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Diffstat (limited to '3765/CH1/EX1.6')
| -rw-r--r-- | 3765/CH1/EX1.6/Ex1_6.sce | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/3765/CH1/EX1.6/Ex1_6.sce b/3765/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..de5edc931 --- /dev/null +++ b/3765/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,32 @@ +clc +// Example 1.6.py +// Consider the flow through a rocket engine nozzle. Assume that the gas flow +// through the nozzle in an isentropic expansion of a calorically perfect gas. +// In the combustion chamber, the gas which results from the combustion of the +// rocket fuel and oxidizer is at a pressure and temperature of 15 atm and +// 2500 K, respectively, the molecular weight and specific heat at constant +// pressure of the combustion gas are 12 and 4157 J/kg K, respectively. The gas +// expands to supersonic speed through the nozzle, with temperature of 1350 K at +// the nozzle exit. Calculate the pressure at the exit. + + +// Variable declaration +pc = 15.0 // pressure combustion chamber (atm) +Tc = 2500.0 // temperature combustion chamber (K) +mol_wt = 12.0 // molecular weight (gm) +cp = 4157.0 // specific heat at constant pressure (J/Kg/K) + +Tn = 1350.0 // temperature at nozzle exit (K) + +// Calculations +R = 8314.0/mol_wt // gas constant = R_prime/mo_wt, R_prime = 8314 J/K +cv = cp - R // specific heat at constant volume (J/Kg/K) +gamma1 = cp/cv // ratio of specific heat + +pn_by_pc = (Tn/Tc** gamma1/(gamma1-1)) // ratio of pressure for isentropic process** pn/pc + +pn = pc * pn_by_pc // pn = pc * pn/pc + +// Result +printf("\n Pressure at the exit is %.3f atm", pn) + |