Presentation on theme: "Lecture# 9 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES"— Presentation transcript:
1 Lecture# 9 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES ENERGY ANALYSIS OF STEADY-FLOW SYSTEMS:A 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 devices.Processes involving such devices can be represented reasonably well by a somewhat idealized process, called the steady-flow process , which was defined in before as a process during which a fluid flows through a control volume steadily. That is, the fluid properties can change from point to point within the control volume, but at any point, they remain constant during the entire process. (Remember, steady means no change with time.)
2 The mass balance for a general steady-flow system was given as: The fluid properties at an inlet or exit remain constant during a steady flow process.The properties may, however, be different at different inlets and exits. They may even vary over the cross section of an inlet or an exit.However, all properties, including the velocity and elevation, must remain constant with time at a fixed point at an inlet or exit.It follows that the mass flow rate of the fluid at an opening must remain constant during a steady flow process.The mass balance for a general steady-flow system was given as:(2-1)The mass balance for a single-stream (one-inlet and one-outlet) steady-flow system was given as:(2-2)
3 Where the subscripts 1 and 2 denote the inlet and the exit states, respectively, ρ is density, V is the average flow velocity in the flow direction, and A is the cross-sectional area normal to flow direction.(2-3)Noting that energy can be transferred by heat, work, and mass only, the energy balance in Eq. 2-3 for a general steady-flow system can also be written more explicitly as:(2-4)
4 Dividing Eq. 2.5 by m gives the energy balance on a unit-mass basis as (9-1)Dividing Eq by m gives the energy balance on a unit-mass basis as(9-2)
5 where q= Q·/m· and w= W·/m· are the heat transfer and work done per unit mass of the working fluid, respectively.When the fluid experiences negligible changes in its kinetic and potential energies (that is, ∆ke ≈ 0, ∆pe ≈ 0), the energy balance equation is reduced further to
6 SOME STEADY-FLOW ENGINEERING DEVICES 1- Nozzles and Diffusers . A nozzle is a device that increases the velocity of a fluid at the expense of pressure.A diffuser is a device that increases the pressure of a fluid by slowing it down.
7 FIGURE 9– 1Nozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies.
8 EXAMPLE 9–1:Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.Determine (a) the mass flow rate of the air and(b) the temperature of the air leaving the diffuser.Solution: Air enters the diffuser of a jet engine steadily at a specified velocity. The mass flow rate of air and the temperature at the diffuser exit are to be determinedFig.(9.2)
9 Assumptions1- This is a steady-flow process since there is no change with time at any point and thus ∆m = 0 and ∆E = 0.2- Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values.3- The potential energy change is zero, ∆pe = 0.4- Heat transfer is negligible.5- Kinetic energy at the diffuser exit is negligible.6- There are no work interactionsAnalysis: We take the diffuser as the system (Fig. 2–6). This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus :.
10 (a) To determine the mass flow rate, we need to find the specific volume of the air first. This is determined from the ideal-gas relation at the inlet conditions.
13 2- Turbines and Compressors In steam, gas, or hydroelectric power plants, the device that drives the electric generator is the turbine. 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. Therefore, compressor involve work inputs.
14 Even though these three devices function similarly, they do differ in the tasks they perform. 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.Note that turbines produce power output whereas compressors, pumps, and fans require power input. Heat transfer from turbines is usually negligible (Q = 0) since they are typically well insulated
15 EXAMPLE 9–2Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressorSolution Air is compressed steadily by a compressor to a specified temperatureand pressure. The power input to the compressor is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus Δm= 0 and ΔE= 0.
18 EXAMPLE 9– 3The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in Fig. 2–8.(a) Compare the magnitudes of h, ke, and pe.(b) Determine the work done per unit mass of the steam flowing through the turbine.(c) Calculate the mass flow rate of the steam
20 Analysis: We take the turbine as the system Analysis: We take the turbine as the system. This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus m.1=m.2. Also, work is done by the system. The inlet and exit velocities and elevations are given, and thus the kinetic and potential energies are to be considered.a) At the inlet, steam is in a superheated vapor state, and its enthalpy isP1 = 2 Mpa ,T1= 4000C ,h1 = kJ/kg (Table A–6)At the turbine exit, we obviously have a saturated liquid–vapor mixture at 15-kPa pressure. The enthalpy at this state ish2= hf + x2hfg = [ (0.9)(2372.3)] kJ/kg = kJ/kg