Patent Publication Number: US-2018045440-A1

Title: Ideal Liquid Compression Refrigeration Cycle

Description:
FIELD OF THE INVENTION 
     The present invention is directed to the mechanical power engineering for refrigeration and heat pumps. 
     BACKGROUND OF THE INVENTION 
     Refrigeration cycles transfer thermal energy from a region of low temperature to one of higher temperature, the reversed Carnot cycle is the perfect model for the refrigeration cycle operating between two fixed temperatures, the most ideal cycle, which has the maximum thermal efficiency, maximum coefficient of performance, and serves as a standard against which actual refrigerator cycles can be compared, reversed Carnot cycle consist of 4 processes, 2 isentropic processes for expansion and compression, and 2 isothermal processes for heat rejection and heat absorption. 
     Now most of the refrigerators and heat pumps are working on the principle of the ideal Vapor compression cycle, that cycle was built on the principals of the reversed Carnot cycle, but this cycle is deviate from the reversed Carnot for the following reasons:
         1—The refrigerant shall enter the compressor at the vapor phase, for the compressor operation.   2—Throttling valve is used in expansion process (constant enthalpy process)   3—The heat rejection and absorption at a constant pressure process, for more practicality.       

     SUMMARY OF INVENTION 
     The intent of this invention is to prove a new ideal refrigerator cycle (the Liquid compression cycle) which has a coefficient of performance higher than the Vapor compression cycle. 
     Technical Problems 
     The coefficient of performance for the ideal Vapor compression refrigeration cycle (VCRC) is lower than the reversed Carnot cycle due to the deviation of its ideal process from the reversed Carnot, this means that the ideal VCRC will consume more electric power than the reversed Carnot cycle at the same refrigeration capacity or when the two cycles are operating at the same maximum and minimum temperatures. 
     Moreover, all issues related to the compressors in the actual VCRC, for example the maintenance, lubrication system, and it&#39;s high cost, etc. 
     Problems Solution 
     The liquid refrigerant pump in the liquid compression refrigeration cycle (LCRC) is acting the same function of the compressor in the VCRC to solve all the above problems. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  showing the LCRC on T-S and T-H diagram. 
         FIG. 2  showing the COP levels for Carnot, LCRC, and VCRC. 
         FIG. 3  showing the VCRC on T-S and T-H diagrams. 
         FIG. 4  showing a simple schematic diagram for the main components. 
     
    
    
     DETAILED DESCRIPTION 
     Liquid compression cycle (LCRC) is a cycles, that can be applied in the refrigeration and heat pumps applications, this cycle has achieved the performance of the reversed Carnot cycle, unlike the vapor compression cycle, where a clear deviation from the reversed Carnot cycle is appeared in it&#39;s ideal case, the deviation is occur due to the compression process where the refrigerant has to be compressed to a temperature higher than the condensing temperature, and the constant enthalpy process in the expansion valve, where energy loss has occurred due to the irreversibility of the process, these deviations from Carnot cycle have been solved in the Liquid Compression Cycle (LCRC) to achieve a thermal efficiency more than the Vapor Compression Cycle (VCRC) efficiency, and we will prove that later. 
     Liquid compression cycle consists of 5 processes, 3 isentropic processes, one isothermal process, and one isobaric process, the cycle (T-H) and (T-S) diagrams are shown in  FIG. 1 . 
     Process (1-2) isentropic compression in a liquid pump 
     Process (2-3) isentropic expansion in a nozzle 
     Process (3-4) isothermal heat absorption in an evaporator 
     Process (4-5) isentropic compression in a diffuser 
     Process (5-1) isobaric heat rejection in a condenser 
     Liquid compression cycle is working between 3 levels of pressure, the refrigerant enter the pump at state 1 as a saturated liquid and compressed from the condenser pressure to a higher level pressure, then the refrigerant enters the expansion nozzle to reach the evaporator pressure, during this expansion process the refrigerant lose a lot of internal energy as well as the pressure is decreasing during the expansion, these amount of energy is converted to kinetic energy at state 3, then the refrigerant is absorbing heat during the isothermal process in the evaporator to reach state 4 in a 2 phase region, then the pressure is regained in the diffuser by converting a part of the kinetic energy again to enthalpy, the refrigerant is isentropic compressed to the condenser pressure at state 5, then the heat is rejected to the ambient at constant pressure to enter the pump again at state 1,  FIG. 4  is showing a schematic diagram for the cycle main components. 
     Example 
     The following example is showing how the Liquid compression cycle has achieved the performance of the reversed Carnot cycle comparing with the Vapor compression cycle at the same levels of condenser and evaporator pressure. 
     As shown in  FIG. 2  a comparison between Carnot, LCRC, and VCRC according to the COP levels 
     Assume refrigerant  134   a  in the Liquid compression cycle is working between the condenser pressure P 1 =1.2 Mpa, and the evaporator pressure P 3 =0.36 Mpa, with refrigerant effect 14 kJ/(kg of refrigerant), now, we can describe and calculate the properties at each state.
         @state 1: saturated liquid phase, P 1 =1.2 Mpa, T 1 =46° C., =117.77 kj/kg, s 1 =0.424 kj/kg. K, v 1 =0.00089 m3/kg.   @state 2: sub-cooled phase, P 2  shall be calculated by applying the energy equation on the total cycle, as follows:       

         w   p   =q   co   −q   ev =( T   1   −T   3 )Δ s  
 
       And, 
       Δ s= 14/(5.8+273)=0.05 KJ/Kg
 
       Hence, 
         w   p =(46−5.8)0.05=2.01 KJ/Kg
 
       But, 
         w   p   =v   1 ( P   2   −P   1 ) 
         P   2 =(2.01/0.00089)+1200=3458 Kpa=3.46 Mpa 
       For isentropic compression,  s   2   =s   1 =0.424 KJ/Kg·K
 
       And, 
         h   2   =v   1 ( P   2   −P   1 )+ h   1 =0.00089(3458−1200)+117.77=119.8 KJ/Kg
         @state 3: P 3 =0.36 Mpa, for isentropic expansion s 2 =s 3 =0.424 KJ/Kg·K, T 3 =5.8° C., x 3 =0.275, h 3 =h f +x 3 h fg =59.72+(0.275×194.08)=113.1 kj/kg.   @state 4: P 4 =P 3 =0.36 Mpa, calculating s 4 =Δs+s 3 =0.424+0.05=0.474 kj/kg·K, x 4 =0.347, calculating h 4 =h f +x 4 h fg =59.72+(0.347×194.08)=127.04 kj/kg   @state 5: P 5 =P 1 =1.2 Mpa, for isentropic compression s 5 =s 4 =0.474 kj/kg·K, x 5 =0.1015, h 5 =h f +x 5 h fg =117.77+(0.102×156.1)=133.61 kj/kg   Assume that the ideal Vapor Compression cycle (VCRC) is working at the same evaporator and condenser pressure as shown in  FIG. 3 :   @state 1: P 1 =1.2 Mpa, h 1 =117.77 kj/kg   @state 2: at throttling process, h 2 =h 1 =117.77 kj/kg   @state 3: P 3 =0.36 Mpa, h 3 =253.81 kj/kg, s 3 =0.9283 kj/kg   @state 4: for isentropic compression s 4 =s 3 =0.928 kj/kg·K, T 4 =50° C., P 4 =1.2 Mpa, h 4 =278.27 kj/kg       

     A—the Coefficient of Performance (COP) for the Rev. Carnot, LCRC, and VCRC: 
       COP carnot   =T   3 /( T   1   −T   3 )=278.8/(46−5.8)=7
 
       COP LCC   =q   ev   /w   p =14/2=7 
       COP VCC   =q   ev   /w   c =( h   3   −h   2 )/( h   4   −h   3 )=136.04/24.46=5.56 
     B— Special Configuration of the Nozzle and Diffuser Devices for the LCRC: 
     In the theoretical study of the liquid compression cycle, special considerations into nozzle and diffuser shall be considered: 
     i. Diffuser Inlet Velocities 
     Defining the relation between the inlet and outlet velocities by applying the energy balancing equation on the diffuser, 
         h   4 +( V   4   2 /2)= h   5 +( V   5   2 /2), 
       ( V   4   2 /2)−( V   5   2 /2)=Δ h   D  
 
     Dividing the two terms by (V 4   2 /2) 
     
       
         
           
             
               
                 
                   
                     
                       1 
                       - 
                       
                         
                           V 
                           5 
                           2 
                         
                         
                           V 
                           4 
                           2 
                         
                       
                     
                     = 
                     
                       
                         2 
                          
                         Δ 
                          
                         
                             
                         
                          
                         
                           h 
                           D 
                         
                       
                       
                         V 
                         4 
                         2 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       
                         V 
                         5 
                       
                       
                         V 
                         4 
                       
                     
                     = 
                     
                       
                         1 
                         - 
                         
                           
                             2 
                              
                             Δ 
                              
                             
                                 
                             
                              
                             
                               h 
                               D 
                             
                           
                           
                             V 
                             4 
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     1 
                      
                     a 
                   
                   ) 
                 
               
             
           
         
       
     
     Defining the relation between the inlet and outlet velocities by applying the mass balancing equation on the diffuser, 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           A 
                           5 
                         
                         · 
                         
                           V 
                           5 
                         
                       
                       
                         v 
                         5 
                       
                     
                     = 
                     
                       
                         
                           A 
                           4 
                         
                         · 
                         
                           V 
                           4 
                         
                       
                       
                         v 
                         4 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       
                         V 
                         5 
                       
                       
                         V 
                         4 
                       
                     
                     = 
                     
                       
                         
                           
                             v 
                             5 
                           
                           
                             v 
                             4 
                           
                         
                         · 
                         
                           
                             A 
                             4 
                           
                           
                             A 
                             5 
                           
                         
                       
                       = 
                     
                   
                 
               
               
                 
                   ( 
                   
                     2 
                      
                     a 
                   
                   ) 
                 
               
             
           
         
       
     
     From equation (1a) and (2a) 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           v 
                           5 
                         
                         
                           v 
                           4 
                         
                       
                       · 
                       
                         
                           A 
                           4 
                         
                         
                           A 
                           5 
                         
                       
                     
                     = 
                     
                       
                         1 
                         - 
                         
                           
                             2 
                              
                             Δ 
                              
                             
                                 
                             
                              
                             
                               h 
                               D 
                             
                           
                           
                             V 
                             4 
                             2 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Then 
                     , 
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     V 
                     4 
                   
                   = 
                   
                     
                       
                         2 
                          
                         Δ 
                          
                         
                             
                         
                          
                         
                           h 
                           D 
                         
                       
                       
                         1 
                         - 
                         
                           
                             
                               v 
                               5 
                               2 
                             
                             
                               v 
                               4 
                               2 
                             
                           
                           · 
                           
                             
                               A 
                               4 
                               2 
                             
                             
                               A 
                               5 
                               2 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     3 
                      
                     a 
                   
                   ) 
                 
               
             
           
         
       
     
     By substituting in equation 3, where, 
     v 4 =0.0202 m3/kg, and v 5 =0.0025 m3/kg (From the previous example) 
     
       
         
           
             
               V 
               4 
             
             = 
             
               
                 
                   2 
                    
                   Δ 
                    
                   
                       
                   
                    
                   
                     h 
                     D 
                   
                 
               
               
                 
                   1 
                   - 
                   
                     0.0153 
                      
                     
                       
                         A 
                         4 
                         2 
                       
                       
                         A 
                         5 
                         2 
                       
                     
                   
                 
               
             
           
         
       
     
     But from the above relation, we found that; 
     
       
         
           
             
               
                 1 
                 - 
                 
                   0.0153 
                    
                   
                     
                       A 
                       4 
                       2 
                     
                     
                       A 
                       5 
                       2 
                     
                   
                 
               
             
             ≈ 
             1 
           
         
       
     
     Hence, 
         V   4 ≈√{square root over (2Δ h   D )}  (4a)
 
     ii. Nozzle Outlet Velocities 
     Defining the relation between the inlet and outlet velocities by applying the energy balancing equation on the diffuser, 
         h   2 +( V   2   2 /2)= h   3 +( V   3   2 /2), 
       ( V   3   2 /2)−( V   2   2 /2)=Δ h   N  
 
     Dividing the two terms by (V 2   2 /2) 
     
       
         
           
             
               
                 
                   
                     
                       1 
                       - 
                       
                         
                           V 
                           2 
                           2 
                         
                         
                           V 
                           3 
                           2 
                         
                       
                     
                     = 
                     
                       
                         2 
                          
                         Δ 
                          
                         
                             
                         
                          
                         
                           h 
                           N 
                         
                       
                       
                         V 
                         3 
                         2 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       
                         V 
                         2 
                       
                       
                         V 
                         3 
                       
                     
                     = 
                     
                       
                         1 
                         - 
                         
                           
                             2 
                              
                             Δ 
                              
                             
                                 
                             
                              
                             
                               h 
                               N 
                             
                           
                           
                             V 
                             3 
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     1 
                      
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     Defining the relation between the inlet and outlet velocities by applying the mass balancing equation on the diffuser, 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           A 
                           3 
                         
                         · 
                         
                           V 
                           3 
                         
                       
                       
                         v 
                         3 
                       
                     
                     = 
                     
                       
                         
                           A 
                           2 
                         
                         · 
                         
                           V 
                           2 
                         
                       
                       
                         v 
                         2 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       
                         V 
                         2 
                       
                       
                         V 
                         3 
                       
                     
                     = 
                     
                       
                         
                           v 
                           2 
                         
                         
                           
                               
                           
                            
                           
                             v 
                             3 
                           
                         
                       
                       · 
                       
                         
                           A 
                           3 
                         
                         
                           A 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     2 
                      
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     From equation (1b) and (2b) 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           v 
                           2 
                         
                         
                           
                               
                           
                            
                           
                             v 
                             3 
                           
                         
                       
                       · 
                       
                         
                           A 
                           3 
                         
                         
                           A 
                           2 
                         
                       
                     
                     = 
                     
                       
                         1 
                         - 
                         
                           
                             2 
                              
                             Δ 
                              
                             
                                 
                             
                              
                             
                               h 
                               N 
                             
                           
                           
                             V 
                             3 
                             2 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Then 
                     , 
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     V 
                     3 
                   
                   = 
                   
                     
                       
                         2 
                          
                         Δ 
                          
                         
                             
                         
                          
                         
                           h 
                           D 
                         
                       
                       
                         1 
                         - 
                         
                           
                             
                               v 
                               2 
                               2 
                             
                             
                               v 
                               3 
                               2 
                             
                           
                           · 
                           
                             
                               A 
                               3 
                               2 
                             
                             
                               A 
                               2 
                               2 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     3 
                      
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     By substituting in equation 3b, 
     Where, 
     v 3 =0.016 m3/kg, and v 2 =0.00089 m3/kg (From the previous example) 
     
       
         
           
             
               V 
               3 
             
             = 
             
               
                 
                   2 
                    
                   Δ 
                    
                   
                       
                   
                    
                   
                     h 
                     N 
                   
                 
               
               
                 
                   1 
                   - 
                   
                     0.003 
                      
                     
                       
                         A 
                         3 
                         2 
                       
                       
                         A 
                         2 
                         2 
                       
                     
                   
                 
               
             
           
         
       
     
     From the above equation, we find that; 
     
       
         
           
             
               
                 1 
                 - 
                 
                   0.003 
                    
                   
                     
                       A 
                       3 
                       2 
                     
                     
                       A 
                       2 
                       2 
                     
                   
                 
               
             
             ≈ 
             1 
           
         
       
     
     Hence, 
         V   3 ≈√{square root over (2Δ h   N )}  (4b)
 
     C— General Configuration on the Actual Liquid Compression Cycle:
         1. As shown in the previous example the higher pressure level is calculated according to the minimum potential work needed for the reversible Liquid compression cycle, in the actual cycle, that pressure shall be increased to overcome all irreversibilities in the cycle.   2. The expansion process occur in the nozzle will be adiabatic irreversible process, where;       

     
       
         
           
             
               
                 
                   
                     η 
                     
                       is 
                       . 
                       N 
                     
                   
                   = 
                     
                    
                   
                     
                       Δ 
                        
                       
                           
                       
                        
                       
                         h 
                         act 
                       
                     
                     
                       Δ 
                        
                       
                           
                       
                        
                       
                         h 
                         is 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         Actual 
                          
                         
                             
                         
                          
                         Kinetic 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         at 
                          
                         
                             
                         
                          
                         exit 
                       
                       ) 
                     
                      
                     
                       / 
                     
                   
                 
               
             
             
               
                 
                     
                    
                   
                     ( 
                     
                       Isentropic 
                        
                       
                           
                       
                        
                       Kinetic 
                        
                       
                           
                       
                        
                       energy 
                        
                       
                           
                       
                        
                       at 
                        
                       
                           
                       
                        
                       exit 
                     
                     ) 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       V 
                       act 
                       2 
                     
                     
                       V 
                       is 
                       2 
                     
                   
                 
               
             
           
         
       
         
         
           
             3. The compression process occur in the diffuser will be adiabatic irreversible process, where; 
           
         
       
    
     
       
         
           
             
               
                 
                   
                     η 
                     
                       is 
                       . 
                       D 
                     
                   
                   = 
                     
                    
                   
                     
                       Δ 
                        
                       
                           
                       
                        
                       
                         h 
                         is 
                       
                     
                     
                       Δ 
                        
                       
                           
                       
                        
                       
                         h 
                         act 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         Isentropic 
                          
                         
                             
                         
                          
                         Kinetic 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         at 
                          
                         
                             
                         
                          
                         exit 
                       
                       ) 
                     
                      
                     
                       / 
                     
                   
                 
               
             
             
               
                 
                     
                    
                   
                     ( 
                     
                       Actual 
                        
                       
                           
                       
                        
                       Kinetic 
                        
                       
                           
                       
                        
                       energy 
                        
                       
                           
                       
                        
                       at 
                        
                       
                           
                       
                        
                       exit 
                     
                     ) 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       V 
                       is 
                       2 
                     
                     
                       V 
                       act 
                       2 
                     
                   
                 
               
             
           
         
       
         
         
           
             4. The compression process occur in the pump will be adiabatic irreversible process. 
             5. A pressure drop shall occur in evaporator and condenser coil, the same as the actual Vapor compression cycle. 
           
         
       
    
     Advantages of the Liquid Compression Cycle on the Vapor Compression Cycle:
         1. The coefficient of performance of LCRC is higher than VCRC.   2. If the refrigerant leaving the condenser in the sub-cooled region, or state 1 is fall in the sub-cooled region, the COP of the LCRC will slightly raised above the reversed Carnot cycle, the limitation for this raise depending on the minimum temperature approach between the refrigerant and the ambient or the cooling medium.   3. The work addition process is occur in the liquid phase, thus the actual process will be close to the isentropic process, unlike the VCRC, the work addition process in the superheat region, where more irreversibility has occur in the actual cycle.   4. The constant enthalpy process in the expansion valve for the VCRC, increasing the irreversibility in the ideal VCRC cycle, as well as, the actual cycle, where there no expansion valve throttling process in the LCRC.   5. The low refrigerant velocity for the vapor line and condenser coil, will decreasing the friction loss in pipes, and hence the irreversibility (or energy loss) will decrease in LCRC comparing with VCRC.   6. The lubrication system challenges in the VCRC are not exist in the LCRC by separating the lubricant from the refrigerant path.   7. The LCRC is more economic than VCRC in maintenance, by replacing the compressor with pump for work addition process.   8. The initial cost of LCRC is lower than VCRC in the reason of using pump instead of the compressor.       

     Disadvantages of the Liquid Compression Cycle on the Vapor Compression Cycle
         1. The required refrigerant mass flow rate is much higher than VCRC for the same evaporator capacity however this increasing in mass flow rate will not affecting on the total volume of the cycle, that because the density of the liquid stat is much higher than density at vapor stat, so the volume of the pump will not increasing as compressors at higher refrigerant mass flow rate, in addition the decreasing in condenser and evaporator effect (heat transfer in KW/Kg of refrigerant mass) will balance the increasing in refrigerant mass flow rate in LCRC, so the total surface area of the condenser and evaporator in LCRC will be close to the VCRC for the same cycle capacity.   2. The high refrigerant velocity for the liquid line and evaporator coil in the LCRC, will increasing the friction loss in pipes, and hence the irreversibility (or energy loss) will increased in LCRC comparing with VCRC, however this energy loss is too low if compared with the VCRC energy loss as discussed above.       

     General Recommendation on the Liquid Compression Cycle:
         1. The positive displacement pump could be more suitable for the higher compression ratio compared to the required mass flow rate.   2. Installing a refrigerant distributor before the evaporator and condenser coils to divide the mass flow rate on a multiple paths, will increase the heat transfer area and decreasing the refrigerant paths, also the nozzle could be a part of the pump casing, or installed directly after the pump to insure that the expansion is occurred suddenly after compression process.   3. For preventing cavitations at the centrifugal pump suction line a pressurized tank shall installed at the pump suction, also the tank for keeping the condenser, and the suction line at a constant pressure.       

     LEGEND 
     
         
         LCRC Liquid Compression Refrigeration Cycle 
         VCRC Vapor Compression Refrigeration Cycle 
         T Temperature 
         P Pressure 
         H enthalpy per unit of refrigerant mass 
         S entropy per unit of refrigerant mass 
         w p  Mechanical pump work per unit of refrigerant mass 
         w c  Mechanical compressor work per unit of refrigerant mass 
         q co  Heat rejected from condenser per unit of refrigerant mass 
         q ev  Heat absorbed to evaporator per unit of refrigerant mass 
         COP Coefficient Of Performance 
         X Mass quality, the ratio of the vapor mass to the total mass of the mixture. 
         Δh N  Total Enthalpy difference in the diversion-conversion nozzle
       Δh N =Δh N2 −Δh N1      
     
         Δh N1  Enthalpy difference in the diversion section of the diversion-conversion nozzle 
         Δh N2  Enthalpy difference in the conversion section of the diversion-conversion nozzle 
         Δh D  Total Enthalpy difference in the diversion-conversion diffuser
       Δh D =Δh D1 −Δh D2      
     
         Δh D1  Enthalpy difference in the diversion section of the diversion-conversion diffuser 
         Δh D2  Enthalpy difference in the conversion section of the diversion-conversion diffuser 
         P Pump. 
         N Nozzle. 
         D Diffuser. 
         CO Condenser coil. 
         EV Evaporator coil. 
         Δh act  Actual enthalpy difference 
         Δh is  Isentropic enthalpy difference 
         V act   2 /2 Actual Kinetic energy 
         V is   2 /2 Isentropic Kinetic energy 
         η is,N  Isentropic efficiency of the nozzle 
         η is,D  Isentropic efficiency of the diffuser