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Control Volumes

Control Volumes. Thermodynamics Professor Lee Carkner Lecture 9. PAL # 8 Specific Heat. Piston of N 2 compressed polytropically Can find work from polytropic eqn. W = mR(T 2 -T 1 )/(n-1) Final pressure: P 1 V n 1 =P 2 V n 2 P 2 = (V 1 /V 2 ) 1.3 (P 1 ) = (2 1.3 )(100) =

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Control Volumes

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  1. Control Volumes Thermodynamics Professor Lee Carkner Lecture 9

  2. PAL # 8 Specific Heat • Piston of N2 compressed polytropically • Can find work from polytropic eqn. • W = mR(T2-T1)/(n-1) • Final pressure: • P1Vn1=P2Vn2 • P2 = (V1/V2)1.3 (P1) = (21.3)(100) = • Final temperature • P1V1/T1 = P2V2/T2 • T2 = (P2/P1)(V2/V1)T1 = (246.2/100)(0.5)(300) =

  3. PAL # 8 Specific Heat • Work is: • W = (0.8)(0.2968)(369.3-300) / (1-1.3) = • Can get DE from cv • Average T = (369+300)/2 = 335 K • DU = mDu = mcvDT • DU = (0.8)(0.744)(369.3-300) = • Q = 54.8 – 41.2 =

  4. Control Volume • For a control volume, mass can flow in and out • Mass flow rate depends on: • Velocity normal to Ac: Vn • Note that: • V = velocity • V =volume

  5. Flow • Mass flowing through a pipe • Not easy to solve • Velocity is not Vavg = V = (1/Ac) ∫ Vn dAc • We can now write:

  6. Volume • The volume flow rate is just: • Related to the mass flow rate just by the density: • In m3/s

  7. Conservation of Mass • We examine: • Mass streams flowing out = mout • So then: • m’in – m’out = dmcv/dt

  8. Common Cases • For the steady flow case: S m’in = S m’out • For systems with just a single stream: r1V1A1 = r2V2A2 • For incompressible fluids, r1 = r2 V’1 = V’2

  9. Flow Work • The amount of energy needed to push fluid into a control volume is the flow work: • Can think of it as a property of the fluid itself

  10. Energy • Our previous result for fluid energy was: • We can add the flow work, but note: • We designate the total fluid energy per unit mass by q: q = h + V2/2 + gz • Now we don’t have to worry about the flow work

  11. Energy and Mass Flow • The total is: E’mass = m’q = m’(h + V2/2 +gz) • If V and z are small: • True for most systems

  12. Steady-Flow Systems • Once they are up and running the properties of the fluid at a given point don’t change with time q – w = Dq = Dh + DV2/2 + gDz • Energy balance for a steady flow system per unit mass

  13. Steady Flow Heat and Work • Heat • This is the net heat Q’in-Q’out • Work • Boundary work = 0, flow work part of enthalpy • Remaining work:

  14. Steady Flow Energies • Enthalpy • Dh is difference between inlet and exit • Kinetic energy • Often very small • Potential energy • 10’s or 100’s of meters

  15. Automotive Cooling System

  16. Example: Radiator • Assume cooling fluid is water • Flow velocity about 1 m/s • Maximum height differential about 1 meter • gz = (9.8)(1) /(1000) = 0.0098 kJ/kg

  17. Next Time • Read: 5.4-5.5 • Homework: Ch 5, P: 17, 21, 36, 75

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