VIEWS: 121 PAGES: 30 POSTED ON: 2/18/2012 Public Domain
Learning Objectives 1. Develop the conservation of mass principle. 2. Apply the conservation of mass principle to various systems including steady-flow control volumes. 3. Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes. 4. Identify the energy carried by a fluid stream crossing a control surface as the sum of internal energy, flow work, kinetic energy, and potential energy of the fluid and to relate the combination of the internal energy and the flow work to the property enthalpy. 5. Solve energy balance problems for common steady-flow devices such as nozzles, compressors, turbines, throttling valves, mixers, heaters, and heat exchangers. 5-1 Conservation of Mass Mass Flow Rate • The amount of mass flowing through a cross section per unit time • Denoted, m General Analysis • The mass flow rate through a differential area dAc is m rVn dAc • r → density of fluid Vn → velocity of fluid normal to the cross-sectional area • Mass flow rate through entire cross- sectional area m rVn dAc Ac 5-1 Conservation of Mass Uniform Flow • In many practical applications, we can make the following approximations: 1. One-dimensional flow 2. All properties are uniform at any cross-section normal to the flow direction 3. Properties have the bulk average values over the cross-section • These assumptions lead to m rVavg Ac • r → density of fluid Vavg → mean velocity of fluid normal to the cross-sectional area Ac → cross-sectional area normal to flow direction 5-1 Conservation of Mass Volumetric Flow Rate • The volume of a fluid flowing through a cross-sectional area per unit time is V Vn dAc Ac V Vavg Ac • The mass and volume flow rates are related by m rV V m v 5-1 Conservation of Mass Conservation of Mass Principle • One of the most fundamental principles in nature • The net mass transfer to or from a control volume (CV) during a process is equal to the net change (increase or decrease) in the total mass of the control volume during that process • Or min mout mCV d • In rate form min mout mCV dt • These expressions are referred to as mass balances • The expressions are valid for any system undergoing any process 5-1 Conservation of Mass Mass Balance Expressions for Multiple Inlets and Outlets m m m in out CV d m m dt mCV in out • When the properties at the inlets and exits as well as within the control volume are not uniform, the mass flow rate can be expressed in the differential form m rVn dAc 5-1 Conservation of Mass Mass Balance for Steady Flow Processes • For a steady flow process, mCV = constant • The conservation of mass principle simplifies to m m in out Incompressible Flow (r = constant) • For an incompressible fluid, the conservation of mass principle simplifies to V V in out 5-2 Flow Work and the Energy of a Flowing Fluid Flow Work • The work required to push a mass into or out of a control volume • Necessary for maintaining a continuous flow through a control volume • Force applied on the fluid element by the imaginary piston F = PA 5-2 Flow Work and the Energy of a Flowing Fluid • To push the entire fluid element into the control volume, this force must act through a distance L • The work done in pushing the fluid element across the boundary (flow work) is then Wflow = FL = PAL = PV • The flow work per unit mass is then wflow = Pv • The flow work is the same whether the fluid is pushed into or out of the control volume 5-2 Flow Work and the Energy of a Flowing Fluid Flow Work (cont.) • Unlike other work quantities, flow work is expressed in terms of properties • For this reason, flow work can be viewed as a combination property, convected energy, or transport energy • As a result, the flow energy can be considered part of the energy of a flowing fluid • This greatly simplifies energy analysis of control volumes 5-2 Flow Work and the Energy of a Flowing Fluid Total Energy of a Flowing Fluid • The total energy of a stationary simple compressible system consists of three parts: internal, kinetic, and potential energies 1 2 e u ke pe u V gz non flowing fluid 2 • A fluid entering or leaving a control volume possesses an additional form of energy, the flow energy, Pv • The total energy of a flowing fluid is then, Pv e Pv u ke pe • This expression can be simplified by recalling that the enthalpy h is given by h = u + Pv 1 h ke pe h V 2 gz flowing fluid 2 5-2 Flow Work and the Energy of a Flowing Fluid Total Energy of a Flowing Fluid (cont.) • By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid, one does not need to be concerned about the flow work • The energy associated with pushing the fluid into or out of a control volume is automatically taken care of by the enthalpy 5-3 The First Law of Thermodynamics Energy Balance Ein Eout Esystem • This relation is referred to as the energy balance • The relationship is valid for any system undergoing any process 5-3 Energy Balance for Steady-Flow Systems Steady-flow process • A process during which a fluid flows through a control volume steadily • No intensive or extensive properties within the control volume change with time • The total volume V, the mass m, and the total energy content E of the control volume remain constant Wb = 0 Q constant W constant 5-3 Energy Balance for Steady-Flow Systems Steady-flow process (cont.) • The boundary work is zero for steady flow systems (VCV = const.) • The total mass or energy entering the control volume must be equal to the total mass or energy leaving it (since, mCV = const. and ECV = const.) • The heat and work interactions between a steady-flow system and its surroundings do not change with time Wb = 0 Q constant W constant 5-3 Energy Balance for Steady-Flow Systems Mass Balance for Steady-Flow Systems m m in out rVA rVA in out For a single stream (one inlet and one outlet), m1 m2 r1V1 A1 r2V2 A2 5-3 Energy Balance for Steady-Flow Systems Energy Balance for Steady-Flow Systems • During a steady-flow process, the total energy content of a control volume remains constant, and thus ECV = 0 • Therefore, the total amount of energy entering a control volume must be equal to the amount of energy leaving it Ein Eout ECV 0 Ein Eout 5-3 Energy Balance for Steady-Flow Systems Energy Balance for Steady-Flow Systems (cont.) Qin Win m Qout Wout m in out h ke pe Qin Win m h 1 V 2 gz Qout Wout m h 1 V 2 gz 2 2 in out Using the formal sign convention, Q W m h 1 V 2 gz m h 1 V 2 gz 2 2 out in For a single stream (one inlet (1) and one outlet (2)), V22 V12 Q W m h2 h1 g z2 z1 2 5-3 Energy Balance for Steady-Flow Systems Energy Balance for Steady-Flow Systems (cont.) Q W m h 1 V 2 gz m h 1 V 2 gz 2 2 out in •Q → Rate of heat transfer between the control volume and its surroundings negative → CV is losing heat adiabatic → Q 0 • W → Power positive → CV is doing work on surroundings • h → Enthalpy of flowing fluid 2 • 1 V → Kinetic energy of flowing fluid 2 • gz → Potential energy of flowing fluid 5-4 Some Steady-Flow Engineering Devices Nozzles • Devices that increase the velocity of a fluid at the expense of pressure Diffusers • Devices that increase the pressure of a fluid by decreasing its velocity General Characteristics • Typically, W 0 • Usually, Q 0 , pe 0 • Large changes in velocity ke 0 5-4 Some Steady-Flow Engineering Devices Turbine • In gas, steam, or hydroelectric power plants, the device that drives the electric generator • A turbine produces work, W 0 Compressors, Pumps, and Fans • Devices used to increase the pressure of a fluid • Work is supplied to these devices, W 0 Fan •Increases the pressure of a gas slightly •Mainly used to mobilize a gas Compressor •Capable of compressing a gas to very high pressures Pumps •Like compressors except that they handle liquids instead of gases 5-4 Some Steady-Flow Engineering Devices General Characteristics of Turbines and Compressors • Usually, Q 0 , pe 0 • Often, ke ≈ 0, except turbines and fans, however often |h| >> |ke| in turbines and fans making ke negligible 5-4 Some Steady-Flow Engineering Devices Throttling Valves • Any kind of flow-restricting device that causes a significant pressure drop in the fluid • The pressure drop is often accompanied by a large drop in temperature • The following assumptions can often be made, q ≈ 0, w ≈ 0, pe ≈ 0, ke ≈ 0 • Based on these assumptions, the conservation of energy equation for a single-stream steady-flow device reduces to h2 ≈ h1 • For this reason, a throttling valve is sometimes called an isenthalpic device 5-4 Some Steady-Flow Engineering Devices Mixing Chamber • The section where a mixing process takes place • The following assumptions can often be made, q ≈ 0, w ≈ 0, pe ≈ 0, ke ≈ 0 5-4 Some Steady-Flow Engineering Devices Heat Exchangers • Devices where two moving fluid streams exchange heat without mixing • Typical characteristics of heat exchangers, w ≈ 0, pe ≈ 0, ke ≈ 0 Double Shell (tube and shell) •Composed of two concentric pipes of different diameters Problem 5-38 Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa and 500 oC with a mass flow rate of 6000 kg/h and leaves at 100 kPa and 450 m/s. The inlet area of the nozzle is 40 cm2. Determine (a) the inlet velocity and (b) the exit temperature. Problem 5-54 Argon gas enters an adiabatic turbine steadily at 900 kPa and 450 oC with a velocity of 80 m/s and leaves at 150 kPa and 150 m/s. The inlet area of the turbine is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon. Problem 5-66 Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant. Problem 5-81 Refrigerant-134a at 1 MPa and 90 oC is to be cooled to 1 MPa and 30 oC in a condenser by air. The air enters at 100 kPa and 27 oC with a volume flow rate of 600 m3/min and leaves at 95 kPa and 60 oC. Determine the mass flow rate of the refrigerant. Problem 5-91 A thin-walled double-pipe counter-flow heat exchanger is used to cool oil (cp = 2.20 kJ/kg · °C) from 150 to 40 °C at a rate of 2 kg/s by water (cp = 4.18 kJ/kg · °C) that enters at 22 °C at a rate of 1.5 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of water.