Question

Air enters a well-insulated nozzle at 400 m/s, 7 kPa, and 417°C and exits at 700 m/s. The nozzle inlet diameter is 0.2 m. Assuming ideal gas behavior, calculate a) the mass flow rate of the air. Then, determine the exit temperature of the air two ways: b) using the air table (A-17, which automatically accounts for the variation of specific heat with temperature); and c) assuming constant specific heats at the inlet temperature of 417°C. i he comatn speche hettn ar roblen?

Answer 1

NOTE: all the tables used in this problems are taken from the cengel thermodynamics book

a) mass flow rate of the air

$density\ of\ air\ at\ inlet:\rho =(P)/(RT)$

$\rho =(7)/(0.287*(417+273))=0.035\,kg/m^3$

$\dot{m}=\rho_{inlet}*A_{inlet}*V_{inlet}=(0.035)*(\pi/4*0.2^2)*(400)$

$\dot{m}=0.444\,kg/s$

b)Exit temperature of air using table:

7 8371 31963 07395 17283 82710 g 7 9 0 2 4 67902 45780 13469 3 3 4 4 4 4 4 4 5 5 5 5 5 56 66666 s* 2 2 2 2 2 22222 22222 22222 2 49491 06673 32344 76829 89386 55705 18656 50516 28518 52974 29751 11009 98887 77666 65555 80511 25334 72912 75208 64321 00901 97420 75318 64208 64106 12345 56788 90122 34567 uk 44444 44444 45555 55555 81806 44636 59082 20575 33233 86814 82830 75323 45678 90134 57802 35793 P-11111 12222 22233 33334 2237 32474 2a 5050 6284-8 k 6 6 778 89901 1 344 56780 89012 3 4 5 7 8 90123 45670 55666 66666 6-777 77778 89012 3 4 5 6 7 8 9012 34568 TK 5 5 6 6 6 66666 66 77 77777

from above table enthalpy of air at T=690 K is 702.52 kJ/kg

$from\,energy\,\,balance:h_1+(V_1^{2}/2000)=h_2+(V_2^{2}/2000)$

$(702.52)+(400^{2}/2000)=h_2+(700^{2}/2000)$

$h_2=537.52\,kJ/kg$

01740 62377 6951 76006 75996 0083 1467 89999 9865 K81357 91 357 9135 01-11 12 2 2 2 23 33 86415 76117 7102 63109 90246 9371 32108 77654 3322 02720 g36937 493 5 2 9 7 4 29 641 9641 1 2 2 3 4 55 678 8901 33333 33333 3344 25528 4114 37446 21387 0660 F-37272 84063 1865 55667 78990 1123 810249 42238 5 79 /k 6 8024 70369 3715 hk- 11222 2333a 45678 9012 5 4 44444 45555 5555 0 TK-45678 90 0 1 2 3 4 5 6 7 44444 45 5 5 5 5 5

for h=537.52 kJ/kg using interpolation between T=530 K and 540 K gives exit temperature T2=533.41 K

c) Exit temperature of air assuming constant specific heat:

$T_1=417+273=690\,K$

Ideal- gas specific heats of various common gase (b) At various temperatures Cy kJ/kg.K kJ/kg.K k Temperature Air 250 300 350 400 450 500 550 600 650 700 750 800 900 1000 1.003 1.005 1.008 1.013 1.020 1.029 1.040 1.051 1.063 1.075 1.087 1.099 1.121 1.142 0.716 1.401 0.718 1.400 0.721 1.398 0.726 1395 0.733 1.391 0.742 1387 0.753 1.381 0.764 1376 0.776 1.370 0.788 1364 0.800 1359 0.812 1.354 0.834 1.344 0.855 1.336

$Using\,\,interpolation\,for\ T=690K: C_P=1.0726 \,kJ/kg.K$

$from\,energy\,\,balance:C_P*T_1+(V_1^{2}/2000)=C_P*T_2+(V_2^{2}/2000)$

$1.0726*690+(400^{2}/2000)=1.0726*T_2+(700^{2}/2000)$

$T_2=536.168\,K$

error in exit temperature by assuming constant specific heat is:

$error=\frac{536.168-533.41}{533.41}*100=0.5\,percent$

As error associated with constant specific heat assumption is very less, it's good approximation for the given problem.