                 8.091632 
               
               
                 300 
                 291.9084 
                 8.091632 
               
               
                 400 
                 391.9084 
                 8.091632 
               
               
                 500 
                 491.9084 
                 8.091632 
               
               
                 600 
                 591.9084 
                 8.091632 
               
               
                 700 
                 691.9084 
                 8.091632 
               
               
                 800 
                 791.9084 
                 8.091632 
               
               
                 900 
                 891.9084 
                 8.091632 
               
               
                 1000 
                 991.9084 
                 8.091632 
               
               
                 1100 
                 1091.908 
                 8.091632 
               
               
                 1200 
                 1191.908 
                 8.091632 
               
               
                 1300 
                 1291.908 
                 8.091632 
               
               
                 1400 
                 1391.908 
                 8.091632 
               
               
                 1500 
                 1491.908 
                 8.091632 
               
               
                 1600 
                 1591.908 
                 8.091632 
               
               
                 1700 
                 1691.908 
                 8.091632 
               
               
                 1800 
                 1791.908 
                 8.091632 
               
               
                 1900 
                 1891.908 
                 8.091632 
               
               
                 2000 
                 1991.908 
                 8.091632 
               
               
                   
               
             
          
         
       
     
         [0029]    A suitable diaphragm  114  may be a rolling seal diaphragm obtained for example from Bellofram Corporation, of Newell, W. Va. Exemplary diaphragms and methods of use are disclosed in U.S. Pat. Nos. 3,137,215 and 3,373,236, each of which is incorporated herein by reference thereto. 
         [0030]    While various descriptions of the present invention are described above, it should be understood that the various features can be used singly or in any combination thereof. Therefore, this invention is not to be limited to only the specifically preferred embodiments depicted herein. 
         [0031]    Further, it should be understood that variations and modifications within the spirit and scope of the invention may occur to those skilled in the art to which the invention pertains. Accordingly, all expedient modifications readily attainable by one versed in the art from the disclosure set forth herein that are within the scope and spirit of the present invention are to be included as further embodiments of the present invention. The scope of the present invention is accordingly defined as set forth in the appended claims.