Energy analysis of Steady-Flow in an open System.

Energy analysis of Steady-Flow in an open System.

1IntroductionA large number of engineering devices such as turbines, compressors, and nozzles operate for long periods of time under the same conditions once the transient start up period is completed and steady operation is established, and they are classified as STEADY-FLOW PROCESS.2Characteristic 1:No property at any given location within the system boundary changes with time. That also means, during an entire steady flow process, the total volume ( ) of the system remains a constant, the total mass( )of the system remains a constant, and that the total energy content Es of the system remains a constant.

3Characteristic 2:Since the system remains unchanged with time during a steady flow process, the system boundary also remains the same.4Characteristic 3:No property at an inlet or at an exit to the open system changes with time. That means during a steady flow process, the mass flow rate, the energy flow rate, pressure, temperature, specific (or molar) volume, specific (or molar) internal energy, specific (or molar) enthalpy, and the velocity of flow at an inlet or at an exit remain constant.5Characteristic 4: Rates at which heat and work are transferred across the boundary of the system remain unchanged.6

Mass Balance for a Steady Flow Process.

Since a steady flow process can be considered as a special process experienced by the open system, we may start from the mass balance for open systems. Characteristic 1 of the steady flow process is that the mass of the open system experiencing a steady flow process remains constant. This is achieved if the mass flow rate at the inlet equals the mass flow rate at the exit.

78The Steady Flow Energy Equation (S.F.E.E)

z1P1, V1, C1z2P2, V2, C2leavingenteringWQEnteringLeavingP1 = PressureP2 = PressureV1 = Specific volumeV2 = Specific volume C1 = velocity C2 = velocity

9Potential energy (PE)Internal Energy (u)PE = mg z (kg m/s2 m) = JPE = gz J/kg (for unit mass)

Kinetic energy (KEJ) Flow Work energy

KE = mv2W = Pv (N/m2 x m3/kg =J/kg)= mc2 (kg x m2/s2 = J)= C2 J/kgThe Steady flow energy equationEntering = leavingPE1 + KE1 + FW1 + IE1 + (Q) = PE2 + KE2 + FW2 + IE2 + W

gz1 + + P1v1+ U1 +Q = gz2 + + P2v2+ U2 + W BUT

FW = flow work energy

gz1 + + h1 + Q = + + h2 + W

In thermodynamics systems the changes in potential energy are mostly small compared to other energies.

+ h1 + Q = + h2 + W

Non-flow energy equation (close system)

gz1 + + P1v1 + U1 + Q = gz2 + + P2v2 + U2 + W

U1 + Q = U2 + WQ = (U2 U1) + WQ = W + U

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Steady Flow Engineering Devices.Nozzles &Diffusers.A nozzle is a device that increases the velocity of a fluid at the expense of pressure and A diffuser is a device that increases the pressure of a fluid by slowing it down.

11SFEE for Nozzles &Diffusers. Since no shaft work is involved in a nozzle or a diffuser, and since the potential energy difference across a nozzle or a diffuser is usually negligible, the steady flow energy equation for flow through a nozzle or a diffuser becomes.

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Reduced SFEE for Nozzles &Diffusers.

The flow through nozzles and diffusers are often considered adiabatic, so that the rate of heat transfer is neglected. Therefore the equation is reduced to;

1314TurbineA turbine is device in which work is produced by a gas passing over and through a set of blades fixed to a shaft which is free to rotate.

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SFEE for Turbines.

Since the fluid flowing through a turbine usually experiences negligible change in elevation, the potential energy term is neglected. Work always leaves the turbine, and therefore the is negative. The steady flow energy equation for flow through a turbine may therefore be written as

16Enthalpy and K.EThe fluid velocities encountered in most turbines are large, and the fluid experiences a significant change in its kinetic energy. However, if this change is small compared to the change in enthalpy then the change in kinetic energy may be neglected. If the fluid flowing through the turbine undergoes an adiabatic process, which is usually the case, then = 0.

17Reduced SFEE for Turbines. which clearly shows that the shaft work delivered by an adiabatic turbine is derived from the enthalpy loss by the fluid flowing through the turbine.

1819Compressors, pumps, and fansMachines developed to make life easier, decrease world anxiety, and provide challenging problems for engineering students.These Machines normally do work on a fluid to raise its pressure, potential, or speed.Mathematical analysis proceeds the same as for turbines, although the signs may differ.20Primary differencesCompressor - used to raise the pressure of a compressible fluid Pump - used to raise pressure or potential of an incompressible fluidFan - primary purpose is to move large amounts of gas, but usually has a small pressure increase21Compressors, pumps, and fans

Axial flow CompressorSide viewEnd viewCentrifugal pump

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Compressors.A compressor is a device used to increase the pressure of a gas flowing through it. The rotating type compressor functions in manner opposite to a turbine. To rotate the shaft of a compressor, work must be supplied from an external source such as a rotating turbine shaft

24SFEE for Compressor.The potential energy difference across a compressor is usually neglected, and the steady flow energy equation for flow through it becomes

25Reduced SFEE for Compressors.The velocities involved in compressor, pump, fan or blower are usually small to cause a significant change in kinetic energy, and often the change in kinetic energy term is neglected. If the compressor, pump, fan or blower is operated under adiabatic conditions, then Qin = 0.2627Reduced SFEE for Compressors.

Note:

which clearly shows that the shaft work provided to an adiabatic compressor, pump, fan or blower is used to increase the enthalpy of the fluid flowing through.

28Sample ProblemAir initially at 103.42kPa and 288.7K with enthalpy of 131.1 KJ/kg is compressed to 517.12kPa and 477.6K with enthalpy of 217.83kJ/kg . The power input to the air is 3.7kW and a heat loss of 4 kJ/kg occurs during the process. Determine the mass flow in kg/s.

29Draw Diagram103.42 kPa288.7K, h1 = 131.1 KJ/kg

517.107 kPa477.6K, h2 = 217.83 KJ/kgq= 4kJ/kg30AssumptionsSteady state flow (SSSF)Neglect potential energy changesNeglect kinetic energy changesAir is ideal gasWhat do we know?INLET T1 = 288.7K P1 = 103.42kPa h1 = 131.1 KJ/kgOUTLET T2 = 477.6K P2 = 517.107kPa h2 = 217.83 KJ/kg

31Heat exchangers.These are devices where two moving fluid streams exchange heat without mixing.Under steady operation, the mass flow rate of each fluid stream flowing through a heat exchanger remains constant.Heat exchangers typically involve no work(w=0) and negligible kinetic and potential energy changes ( ) and ( ) for each fluid stream.

3233Heat exchangers are used in a variety of industriesAutomotive - radiatorRefrigeration - evaporators/condensersPower production - boilers/condensersPower electronics - heat sinksChemical/petroleum industry- mixing processes34Heat exchangers can take a variety of shapes

35Condenser/evaporator for heat pump

36Cooling towers are a type of heat exchanger.

37Something a little closer to home..

38With heat exchangers, we have to deal with multiple inlets and outlets

39At steady flow, what is the relationship between

40What assumptions to make?Ask yourself:See any devices producing/using shaft work?What about potential energy effects?What about kinetic energy changes?Can we neglect heat transfer?41Apply conservation of mass on both streams...

If we have steady flow, then:And

Fluid AFluid B42Conservation of energy can be a little more complicated...

There is a control volume around the whole heat exchanger.Implications:No heat transfer from the control volume.Fluid AFluid B43Conservation of energy looks pretty complicated:

We know from conservation of mass:44Conservation of energy equation for the heat exchanger

Apply what we know about the mass flow relationships:45Heat ExchangersGenerally, there is no heat transfer from or to the heat exchanger, except for that leaving or entering through the inlets and exits.So,And, because the device does no work,

Also, potential and sometimes kinetic energy changes are negligible.

46Heat Exchangers - apply assumptions

00000047Heat Exchangers

After throwing away a bunch of terms, were left with:The energy change of fluid A is equal to the negative of the energy change in fluid B.48Question.How would the energy equation differ if we drew the boundary of the control volume around each of the fluids?49Heat ExchangersNow if we want the energy lost or gained by either fluid we must let that fluid be the control volume, indicated by the red.

4950Heat Exchangers

The energy equation for one side:Or dividing through by the mass flow:50Examples.A turbine receives a steady flow of steam under the following conditions: inlet enthalpy 2800 KJ/kg; exit enthalpy 2550kJ/kg; inlet velocity 50m/s; outlet velocity 110m/s; heat lost to surroundings 25kJ/s; mass of steam 9.1kg/s. Calculate the power developed by the turbine. 2. A gas flows at a steady rate of 4 kg/s. It enters a machine at a pressure of 620 kPa, with a velocity of 300 m/s, a specific internal energy of 2100 kJ/kg and a specific volume of 0.37 m3/kg. It leaves at a pressure of 130 kPa, with a velocity of 150 m/s, a specific internal energy of 1500 kJ/kg and a specific volume of 1.2 m3/kg. Whilst flowing through the machine heat loss from the gas to the surroundings is 30 kJ/kg. Calculate:(a)Density at entry and exit.[2.7 kg/m3, 0.833 kg/m3] (b)Areas of flow at entry and exit.[0.004933 m2, 0.032 m2](c)Specific enthalpy at entry and exit[2329 kJ/kg, 1656 kJ/kg](d)Power produced.[686 kW];51