Question

air is stored in a tank 0.037661m

Problem 25. Air is stored in a tank 0,037661m in at an initial pressure of volume temperature of 321.4K. The gas is discharged through a converging P: 5,760.6 kPa and a nozzle with an exit area of 3.167x10 m2. For a back-pressure of 101 kPa, assuming spatially lumped polytropic process in the tank, i.e., pv constant, and isentropic flow in the nozzle, i.e., pv = constant, compare predicted tank pressures to the measured values contained the following table. Try various values of the polytropic exponent, n. from 1.0 (isothermal) to 1.4 (isentropic). Perform only a Stage I analysis, i.e., the nozzle is a 68choked. time, 11.0 13.0 15.0 17.0 19.0 7.0 9,0 4.0 6.0 5.0 3.0 0,0 1.0 2.0 sec 1.000 0.7170.551 0.448 0.358 0.281 0.241 0.199 0.142 0.104 0.078 0,059|0.044 0.033 Po/Pot Now from the continuity equation dmme-PeAcVe dt For polytropic expansion within the tank 52

Answer 1

Pol So Po Po Pol Pol And for isentropic expansion in the nozzle Po_Pe So Po Pe Po Pol Y For a choked flow: M 1, Ve = a. = vyp/pe and YY-1) Y-1 Po 2 2 Pe So, Pe Po 2 Therefore, 1 y1-Po Pe Pol 2 Pol Now Pe AeVYPePe Pe mePeAeVe PeAae 1 1 Y+121-y)1/2( Y+12(1-y) Po2n = VYAgpp Pol Pol 2 2 Pol

VYPol/Poi the mass flow rate at the exit can be written as Using ao 1 n+1 2n 2n me Aeal Pol +12(1-y)1 1/2 2 Pol within the tank is given by Now the time rate of change of the mass (1-n Poiv dpo Po , dpo dm 1 dt dt dt Equating this to the exiting flow rate gives 1 n+1 Y+1 Pol 12(1-y) (1-n) dpo-me =-Aeaol /2 Pol Po 1/n npol 1 2n 2n 11 Po dt 2 Pol Pol or 1-n 1-n n+1 Y+1 -nAeaolY+1 2(1-y), 1/n-1/2-1/2n Pol 1-3n dp, 2n 2n n+1 dt dt dt 2n Y+1 1-n -nAeaol Y1| 2(1-y) 'Po1 2n 2 Integration yields, (note 1-3n 1 -1 2n 2n Y+1 1-n 1-n 1-n Aea ol 1 2(1-) 2n Pol n -1 2n - 2 V 2 Rearrangement brings

2n Y+1 1-n Aea ol 12(1-y) n-1 + Ро t 2 2 Pol Note this is not valid for n -1, the isothermal case which must be treated separately. For n 1 Y+1 Pol Y121-y) Ро me Aea ol Pol 2 Po Po Pol Pol Y+1 ,dpo Pol dpo-ne =-Aea/ Pol( Y+1\2(1-Y), dm Po dt dt 2 Pol dt Pol Canceling, separating variables, integrating and rearranging yields, Y+1 Aeaol Y+l )2(1-Y)t Ро . 2 e Pol A spreadsheet program was written and run for various n. A table of the results is as follows 1.0 1.1 1.2 1.3 1.4 n = Ip./pa (ехp) Р. р. Р.Рa Р.рa t P/Pol P/Pol 0.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.000 0.7990 1.0 0.8396 0.8257 0.8122 0.7862 0.717 0.6421 0.6230 2.0 0.7049 0.6830 0.6620 0.551 3.0 0.5918 0.5658 0.5415 0.5187 0.4974 0.448 0.4695 4.0 0.4969 0.4443 0.4212 0.3999 0.358 5.0 0.281 0.4172 0.3902 0.3658 0.3437 0.3237 6.0 0.3503 0.3247 0.3021 0.2818 0.2636 0.241 0.199 7.0 0.2941 0.2707 0.2502 0.2321 0.2159 0.2073 0.1890 9.0 0.1731 0.1594 0.1473 0.142 0.1327 11.0 0.1461 0.1211 0.1112 0.1025 0.104 13.0 0.1030 0.0937 0.0856 0.0787 0.0726 0.078 15.0 0.0726 0.0665 0.0611 0.0564 0.0522 0.059 17.0 0.0512 0.0474 0.0440 0.0410 0.0382 0.044 0.0340 19.0 0.0361 0.0320 0.0301 0.0283 0.033