Mass and Energy Analysis of Control Volumes Chapter 5 Mass and Energy Analysis of Control Volumes

The First Law of Thermodynamics Open System (Control Volume)

Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steady- and unsteady-flow control volumes. Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes. 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. Solve energy balance problems for common steady-flow devices such as nozzles, compressors, turbines, throttling valves, mixers, heaters, and heat exchangers. Apply the energy balance to general unsteady-flow processes with particular emphasis on the uniform-flow process as the model for commonly encountered charging and discharging processes.

How does energy cross an open system boundary? The energy content of a control volume can be changed by mass flow as well as heat and work interactions

Mass Flow Rate The amount of mass flowing through a cross section per unit time is called the mass flow rate. It is given as  = density of fluid, Kg/m3 Vm = mean velocity normal to A A= cross sectional area of pipe. Usually we drop the subscript m The unit is Kg/s

Mass Flow Rate (V=F(Area)) The mass flow rate through the entire cross-sectional area of a pipe Where Vn is the component of the flow velocity normal to the cross sectional area Ac Average Velocity =

Volume Flow Rate Volume flow rate is the volume of fluid flowing through a cross section per unit of time It is given as Vm = mean velocity normal to A A= cross sectional area of pipe. Usually we drop the subscript m The unit is m3/s

Mass and Volume Flow rate The mass and volume flow rates are related by

Conservation of mass principle _ Total mass entering the CV during Dt Total mass leaving the CV during Dt Net change in mass within the CV during Dt = On a rate basis,

Conservation of mass principle Total mass within the CV: Rate of change of mass within the CV:

Conservation of mass principle Normal component of velocity: Net mass flow rate: General conservation of mass:

General Conservation of mass for multi-inlet and outlet

Conservation of mass principle (continued) For many inlets and exits, On a rate basis,

For steady process, no accumulation of mass in CV For single stream,

Flow rate for Steady compressible flow ( = not const.) Although, in a steady flow process, Volume Flow Rates are not Necessarily the same This is because

Special Case: Flow rate for Steady Incompressible Flow ( =constant) For single stream

Example: 5-8

Flow Work (Flow Energy) or Energy of Flowing Fluid Some people call it “flow energy” Flow work is defined as the work required to push a unit mass of fluid into or out of a control volume. It is equal to Pv

Flow Work derivation The force applied on the fluid by the imaginary piston is Wf per unit mass is

Total Energy of a Flowing Fluid Recall that The total energy of a system is For control volume, the fluid entering or leaving possesses an additional form of energy (the flow work or the flow energy) The total energy of a flowing fluid

Observations Using h instead of u to represent the energy of a flowing fluid, results in Taking care of the energy required to push the fluid in or out of the control volume. In fact, this is the main reason behind introducing the enthalpy.

The Total Energy of a Flowing and non-Flowing Fluid Closed System Control Volume

Energy Transport by Mass The total energy of a flowing fluid of mass m with uniform properties is (Amount of energy transport) The rate of energy transport by a fluid with a mass flow rate of is (Rate of energy transport) When the KE and PE of a flow stream are negligible, then

Energy Transport by Mass the total energy transported by mass through an inlet or exit

Example 3-14: Energy transport by mass Steam is leaving a 4-L pressure cooker whose operating pressure is 150 Kpa. The liquid in the cooker decreased 0.6 L in 40 min. The cross sectional area of the exit is 8 mm2. Determine: The mass flow rate of steam The exit velocity The total and flow energies of the steam per unit mass. The rate at which the energy is leaving the cooker by steam.

Solution: Energy transport by mass Assumptions: The flow exiting the cooker is steady. KE and PE are negligible. Saturation conditions exist all times. Thus: (1) The steam leaving the cooker is saturated vapor (x=1) at the operating pressure. (2) The liquid remaining is saturated liquid (x-0) at the operating pressure.

Steady: the decrease in Liquid mass = the increase in vapor mass Note: the negligible K.E. Differentiate between the flow and total Energy