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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /572/CH9/EX9.14 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '572/CH9/EX9.14')
| -rwxr-xr-x | 572/CH9/EX9.14/c9_14.sce | 51 |
1 files changed, 51 insertions, 0 deletions
diff --git a/572/CH9/EX9.14/c9_14.sce b/572/CH9/EX9.14/c9_14.sce new file mode 100755 index 000000000..c7dfc4f1d --- /dev/null +++ b/572/CH9/EX9.14/c9_14.sce @@ -0,0 +1,51 @@ +//(9.14) A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1.0 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4, determine the mass flow rate, in kg/s, and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
+
+//solution
+
+//variable initialization
+Tnot = 360 //in kelvin
+pnot = 1 //in MPa
+A2 = .001 //in m^2
+k = 1.4
+
+pstarbypnot = (1+(k-1)/2)^(k/(1-k))
+pstar = pstarbypnot*pnot
+//part(a)
+//since back pressure of 500 kpa is less than critical pressure pstar(528kpa in this case) found above, the nozzle is choked
+//at the exit
+M = 1
+p2 = pstar //in MPa
+printf('the exit mach number for back pressure of 500kpa is: %f',M)
+T2 = Tnot/(1+((k-1)/2)*(M^2)) //exit temperature in kelvin
+R = 8.314 //universal gas constant, in SI units
+M = 28.97 //molar mass of air in grams
+V2 = sqrt(k*(R/M)*T2*10^3) //exit velocity in m/s
+mdot = (p2/((R/M)*T2))*A2*V2*10^3 //mass flow rate in kg/s
+printf('\nthe mass flow rate in kg/s for back pressure of 500kpa is: %f',mdot)
+
+//part(b)
+//since the back pressure of 784kpa is greater than critical pressure of pstar determined above,the flow throughout the nozzle is subsonic and the exit pressure equals the back pressure,
+p2 = 784 //exit pressure in kpa
+M2 = {(2/(k-1))*[(pnot*10^3/p2)^((k-1)/k)-1]}^.5 //exit mach number
+T2 = Tnot/(1+((k-1)/2)*(M2^2)) //exit temperature in kelvin
+V2 = M2*sqrt(k*(R/M)*10^3*T2) //exit velocity in m/s
+mdot2 = (p2/((R/M)*T2))*A2*V2 //mass flow rate in kg/s
+printf('\n\nthe mass flow rate at the exit in kg/s for back pressure of 784kpa is: %f',mdot2)
+printf('\nthe exit mach number for back pressure of 784 kpa is: %f',M2)
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