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Flow Devices Thermodynamics Professor Lee Carkner Lecture 10
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PAL # 9 Control Volumes Air flow through pipe Find diameter from area, find area from mass flow rate (= AV) v = RT/P = (0.287)(350)/(300) = A = m’ v /V = (18/60)(0.3349)/(25) = 0.004018 m 2 D = (4A/ ) ½ = (4(0.004018) / ) ½ =
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PAL # 9 Control Volumes Rate of flow energy = m’P v W’ flow = (18/60)(300)(0.3349) = Energy transport by mass: E’ mass = m’(h+ke) = m’(c p T + ½V 2 ) E’ mass = (18/60)[(1.008)(350) + (½)(25) 2 (1/1000)] = Neglecting ke E’ mass = m’h = m’c p T = (18/60)(1.005)(350) = 105.84 kW
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Steady-Flow Devices E mass,in = E mass,out W’ in + Q’ in + m’ in (h + V 2 /2 + gz) in = W’ out + Q’ out + m’ out (h + V 2 /2 + gz) out but mass and energy are conserved in each
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Nozzles V’ 1 = V’ 2 V 1 A 1 = V 2 A 2 V 1 /V 2 = A 2 /A 1 Velocity changes opposite to pipe diameter change Small to large opening, velocity decreases, diffuser But basic idea holds for gases
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Working With Nozzles Nozzles increase velocity at the expense of pressure Other energies usually negligible Can often use ideal gas law and enthalpy tables to solve problems n.b. Simplified nozzle equation h 1 +½V 2 1 = h 2 +½V 2 2
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Turbines The flow work pushes the fluid into a control volume Change enthalpy into work Change work into enthalpy Q ~ PE ~ 0
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Working with Turbines but the change in kinetic energy is normally small compared to enthalpy change Simplified turbine equation: W’ = m’( h + ke)
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Throttling Valve A throttling valve restricts flow causing a pressure drop So conservation of energy gives us h 1 = h 2 A throttling process shifts energy between internal and flow energies Simplified throttling equation: u 1 + P 1 v 1 = u 2 + P 2 v 2
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Mixing Chamber Usually Q ~ W ~ KE ~ PE ~ 0 Simplified mixing chamber equations: m’ 1 + m’ 2 = m’ 3 m’ 1 h 1 + m’ 2 h 2 = m’ 3 h 3
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Heat Exchanger Generally, W ~ PE ~ KE ~ 0 This means that mass flow and enthalpy are the only important variables
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Working With Heat Exchangers Fluid B enters at 3 and exits at 4 Mass conservation: m’ 3 = m’ 4 = m’ B Energy conservation Simplified heat exchanger equation: m’ A (h 1 -h 2 ) = m’ B (h 4 -h 3 )
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Unsteady Flow Unsteady flow involves changes within the control volume Example: Mass within the flow volume increases Deal with changes in properties with time rather than rates
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Uniform Flow If the input and output flow is steady and only the properties of the control volume change, it is a uniform flow process Q-W = (mh) out – (mh) in + (m 2 u 2 -m 1 u 1 ) system Where: mh is the sum of all the enthalpies in or out
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Next Time No class Friday For Monday: Read: 6.1-6.4 Homework: tba
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