Chapter 4 The First Law of Thermodynamics. Work energy can be thought of as the energy expended to lift a weight. Closed System First Law

Transcription

1 Chapter 4 The First Law of Thermodynamics The first law of thermodynamics is an expression of the conservation of energy prciple. Energy can cross the boundaries of a closed system the form of heat or work. Energy transfer across a system boundary due solely to the temperature difference between a system and its surroundgs is called heat. Work energy can be thought of as the energy expended to lift a weight. Closed System First Law A closed system movg relative to a reference plane is shown below where z is the elevation of the center of mass above the reference plane and V is the velocity of the center of mass. Heat Closed System CM z V Work For the closed system shown above, the conservation of energy prciple or the first law of thermodynamics is expressed as or F HG Reference Plane, z 0 I Total energy Total energy The change total enterg the systemkj F H G I leavg the system K J F H G I energy of the system K J E E E system Chapter 4 -

2 Accordg to classical thermodynamics, we consider the energy added to be net heat transfer to the closed system and the energy leavg the closed system to be net work done by the closed system. So Q W E net net system Normally the stored energy, or total energy, of a system is expressed as the sum of three separate energies. The total energy of the system, E system, is given as E Internal energy + Ketic energy + Potential energy E U + KE + PE Recall that U is the sum of the energy contaed with the molecules of the system other than the ketic and potential energies of the system as a whole and is called the ternal energy. The ternal energy U is dependent on the state of the system and the mass of the system. For a system movg relative to a reference plane, the ketic energy KE and the potential energy PE are given by KE mv dv V V 0 z z 0 mv PE mg dz mgz The change stored energy for the system is E U + KE + PE Chapter 4 -

3 Now the conservation of energy prciple, or the first law of thermodynamics for closed systems, is written as Q W U + KE + PE net net If the system does not move with a velocity and has no change elevation, the conservation of energy equation reduces to Q W U net We will fd that this is the most commonly used form of the fist law. Closed System First Law for a Cycle Sce a thermodynamic cycle is composed of processes that cause the workg fluid to undergo a series of state changes through a series of processes such that the fal and itial states are identical, the change ternal energy of the workg fluid is zero for whole numbers of cycles. The first law for a closed system operatg a thermodynamic cycle becomes 0 net Q W U net net cycle Q net W net Chapter 4-3

4 Example 4- Complete the table given below for a closed system under gog a cycle. Process Q net W net U U Cycle (Answer to above problem) Row : +0, Row : +0, Row 3: -0, -5 Row 4: +5, +5, 0 Example 4- Fd the required heat transfer to the water Example 3-5. Review the solution procedure of Example 3-5 and then apply the first law to the process. Conservation of Energy: E E E Q W U net,4 net,4 4 Chapter 4-4

5 In Example 3-5 we found that W net, 4. The heat transfer is obtaed from the first law as Q W + U net,4 net,4 4 where U U U m( u u ) At state, T 00 C, v m 3 /kg and v f < v < v g at T. The quality at state is v v + xv f x fg v vf v fg u u + xu f fg ( )( ) kg Because state 4 is a saturated vapor state and v m 3 /kg, terpolatg either the saturation pressure table or saturation temperature table at v 4 v g gives Chapter 4-5

6 u kg So U m( u u ) 4 4 ( kg)( ) 07. kg The heat transfer is Q W + U net, 4 net, Heat the amount of 07.4 is added to the water. Review: Specific Heats and Changes Internal Energy and Enthalpy for Ideal Gases Before the first law of thermodynamics can be applied to systems, let s review the ways to calculate the change ternal energy and enthalpy of the substance enclosed by the system boundary. For real substances like water, the property tables are used to fd the ternal energy and enthalpy change. For ideal gases we learned Chapter that the ternal energy and enthalpy are found by knowg the specific heats. Chapter 4-6

7 C V u du h dh and C T dt T dt P v ideal gas P ideal gas The change ternal energy and enthalpy of ideal gases can be expressed as u u u C ( T) dt C ( T T) z z V V, ave h h h C ( T) dt C ( T T) P P, ave where C V,ave and C P,ave are average or constant values of the specific heats over the temperature range. We will drop the ave subscript shortly. P a b c T T P-V DIAGRAM FOR SEVERAL PROCESSES FOR AN IDEAL GAS. In the above figure an ideal gas undergoes three different process between the same two temperatures. Process -a: Constant volume Process -b: P a + bv, a lear relationship Process -c: Constant pressure These ideal gas processes have the same change ternal energy and enthalpy because the processes occur between the same temperature limits. V Chapter 4-7

8 z z a b c V u u u C ( T) dt h h h C ( T) dt a b c P To fd u and h we often use average, or constant, values of the specific heats. Some ways to determe these values are as follows:. The best average value (the one that gives the exact results) See Table A-(c) z for variable specific data. CV T dt C ( ) CP ( T) dt and CPave T T T vave,, T. Good average values are z C and CV( T) + CV( T) CP( T) + CP( T) and CPave vave,, Cv, ave CV ( Tave) and CP, ave CP ( Tave) where T ave T + T 3. Sometimes adequate (and most often used) values are the ones evaluated at 300 K and are given Table A-(a). C C ( 300K) and C C ( 300K) vave, V Pave, P Chapter 4-8

9 Let's take a second look at the defition of u and h for ideal gases. Just consider the enthalpy for now. z h h h CP ( T) dt Let's perform the tegral relative to a reference state where h h ref at T T ref. T T z z + ref h h h C ( T ) dt C ( T ) dt T P T zt T or zt h h h CP ( T ) dt CP ( T ) dt Tref ref ( h h ) ( h h ) ref At any temperature, we can calculate the enthalpy relative to the reference state as zt h href CP ( T ) dt Tref or ref h h + C ( T ) dt ref zt T ref This result is the basis of Table A-7 for air and Tables A-8 through A-5 for other gases. A similar result is found for the change ternal energy. P ref P Chapter 4-9

10 The Systematic Thermodynamics Solution Procedure When we apply a methodical solution procedure, thermodynamics problems are relatively easy to solve. Each thermodynamics problem is approached the same way as shown the followg, which is a modification of the procedure given the text: Thermodynamics Solution Method. Sketch the system and show energy teractions across the boundaries.. Determe the property relation. Is the workg substance an ideal gas or a real substance? Beg to set up and fill a property table. 3. Determe the process and sketch the process diagram. Contue to fill the property table. 4. Apply conservation of mass and conservation of energy prciples. 5. Brg other formation from the problem statement, called physical constrats, such as the volume doubles or the pressure is halved durg the process. 6. Develop enough equations for the unknowns and solve. Chapter 4-0

11 Example 4-3 A tank contas nitrogen at 7 C. The temperature rises to 7 C by heat transfer to the system. Fd the heat transfer and the ratio of the fal pressure to the itial pressure. System: Nitrogen the tank. System boundary NITROGEN GAS P T 7 C T 7 C V P-V diagram for a constant volume process Property Relation: Nitrogen is an ideal gas. The ideal gas property relations apply. Let s assume constant specific heats. (You are encouraged to rework this problem usg variable specific heat data.) Process: Tanks are rigid vessels; therefore, the process is constant volume. Conservation of Mass: m m Usg the combed ideal gas equation of state, PV T PV T Sce R is the particular gas constant, and the process is constant volume, Chapter 4 -

12 V P P V T T ( ) K ( ) K 333. Conservation of Energy: The first law closed system is E E E net Q W U net For nitrogen undergog a constant volume process (dv 0), the net work is (W other 0) z net, + b, W 0 W PdV 0 Usg the ideal gas relations with W net 0, the first law becomes (constant specific heats) z Qnet 0 U m CV dt mcv ( T T) The heat transfer per unit mass is Chapter 4 -

13 q net Qnet CV ( T T) m b kg K kg g K Example 4-4 Air is expanded isothermally at 00 C from 0.4 MPa to 0. MPa. Fd the ratio of the fal to the itial volume, the heat transfer, and work. System: Air contaed a piston-cylder device, a closed system Process: Constant temperature System boundary P AIR W B T const. V P-V diagram for T constant Property Relation: Assume air is an ideal gas and use the ideal gas property relations with constant specific heats. Chapter 4-3

14 Conservation of Energy: PV mrt u C ( T T) V E E E Q W U net The system mass is constant but is not given and cannot be calculated; therefore, let s fd the work and heat transfer per unit mass. Work Calculation: net b g + W W W net, net other, b, mrt Wb, P dv dv V V mrt ln V Conservation of Mass: For an ideal gas a closed system (mass constant), we have Sce the R's cancel and T T m PV RT m PV RT Chapter 4-4

15 V V P 04. MPa 4 P 0. MPa Then the work expression per unit mass becomes w b, F Wb, V RT ln m H G V F wb, HG KJ ( ) K ln 4 kg K kg The net work per unit mass is I I KJ b g w net, 0+ wb, 484. kg Now to contue with the conservation of energy to fd the heat transfer. Sce T T constant, U m u mcv ( T T ) 0 So the heat transfer per unit mass is Chapter 4-5

16 Qnet qnet m qnet wnet u 0 q w net net kg The heat transferred to the air durg an isothermal expansion process equals the work done. Examples Usg Variable Specific Heats Review the solutions Chapter 4 to the ideal gas examples where the variable specific heat data are used to determe the changes ternal energy and enthalpy. Extra Problem for You to Try: An ideal gas, contaed a piston-cylder device, undergoes a polytropic process which the polytropic exponent n is equal to k, the ratio of specific heats. Show that this process is adiabatic. When we get to Chapter 6 you will fd that this is an important ideal gas process. Chapter 4-6

17 Example 4-5 Incompressible Liquid A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage is 35 C, and you always drk your beverage at 0 C. a. How much heat energy must be removed from your two liters of beverage? b. You are havg a party and need to cool 0 of these two-liter bottles one-half hour. What rate of heat removal, kw, is required? Assumg that your refrigerator can accomplish this and that electricity costs 8.5 cents per kw-hr, how much will it cost to cool these 0 bottles? System: The liquid the constant volume, closed system contaer System boundary My beverage Q The heat removed Property Relation: Incompressible liquid relations, let s assume that the beverage is mostly water and takes on the properties of liquid water. The specific volume is 0.00 m 3 /kg, C 4.8 /kg K. Process: Constant volume V V Chapter 4-7

18 Conservation of Mass: m m m m V v L 3 m kg F H G Conservation of Energy: The first law closed system is 3 m 000 E E E L I KJ kg Sce the contaer is constant volume and there is no other work done on the contaer durg the coolg process, we have b g 0 W W + W net net other b The only energy crossg the boundary is the heat transfer leavg the contaer. Assumg the contaer to be stationary, the conservation of energy becomes E E Q U mc T Chapter 4-8

19 Q ( kg)(4.8 )(0 35) K kg K Q 09. Q 09. The heat transfer rate to cool the 0 bottles one-half hour is Q ( 0bottles)( 09. ) bottle 05. hr F hr H G I F s K JG GG 3600 H 6. kw kw s I J JJ K Cost (. kw)(. hr) $ kw hr $0. 05 Chapter 4-9

20 Conservation of Energy for Control volumes The conservation of mass and the conservation of energy prciples for open systems or control volumes apply to systems havg mass crossg the system boundary or control surface. In addition to the heat transfer and work crossg the system boundaries, mass carries energy with it as it crosses the system boundaries. Thus, the mass and energy content of the open system may change when mass enters or leaves the control volume. W net m, V i i V CM Control surface Z i Q net Z CM Z e m, V e e Reference plane Typical control volume or open system Thermodynamic processes volvg control volumes can be considered two groups: steady-flow processes and unsteady-flow processes. Durg a steady-flow process, the fluid flows through the control volume steadily, experiencg no change with time at a fixed position. Let s review the concepts of mass flow rate and energy transport by mass. Mass Flow Rate Mass flow through a cross-sectional area per unit time is called the mass flow rate m. If the fluid density and velocity are constant over the flow cross-sectional area, the mass flow rate is m V A V A av ρ av v where ρ is the density, kg/m 3 ( /v), A is the cross-sectional area, m ; and V av is the average fluid velocity normal to the area, m/s. Chapter 4-0

21 The fluid volume flowg through a cross-section per unit time is called the volume flow rate V. V VA 3 ( m / s) The mass and volume flow rate are related by m ρv V v ( kg / s) Conservation of Mass for General Control Volume The conservation of mass prciple for the open system or control volume is expressed as L NM Sumof rate of massflowg tocontrol volume O L QP NM Sumof rate of massflowg fromcontrol volume O L QP NM Time ratechange of massside control volume O QP or m m m ( kg/ s) system Conservation of Energy for General Control Volume The conservation of energy prciple for the control volume or open system has the same word defition as the first law for the closed system. Expressg the energy transfers on a rate basis, the control volume first law is Chapter 4 -

22 L NM Sumof rate of energy flowg tocontrol volume or O L QP NM Sumof rate of energy flowg fromcontrol volume O L QP NM Time ratechange of energy side control volume E E E system ( kw) Rate of net energy transfer by heat, work, and mass Rate change ternal, ketic, potential, etc., energies Considerg that energy flows to and from the control volume with the mass, energy enters because net heat is transferred to the control volume, and energy leaves because the control volume does net work on its surroundgs, the open system, or control volume, the first law becomes decv Qnet + m4 iθi Wnet m: eθe ( kw) dt for each let for each exit O QP where θ is the energy per unit mass flowg to or from the control volume. The energy per unit mass, θ, flowg across the control surface that defes the control volume is composed of four terms: the ternal energy, the ketic energy, the potential energy, and the flow work. The total energy carried by a unit of mass as it crosses the control surface is V θ u+ Pv+ + gz h + V + gz Here we have used the defition of enthalpy, h u + Pv. Chapter 4 -

23 E E ECV Q F V I m h gz W F i V I e m h gz E net + i i + + i net e e e HG KJ HG + + KJ for each let for each exit Where the time rate change of the energy of the control volume has been written as E CV. Steady-State, Steady-Flow Processes Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossg the control surface are constant with time. The states of the mass streams crossg the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time. CV dm dt de dt CV CV m 0 CV E 0 CV Steady-state, Steady-Flow Conservation of Mass: Sce the mass of the control volume is constant with time durg the steady-state, steady-flow process, the conservation of mass prciple becomes L NM Sumof rate of massflowg tocontrol volume O L QP NM Sumof rate of massflowg fromcontrol volume O QP or m m ( kg/ s) Chapter 4-3

24 Steady-state, steady-flow conservation of energy Sce the energy of the control volume is constant with time durg the steady-state, steady-flow process, the conservation of energy prciple becomes or L NM Sumof rate of energy flowg tocontrol volume O L QP NM Sumof rate of energy flowg fromcontrol volume O QP E E E system ( kw) Rate of net energy transfer by heat, work, and mass or 0 Rate change ternal, ketic, potential, etc., energies E E : Rate of net energy transfer by heat, work, and mass to the system Rate of energy transfer by heat, work, and mass from the system Considerg that energy flows to and from the control volume with the mass, energy enters because heat is transferred to the control volume, and energy leaves because the control volume does work on its surroundgs, the steady-state, steady-flow first law becomes F I Q W Vi m h gz Q W Ve + + i i + + i m e he gze HG KJ + + HG + + KJ for each let F for each exit I Chapter 4-4

25 Most often we write this result as F Q W Ve Vi net net me he + + gze mi hi gzi HG KJ HG + + KJ for each exit I F for each let I where Q Q Q net W W W net Steady-state, steady-flow for one entrance and one exit A number of thermodynamic devices such as pumps, fans, compressors, turbes, nozzles, diffusers, and heaters operate with one entrance and one exit. The steady-state, steady-flow conservation of mass and first law of thermodynamics for these systems reduce to m m ( kg/ s) v VA v VA Q W L V V O m h h+ + gz ( z) ( kw) NM QP where the entrance is state and the exit is state and m is the mass flow rate through the device. When can we neglect the ketic and potential energy terms the first law? Consider the ketic and potential energies per unit mass. Chapter 4-5

26 For V 45 m s V40 m s ke V ( 45m / s) / kg ke 000m / s kg ( 40m / s) / kg ke 0 000m / s kg pe gz m / kg For z 00m pe m 098. s 000m / s kg m / kg z 000m pe m 98. s 000m / s kg When compared to the enthalpy of steam (h 000 to 3000 /kg) and the enthalpy of air (h 00 to 6000 /kg), the ketic and potential energies are often neglected. When the ketic and potential energies can be neglected, the conservation of energy equation becomes Q W m ( h h ) ( kw) We often write this last result per unit mass flow as q w ( h h ) ( / kg) where q Q m and w W m. Chapter 4-6

27 Some Steady-Flow Engeerg Devices Below are some engeerg devices that operate essentially as steady-state, steady-flow control volumes. Nozzles and Diffusers V V >> V V V << V For flow through nozzles, the heat transfer, work, and potential energy are normally neglected, and nozzles have one entrance and one exit. The conservation of energy becomes Chapter 4-7

28 m m m m m E E Q F V I m h gz W F i V I e net + i i + + i net m e he gze HG KJ + HG + + KJ for each let for each exit F V I F V I mh + mh HG KJ HG + KJ Solvg for V V ( h h ) + V Example 4-6 Steam at 0.4 MPa, 300 o C, enters an adiabatic nozzle with a low velocity and leaves at 0. MPa with a quality of 90%. Fd the exit velocity, m/s. Control Volume: The nozzle Property Relation: Steam tables Process: Assume adiabatic, steady-flow Conservation Prciples: Conservation of mass: For one entrance, one exit, the conservation of mass becomes Chapter 4-8

29 m m m m m Conservation of energy: Accordg to the sketched control volume, mass crosses the control surface, but no work or heat transfer crosses the control surface. Neglectg the potential energies, we have F HG mh E E I V V + mh F KJ HG + I KJ Neglectg the let ketic energy, the exit velocity is V ( h h ) Now, we need to fd the enthalpies from the steam tables. o T 300 C P MPa h kg U V W P x At 0. MPa h f /kg and h fg 0.9 /kg. 0. MPa 090. U V W h h h + x h f fg ( 0. 90)( 09. ) kg Chapter 4-9

30 V ( ) m s kg 000m / s / kg Turbes m Control surface Turbe control volume W m If we neglect the changes ketic and potential energies as fluid flows through an adiabatic turbe havg one entrance and one exit, the conservation of mass and the steady-state, steady-flow first law becomes Chapter 4-30

31 m m m m m E E F I F I Q Vi m h gz W Ve net + HG i i + + KJ + i net m HG e h + + e gz KJ e for each let mh mh + W W m ( h h ) for each exit Example 4-7 High pressure air at 300 K flows to an aircraft gas turbe and undergoes a steady-state, steady-flow, adiabatic process to the turbe exit at 660 K. Calculate the work done per unit mass of air flowg through the turbe when (a) Temperature-dependent data are used. Chapter 4-3

32 (b) C p,ave at the average temperature is used. (c) C p at 300 K is used. Control Volume: The turbe Property Relation: Assume air is an ideal gas and use ideal gas relations Process: Steady-state, steady-flow, adiabatic process Conservation Prciples: Conservation of mass: m m m m m Conservation of energy: F I Q Vi m h gz W Ve + i i + + i m e he gze HG KJ + HG + + KJ for each let F for each exit I Accordg to the sketched control volume, mass and work cross the control surface. Neglectg ketic and potential energies and notg the process is adiabatic, we have + mh W + mh W m ( h h ) 0 The work done by the air per unit mass flow is Chapter 4-3

33 W w h h m Notice that the work done by a fluid flowg through a turbe is equal to the enthalpy decrease of the fluid. (a) Usg the air tables, Table A-7 at T 300 K, h /kg at T 660 K, h /kg w h h ( ) 755. kg kg (b) Usg Table A-(c) at T ave 980 K, C p, ave.38 /kg K w h h Cp, ave( T T) 38. ( ) K kg K kg c. Usg Table A-(a) at T 300 K, C p.005 /kg K w h h C ( T T ) p 005. ( ) K kg K 643. kg Chapter 4-33

34 Compressors and fans m W Steady-Flow Compressor m Compressors and fans are essentially the same devices. However, compressors operate over larger pressure ratios than fans. If we neglect the changes ketic and potential energies as fluid flows through an adiabatic compressor havg one entrance and one exit, the steady-state, steady-flow first law or the conservation of energy equation becomes Chapter 4-34

35 F I F I Q Vi m h gz W Ve net + HG i i + + KJ + i net m HG e h + + e gz KJ e for each let W m ( h h ) net ( W ) m ( h h) W m ( h h ) for each exit Example 4-8 Nitrogen gas is compressed a steady-state, steady-flow, adiabatic process from 0. MPa, 5 o C. Durg the compression process the temperature becomes 5 o C. If the mass flow rate is 0. kg/s, determe the work done on the nitrogen, kw. Control Volume: The compressor (see the compressor sketched above) Property Relation: Assume nitrogen is an ideal gas and use ideal gas relations Process: Adiabatic, steady-flow Chapter 4-35

36 Conservation Prciples: Conservation of mass: m m m m m Conservation of energy: F I Q Vi m h gz W Ve net + i i + + i net m e he gze HG KJ + HG + + KJ for each let F for each exit I Accordg to the sketched control volume, mass and work cross the control surface. Neglectg ketic and potential energies and notg the process is adiabatic, we have for one entrance and one exit 0+ m ( h+ 0+ 0) ( W ) + m ( h ) W m ( h h ) The work done on the nitrogen is related to the enthalpy rise of the nitrogen as it flows through the compressor. The work done on the nitrogen per unit mass flow is w W m h h Assumg constant specific heats at 300 K from Table A-(a), we write the work as Chapter 4-36

37 w C ( T T) p 039. ( 5 5) K kg K kg F kg W mw 0.. s H G kw s kg I KJ Throttlg devices Consider fluid flowg through a one-entrance, one-exit porous plug. The fluid experiences a pressure drop as it flows through the plug. No net work is done by the fluid. Assume the process is adiabatic and that the ketic and potential energies are neglected; then the conservation of mass and energy equations become Chapter 4-37

38 mi me F I F I Q Vi m h gz W Ve net + HG i i + + KJ + i net m HG e h + + e gz KJ e for each let mh mh i i e e for each exit h i h e This process is called a throttlg process. What happens when an ideal gas is throttled? ze i C p h e hi h h 0 ( T) dt T i e e or T When throttlg an ideal gas, the temperature does not change. We will see later Chapter 0 that the throttlg process is an important process the refrigeration cycle. 0 i A throttlg device may be used to determe the enthalpy of saturated steam. The steam is throttled from the pressure the pipe to ambient pressure the calorimeter. The pressure drop is sufficient to superheat the steam the calorimeter. Thus, the temperature and pressure the calorimeter will specify the enthalpy of the steam the pipe. Example 4-9 Chapter 4-38

39 One way to determe 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. MPa, 00 o C. Determe the quality of the steam at 0.4 MPa. Throttlg orifice Control Volume: The throttle Control Surface Property Relation: The steam tables Process: Steady-state, steady-flow, no work, no heat transfer, neglect ketic and potential energies, one entrance, one exit Conservation Prciples: Conservation of mass: Conservation of energy: m m m m m E E F I F I Q Vi m h gz W Ve net + HG i i + + KJ + i net m HG e h + + e gz KJ e for each let for each exit Accordg to the sketched control volume, mass crosses the control surface. Neglectg ketic and potential energies and notg the process is adiabatic with no work, we have for one entrance and one exit Chapter 4-39

40 Therefore, 0+ m ( h+ 0+ 0) 0+ m ( h ) mh mh h h o T 00 C P MPa h kg h U V W h 676. h + xh kg d i f P 04. MPa x h h h fg f Mixg chambers The mixg of two fluids occurs frequently engeerg applications. The section where the mixg process takes place is called a mixg chamber. The ordary shower is an example of a mixg chamber. Chapter 4-40

41 Example 4-0 Steam at 0. MPa, 300 o C, enters a mixg 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? Steam Cold water Mixg chamber Saturated water 3 Control surface Control Volume: The mixg chamber Property Relation: Steam tables Process: Assume steady-flow, adiabatic mixg, with no work Conservation Prciples: Conservation of mass: m m + m m 3 m m m m 3 Chapter 4-4

42 Conservation of energy: E E F I F I Q Vi m h gz W Ve net + HG i i + + KJ + i net m HG e h + + e gz KJ e for each let for each exit Accordg to the sketched control volume, mass crosses the control surface. Neglectg ketic and potential energies and notg the process is adiabatic with no work, we have for two entrances and one exit mh + mh mh 3 3 mh + ( m3 m ) h mh 3 3 m ( h h) m 3( h3 h) m m ( h h ) 3 3 ( h h ) Now, we use the steam tables to fd the enthalpies: o T 300 C P MPa h kg U V W o 0 C U V o C 0. W 0. MPa U VW h h T P MPa h h kg P3 Sat. liquid 3 0. MPa kg Chapter 4-4

43 m m ( h h ) 3 3 ( h h) kg ( ) / kg 0 s ( ) / kg kg 8. s m m m 3 ( 0. 8) 78. kg s kg s Heat exchangers Heat exchangers are normally well-sulated devices that allow energy exchange between hot and cold fluids with mixg the fluids. The pumps, fans, and blowers causg the fluids to flow across the control surface are normally located side the control surface. Chapter 4-43

44 Example 4- Air is heated a heat exchanger by hot water. The water enters the heat exchanger at 45 o C and experiences a 0 o C drop temperature. As the air passes through the heat exchanger, its temperature is creased by 5 o C. Determe the ratio of mass flow rate of the air to mass flow rate of the water. Air let Water let Control surface Air exit Water exit Control Volume: The heat exchanger Property Relation: Air: ideal gas relations Chapter 4-44

45 Process: Assume adiabatic, steady-flow Conservation Prciples: Conservation of mass: Water: steam tables or compressible liquid results m m m ( kg/ s) system 0(steady) For two entrances, two exits, the conservation of mass becomes m m m + m m + m air, w, air, w, For two fluid streams that exchange energy but do not mix, it is better to conserve the mass for the fluid streams separately. m m m air, air, air m m m w, w, w Conservation of energy: Accordg to the sketched control volume, mass crosses the control surface, but no work or heat transfer crosses the control surface. Neglectg the ketic and potential energies, we have for steady-flow E E E system ( kw) Rate of net energy transfer by heat, work, and mass 0(steady) Rate change ternal, ketic, potential, etc., energies Chapter 4-45

46 E E m h + m h m h + m h air, air, w, w, air, air, w, w, m air ( hair, hair, ) m w( hw, hw, ) m ( h, h, ) air w w m ( h h ) w air, air, We assume that the air has constant specific heats at 300 K, Table A-(a) (we don't know the actual temperatures, just the temperature difference). Because we know the itial and fal temperatures for the water, we can use either the compressible fluid result or the steam tables for its properties. Usg the compressible fluid approach for the water, Table A-3, C p, w 4.8 /kg K. m m air w Cpw, ( Tw, Tw, ) C ( T T ) p, air air, air, 48. 0K kgw K b K kgair K kgair / s 333. kg / s w b g g Chapter 4-46

47 A second solution to this problem is obtaed by determg the heat transfer rate from the hot water and notg that this is the heat transfer rate to the air. Considerg each fluid separately for steady-flow, one entrance, and one exit, and neglectg the ketic and potential energies, the first law, or conservation of energy, equations become E E air : m h + Q m h water : mw, hw, Q, w + mw, hw, Q Q, air, w air, air,, air air, air, Pipe and duct flow The flow of fluids through pipes and ducts is often a steady-state, steadyflow process. We normally neglect the ketic and potential energies; however, dependg on the flow situation, the work and heat transfer may or may not be zero. Example 4- In a simple steam power plant, steam leaves a boiler at 3 MPa, 600 o C, and enters a turbe at MPa, 500 o C. Determe the -le heat transfer from the steam per kilogram mass flowg the pipe between the boiler and the turbe. Steam from boiler Q Steam to turbe Control surface Chapter 4-47

48 Control Volume: Pipe section which the heat loss occurs. Property Relation: Steam tables Process: Steady-flow Conservation Prciples: Conservation of mass: m m m ( kg/ s) system For one entrance, one exit, the conservation of mass becomes m m m m m 0(steady) Conservation of energy: Accordg to the sketched control volume, heat transfer and mass cross the control surface, but no work crosses the control surface. Neglectg the ketic and potential energies, we have for steady-flow 0(steady) E E E system ( kw) Rate of net energy transfer Rate change ternal, ketic, by heat, work, and mass potential, etc., energies We determe the heat transfer rate per unit mass of flowg steam as Chapter 4-48

49 mh m h + Q Q m ( h h ) q Q h h m We use the steam tables to determe the enthalpies at the two states as o T 600 C h P 3MPa kg o T 500 C h P MPa kg q h h ( ) 4.7 kg kg Example 4-3 Air at 00 o C, 0.5 MPa, 40 m/s, flows through a convergg duct with a mass flow rate of 0. kg/s. The air leaves the duct at 0. MPa, 3.6 m/s. The exit-to-let duct area ratio is 0.5. Fd the required rate of heat transfer to the air when no work is done by the air. Q Air let Chapter 4-49 Air exit Control surface

50 Control Volume: The convergg duct Property Relation: Assume air is an ideal gas and use ideal gas relations Process: Steady-flow Conservation Prciples: Conservation of mass: m m m ( kg/ s) system For one entrance, one exit, the conservation of mass becomes m m m m m 0(steady) Conservation of energy: Accordg to the sketched control volume, heat transfer and mass cross the control surface, but no work crosses the control surface. Here keep the ketic energy and still neglect the potential energies, we have for steadystate, steady-flow process 0(steady) Chapter 4-50

51 E E E system ( kw) Rate of net energy transfer Rate change ternal, ketic, by heat, work, and mass potential, etc., energies F I F I V m HG h Q V + KJ m h + HG + KJ F Q V m ( h h ) + HG V I KJ In the first law equation, the followg are known: P, T (and h ), V, V, m, and A /A. The unknowns are Q, and h (or T ). We use the first law and the conservation of mass equation to solve for the two unknowns. m m ( kg/ s) VA VA v v P P VA VA RT RT Solvg for T Chapter 4-5

52 T T P P A A V V F 0. MPa ( ) K H G 05. MPa 353. K I F KJ b ghg m s /. 40m/ s I KJ Assumg C p constant, h - h C p (T - T ) F V Q m C ( T T ) + HG p V kg 0. ( 005. ( ) K s kg K ( ) m / s / kg + 000m / s kw s I KJ ) Looks like we made the wrong assumption for the direction of the heat transfer. The heat is really leavg the flow duct. (What type of device is this anyway?) Q Q 87. kw Liquid pumps Chapter 4-5

53 Fluid exit Pump W z Fluid let Liquid flow through a pump The work required when pumpg an compressible liquid an adiabatic steady-state, steady-flow process is given by L NM V V Q W m h h+ + gz ( z) ( kw) The enthalpy difference can be written as h h ( u u ) + ( Pv) ( Pv) For compressible liquids we assume that the density and specific volume are constant. The pumpg process for an compressible liquid is essentially isothermal, and the ternal energy change is approximately zero (we will see this more clearly after troducg the second law). Thus, the enthalpy difference reduces to the difference the pressure-specific volume products. Sce v v v the work put to the pump becomes L NM V V W m v( P P) + + gz ( z) ( kw) O QP O QP W is the net work done by the control volume, and it is noted that work is put to the pump; so, W,. W pump Chapter 4-53

54 If we neglect the changes ketic and potential energies, the pump work becomes ( W, pump) m v( P P) ( kw) W m v( P P), pump We use this result to calculate the work supplied to boiler feedwater pumps steam power plants. Uniform-State, Uniform-Flow Problems Durg unsteady energy transfer to or from open systems or control volumes, the system may have a change the stored energy and mass. Several unsteady thermodynamic problems may be treated as uniform-state, uniform-flow problems. The assumptions for uniform-state, uniform-flow are The process takes place over a specified time period. The state of the mass with the control volume is uniform at any stant of time but may vary with time. The state of mass crossg the control surface is uniform and steady. The mass flow may be different at different control surface locations. To fd the amount of mass crossg the control surface at a given location, we tegrate the mass flow rate over the time period. z z t t Inlets: mi m idt Exits: me m edt 0 0 The change mass of the control volume the time period is CV z0 ( m m ) t dm dt CV dt Chapter 4-54

55 The uniform-state, uniform-flow conservation of mass becomes m m ( m m) i e CV The change ternal energy for the control volume durg the time period is ( mu mu) CV z0 t du dt CV dt The heat transferred and work done the time period are z z t t 0 0 Q Qdt and W W dt The energy crossg the control surface with the mass the time period is z j j j j 0 F t Vj mθ m h + + gz HG j I KJ dt where j i, for lets e, for exits The first law for uniform-state, uniform-flow becomes Chapter 4-55

56 E E ECV F V I F V I e i Q W me he + + gze mi hi gzi m e m e HG KJ HG + + KJ + b gcv When the ketic and potential energy changes associated with the control volume and the fluid streams are negligible, it simplifies to b g ( ) e e i i CV Q W mh mh + m u mu Example 4-4 Consider an evacuated, sulated, rigid tank connected through a closed valve to a high-pressure le. The valve is opened and the tank is filled with the fluid the le. If the fluid is an ideal gas, determe the fal temperature the tank when the tank pressure equals that of the le. Control Volume: The tank Property Relation: Ideal gas relations Process: Assume uniform-state, uniform-flow Chapter 4-56

57 Conservation Prciples: Conservation of mass: m m ( m m) i e CV Or, for one entrance, no exit, and itial mass of zero, this becomes m i ( m ) CV Conservation of energy: For an sulated tank Q is zero and for a rigid tank with no shaft work W is zero. For a one-let mass stream and no-exit mass stream and neglectg changes ketic and potential energies, the uniform-state, uniform-flow conservation of energy reduces to Q W mh e e mh i i+ m u mu CV 0 mh + ( m u ) i i CV b g ( ) or mh h ( m u ) i i CV u u + Pv u u u Pv C ( T T) Pv i i i i i i i v i i i Chapter 4-57

58 C ( T T) RT v i i T Cv + C kt i v R T i Cp C T i v If the fluid is air, k.4 and the absolute temperature the tank at the fal state is 40 percent higher than the fluid absolute temperature the supply le. The ternal energy the full tank differs from the ternal energy of the supply le by the amount of flow work done to push the fluid from the le to the tank. Extra Assignment Rework the above problem for a 0 m 3 tank itially open to the atmosphere at 5 o C and beg filled from an air supply le at 90 psig, 5 o C, until the pressure side the tank is 70 psig. Chapter 4-58