MASS AND ENERGY ANALYSIS OF CONTROL VOLUMESPradeep SahooRoom No: 135Dept of Mech. and Ind. EngineeringIIT Roorkee

*CONSERVATION OF MASSMass is conserved even during chemical reactions.Conservation of mass: Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume.

*Conservation of Mass PrincipleConservation of mass principle for an ordinary bathtub.The conservation of mass principle for a control volume: The net mass transfer to or from a control volume during a time interval t is equal to the net change (increase or decrease) in the total mass within the control volume during t.General conservation of mass in rate form

*Mass Balance for Steady-Flow ProcessesConservation of mass principle for a two-inletone-outlet steady-flow system.During a steady-flow process, the total amount of mass contained within a control volume does not change with time (mCV = constant). For steady-flow processes, we are interested in the amount of mass flowing per unit time, that is, the mass flow rate.Multiple inlets and exitsSingle stream

*Special Case: Incompressible FlowThe conservation of mass relations can be simplified even further when the fluid is incompressible, which is usually the case for liquids.Steady, incompressibleSteady, incompressible flow (single stream)There is no such thing as a conservation of volume principle. However, for steady flow of liquids, the volume flow rates, as well as the mass flow rates, remain constant since liquids are essentially incompressible substances.

*During a steady-flow process, volume flow rates are not necessarily conserved although mass flow rates are.

*FLOW WORK AND THE ENERGY OF A FLOWING FLUIDSchematic for flow work.Flow work, or flow energy: The work (or energy) required to push the mass into or out of the control volume. This work is necessary for maintaining a continuous flow through a control volume.

*Total Energy of a Flowing FluidWhere h = u + Pv

*The total energy consists of three parts for a non-flowing fluid and four parts for a flowing fluid.The flow energy is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy.

*Energy Transport by MassWhen the kinetic and potential energies of a fluid stream are negligibleWhen the properties of the mass at each inlet or exit change with time as well as over the cross section

*ENERGY ANALYSIS OF STEADY-FLOW SYSTEMSMany engineering systems such as power plants operate under steady conditions.Under steady-flow conditions, the mass and energy contents of a control volume remain constant.Under steady-flow conditions, the fluid properties at an inlet or exit remain constant (do not change with time).

*Mass and Energy balances for a steady-flow processMass balanceEnergy balance

*Energy balance relations with sign conventions (i.e., heat input and work output are positive)when kinetic and potential energy changes are negligible

*Nozzles and DiffusersNozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies.Energy balance for a nozzle or diffuser:

*Example: Steam at 0.4 MPa, 300oC, enters an adiabatic nozzle with a low velocity and leaves at 0.2 MPa with a quality of 90%. Find the exit velocity, in m/s.Conservation of mass: For one entrance, one exit, the conservation of mass becomesConservation of energy:

*Neglecting the inlet kinetic energy, the exit velocity isAt 0.2 MPa hf = 504.7 kJ/kg and hfg = 2201.6 kJ/kg.

*Turbines and CompressorsEnergy balance for the compressor in this figure:

*Example: High pressure air at 1300 K flows into an aircraft gas turbine and undergoes a steady-state, steady-flow, adiabatic process to the turbine exit at 660 K. Calculate the work done per unit mass of air flowing through the turbine. when (a) Temperature-dependent data are used. (b) Cp,ave at the average temperature is used. (c) Cp at 300 K is used. Solution:Conservation of energy:

*According to the sketched control volume, mass and work cross the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic, we haveThe work done by the air per unit mass flow is

*(a) Using the air tables, Table A-17 at T1 = 1300 K, h1 = 1395.97 kJ/kgat T2 = 660 K, h2 = 670.47 kJ/kgw = h1-h2 = 725.5 kJ/kg (b) Using Table A-2(c) at Tave = 980 K, Cp, ave = 1.138 kJ/kgK w = h1 h2 = Cp(T1-T2) = 1.138(1300-660) = 728.3 kJ/kg (c) Using Table A-2(a) at T = 300 K, Cp = 1.005 kJ/kg Kw = h1 h2 = Cp(T1-T2) = 1.005(1300-660) = 643.2 kJ/kg

*Turbine drives the electric generator In steam, gas, or hydroelectric power plants.As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work.Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft.A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. A compressor is capable of compressing the gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases.

*Example: Nitrogen gas is compressed in a steady-state, steady-flow, adiabatic process from 0.1 MPa, 25C. During the compression process the temperature becomes 125C. If the mass flow rate is 0.2 kg/s, determine the work done on the nitrogen, in kW.

*Throttling valvesThrottling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. What is the difference between a turbine and a throttling valve? The pressure drop in the fluid in throttling is often accompanied by a large drop in temperature instead of work production, and for that reason throttling devices are commonly used in refrigeration and air-conditioning applications.Energy balance

*The temperature of an ideal gas does not change during a throttling (h = constant) process since h = h(T).During a throttling process, the enthalpy of a fluid remains constant. But internal and flow energies may be converted to each other.

*Example: One way to determine the quality of saturated steam is to throttle the steam to a low enough pressure that it exists as a superheated vapor. Saturated steam at 0.4 MPa is throttled to 0.1 MPa, 100oC. Determine the quality of the steam at 0.4 MPa. Solution:

*Mixing chambersIn engineering applications, the section where the mixing process takes place is commonly referred to as a mixing chamber.The T-elbow of an ordinary shower serves as the mixing chamber for the hot- and the cold-water streams.

*Energy balance for the adiabatic mixing chamber in the figure is:

*Example: Steam at 0.2 MPa, 300C, enters a mixing chamber and is mixed with cold water at 20C, 0.2 MPa, to produce 20 kg/s of saturated liquid water at 0.2 MPa. What are the required steam and cold water flow rates?

*Heat exchangersHeat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchangers are widely used in various industries, and they come in various designs.A heat exchanger can be as simple as two concentric pipes.

*Mass and energy balances for the adiabatic heat exchanger in the figure is:Continuity Energy balance

*Example: Air is heated in a heat exchanger by hot water. The water enters the heat exchanger at 45C and experiences a 20C drop in temperature. As the air passes through the heat exchanger, its temperature is increased by 25C. Determine the ratio of mass flow rate of the air to mass flow rate of the water.

*ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSESMany processes of interest, however, involve changes within the control volume with time. Such processes are called unsteady-flow, or transient-flow, processes.Most unsteady-flow processes can be represented reasonably well by the uniform-flow process.Uniform-flow process: The fluid flow at any inlet or exit is uniform and steady, and thus the fluid properties do not change with time or position over the cross section of an inlet or exit. If they do, they are averaged and treated as constants for the entire process.

*Charging of a rigid tank from a supply line is an unsteady-flow process since it involves changes within the control volume.The shape and size of a control volume may change during an unsteady-flow process.

*Mass balanceEnergy balance

*The energy equation of a uniform-flow system reduces to that of a closed system when all the inlets and exits are closed.

*SummaryConservation of massMass and volume flow ratesMass balance for a steady-flow processMass balance for incompressible flowFlow work and the energy of a flowing fluidEnergy transport by massEnergy analysis of steady-flow systemsSome steady-flow engineering devicesNozzles and DiffusersTurbines and CompressorsThrottling valves Mixing chambers and Heat exchangersEnergy analysis of unsteady-flow processes

*Example: Consider a water cooled condenser of a large refrigeration system in which R-134a is the refrigerant. The refrigerant enters the condenser at 1.0 MPa and 600C, at the rate of 0.2 kg/s and exits as a liquid at 0.95 MPa, and 350C. Cooling water enters the condenser at 100C and leaves at 200C. Determine the cooling water flow rate.hw,i = 42.0 kJ/kghw,e = 84.0 kJ/kghr,i = 441.5 kJ/kghr.,e = 249.0

*Example 2: Consider a simple power shown below. The pump work is 4 kJ/kg. Determine the following: Heat transfer in line between boiler and turbine Turbine work Heat transfer in condenserHeat transfer in the boiler

*The quality of vapour at the evaporator inlet, The rate heat transfer in the evaporator, and The rate of heat transfer from the compressor.h1 = 378.2, h2 = 435.1h4 = h3 = 241.8,h(f,4) = 167.4H(fg,4) = 215.6