ME 131B Fluid Mechanics Solutions to Week Three Problem Session: Isentropic Flow II (1/26/98) 1. From an energy view point, (a) a nozzle is a device that converts static enthalpy into kinetic energy. (b) a diuser is a device that converts kinetic energy into static enthalpy. In the absence of heat transfer and non-ow work, the stagnation enthalpy of the ow is constant. 2. For a steady, quasi-one-dimensional, adiabatic ow without wall friction, what do the following principles simplify to: (a) Conservation of Mass: V A = constant Remarks: The mass ow rate (product of density, velocity and area) is required to be a constant to conserve mass. In subsonic ow, the change in density is not so drastic. Velocity varies in a way which is consistent with our everyday experience. But the situation is so much dierent in the supersonic regime. Density change becomes very appreciable. Take the case of a supersonic ow in a converging passage: the density increase outweighs the area decrease and forces velocity to go down in order to conserve mass ow. Similar behavior is found in a supersonic ow in a diverging passage but opposite eects are observed. In summary, the \strange" behavior of supersonic ow is caused by the appreciable density change. It seems counter-intuitive because the world we encounter with on a daily basis operates mostly in the incompressible regime. (b) Momentum Equation: dp = V dv Remarks: Pressure and velocity change in opposite direction to each other in both subsonic and supersonic regimes. 1
(c) Conservation of Energy: h 0 = h + V 2 2 = constant Remarks: When the ow speeds up, the uid cools down and vice versa. This interchange between static enthalpy and kinetic energy is fundamental in understanding an adiabatic ow. (d) Second Law of Thermodynamics: s = constant 3. We have discussed how to locate the stagnation state of a given ow state (T 1 ; P 1 ; M 1 ) last week. How about its sonic ( ) state? Can you locate it on the T s diagram? By denition, the sonic ( ) state should have the same specic entropy as its static state. Hence, they should both be on the same vertical line on the T s diagram. By denition, the Mach number of the sonic state is unity. Hence, its location relative to that of its static state depends on the ow Mach number. If the ow is subsonic (M < 1), the sonic state will be below its static state. If the ow is supersonic (M > 1), the sonic state will be above its static state. Based on the above conclusions, we can locate the sonic state relative to its static state on a T s diagram in the following gures: Subsonic Case Supersonic Case T P 1 P * 1 T P * 1 T 1 P 1 T * 1 T * 1 T 1 s 1 s s 1 s 2
4. Complete the following table with increases, decreases, remains constant for an isentropic ow: Subsonic Flow: Converging Channel Diverging Channel P decreases increases decreases increases T decreases increases V increases decreases c decreases increases M increases decreases P 0 remains constant remains constant 0 remains constant remains constant T 0 remains constant remains constant A remains constant remains constant P remains constant remains constant remains constant remains constant T remains constant remains constant Remarks: Supersonic Flow: Converging Channel Diverging Channel P increases decreases increases decreases T increases decreases V decreases increases c increases decreases M decreases increases P 0 remains constant remains constant 0 remains constant remains constant T 0 remains constant remains constant A remains constant remains constant P remains constant remains constant remains constant remains constant T remains constant remains constant Both the stagnation state and the sonic state are constant in an isentropic ow. They serve as convenient reference states for the ow. 3
Trace each process path on a T s and a P diagram. Subsonic Case Supersonic Case T T Diverging T * 1 T 1 Converging T 1 Converging Diverging T * 1 s 1 s s 1 s P P ρ k = constant P P ρ k = constant P * 1 P 1 Diverging P 1 P * 1 ρ * 1 Converging ρ 1 ρ Diverging ρ 1 Converging ρ * 1 ρ 5. Choose the best answer in the following questions which concern the sonic state in an adiabatic, non-isentropic ow: (a) T remains constant in the ow. For any specic gas, the ratio of stagnation temperature to sonic temperature is a constant: T 0 T = 1 + k 1 2 Since the stagnation temperature of an adiabatic ow is constant, so is the sonic temperature. (b) P decreases in the ow. For any specic gas, the ratio of stagnation pressure to sonic pressure is a constant:! P 0 P = 1 + k 1 k k 1 2 4
Since the stagnation pressure of an adiabatic, non-isentropic ow decreases in the ow direction, so does the sonic pressure. (c) A increases in the ow. To see this point clearly, we can evaluate the mass ow rate at the sonic point: _m = A c where c = p k R T : We know from previous results that T ; c remain constant but P decreases in the ow. This leads us to conclude that decreases in the ow also, from the ideal gas equation. Hence, A has to increase to conserve the same mass ow rate. 6. A large supply chamber containing air at 6.0 atm and 300 K is connected to a converging nozzle on the left side and a converging-diverging (C-D) nozzle on the right side. Both nozzles share the same minimum passage area of 100 cm 2. The C-D nozzle has an exit-to-throat area ratio of 1.2. B P = 6.0 atm T = 300 K D C A Converging Nozzle C D Nozzle (a) Let us consider the converging nozzle on the left. i. Compare the pressure level at Point A, B, C and D. The main point of this part is to visualize the pressure variation and uid acceleration within the supply chamber. When the uid \senses" the pressure dierential between the inner chamber, P 0, and the surrounding ambient,, it accelerates from negligible velocity at chamber pressure to some nite velocity closed to the nozzle inlet. Associate with this ow acceleration, there is a corresponding pressure drop. We can treat the ow going through an \imaginary" converging passage from the inner chamber to the nozzle inlet. 5
Between the inlet and exit, the ow continues to accelerate and pressure continues to drop. We can conclude that P A > P C > P D The cause of pressure dierence between Point B and C is apparent after we draw the streamlines around the inlet of the converging nozzle. Since the streamlines curve around the corner, there is a positive pressure gradient developed in the normal (to the streamline) direction. Hence, P C > P B The comparison between pressure level at Point B and D depends on the exact nozzle geometry and requires further quantitative analysis. ii. If the ambient pressure is reduced to 5.0 atm, what is the mass ow rate in the nozzle? In this type of problem, we always need to check if the converging nozzle is choked at = 5:0 atm. For a converging nozzle, we learned that the ambient pressure has to be lower than 52.8 % of the chamber pressure before choking occurs. In this case, = 5:0 = 0:833 > 0:528 P 0 6:0 Hence, the nozzle is not choked. Furthermore, we can conclude that the pressure at the exit plane is the same as the ambient value. For the given pressure ratio P exit P 0 = 0:833 We can nd out from the isentropic ow table that And the temperature ratio is M exit = 0:517 T exit T 0 = 0:94924 which gives an exit temperature of T exit = 284:8 K. Using the thermal equation of state for an ideal gas P = R T we obtain an exit density of exit = 6:201 kg / m 3. 6
The mass ow rate can then be computed by _m = exit V exit A exit = exit M exit qk R T exit A exit ) _m = 10:8 kg/sec iii. How much do we need to lower the ambient pressure (relative to the chamber pressure) to reach the choking point of this converging nozzle? For a converging nozzle, the ambient pressure has to be lower than 52.8% of the chamber pressure to choke the converging nozzle. This corresponds to an ambient pressure of 3:17 atm If is lower than 3.17 atm, the exit plane pressure will not be the same as the ambient value (pressure mismatch). will keep staying at 3.17 atm. This is because no downstream pressure information can propagate upstream past the sonic point (exit plane). The ow within the nozzle becomes invariant once the sonic condition is attained at the exit. iv. What is the corresponding mass ow rate at the choking condition? When = 3:17 atm, the Mach number at the exit plane just reaches unity. Pressure at the exit plane equals to the ambient pressure = 3:17 atm From the isentropic ow table, we obtain T exit T 0 = 0:8333 ) T exit = 250 K Using the ideal gas equation, we obtain The mass ow rate is (b) Let us consider the C-D nozzle on the right. exit = 4:478 kg/m 3 _m = 14:2 kg/m 3 i. If the ambient pressure is set at 5.0 atm, do you expect the mass ow rate in the C-D nozzle to be the same as that in the converging nozzle computed before? 7
For this C-D nozzle case, we also need to check if the nozzle is choked at = 5:0 atm. The main dierence between the C-D nozzle and the converging nozzle is that the choking pressure ratio is dependent on the exit-to-throat area ratio (not a universal constant anymore). With an area ratio of 1.20, we nd from the isentropic ow table that the subsonic solution gives a pressure ratio P = 0:78997 < 5:0 P 0 6:0 Hence, we conclude that { the ambient pressure is high enough that the ow is not choked { the ow remains subsonic within the C-D nozzle { exit ; M exit ; T exit are the same as those in the converging nozzle case Since the exit area is 1.20 times as large as that of the converging nozzle, we expect a 20 % increase in the mass ow rate. Hence, _m = 13:0 kg/m 3 ii. How much do we need to lower the ambient pressure for the nozzle to operate at its rst critical point? The rst critical point corresponds to an isentropic, subsonic solution with Mach 1.0 ow at the throat. We obtain from the isentropic ow table that P 0 = 0:78997 ) = 4:74 atm iii. What is the corresponding mass ow rate at the rst critical point? Once this converging-diverging nozzle is choked at its rst critical point, we know that Mach 1.0 is achieved at its minimum ow area, i.e. at the throat. Furthermore, P throat ; T throat ; M throat are the same as those of the converging nozzle choked case. Hence, we expect the same mass ow rate as that of the converging nozzle choked case _m = 14:2 kg/m 3 iv. At the design point (third critical), A. what is the ambient pressure? The third critical point corresponds to an isentropic, supersonic solution in the C-D nozzle. 8
For an area ratio of 1.20, we obtain a supersonic solution from the isentropic ow table M exit = 1:534 This solution gives a pressure ratio of P 0 = 0:25922 ) = 1:55 atm B. determine the density and velocity at the exit plane. For the M exit = 1:534 solution, we obtain a temperature ratio of T exit T 0 = 0:67995 ) T exit = 204 K Using the ideal gas model, we obtain (c) Look back to your calculations, exit = 2:693 kg/m 3 and V exit = 439 m/sec i. How do you compare the ambient pressure which is required to choke the converging and C-D nozzle? Which one is higher? Can you explain it? For the converging nozzle, P choke = 3:17 atm. For the C-D nozzle, P choke = 4:74 atm. We conclude that the C-D nozzle is choked at a higher ambient pressure than the converging nozzle. This conclusion can be explained by the following pressure plot for C-D nozzle operation: / P 0 0.790 First Critical (C D nozzle) 0.528 Choking Point (Converging nozzle) Third Critical (C D nozzle) x Due to the pressure recovery in the diverging section of a C-D nozzle (subsonic ow), the C-D nozzle is choked at a higher back-to-plenum pressure ratio. The exact value of the this pressure ratio depends only on the exit-to-throat area ratio. 9
In summary, there are three operating regimes: A. =P 0 > 0:790, both nozzles are not choked. B. 0:790 > =P 0 > 0:528, only the C-D nozzle is choked. C. =P 0 < 0:528, both nozzles are choked. ii. How do you compare the mass ow rate between the two nozzles: A. before choking? Before any choking occurs, the C-D nozzle has a higher mass ow rate (20 % higher) than the converging nozzle simply because the exit area of the C-D nozzle is 20 % larger than that of the converging nozzle. As the ambient pressure is reduced, the C-D nozzle gets choked rst. Once it is choked, its mass ow rate is not aected by the ambient pressure anymore. Meanwhile the mass ow rate of the converging nozzle keeps increasing as the ambient pressure is reduced. B. after choking? The mass ow rate is the same in both nozzles after they are both choked. Graphically, the mass ow rate of the two nozzles can be compared as follows:. m 14.2 kg/sec C D Nozzle Converging Nozzle 0.528 0.790 / P 0 7. You are asked to build a supersonic wind tunnel with operating Mach number of 2.0 in the test section. The plenum conditions are constantly kept at 300 K and 10.0 bars. Due to cost factor, air ow is delivered at a rate of 1 kg/sec. (a) If the ow is treated as isentropic, what is the downstream cross-sectional area? To achieve a supersonic ow in the test section, we need a C-D nozzle connecting the plenum and the test section, with a Mach 1.0 ow (sonic state) right at the minimum throat area. 10
From the isentropic ow table, we obtain P P 0 = 0:52828 ) P = 5:283 bar T T 0 = 0:83333 ) T = 250 K The ideal gas equation further gives us the density at the throat: = 7:365 kg/m 3 The size of the throat can be found from the mass ow rate equation _m = V A throat = p k R T A throat ) A throat = 4:285 cm 2 To nd out the cross-sectional area at the test section, we need to relate the Mach 2.0 ow in the test section with the sonic state at the throat. From the isentropic ow table, M section = 2:0 gives A section A throat = 1:68750 ) A section = 7:230 cm 2 (b) If the entropy change between the plenum and the test section is 40 J/kg-K, what will be the cross-sectional area in the test section? Compare the result with Part (a) and label the two states on the same T s diagram. The ow remains to be adiabatic. Hence, T 0 ; T remains constant even in this non-isentropic ow. However, the entropy increase in the nozzle causes a drop in the stagnation pressure. (Take State 1 to be the plenum state and State 2 to be the test section state in the following analysis.)!! s 2 s 1 = C p log T 0;2 R log P 0;2 T 0;1 P 0;1 ) P 0;2 P 0;1 = exp = 0:870 ) P 0;2 = 8:70 bar From the isentropic ow table, M 2 s2 s 1 R = 2:0 gives P 2 P 0;2 = 0:12780 ) P 2 = 1:11 bar T 2 T 0;2 = 0:55556 ) T 2 = 167 K 11
The ideal gas equation further gives us the density in the test section: 2 = 2:32 kg/m 3 The cross-sectional area in the test section can be found from the mass ow rate equation _m = 2 V 2 A 2 = 2 M 2 qk R T 2 A 2 ) A 2 = 8:31 cm 2 Comparing this result with that of Part (a), we conclude that a larger test section area is necessary when irreversible eects are taken into account. It is also interesting to point out that the ratio between the area obtained in Part (a) and Part (b): A section;a A section;b = 7:230 cm2 8:310 cm 2 = 0:870 is the same ratio as the stagnation pressure loss. Hence, we conclude from this observation that P 0 A = constant for an adiabatic ow. When stagnation pressure drops, the sonic area increases. Both the isentropic and non-isentropic solutions are shown in the following T s diagram for reference: T P 2,a P 2,b T * = T* 2,a 2,b T = T 2,a 2,b Isentropic Solution Non Isentropic Solution s 2,a s 2,b s Comments: The ridiculously small test section area is not reasonable for conducting wind tunnel experiments. We can increase its size by { reducing the plenum pressure { increasing the plenum temperature { paying a higher cost to allow a higher mass ow rate 12