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Chapter 4 The First Law of Thermodynamics: Control Volumes

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4-1-1 Conservation of Mass principle 4-1 Thermodynamic analysis of Control Volume 4-1-2 Conservation of Energy principle for Control Volume

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The net energy change in energy of CV Total energy crossing boundary + Total energy of mass entering CV = - Total energy of mass leaving CV

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4-1-3 Flow Work

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4-1-4 Total Energy of a Flowing Fluid On a unit-mass basis, the energy stored in fluid: Considering the flow work, we have The total energy of a flowing fluid (denoted θ):

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4-1-5 Enthalpy Internal energy and flow work usually transferred at the same time in flowing fluid, so we define Professor J. Kestin proposed in 1966 that the term θ be called methalpy (from metaenthalpy, which means beyond enthalpy)

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4-2 The Steady-Flow Process 4-2-1 Definition of Steady-Flow Process A process during which a fluid flows through a control volume steadily 4-2-2 Characteristics of Steady-Flow Process 1.No properties in the control volume change with time 2. No properties changes at the boundaries of the control volume with time

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3. The heat and work interactions between a steady flow system and its surroundings do not change with time 4-2-3 Energy Equation of Steady-Flow Process From the above discussion we can conclude that: 2. The net change in energy of CV is Zero

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Energy Equation of Steady-Flow Total energy crossing boundary per unit time = Total energy of mass leaving CV per unit time - Total energy of mass entering CV per unit time

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Using subscript 1and subscript 2 for denoting inlet and exit states Dividing the equation byyields

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Work

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3-3 Some Steady-Flow Engineering devices 3-3-1 Nozzles and Diffusers q = Δh + ΔV 2 / 2 + gΔz + w q = 0 Δz = 0 w = 0

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3-3-2 Steam Turbine from q = Δh + ΔV 2 / 2 + gΔz + w s if q = 0; Δc = 0; Δz = 0 then w s = -Δh = h 1 - h 2

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3-3-3 Throttling Valves From q = Δh + ΔV 2 / 2 + gΔz + w s as q = 0; w s = 0; z = 0 ; ΔV 2 / 2 =0 then Δh =0 That is h 1 = h 2

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3-3-4 Mixing Chambers From q = Δh + ΔV 2 / 2 + gΔz + w s as q = 0; w s = 0; z = 0 ; ΔV 2 / 2 =0 then Δh =0 That is ∑h in = ∑ h exit

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3-3-5 Heat exchangers From q = Δh + ΔV 2 / 2 + gΔz + w s as q = 0; w s = 0; Δ z = 0;ΔV 2 / 2 =0 Δh =0 That is ∑h in = ∑ h exit

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3-4 Unsteady-Flow Processes 3-4-1 Model 1. The net energy transferred to the system: dQ ， e 1 dm 1 ， p 1 v 1 dm 1 2. The net energy transferred out of the system: dWs ， e 2 dm 2 ， p 2 v 2 dm 2

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3-4-2 Conversation of Energy

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Divide dτ on each side

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3-5 Uniform-Flow Processes 1. At any instant during the process, the state of control volume is uniform 2. The fluid properties may differ from one inlet or exit to another, but the fluid flow at an inlet or exit is uniform and steady

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The end of this chapter Thanks for your attention!

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Lecture# 9 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

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