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Choking Due To Friction The theory here predicts that for adiabatic frictional flow in a constant area duct, no matter what the inlet Mach number M 1 is, the flow downstream tends toward the sonic point. There is a certain duct length L*(M 1 ) for which the exit Mach number will be exactly unity. But what if the actual duct length L is greater than the predict- ed “maximum” length L* ? Then the flow condition must change and there are two classifications. Subsonic Inlet If L ˃ L*(M 1 ), the flow slows down until an inlet Mach number M 2 is reached such that L = L*(M 2 ). The exit flow is sonic and the mass flow has been reduced by frict- ional choking. Further increases in duct length will continue to decrease the inlet Mach number M and mass flux. Prof. Dr. MOHSEN OSMAN1

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Supersonic Inlet From Table we see that friction has a very large effect on supersonic duct flow. Even an infinite inlet Mach number will be reduced to sonic conditions in only 41 diameters M = M = ∞ Fig. (4.3) Behavior of duct flow with a nominal supersonic inlet condition M = 3.0 (a) L/D ≤ 26, flow is supersonic throughout duct; (b) L/D = 40 ˃ L*/D; normal shock at M = 2.0 with subsonic flow then accelerating to sonic exit point. Prof. Dr. MOHSEN OSMAN2

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(c) L/D = 53, shock must now occur at M = 2.5; (d) L/D ˃ 63, flow must be entirely subsonic and choked at exit Some typical numerical values are shown in Fig. (4.3), assuming an inlet M = 3.0 and. For this condition L* = 26 diam. If L is increased beyond 26 D, the flow will not choke but a normal shock will form at just the right place for the subsequent subsonic frictional flow to become sonic exactly at the exit. Figure (4.3) shows two examples, for As length increases, the required shock moves upstream until, for Fig.(4.3), the shock is at the inlet for Further increase in L causes the shock to move upstream of the inlet into the supersonic nozzle feeding the duct. Yet the mass flux is still the same as for the very short duct, because presumably the feed nozzle still has a sonic throat. Prof. Dr. MOHSEN OSMAN3

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Eventually, a very long duct will cause the feed-nozzle throat to become choked, thus reducing the duct mass flux. Thus supersonic friction changes the flow pattern if L ˃ L* but does not choke the flow until L is much larger than L*. Example 4.3 Air enters a 3-cm-diameter duct at and V 1 = 100 m/s. The friction factor is Compute (a) the maximum duct length for this condition, (b) the mass flux if the duct length is 15 m, and (c) the reduced mass flux if L = 30 m. Required: D = 3 cm & f = 0.02 (a) L max for given condition (b) for L = 15 m P o =200 kPa (c) for L = 30 m T o =500 K V 1 = 100 m/s Prof. Dr. MOHSEN OSMAN4

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Solution (a) Apply energy equation between stagnation and inlet states C P T o = C P T 1 + ½ V 1 2 Interpolating into Fanno-Line Table across, from M=M 1 =0.225 we get Prof. Dr. MOHSEN OSMAN5

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(b) The given L=15 m is less than L max, and so is not choked and the mass flux follows from inlet conditions; Prof. Dr. MOHSEN OSMAN6

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(c) Since L = 30 m is greater than L max, the duct must choke back until L = L max, corresponding to a lower inlet M 1 Interpolation in Fanno-Line Table we find that this value of 20 corresponds to M 1, choked = (23% less) From Isentropic Table : Prof. Dr. MOHSEN OSMAN7

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Isothermal Flow with Friction The adiabatic frictional-flow assumption is appropriate to high-speed flow in short ducts. For flow in long ducts, e.g., natural-gas pipelines, the gas state more closely approximates an isothermal flow. The analysis is the same except that the iso-energetic energy equation (4.1c) is replaced by the simple relation. T = constant dT = 0.0 (4.11) Again it is possible to write all property changes in terms of the Mach number. Integration of the Mach number versus friction relation yields. which is the isothermal analog of equation (4.7) for adiabatic flow. Prof. Dr. MOHSEN OSMAN8

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This friction relation has the interesting result that L max becomes zero not at the sonic point but at a. If the tube length L is greater than L max from Eq. (4.12), a sub- sonic flow will choke back to a smaller M 1 and mass flux and a supersonic flow will experience a normal-shock adjustment to Fig. (4.3). The exit isothermal choked flow is not sonic, and so the use of the asterisk (*) is approximate. Let represent properties at the choking point L=L max. Then the isothermal analysis leads to the following Mach number relations for the flow properties:. Prof. Dr. MOHSEN OSMAN9

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An interesting by-product of the isothermal analysis is an exact relation between pressure drop and duct mass flux. In contrast, it is practically impossible in adiabatic flow to eliminate the Mach number between eqns. (4.7) and (4.9a) to get a direct relation between pressure and friction. Consequently, the common problem of predicting adiabatic duct mass flux for a given pressure drop can only be solved by a laborious Mach-number-iteration procedure. For isothermal flow, we can substitute in Eq.(4.1b) to obtain. But from Eqs.(4.1a) and (4.2) with dT = 0.0, we have. so that Eq. (4.14) becomes. Prof. Dr. MOHSEN OSMAN10

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Since G 2 RT is constant for isothermal flow, these are exact differentials and can be integrated from giving:. Thus we have an explicit expression for isothermal mass flux as a function of the pressure drop in the duct. If you don’t think eqn (4.16) is beautiful, try solving the same problem for adiabatic flow with the Mach-number relations or Table. Eqn (4.16) has one flow: with the Mach number eliminated, it cannot recognize the choking phenomenon. Therefore one should check the physical realism of a solution using Eq(4.16) by computing the exit Mach number M 2 to ensure that it is not greater than for a subsonic inlet flow. Prof. Dr. MOHSEN OSMAN11

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Example 4.4 Air enters a pipe of 1 in diameter at subsonic velocity and P 1 = 30 psia, T 1 = 550 o R. If the pipe is 10 ft long, and the exit pressure P 2 =20 psia, compute the mass flux for (a) isothermal flow and (b) adiabatic flow Solution 1 f = A) Isothermal Flow. D = 1inch..... L pipe = 10 inches P 2 =20 psia P 1 = 30 psia T 2 =T 1 = 550 o R T 1 = 550 o R For isothermal flow Eq. (4.16) applies. Compute. Prof. Dr. MOHSEN OSMAN12

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Accordingly, G = 1.7 lb f.s/ft 3 = 1.7 slugs/(s.ft 2 ) (Note: mass unit = slug = (weight unit / g) ) Check the Mach number at inlet and exit. Prof. Dr. MOHSEN OSMAN13

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Then & Since these are well below choking, the solution is accurate. B) Adiabatic Flow L max1 – L max2 = 10 f = P 1 = 30 psia P 2 = 20 psia T 1 = 550 o R For adiabatic flow, the only solution you can think of is to guess M 1 Compute read M 2, finally compute which is the ratio of the given pressures Prof. Dr. MOHSEN OSMAN14

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If so, you have found the correct solution. If not try another M 1 and iterate to find the proper solution. The isothermal relation provides a guess for The last value is close enough : Prof. Dr. MOHSEN OSMAN15 M1M1 M2M2 P 1 / P – 3 = – 3 = – 3 = – 3 =

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Example 4.5 At the inlet of 0.1 m diameter duct, the initial Mach no. is 0.2 and the static pressure and temperature are 200 kPa and 300K respectively. If the friction factor is 0.02, find the Mach number and properties after the flow has traversed 50 m of duct. Note: M 1 = 0.2, P 1 = 200 kPa T 1 = 300 K (1)________________________(2) (1)_______________(2) P*,T* f = 0.02 L = 50 m D = 0.1 m f = 0.02 L = 50 m D = 0.1m _________________________ ______________ (a) Problem Statement (b) Sonic position indicated Prof. Dr. MOHSEN OSMAN16

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From Isentropic Table From Fanno line M 1 = 0.2 Interpolating into Fanno Line Table across from Prof. Dr. MOHSEN OSMAN17

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Example 4.6 A converging-diverging nozzle / constant-area duct system is connected to a reservoir containing air with a stagnation pressure of 700 kPa. The exit to throat area ratio of the nozzle is 3; the constant area duct is 6 m long and 0.3 m in diameter with a friction coefficient of Determine the range of back pressures for which a normal shock wave will stand in the duct SOLUTION Using Isentropic Table for we get:.M e = 2.64, If a normal shock wave will stand at inlet to the duct; use N.S. Table for M x = M e = 2.64, we get kPa Using Fanno Line Table for M = M y = , we get: Prof. Dr. MOHSEN OSMAN18

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Flow parameters If a normal shock wave will stand at exit of the duct; use Fanno Line Table for M 1 = M e = 2.64, we get: Using N.S. Table for M x = M 2 = , we get: kPa ˂ P b ˂ kPa Prof. Dr. MOHSEN OSMAN19

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