Control Volumes

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
• P1Vn1P2Vn2
• 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 (369300)/2 335 K
• DU mDu mcvDT
• DU (0.8)(0.744)(369.3-300)
• Q 54.8 41.2

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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

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Volume

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

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Conservation of Mass

• We examine
• Mass streams flowing out mout
• So then
• min mout dmcv/dt

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Common Cases

• For the steady flow case
• S min S mout
• For systems with just a single stream
• r1V1A1 r2V2A2
• For incompressible fluids, r1 r2
• V1 V2

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

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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 dont have to worry about the flow work

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Energy and Mass Flow

• The total is
• Emass mq m(h V2/2 gz)
• If V and z are small
• True for most systems

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Steady-Flow Systems

• Once they are up and running the properties of
  the fluid at a given point dont change with time
• q w Dq Dh DV2/2 gDz
• Energy balance for a steady flow system per unit
  mass

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Steady Flow Heat and Work

• Heat
• This is the net heat Qin-Qout
• Work
• Boundary work 0, flow work part of enthalpy
• Remaining work

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Steady Flow Energies

• Enthalpy
• Dh is difference between inlet and exit
• Kinetic energy
• Often very small
• Potential energy
• 10s or 100s of meters

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Automotive Cooling System
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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

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Next Time

• Read 5.4-5.5
• Homework Ch 5, P 17, 21, 36, 75