im asked to know the derivation of daynamic and thermo dynamic and fluid equtions for them
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THERE IS A BOOK CALLED "TURBINES,COMPRESSORS AND FANS :BY S .M YAHAYA"IT WILL ANSWER ALL UR QUERIES.
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Its so easy to do a search\n\nhttp://www.taftan.com/thermodynamics/STURBINE.HTM\n\nhttp://en.wikipedia.org/wiki/Steam_turbine
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OK...You can start with the equation of DG (I'm using D instead of writing "delta").\n\nDG = DH -TDS  You've seen that equation, right?  OK..Since the boiling point of water is 100 degreed C at 1 atm pressure, we are really talking about an equilibrium situation.  So, DG= 0.\n\nConsequently, DH = TDS.  \n\nSince you know DH and you know T (in Kelvin), you can calculate DS.  Make sure that you convert DH into Joules (instead of kilojoules) before you finish the problem.
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USE PRINCIPLE OF CONSERVATION OF linear MOMENTUM.\n\nearth gas lower KE because it has huge mass and huge inertia
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I'm going to assume the dynamic pressure of the outflowing water equals the static pressure just inside the tank:\n\n0.5ρV²  =  ρgh → V = √(2gh) = √(2*9.8*10) = 14 m/sec\n\nFlowrate = A*V → A = F/V = 2.9E-3/14 = 2.071E-4 m² and\n\nd = √(4A/π) = 1.624E-2 m = 16.24 mm
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http://www.thermo.ae/thermo/default.htm
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This is a problem in open channel flow in Fluid Mechanics, not a pre-calc problem.  A proper solution needs calculus and a better description of the harbor bottom.\n\nAn intuitive set-up based on the information you've given would be \nh = (e^-kt)[(10 - 9) - 9sin(2πt/15)]\nwhere h is in feet, t is in minutes, and k is a damping constant, probably on the order of 1/15.
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depends on the transit time to thermal transfer rate ratio.\n1) Adiabatic if the transit time is much less than the thermal equalization time. \n2) PV = nRT if the transit time is much greater than the thermal equalization time. \n\nSo for the sake of simplicity, for option 2\nV = nRT/P\nV1/V2 = T1P2/T2P1\n\nT1/T2 = (273 + 7)/(273 + 20) = 0.9556\nP2/P1 = 33/49 =0.6735 [33 ft depth is +1 atmospheres]\n\nV1/V2 = 0.9556 * 0.6735 = 0.6436\n\nV2 = V1/0.6436 = 3.12 cm3\n\nAs the transit time becomes less than the thermal equilibrium time, the air bubble is cooler so the volume is less than the calculated value.
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let T1 and V1 be velocity and time to lake\nand T2 and V2 be velocity and time to ocean\n \nT1 x V1 = 600\nT2 x V2 = 240\nalso\nT1 = 2T2\nV1= V2 + 6\nhence\n2T2 x V1 = 600\nsubstituting again V1 = V2 + 6\n2T2 x(V2+6) = 600\nT2 x V2     = 240\nor T2       = 240/V2\nsubstituing in equation:\n2 x (240/V2) ( V2 + 6) = 600\nx V2 both sides\n2x 240 (V2+6)= 600 V2\n480 V2 + (6x480) = 600 V2\nsubtract 480 V2 from both sides\n6x 480 = 120 V2\ndivide by 120 each side\n6 x 480/120 = V2\nV2 = 24\nV1 = 30 \nT2 = 10 hours\nt1 = 20 hours \nchecked in top equations
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Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
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The link below describe Navier-Stokes Equation in detail.  It can tell you much more than I can write over here!  As of now the solution to the equation does not exist.  If you can solve it let me know.
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The following link is a pdf slide show that gives examples of buoyancy calculations and theory of floating objects.\n\nhttp://www.colorado.edu/physics/phys1010/phys1010_fa05/lectures/class12-09-29.pdf
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http://www.seafriends.org.nz/oceano/waves.htm\n\nIs a good place to start.\n\nhttp://www.fluidmech.net/tutorials/ocean/w_waves.htm\nIs more advanced\n\nYou will need to go to a libray and get a book to see equations and solutions.
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a=(delta V)/(delta t)\n\ndelta V = V_final - V_initial = 0 - 3.2\ndelta t = 0.01 sec\n\na=3.2/0.01=320 m/s2\n\nforce=mxa=0.084x320=26.88 N
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A good place to start will be to look at the following websites \n\nThermal Energy and States of Matter\nhttp://www.pschweigerphysics.com/thermenergy.html\n\nThermal Energy\nhttp://sol.sci.uop.edu/~jfalward/physics17/chapter6/chapter6.html\n\nThermodynamics and Thermal Physics guides\nhttp://physics.about.com/cs/thermodynamics/\n\nIf your doing a project you might look into "Ocean Thermal energy conservation" as a topic.\n\nOcean Thermal energy conservation is used to generate electricity using the temperature difference of seawater at different depths
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Entropy is the change in heat energy, Q, over a given temperature, T.\n\nIf you are familure with calculus, the ideal gas law, and the first law of thermodynamics then you can derive it.\n\ndS = dQ/ T  (change in entropy per change in heat)\n\nSince the tempurature is not constant you will have to use the first law of thermodynamics\n\ndQ - dW = dEint   \n\nUsing the ideal gas law \n\nPV = nRT \n\nand the fact work is \n\ndW = PdV \n\nand \n\ndEint = nCvdT \n\nyou can write an equation for a non-isothermal and non-isochoric situation:\n\nS = nR ln (Vf/Vi) + nCvln(Tf/Ti)\n\nWhere\nS - change in entropy\nn - number of moles\nR - gas constant\nVf - Final Volume\nVi - Initial Volume\nCv - specific heat\nTf - Final temperature\nTi - Initial temperature
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thermal eneergy or heat