Answers assignment 4 real fluids-fluid mechanics


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Answers assignment 4 real fluids-fluid mechanics

  1. 1. 5PHYS207 Assignment 4 Real FluidsQ1. A 1:100 model submarine is 1.25 m long and is towed underwater at a velocity of 4.4 m/s. Atwhat velocity will the full scale submarine operate for dynamically similar forces? (Note: this is not adesign target but a laboratory examination of similitude eg similar levels of turbulence).This is a submerged situation, so the key dimensionless number is Reynolds. Using the model data,the Reynolds number is:Re = VD / = 4.4(1.25)1000/0.001 = 5.5x106(We can use any characteristic length, so long as we use the same length in model and prototype. Inthis case we have used the hull length. The fluid data states only that it is water, so we can assumerough figures of 1000 kg/m3 and 0.001 Pa.s for the density and viscosity.)The prototype will have a similar Reynolds number for similar forces:Re = VD / = V(125)1000/0.001 = 5.5x106The remaining unknown is the velocity:V = 0.001(5.5x106)/(125(1000)) = 0.044 m/sThis agrees with the point that submerged models at high speed are dynamically similar to submergedprototypes at low speeds. But is it realistic to assume that the full scale submarine will behave thisway? 0.044 m/s is a fairly low speed…Q2. The towing force on the model submarine of the previous question was measured as 7.4 N. Whatkinematically similar force would be applied to the prototype to attain the calculated speed?For similitude of forces we need to use the Euler number, which in Reynolds cases leads to a forceratio ofFr = r (p. 121 study guide)The fluid density is the same for the model and the prototype, so Fr = r = 1This means that the forces on the prototype are the same as on the model for kinematically similarspeeds.Fp = 7.4 NNo wonder the prototype velocity is so slow! This motive force is tiny compared to the actual powerneeded for even a small submarine. The point is that dynamic similitude, while providing a usefulpoint of comparison between model and prototype, is not realistic as a description of actualperformance.
  2. 2. 5Q3. The submarine of the previous example is also tested in a high float surface propulsion mode. Fora towing force of 6.2 N a slightly higher velocity of 4.8 m/s is obtained. What are the correspondingprototype velocity and propulsion force?Surface propulsion is a Froude number situation: We can work out the Froude number for the modeland use it to work out the prototype velocity:gyVFrFor the model, Fr = 4.8/√(9.8(1.25)) = 1.37(Again we can use any characteristic length, so long as we use the same length in model andprototype)For the prototype, Fr = V/√(9.8(125)) = 1.37This leaves only V: V = 1.37√(9.8(125)) = 48 m/sThis seems more realistic but in fact it only represents a design point for similar dynamic effectscompared to the model.The propulsion force ratio is similarly determined from the Euler and Froude numbers:Fr = rLr3( study guide)= Lr3(same density for model and prototype)That is Fm/Fp = (1/100)3And Fp = Fm(1003) = 6.2(1003) = 6.2 MN thrust.At 48 m/s this would require a power of about 300MW, a very unrealistic design target but notunacceptable for establishing wave friction loads for the experimental design process.Q4. A reservoir’s water surface is at an elevation of 80 m. The reservoir drains via a cast iron pipewhich is 250 m long and 400 mm in diameter into a pre-treatment pond. If the pond and outfall are atan elevation of 40 m, what flowrate can the pipeline deliver?It helps to draw a diagram:80 m40 m250 mHGLEntry losseslosses
  3. 3. 5The hydraulic gradeline loses a small incremental amount due to entry turbulence and then a largergradual amount due to pipe friction.The entry losses are he = ke V2/2g where ke is about 0.5The pipe losses are hf = fLV2/2gD where f is an as yet unknown friction factor.The two losses add up to the total head difference between the two water surfaces:he + hf = 40 mLet’s assume a value of f = 0.02 and check that later:V2(ke/2g + fL/2gD) = V2(0.5/(2(9.8)) + 0.02(250)/(2(9.8)0.4) = V2(0.663) = 40 mSo V = √(40/0.663) = 7.7 m/sThis gives a Reynolds number of Re = 7.7(0.4)1000/0.001 = 3.1x106Using the moody diagram for cast iron this gives a friction factor of 0.012. Repeat the calculation withthis value:V2(0.5/(2(9.8)) + 0.012(250)/(2(9.8)0.4) = V2(0.408) = 40 mV = √(40/0.408) = 9.9 m/sRe = 9.9(0.4)1000/0.001 = 4.0x106The value of f does not change for this small difference in Re.Hence V = 9.9 m/s and Q = VA = V( D2/4) = 9.9( 0.42/4) = 1.24 m3/sQ5. A pipeline needs to be designed to carry water from the pre-treatment pond of the previousexample to the treatment works 200 m away. The treatment works are at an elevation of 30 m andhave an internal working head of 15 m. A pump of design head 25 m is specified to supply 300 l/sfrom the pond to the treatment works. What diameter pipe is required to carry this amount?Let’s do a diagram again:The total headloss between the pump and the pressurised works is (40 + 25) – (30 + 15) = 20 m. Thisis shared between the pipe losses and an exit loss into the works tank:P40 m30 m45 m+ 15 m+ 25 m200 m
  4. 4. 5ht = he + hf = keV2/2g + fLV2/2gD = 20 mke is an exit coefficient, ke ~ 1.0 and f is the friction factor, again guessed initially as 0.02.We don’t know the velocity or the diameter of the pipe, but we do know the flowrateQ = VA = V D2/4Q2= V2 2D4/16 = 0.32= 0.09So V2= 0.09(16)/( 2D4) = 0.146/D4Putting in the other numerical values we can express the headloss as a function of D:ht = he + hf = 0.0074/D4+ 0.03/D5= 20 mWe can ignore the first (small) term and evaluate D = 5√(0.03/20) = 0.272 mV is then V = √0.146/D4= 5.1 m/sWe need to check if our assumed f is correct: Calculate a Reynolds number:Re = 5.1(0.272)1000/0.001 = 1.374x106Assuming cast iron pipe, this is f = 0.013The solution amounts to correcting this factor:D = 5√(0.03x(0.013/0.02)/20) = 0.250 mV = √0.146/D4= 6.1 m/sRe = 6.1(0.250)1000/0.001 = 1.525x106The friction factor does not change much for this small change in Reynolds number:D = 250 mm or next standard size upQ6. An airfoil of 37 m2area (1 wing only) has an angle of attack of 6oand is travelling at 25 m/s. Theaircraft has a total profile area of 4.2 m2in the direction of travel. If the coefficient of drag varieslinearly from 0.04 at 4oAoA to 0.12 at 14oAoA, what power is required to maintain this velocity at4oC temperature and 0.9x105Pa absolute pressure? If the lift coefficient is inversely proportional tothe square root of the drag coefficient, what is the lift on the wing at this speed? Corrections addedvia Moodle discussion.Start by looking at the drag equation:pD AUCDrag22Total Drag ForceWe can work out the drag coefficient CD by interpolating the data givenWe can work out the density using the absolute temperature and pressure
  5. 5. 5The slope of the CD profile is k = (0.12 – 0.04) / (14 – 4) = 0.008The value of CD at 6ois thus CD(6o) = 0.04 + 0.008(6 – 4) = 0.056The density of air at 4oC (277 K) and 0.9x105Pa is= P/RT = 0.9x105/(287(277)) = 1.132 kg/m3The total profile area must increase as the angle of attack increases. In the absence of blueprints, let’sassume it is approximately the base profile area + the wing area x the sine of the angle of attack:Ap = 4.2 + 2(37)sin(6) = 11.94 m2.So we can work out the drag force on the wing:pD AUCDrag22= 0.056(1.132)252(11.94)/2 = 237 NThe power required to maintain that velocity isPower = F.V = 237(25) = 5.9 kWThe lift coefficient is inversely proportional to the square root of the drag coefficient:CL = CD-1/2= 0.056-1/2= 4.2And the lift force isWL AUCLift22= 4.2(1.132)252(37)/2 = 55 kNAoA14o4o6o0.040.12