# Introduction n aerodynamics the air flows over (aircraft wing) or inside (pipes and nozzles) solid bodies. Low speed flows are usually treated as 'incompressible', that is, changes in density of air as it flows are ignored (density is assumed constant). However, that is only an assumption and not a reality; in fact Anderson 1 calls it a myth. The analysis of flows as incompressible will always be in error, howsoever small, but the results may be in error by a small amount which may be acceptable in practical life. It would be preferable if the analysis can be carried out as compressible flow provided the mathematics is not too involved. That is an aim of this communication. # II. # Analysis of Normal Shock Wave Fig, 1 shows the sketch of a normal shock wave. V 1 , P 1 , T 1 and ? 1 are the velocity, pressure, temperature and density of air just before the shock The parameters are related to each other by the following equations: Continuity: ? 1 .V 1 = ? 2 .V 2 (1) Momentum: P 1 + ? 1 .V 1 2 = P 2 + ? 2 .V 2 2 (2) Gas law: P 1 = ? 1 .R.T 1 and P 2 = ? 2 .R.T 2 (3) Where R is the gas constant. R = 287 J/kg.K for air. Energy equation: (4) Dividing eqn 2 by eqn 1: (5) Using gas law (eqn.3), eqn 5 can be simplified to: With V 2 known, T 2 can be calculated from eqn 7. Eqn 5 can be simplified with the help of eqn 1: (10) III. # Solution of Equations # I . (India). e-mail: bsgill.fet@mriu.edu.in V 2 can be determined from eqn 9. It is, however, not easy to solve the equation. The 'goal seek' facility of spreadsheet (like Excel of Microsoft Office) is used for this purpose. Once V 2 is known, temperature, pressure and density downstream of the shockwave are easily calculated. Downstream pressure P 2 can be calculated from eqn 10. Downstream density ? 2 can be calculated from eqn 3 as all parameters are known. wave. V 2 , P 2 , T 2 and ? 2 are the parameters just after the shock wave. It may be noted that V 2 can be determined from It can thus be seen that it is not necessary to read Mach number from any table for solving the normal shock wave problem. The sonic velocity and Mach number at both upstream and downstream regions of the shock wave can be easily calculated. Procedure for arriving at the solutions is described below. In this example V 1 = 680 m/s which means the flow is supersonic before the shock wave. The flow will then be subsonic after the shock wave. This means there will be two solutions of the equations: one where there is no shock and the second where a shock wave occurs. To arrive at the first solution (case 1), assume an arbitrary value of V 2 greater than 680 m/s, say 800 m/s. In this case LHS is 1. 403 (cell C15). Click on Data and then What-if Analysis in the spreadsheet. Then select Goal Seek. In Set Cell enter C15, the cell that contains value of LHS. In 'To value' enter 2.251, the value of RHS in cell C11. In 'By changing cell' enter C14, the cell that contains the value of V 2 . Click OK. The cell C14 will now show value close to 680 which is the same as value of V 1 . This means that no shock has occurred in this case. This shows that the Excel has been properly filled. Click OK. The values of T 2 , P 2 , ? 2 , c 2 , M 2 and P0 2 in cells C19, C20, C22, C24, C26 and C28 respectively will automatically update. If a shock has occurred then, downstream of the shock wave, the velocity will be subsonic. Refer to values in column D (case 2). Enter a small value, say 50, in cell D1. Values in subsequent cells will change. Use Data-What if analysis-Goal Seek as before and enter D15 in Set Cell, 2.251 in 'To Value' and D14 in 'By changing cell' and click OK. The value of V 2 in cell D14 is now 255.139 m/s. The values of T 2 , P 2 , ? 2 , c 2 , M 2 and P0 2 are automatically updated. It can be noted that P 2 = 4.495 atm and M 2 = 0.578. A complete solution of the problem is thus in the Spreadsheet. It can be observed that there was no need to read Mach number value from charts; the problem could be easily solved in a normal mathematical fashion. However, if desired values of sonic velocity and mach number before and after the shock could be easily calculated as shown in Table 1. Even total pressure values upstream and downstream of the normal shock could be calculated. # IV. # Pitot Tube Analysis Pitot static tubes are used to measure flow velocities. There are three cases, viz. incompressible subsonic, compressible subsonic and compressible supersonic. These cases are analyzed below. # a) Incompressible subsonic flow The instrument used to measure velocity of fluid at any location is called Pitot-static tube. The Pitot has two openings, one which faces opposite to the flow direction and another which is perpendicular to the flow direction. The former measures the total or stagnation pressure at the location of the Pitot tube and the latter measures the static pressure. No flow takes place through any of these two openings. The stagnation pressure represents a sum of the static pressure and the dynamic pressure. The difference between the stagnation pressure and the static pressure is measured by appropriate means. Bernoulli's equation is applied to the two readings. The stagnation pressure P0 is related to the static pressure P1 by the equation: # b) Compressible subsonic flow The Pitot tube can also be used to determine flow velocity in high speed but subsonic flow. The formula for velocity V1 in a compressible subsonic flow is: For the case with P0 = 1104326 Pa, P1 = 101325 Pa, ? = 1.4 and ? = 1.225 kg/m3, V 1 = 69.61 m/s. It may be noted that the difference between incompressible velocity and compressible velocity is hardly 0.39 m/s or 0.56%. In case P 0 = 140000 Pa, the incompressible velocity will be 251.28 m/s whereas compressible velocity will be 236.63 m/s. The difference is 14.65 m/s or 6.19% which may be difficult to ignore. # c) Supersonic flow When a Pitot tube is used to measure flow velocity in supersonic flow, it acts as an obstruction to the flow and velocity is brought down to zero in front of it. But transition from supersonic flow to subsonic flow means a shock has occurred somewhere on the way. It is apparent that the Pitot tube will measure total pressure behind the shock; hence it is necessary to determine relationship between dynamic or total pressure behind the shock and flow velocity before the shock. Table 2 shows the calculation. Column C gives data for a random case in supersonic flow. In the case shown, flow velocity of 350 m/s (cell C7) will give rise to Pitot tube reading of 198845 Pa (cell C26). Consider a Pitot tube reading of 275000 Pa for which we need to determine flow velocity. A two or three step iteration will be needed. A B C D E F 1 2 Table 2: Supersonic Pitot Tube 3 Step 1 Step 2 Step 2 depicts values that one will see in column C). The value in column C26 will now read 280029 which is somewhat different from the desired value of 275000. Repeat the iteration and we get the value of 274774 in cell C26 (as depicted in cell E26 in Table 2). One more iteration gives value of 275000 (the desired figure) in cell C26 and the corresponding figure of velocity V 1 (441.5 m/s) in cell C7 (as depicted in cell F7 in Table 2). Thus a reading Pa (equivalent to 2.714 atm) corresponds to V 1 = 441.5 m/s. Sonic velocity upstream shock is calculated as 339.6 m/s and downstream of shock as 370.6 m/s. Mach numbers upstream and downstream of shock are 1.300 and 0.786 respectively. Another example is shown in Table 3. It can be seen that Pitot tube reading of 1221980 Pa (12.06 atm) represents a velocity upstream of shock as 1018.7 m/s. Sonic velocity upstream of shock is calculated as 339.6 m/s and downstream of shock as 555.8 m/s. Mach numbers upstream and downstream of shock are 3.0 and 0.475 respectively. Complete information about the example like downstream velocity, pressure, temperature and density is available in the spreadsheet. There never was a need to refer to any tables. V. A B C D E F 1 # Flow Through Nozzles The spreadsheet solution of compressible flow # Continuity equation: ? 1 A 1 V 1 = ? 2 A 2 V 2 or ? 1 /? 2 ) = (A 2 /A 1 ) (V 2 /V 1 ) (A 1 and A 2 are cross sectional areas of two locations in the nozzle and V 1 and V 2 are velocities at these locations respectively. Location 2 could be anywhere in the nozzle between inlet and exit). # From equation of state: Steady state energy equation (neglecting potential energy): Or where Z = 2?R/ (?-1) =2009 for air. For isentropic flow: # Or From equations 1, 2 and 3, Velocity V 2 at the downstream location can be calculated from value of A 2 since all other information pertains to upstream and is known. Pressure P 2 at the downstream location can be calculated from eqn (17) and temperature T 2 can be calculated from eqn (15). All the information is available to determine the density at the downstream location. Table 4 is a sample of the nozzle flow analysis. It may be noted that column C is for parameters at the first section and column D is for second section where areas A 1 = 0.00785 sq m and A 2 = 0.00385 sq m. Also this table represents data for the subsonic solution of the problem. A second solution exists for the supersonic section of the problem. To arrive at the supersonic section of the solution, give a large value (say 1000) of V 2 (row 14) as the initial guess of V 2 and solve the problem through 'What-if-Analysis'. # VI. # Conclusion It is shown that compressible flow can be analyzed without going through the step of referring to pre-calculated tables. Use of spreadsheet makes it possible. In general, the reading of Mach number M may lie between two values and interpolation is required; interpolated value could be in error. Also the complete problem is solved in one step, i.e. all parameters are determined in one step. ![Fig. 1: Sketch of a normal shock wave Knowing that energy equation (4) can be simplified to:](image-2.png "") ![from which velocity V 1 is determined as:As an example, if P 0 = 104326 Pa, P 1 = 101325 Pa and ? = 1.223 kg/m 3 ,](image-3.png "") 2![Fig. 2: Flow velocity as a function of Pitot tube reading Fig. 2 shows the variation of flow velocity as a function of Pitot tube reading for both subsonic and supersonic regions.](image-4.png "Fig. 2 :") 1gives one example of solution of anormal shock wave problem.ABCD1problem2Case 1Case 23?1.41.44R, J/kg.K2872875V1, m/s6806806T1, K2882887P1,Pa1013201013208P1,atm119Ro1, kg/m31.2261.22610Z2009.0002009.00011RHS2.2512.25112a-6.000-6.00013b0.012134236 0.01213423614V2680.042255.13915LHS2.2512.25116c288.018108.05917d1611.351226.81618e1611.250604.51219T2, K287.917485.75520P2, Pa101284.592455460.90321P2, atm1.0004.49522Ro2, kg/m31.2263.26723c1340.174340.17424c2340.125441.78825M11.9991.99926M21.9990.57827P01, atm3.7973.79728P02, atm3.7975.545 1( ) Volume Xx X Issue I V ersion I Dof Researches in EngineeringGlobal Journal Compressible Flow Analysis through SpreadsheetUnder Data-(13)(14)Year 2020( ) Volume Xx X Issue I V ersion I D4 5 6 7 8 9cp, J/kg.K ? R, J/kg.K V1, m/s T1, K P1,Pa1004.5 1.4 287 350.0 287 1013201004.5 1.4 287 446.8 287 1013201004.5 1.4 287 441.3 287 1013203 1004.5 1.4 287 441.5 287 101320of Researches in Engineering10 11 12 13 14 15 16 17 18 19P1,atm Ro1, kg/m3 Z RHS a b V2, m/s LHS c d1 1.230 2009 5.707 -6.000 0.033448 332.9 5.707 272.959 386.0881 1.230 2009 3.889 -6.000 0.022135 289.7 3.890 186.133 292.5121 1.230 2009 3.961 -6.000 0.022572 291.4 3.961 189.493 295.7941 1.230 2009 3.958 -6.000 0.022552 291.4 3.958 189.387 295.794Global Journal20 21 22 23 24e T2, K P2, Pa P2, atm Ro2, kg/m3405.948 292.8 108692 1.073 1.293451.028 344.6 187607 1.852 1.897448.000 341.7 182703 1.803 1.863448.252 341.8 182884 1.805 1.86425T02, K348.0386.4384.0384.126P02, Pa19884528002927477427500027P02,atm1.9622.7642.7122.71428c1, m/s339.6339.6339.6339.629c2, m/s343.0372.1370.5370.630M11.0311.3161.3001.30031M20.9700.7790.7860.786© 2020 Global Journals 32Step 1Step 2Step 33cp, J/kg.K1004.51004.51004.51004.54?1.41.41.41.45R, J/kg.K2872872872876V1, m/s350.01034.31019.01018.77T1, K2872872872878P1,Pa1013201013201013201013209P1,atm111110Ro1, kg/m31.2301.2301.2301.23011Z200920092009200912RHS5.7071.5391.5551.55613a-6.000-6.000-6.000-6.00014b0.033448 0.007289 0.007414 0.00741715V2, m/s332.9265.4264.1264.116LHS5.7071.5401.5551.55517c272.95973.64974.37774.40018d386.088245.443243.002243.00219e405.948956.451937.674937.39320T2, K292.8784.7769.0768.821P2, Pa10869210794581047631104696422P2, atm1.07310.65410.34010.33323Ro2, kg/m31.2934.7934.7464.74524T02, K348.0819.7803.8803.525P02, Pa19884512579291222696122198026P02,atm1.96212.41512.06712.06027c1, m/s339.6339.6339.6339.628c2, m/s343.0561.5555.9555.829M11.0313.0463.0013.00030M20.9700.4730.4750.475 423cp1004.51004.54?1.41.45R2872876A10.007854 0.007853987P1, atm118P1, Pa1.01E+051.01E+059T1, K31331310V1, m/s10010011Ro1, kg/m31.13E+001.13E+0012Z2009200913RHS63.881763.881714V2, m/s 99.999787 258.77258715V2/V10.9999979 2.5877258716A20.007854 0.0038484517A1/A20.9999977 2.0408163318LHS63.88169 63.880852519P2, Pa1.01E+057.27E+0420P2, atm1.00E+007.17E-0121P1/P21.00E+001.39E+0022T2, K312.99997 284.64174923T1/T21.0000001 1.0996278724Ro2, kg/m31.13E+008.90E-0125Ro1/Ro2 1.00E+001.27E+0026c1, m/s 354.63136 354.63135827c2, m/s 354.63134 338.18494128M10.281983 0.2819829629M20.2819824 0.76518069 © 2020 Global Journals * JohnDAnderson McgrawHill * Equations for isentropic compressible flow through variable area ducts BSGill International Journal of Aerospace and Mechanical Engineering 7 2 May 2020