ENT 254: Applied Thermodynamics

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1 ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP)

2 Chapter 4 The 1st Law of Thermodynamics

3 Chapter Summary Mass balance for control volume, volume and mass flow rates Energy balance for control volume, flow work etc. Steady flow processes; nozzles & diffusers, turbines and compressors, throttling valves, mixing chambers, pipe, etc. Unsteady flow processes Uniform flow processes

4 Control olume A control volume differs from a closed system in that it involves mass transfer. Mass carries energy with it, and thus the mass and energy content of a system change when mass enters or leaves. Control volume boundary may be physical or imaginary surfaces In most cases the shape and size of the boundary is fix Example of cases that usually analyze using control volume are compressors, turbines, nozzles, heat exchanger etc.

5 Control olumes vs. Control Mass The energy content of a control volume can be changed by mass flow as well as heat and work interactions 3 modes of energy interaction ~ heat, work and mass transfer

6 Conservation of Mass (mass balance) Net mass transfer to or from a system during a process is equal to the net change in the total mass of the system during the process or in rate form Mass Flow Rate Rate of change of mass

7 Conservation of Mass (mass balance) m in m out m in Δm m out Δm in mass flow rate

8 Mass and olume Flow Rates Mass flow through a cross section per unit time is called the mass flow rate and it is expressed as. The fluid volume flowing through a cross section per unit time is called the volume flow rate. It is given by The mass and volume flow rates are related by

9 olume and Mass Flow Rates vda m Defining : - v av ρ vda vda A m v av ρv A av A

10 Conservation of Energy (Energy Balance) The net change in the total energy of a system during a process is equal to the difference between the total energy entering and total energy leaving the system during that process In rate form : -

11 First Law for Control olume Considering that : - energy flows into and from the control volume with the mass energy enters because net heat is transferred to the control volume energy leaves because the control volume does net work on its surroundings the open system, or control volume, first law becomes Q m i θi W mo θo de dt C θ is the energy per unit mass flowing into or from the control volume. θ composed of four terms, the internal energy, kinetic energy, potential energy, and the flow work.

12 Work Transfer in Control olume Control olume may involve Boundary, Electrical, and Shaft Work W W W b e W s

13 Mass Transfer ~ Flow Work Work is required to push the mass into and out of a control volume. This work is known as flow work or flow energy. This is considered as part of energy transported, θ Work done by surrounding on system Work done by system on surrounding A δw flow FdL FdL A Pd Pvδm δw δw δ m flow flow Pv

14 Net Mass Transfer in Control olume Recall ~ θ composed of four terms, the internal energy, kinetic energy, potential energy, and the flow work. Chap. 4: 1st Law of Thermodynamics : Control olumes gz h pv gz u w pe ke u flow θ θ δ θ Thus, general 1st Law Equation becomes : - Δ C o o b e s i i E gz h m W W W gz h m Q

15 Steady Flow Processes Thermodynamic processes involving control volumes can be considered in two groups: steady-flow processes unsteady-flow processes. During a steady-flow process, at a fix position, there is no change in fluid properties with time. The mass and energy content of the control volume remain constant. dm dt C m C 0 de dt and C E C 0

16 Steady Flow Processes ( cont..) Steady Flow Mass Conservation Equation (SFMCE) : - sum of rate of mass sum of rate of mass flowing into cv flowing out of cv min mout Steady Flow Energy Equation (SFEE) sum of rate of energy sum of rate of energy flowing into cv flowing out of cv E in E out

17 Steady Flow Processes ( cont..) Chap. 4: 1st Law of Thermodynamics : Control olumes Steady Flow Energy Equation (SFEE) (cont ) o o out out i i in in gz h m W Q gz h m W Q which is commonly written as :~ i i o o net net gz h m gz h m W Q where out in net Q Q Q in out net W W W

18 Single Stream Systems System with one inlet and one exit. The steady flow equations become : ~ Chap. 4: 1st Law of Thermodynamics : Control olumes m m m 1 ( ) ( ) z z g h h m W Q net net In term of unit mass : ~ ( ) ( ) z z g h h w q net net If KE and PE negligible : ~ h h 1 w q net net

19 Steady Flow Engineering Devices Many engineering devices operate essentially under the same conditions for long periods of time, thus can be considered as steady flow devices Nozzle Diffuser Turbine Compressor Throttling valve Mixing chamber Heat Exchangers Pipe flows

20 Nozzles and Diffusers m1 m m 1 >> 1 q 0 w 0 z z 1 1 << 1 1 h 1 h Nozzles and Diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies

21 Turbines and Compressors Turbines ~ In turbines, as the fluid passes through the turbines, work is done against the blades, which are attached to a shaft. The shaft rotates and turbine produces work. Compressor (as well as pump and fan) ~ Device to increase the pressure of a fluid. Work is supplied to the devices from external source through a rotating shaft. Work is -ve. In general :~ m1 m m q 0 w 0 z z Δke 0 1 w ( ) m h 1 h

22 Throttling alves Any kind of flow restricting devices that cause a significant pressure drop in fluid q 0 w 0 z1 h1 h z Δke 0 Thus u 1 Pv 1 1 u Pv h h T T 1 T For ideal gas :~ ( ) Commonly used in refrigeration and air conditioning to cause large temperature drop

23 Mixing Chamber Mixing of two stream of fluids is common in engineering q 0 w 0 Δke 0 Δpe 0 m 1 m m3 m 1h1 mh m3h3 T-elbow of ordinary shower

24 Heat Exchangers Heat Exchangers are devices where moving fluid streams exchange heat without mixing w 0 Δke 0 Δpe 0 The heat transfer associated with a heat exchanger may be zero or non-zero depending on how the system is selected m a ( h h ) m ( h ) 1 b 4 h3

25 Pipe and Duct Flow. w 0 q 0 Δke 0 Δpe 0 The flow of fluids through pipes and ducts is often steady-state, steadyflow. We normally neglect the kinetic and potential energies; however, depending on the flow situation the work and heat transfer may or may not be zero. Eg. Pumping of fluid from well

26 Unsteady Flow Processes Unsteady or Transient flow process involve changes in properties within the control volume with time. The process start and end over some finite time period instead of continuing indefinitely General mass balance : ~ Q W m i Δ mc o m General energy balance : ~ mo h gz mi h o gz i Δ E C

27 Uniform Flow Process Uniform flow process is a special case of unsteady flow process with the following idealization : ~ At any instant during the process, the state of the control volume is uniform The fluid flow at an inlet or exit is uniform and steady The energy balance for uniform flow process reduced to : ~ When the KE and PE changes associated with the control volume and the fluid streams are negligible : ~

enters a tank as a result of flow energy being converted to internal energy o o In a pressure

28 Uniform Flow Process Rigid tank charging from a supply line is an unsteady flow process since it involves changing in the control volume The Temperature of Steam rises from 300 to 456 C as it enters a tank as a result of flow energy being converted to internal energy o o In a pressure cooker, the enthalpy of the existing steam is H g@p (enthalpy of the saturated vapor at the given pressure)

29 Extra Examples Example 1 : Steam at 0.4 MPa, 300 o C enters an adiabatic nozzle with a low velocity and leaves at 10 kpa with a quality of 90%. Find the exit velocity, in m/s. (Nozzles) 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 (Turbines) (a) Temperature dependent data is used. (b) C p;av at the average temperature is used. (c) C p at 300 K is used. Example 3 : Nitrogen gas is compressed in a steady-state, steady-flow, adiabatic process from 0.1 MPa, 5 o C. During the compression process the temperature becomes 15 o C. If the mass flow rate is 0. kg/s, determine the work done on the nitrogen, in kw. (Compressor) Example 4 : 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, 100 o C. Determine the quality of the steam at 0.4 MPa. (Throttling)

30 Extra Examples Example 5 : Steam at 0. MPa, 300 o C enters a mixing chamber and is mixed with cold water at 0 o C, 0. MPa to produce 0 kg/s of saturated liquid water at 0. MPa. What are the required steam and cold water flow rates? (mixing chamber) Example 6 : Air is heated in a heat exchanger by hot water. The water enters the heat exchanger at 45 o C and experiences a 0 o C drop in temperature. As the air passes through the heat exchanger its temperature is increased by 0 o C. Determine the ratio of mass flow rate of the air to mass flow rate of the water. (heat exchanger) Example 7 : In a simple steam power plant, steam leaves a boiler at 3 MPa, 600 o C and enters a turbine at MPa, 500 o C. Determine the in-line heat transfer from the steam per kilogram mass flowing in the pipe between the boiler and the turbine. (pipe) Example 8 : Air at 100 o C, 0.15 MPa, 40 m/s flows through a converging duct with a mass flow rate of 0. kg/s. The air leaves the duct with at 0.1 MPa, m/s. The exit-to-inlet duct area ratio is 0.5. Find the required rate of heat transfer to the air when no work is done by the air. (pipe)

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 5. Mass and Energy Analysis of Control Volumes Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)