In which of the following scenarios is applying the following form of Bernoulli s equation: steady, inviscid, uniform stream of water. Ma = 0.


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1 bernoulli_11 In which of the following scenarios is applying the following form of Bernoulli s equation: p V z constant! g + g + = from point 1 to point valid? a. 1 stagnant column of water steady, inviscid, uniform stream of water b. aircraft Ma = c. pump 1 d. boundary layer 1 e. 1 oscillating Utube manometer containing an incompressible, inviscid fluid Page 1 of
2 bernoulli_11 SOLUTION: Bernoulli s equation, as written in the problem statement, can be used in NONE of the scenarios presented. a. The flow is rotational at the interface between the vertical and horizontal channels and, hence, Bernoulli s equation cannot be applied across the flow streamlines. b. Since Ma > 0.3, the flow should be considered compressible. The given form of Bernoulli s equation is valid only for incompressible flows. An alternate form of Bernoulli s equation that takes compressibility effects into account could be used, however. c. The pump between points 1 and adds energy to the flow and, hence, the constant in Bernoulli s equation changes across the pump. The Extended Bernoulli s Equation (aka energy equation) could be used in this scenario instead of the given form of Bernoulli s equation. d. Bernoulli s equation assumes inviscid flow. Viscous effects are significant in boundary layers and thus Bernoulli s equation may not be used. e. The given form of Bernoulli s equation assumes steady flow. The oscillating Utube is unsteady and the given Bernoulli s equation cannot be used. Note that it is possible to derive an unsteady form of Bernoulli s equation that could be used in the given situation. Page of
3 Water is flowing in a horizontal flat channel with a 90 bend as shown. The width of the channel is constant. The inner radius of the bend is r 1. The flow velocity is V.!!What is the correct description for pressures at locations A, B, and C? Circle your selection.! i. P A = P B = P C ii. P A > P B > P C iii. P A < P B < P C iv. P A = P C > P B v. P A = P C < P B Two velocity profiles are shown for flows in straight ducts. Circle the correct statement about the two flow fields or fluids.! i. (left) steady flow (right) unsteady flow ii. (left) Newtonian fluid (right) nonnewtonian fluid iii. (left) incompressible flow (right) compressible flow iv. (left) inviscid flow (right) viscous flow
4 !"%()*+,.*/*01+,,3+*3.,4*/51.*/67+,/%418*++1* % % % % %!"!!"!!"!!"!!"! ()*+,./,01!"! (* /81!"!!"! (9*:7;,0,<=/;=./81 (9*)>70?7;,0,<=/;=./81
5 normalshock_6 Consider the supersonic wind tunnel shown in the following schematic. Air is the working fluid and the test section area is constant. stagnation conditions, 100 kpa (abs), 300 K discharge to a back pressure throat area = 0.1 m test section area = exit area = m p/p p * /p throat test section exit x a. What is the design Mach number of the test section? SOLUTION: The test section design Mach number may be found using the isentropic sonic area ratio and choosing the supersonic test section Mach number (case 4 in the diagram above), A TS A T = m 0.1 m =.637 = A TS A * = 1 " 1+ k!1 Ma TS Ma TS 1+ k!1 % k+1 ( k!1) Note that at design conditions, the throat Mach number is one.! Ma TS =.50. (1) b. What is the mass flow rate through the wind tunnel at design conditions? SOLUTION: The flow through the wind tunnel will be choked at design conditions, with a mass flow rate of, "!m choked = 1+ k!1 % where A * = A T. k+1 1!k ( ) p0 k RT 0 A *!!m = 3.3 kg/s, () Page 1 of 3
6 normalshock_6 c. What is the maximum back pressure at which the throat will reach sonic conditions? SOLUTION: When the throat just reaches sonic conditions (case 3 in the diagram above), the throat area will equal the sonic area (A * = A T ) and the exit Mach number may be found using the isentropic sonic area ratio since the flow through the entire convergingdiverging nozzle will be subsonic (no shock waves), A E A T = m =.637 = A E 0.1 m A = 1 " 1+ k!1 * Ma E Ma E 1+ k!1 % k+1 ( k!1)! Ma E = (3) The exit pressure may be found from this Mach number using the isentropic stagnation pressure ratio, p E " = 1+ k!1 p 0 Ma % E k 1!k! p E /p 0 = ! p E = 96.5 kpa (abs), (4) using p 0 = 100 kpa (abs). Since the exit Mach number is subsonic, the exit and back pressures are equal. Hence, p B = p E = 96.5 kpa (abs). (5) d. Assume a shock wave stands at the exit of the convergingdiverging nozzle. What is the back pressure at these conditions? SOLUTION: The Mach number just upstream of the shock wave may be found using the isentropic sonic area ratio since the flow leading up to the shock wave is isentropic and the throat area is at sonic conditions (case 6 in the diagram shown above), A E m = =.637 = A E1 A T 0.1 m A = 1 * The pressure ratio across the shock is, " 1+ k!1 Ma E1 Ma E1 1+ k!1 % k+1 ( k!1)! Ma E1 =.50. (6) p E = k p E1 k +1 Ma E1! k!1 k +1! p E/p E1 = (7) The pressure just upstream of the shock wave is, p E1 " = 1+ k!1 p 01 Ma % E1 k 1!k! p E1 /p 01 = (8) The pressure just downstream of the shock is,! p E = p! E p E1 " p E1 % " p 01 % p 01 = (7.15)(0.0585)(100 kpa (abs)) = 41.7 kpa (abs) (9) Since the exit pressure just downstream of the shock is subsonic, the exit pressure and back pressure will be the same. Hence, p B = p E = 41.7 kpa (abs) (10) An alternate approach is to calculate the stagnation pressure ratio across the shock is, k 1 p " 0 ( k +1)Ma % ( k!1 ) " = E1 k +1 % ( k!1 )! p p 01 + ( k!1)ma E1 kma 0 /p 01 = (11) E1! ( k!1) The Mach number just downstream of the shock wave is, ( Ma E = k!1 )Ma E1 +! Ma kma E = (1) E1! ( k!1) Page of 3
7 normalshock_6 The exit pressure just downstream of the shock may be found using the isentropic relations and the Mach number just downstream of the shock, p E " = 1+ k!1 p 0 Ma % E k 1!k! p E /p 0 = (13) Accounting for the change in stagnation pressure ratio across the shock,! p E = p! E p 0 " % " % p 01 = (0.8357)(0.4990)(100 kpa (abs)) = 41.7 kpa (abs), (14) p 0 p 01 which is the same result found previously. Page 3 of 3
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