first law of ThermodyNamics

Transcription

1 first law of ThermodyNamics

2 First law of thermodynamics - Principle of conservation of energy - Energy can be neither created nor destroyed

3 Basic statement When any closed system is taken through a cycle, the net work delivered to the surroundings is proportional to the net heat taken from the surroundings. Q net The symbol W net is called the cyclic integral of the heat transfer, represents the net heat transfer during the cycle J is the cyclic integral of the work, represents the net work during the cycle is a proportionality factor that depends on the units used for work and heat.

4 Because these units are equivalent, it is not necessary to include the factor J explicitly in the Eq. but simply to recognize that for any system of units, each equation must have consistent units throughout. Therefore, we may write Net heat = Net work or, Q in Q out = W out W in or, Q = W or, W in Q in Q out W out

5 First Law of Thermodynamics for a Change in State of a System Consider the first law of thermodynamics for a system (control mass) that undergoes a cycle in which it changes from state 1 to state 2 by process A and returns from state 2 to state 1 by process B. This cycle is shown in Fig. on a pressure (or other intensive property) volume (or other extensive property) diagram. From the first law of thermodynamics, we have (1) Considering the two separate processes (Cycle 1-A-2-B-1) Now consider another cycle (Cycle 1-C-2-B-1) (2) (3)

6 Subtracting (3) from (2), (4) Since A and C represent arbitrary processes, both between states 1 and 2, the quantity is the same for all processes between states 1 and 2. Therefore, the quantity (δq δw ) between state 1 and 2 - depends only on the initial and final states and not on the path followed between the two states. - is a point function - is the differential of a property of the system (control mass). This property is the energy of the system (control mass) and is given the symbol E. de = δq δw or, δq = de + δw (5) Because E is a property, its derivative is written de.

7 By integrating from an initial state 1 to a final state 2, we have where E 1 and E 2 1Q 2 1W 2 1Q 2 = E 2 E W (6) are the initial and final values of the energy E of the control mass, is the heat transferred to the control mass during the process from state 1 to state 2, and is the work done by the control mass during the process between state 1 to state 2 The physical significance of E The physical significance of the property E is that it represents all the energy of the system in the given state. This energy might be present in a variety of forms, such as kinetic energy (KE) - energy associated with the motion potential energy (PE) - energy associated with a mass that is located at a specified position in a force field.

8 Internal Energy (U) - some forms of energy, e.g., chemical, nuclear, magnetic, electrical, and thermal depend in some way on the molecular structure of the substance that is being considered, and these energies are grouped as the internal energy of a system, U. KE & PE are external forms of energy as these are independent of the molecular structure of matter. These are associated with the coordinate frame that we select and can be specified by the macroscopic parameters of mass, velocity & elevation. The internal energy U includes all other forms of energy of the control mass and is associated with the thermodynamic state of the system. Internal energy, like kinetic and potential energy, has no natural zero value. Therefore, it is necessary to arbitrarily define the specific internal energy of a substance to be zero at some state that is referred to as the reference state.

9 Thus, E = Internal energy + kinetic energy + potential energy or, E = U + KE + PE Since the terms comprising E are point functions, we can write de = du + d(ke) + d(pe) (7) The first law of thermodynamics for a change of state may therefore be written de = du + d(ke) + d(pe) = δq δw or δq = du + d(ke) + d(pe) + δw (8) In words, this equation states that as a control mass undergoes a change of state, energy may cross the boundary as either heat or work, and each may be positive or negative. The net change in the energy of the system will be exactly equal to the net energy that crosses the boundary of the system. The energy of the system may change in any of three ways by a change in internal energy, in kinetic energy, or in potential energy.

10 Consider a system is initially at rest. Let the system be acted on by an external horizontal force F that moves the system a distance dx in the direction of force Here, d(pe) = 0 (no change in PE) δq = 0 (no heat transfer) du = 0 (no change in internal energy) Now from the first law, equ (7): From Newton s 2nd law of motion; Integrating, we obtain The expression of KE Then, 0 = 0 + d(ke) (-Fdx) δ W = F dx = dke

11 Consider a control mass that is initially at rest and at the elevation of some reference level. Let this mass be acted on by a vertical force F of such magnitude that it raises (in elevation) the mass with constant velocity an amount dz. Let the acceleration due to gravity at this point be g. From the first law, Eq. 7, we have Then Integrating gives δq = 0, d(ke)=0, du=0 δw = F dz = d PE F = ma = mg d PE = F dz = mg dz (work is done on the system) PE 2 PE 1 = mg(z 2 Z 1 )

12 Substituting the expressions for kinetic and potential energy into Eq. 7 and 8, we have de = du + mv dv + mg dz δq=du + mvdv + mgdz + δw (9a) (9b) Integrating for a change of state from state 1 to state 2 with constant g, (10a) 1Q 2 = + 1 W (10b) This is the integral from of First Law of Thermodynamics. In the absence of ke & PE non-flow energy eqn. Eqs. 9b and 10b are in effect a statement of the conservation of energy. The net change of the energy of the control mass is always equal to the net transfer of energy across the boundary as heat and work.

13 PROBLEM- 1: A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kj. The heat transfer from the tank is 1500 kj. Consider the tank and the fluid inside a control surface and determine the change in internal energy of this control mass.

14 PROBLEM- 1: A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kj. The heat transfer from the tank is 1500 kj. Consider the tank and the fluid inside a control surface and determine the change in internal energy of this control mass. Solution: The first law of thermodynamics is 1Q 2 = + 1 W (10b) Since there is no change in kinetic and potential energy, this reduces to U 2 U 1 = 1 Q 2 1 W 2 = 1500 ( 5090) kj = 3590 kj

15 Internal Energy (U) A Thermodynamic Property Internal energy is an extensive property because it depends on the mass of the system. The symbol U designates the total internal energy of a given mass of a substance. The symbol u designates the internal energy per unit mass (specific internal energy) In tables of thermodynamic properties such as the steam tables, the value of internal energy can be tabulated along with other thermodynamic properties. In the liquid vapor saturation region, U = U liq + U vap Dividing by m and introducing the quality x gives or, mu = m liq u f + m vap u g u = (1 x)u f + xu g u = u f + xu f g

16 Problem -2: A vessel having a volume of 5 m 3 contains 0.05 m 3 of saturated liquid water and 4.95 m 3 of saturated water vapor at 0.1 MPa. Heat is transferred until the vessel is filled with saturated vapor. Determine the heat transfer for this process. Control mass: All the water inside the vessel. Process: Constant volume and mass; therefore, constant specific volume From the first law we have From examining the control surface for various work modes, we conclude that the work for this process is zero. Furthermore, the system is not moving, so there is no change in kinetic energy. There is a small change in the center of mass of the system, but we will assume that the corresponding change in potential energy (in kilojoules) is negligible.

17 Therefore, 1Q 2 = U 2 U 1 The solution proceeds as follows:

18 Then U 1 = m 1 iq u 1 liq + m 1 vap u 1 vap = 47.94*(417.36) *(2506.1) kj = kj The properties for the final state are the quality, x = 100%, and v 2, the final specific volume, m = m 1 liq + m 1 vap = kg = kg v 1 = From the saturated liquid and saturated vapor table: by interpolation, we get that at a pressure of 2.03 MPa, v g = m 3 /kg. The final pressure of the steam is therefore 2.03 MPa. u 2 = kj/kg U 2 = mu 2 = 50.86*(2600.5) = kj 1Q 2 = U 2 U 1 = = kj

19

20 NoN-flow Processes (First Law Analysis for A Control Mass)

21 we can write de = du + d(ke) + d(pe) The first law of thermodynamics for a change of state may therefore (7) be written de = du + d(ke) + d(pe) = δq δw or δq = du + d(ke) + d(pe) + δw (8) In words, this equation states that as a control mass undergoes a change of state, energy may cross the boundary as either heat or work, and each may be positive or negative. The net change in the energy of the system will be exactly equal to the net energy that crosses the boundary of the system. The energy of the system may change in any of three ways by a change in internal energy, in kinetic energy, or in potential energy.

22 Equation (8) for state1 to state 2 is 1Q 2 = + 1 W 2 And the non-flow energy equation (in absence of KE and PE) is presented in the form non-flow energy eqn.

23 ENTHALPY - A thermodynamic property consider a control mass undergoing a quasi-equilibrium constant-pressure process, as shown in Fig. Assumption There is no change in kinetic There is no change potential energy The only work done during the process is that associated with the boundary movement. Taking the gas as control mass and applying the first law, we have, in terms of Q, Fig: The constant-pressure quasi-equilibrium process. First Law of Thermodynamics : 1Q 2 = + 1 W 2 or, or,

24 The work done can be calculated from the relation Since the pressure is constant, Therefore, The heat transfer during the process is given in terms of the change in the quantity U + PV between the initial and final states. Because all these quantities are thermodynamic properties, that is, functions only of the state of the system, their combination must also have these same characteristics. We find U +PV convenient to define a new extensive property, the enthalpy, H = U + PV or, per unit mass, h = u + Pv h, is specific enthalpy,, and H. total enthalpy.

25 Enthalpy Thermodynamic property Extensive property Independent of process Combination of energy of the system due to temperature, pressure, and volume A property of the system that measures its heat content. The heat transfer in a constant-pressure quasi-equilibrium process is equal to the change in enthalpy, which includes both the change in internal energy and the work for this particular process. This is by no means a general result. It is valid for this special case only because the work done during the process is equal to the difference in the PV product for the final and initial states. This would not be true if the pressure had not remained constant during the process. Internal energy can be calculate at a state: u = h - Pv

26 If a system processes that do not occur at constant pressure, for which enthalpy has no physical significance, then What to do? Ans: Being a property, is a state or point function, and its use in calculating internal energy at the same state is not related to, or dependent on, any process that may be taking place. When enthalpy and internal energy are given values relative to the same reference state, as they are in essentially all thermodynamic tables, the difference between internal energy and enthalpy at the reference state is equal to Pv. In many thermodynamic tables, values of the specific internal energy u are not given. Since these values can be readily calculated from the relation u = h Pv, Example: let us calculate the internal energy u of superheated R-134a at 0.4 MPa, 70 C.

27 For substances for which compressed-liquid tables are not available, the enthalpy is taken as that of saturated liquid at the same temperature. The enthalpy of saturated liquid has the symbol h f, saturated vapor h g, and the increase in enthalpy during vaporization h fg. For a saturation state, the enthalpy can be calculated by one of the following relations:

28 Problem - 3 A cylinder fitted with a piston has a volume of 0.1 m 3 and contains 0.5 kg of steam at 0.4 MPa. Heat is transferred to the steam until the temperature is 300 C, while the pressure remains constant. Determine the heat transfer and the work for this process. Solution Control mass: Process: Steam inside cylinder Constant pressure.. There is no change in kinetic energy (KE = 0) There is no change in potential energy (PE = 0). Work is done by movement at the boundary. Assumption The process to be quasi-equilibrium.

29 Since the pressure is constant, we have Therefore, the first law is, in terms of Q,

30 v = v f + xv f g State 1 is saturated liquid-vapor mixture region At 0.4Mpa, T sat = C. But T 2 = C > T sat so state 2 is superheated region. Therefore,

31

32

33 Specific Heats (Heat Capacity) A Thermodynamic property The amount of heat required per unit mass to raise the temperature by one degree. It is an intensive property. Consider a homogeneous phase of a substance of constant composition. This phase may be a solid, a liquid, or a gas, but no change of phase will occur. Assumptions There is no changes in kinetic and potential energies, The substance is compressible and The substance goes through a quasi-equilibrium process, This expression can be evaluated for two separate special cases: 1.Constant volume, for which the work term (PdV) is zero, so that the specific heat (at constant volume) is

34 2. Constant pressure, for which the work term can be integrated and the resulting PV terms at the initial and final states can be associated with the internal energy terms, The corresponding specific heat (at constant pressure) is Special case 1. Consider either a solid or a liquid. Since both of these phases are nearly incompressible, For both of these phases, the specific volume is very small, so in many cases where C is either the constant-volume or the constant-pressure specific heat 2. In many processes involving a solid or a liquid, specific heat is assumed constant (unless the process occurs at low temperature or over a wide range of temperatures).

35 Constant Volume Process (V = constant) the work term W = (PdV) = 0 Q 12 = U 2 U 1 = m (u 2 u 1 ) non-flow energy eqn. That is, the quantity of heat is equal to the change of internal energy.

36 Polytropic Process The constant volume and constant pressure processes can be regarded as limiting cases of a more general type of process in which both the volune and pressure change, but in a certain specified manner. The states during an expansion or compression can be described approximately by a relation of the form, Work done per unit mass where, n is a constant called the index of expansion or compression, and p and v are the average values of the pressure and the specific volume for the system.

37 The area under the curve or the line represents the magnitude of the work done in each case.

38 Since For initial, final and any intermediate state of the process = constant The work done per unit mass during a change from state 1 to state 2 may then be found by integration as folows.

39 Adiabatic Process (Q = 0) Any process during which heat is prevented from crossing the boundary of the system (no heat transfer) is called adiabatic process In this case the system is thermally insulated from its surroundings. For adiabatic non-flow process, the energy equation reduces to W + U 2 U 1 = 0

40

41 Isothermal Process (T = constant) When the quantities of heat and work are so proportional during an expansion or compression that the temperature of the fluid remains constant, the process is said to be isothermal.

42 flow Processes (First Law Analysis for A Control Volume)

43 First Law of Thermodynamics for state change 1 to 2 is Q-W= E, if steady then ( E/ t)=0 and m 1 =m 2 1Q 2 = + 1 W 2 Add flow work W f = p( m)v, here v is specific vol. Q= m(u 2 -u 1 )+ m(v 22 -v 12 )/2+ mg(z 2 -z 1 )+ (W s +W f2 -W f1 ) Q=(u 2 -u 1 )+(v 22 -v 12 )/2+g(z 2 -z 1 )+(W s +W f2 -W f1 ) Q=(u 2 -u 1 )+(v 22 - v 12 )/2+g(z 2 -z 1 )+ W s +(p 2 v 2 -p 1 v 1 )

44 First Law Analysis for Open System (Control Volume, CV) Mass Flow Rate and Mass Balance The conservation of mass law The physical law says that we cannot create or destroy mass.

45 The conservation of mass principle states In general, there may be several locations on the boundary through which mass enters or exits. This can be accounted for by summing, as follows This is the continuity equation Consider several contributions to the mass as Assume that a fluid is flowing in a pipe or duct as shown in the figure

46 The flow across the control volume surface can be indicated with an average velocity shown to the left of the valve or with a distributed velocity over the cross section, as shown to the right of the valve. The volume flow rate is so the mass flow rate becomes Where, V is the fluid flow velocity A is the perpendicular area to the direction of flow velocity where often the average velocity is used. Note that the above equation involves summations over the inlets and exits of the control volume. Each individual term in these sums applies to a particular inlet or exit. The area, velocity, and specific volume appearing in each term refers only to the corresponding inlet or exit.

47 Forms of CV Energy Rate Balance The fluid flowing across the control surface enters or leaves with an amount of energy per unit mass as Whenever a fluid mass enters a control volume at state i or exits at state e, there is a boundary movement work associated with that process. They are commonly referred to as flow work or flow energy or Pressure energy.

48

49 For the general control volume we may have several entering or leaving mass flow rates, so we may write the first law of thermodynamics for general CV, The equation expresses that the rate of change of energy inside the control volume is due to a net rate of heat transfer, a net rate of work (measured positive out), and the summation of energy fluxes due to mass flows into and out of the control volume.

50 Steady-state, steady-flow Process (SSSF) Let us consider a certain set of assumptions that lead to a reasonable model for this type of process, which we refer to as the steady-state process. 1. The control volume is fixed i.e. does not move relative to the coordinate frame. (all velocities measured relative to the coordinate frame are also velocities relative to the control surface, and there is no work associated with the acceleration of the control volume.) 2. The state of the mass at each point in the control volume does not vary with time. (No accumulation of mass within the CV.) It requires that Therefore, we conclude that for the steady-state process. Continuity equation: For various mass flow (many inlet and outlet) For one flow stream

51 3. As for the mass that flows across the control surface, the mass flux and the state of this mass at each discrete area of flow on the control surface do not vary with time. The rates at which heat and work cross the control surface remain constant. At steady state: First law: -- For the various mass flows -- For one flow stream -- This is the equation for a Steady State Steady Flow (SSSF) Process. -- Used in the analysis of Heat exchanger, Nozzle, Difffuser, Compressor, Pump, Power plants, Refrigerator, etc.

52 Examples of SSSF Processes Heat Exchanger A steady-state heat exchanger is a simple fluid flow through a pipe or system of pipes, where heat is transferred to or from the fluid. The fluid may be heated or cooled (remaining at the same phase), And may or may not boil, changing from liquid to vapor, or condense, changing from vapor to liquid. Example is the condenser in an R-134a refrigeration system. Superheated vapor enters the condenser and liquid exits. Fig: A refrigeration system condenser.

53 Assumptions: The process tends to occur at constant pressure, since a fluid flowing in a pipe usually undergoes only a small pressure drop because of fluid friction at the walls. The pressure drop may or may not be taken into account in a particular analysis. There is no means for doing any work (shaft work, electrical work, etc.), = 0 Changes in kinetic and potential energies are commonly negligibly small. ΔKE = 0 and ΔPE = 0. (One exception may be a boiler tube in which liquid enters and vapor exits at a much larger specific volume. In such a case, it may be necessary to check the exit velocity.) The heat transfer in most heat exchangers is then found from SSSF equation as the change in enthalpy of the fluid. SSSF equation :

54 Problem: Consider a water-cooled condenser in a large refrigeration system in which R-134a is the refrigerant fluid. The refrigerant enters the condenser at 1.0 MPa and 60 0 C, at the rate of 0.2 kg/s, and exits as a liquid at 0.95 MPa and 35 0 C. Cooling water enters the condenser at 10 C and exits at 20 C. Determine the rate at which cooling water flows through the condenser. Solution Assumptions: Both kinetic and potential energy changes are negligible. ΔKE = 0 and ΔPE = 0. There is no means for doing any work (shaft work, electrical work, etc.). = 0 There is no heat transfer across the control surface. = 0

55 SSSF equation : Applying the assumptions it becomes: Subscript r for refrigerant and w for water For refrigerant R-134a: At inlet: The flow rate, = 0.2 kg/s P = 1.0 Mpa and T = 60 0 C but at P = 1.0 Mpa, T sat = C Since T > T sat, the refrigerant is in superheated region. So, For Water: At inlet T = 10 0 C, for saturated liquid water, T = 20 0 C, for saturated liquid water, At outlet : P = 0.95 MPa and T = 35 0 C But at T = 35 0 C, P sat = Mpa. Since P > P sat, refrigerant is in compressed liquid region.

56

57

58

59 Nozzle and Diffuser A nozzle is a steady-state device whose purpose is to create a high-velocity fluid stream at the expense of the fluid s pressure. It is contoured in an appropriate manner to expand a flowing fluid smoothly to a lower pressure, thereby increasing its velocity. The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers. A steady-state diffuser is a device constructed to decelerate a high-velocity fluid in a manner that results in an increase in pressure of the fluid.

60 In essence, diffuser is the exact opposite of a nozzle, and it may be thought of as a fluid flowing in the opposite direction through a nozzle, with the opposite effects. Assumptions There is no means to do any work there are no moving parts. There is little or no change in potential energy, Δ PE = 0. The rate of heat transfer between the fluid flowing through a nozzle or a diffuser and the surroundings is usually very small (Q= 0) (usually insulated or since the fluid has high velocities, and thus it does not spend enough time in the device for any significant heat transfer to take place.) As a fluid passes through a nozzle or diffuser, it experiences large changes in its velocity. Therefore, the kinetic energy changes must be accounted for in analyzing the flow through these devices (ΔKE 0). SSSF equation : = 0

61 Problem Steam at 0.6 MPa and C enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a velocity of 600 m/s. Determine the final temperature if the steam is superheated in the final state and the quality if it is saturated. For the nozzle the SSSF equation is:

62 At inlet: P = 0.6 MPa, T = C and V i = 50 m/s But at P = 0.6 MPa, T sat = C Since T >T sat, so the steam is in superheated region. From superheated Table: h i = kj/kg At outlet: P = 0.15 MPa and V e = 600 m/s

63 From the saturated liquid-satured vapor table: at P = 0.15 MPa, we see that h f < h e (= kj/kg) < h g So, the fluid at the exit of the nozzle is in the saturated liquid-satured vapor mixture region. The exit temperature, T e = T sat (at 0.15 MPa) = C and the quality is

64 Throttlin Process A throttling process occurs when a significant reduction in pressure can be achieved for a fluid flowing in a line suddenly encounters a restriction in the flow passage. This may be a plate with a small hole in it, or may be a partially closed valve protruding into the flow passage, or it may be a change to a tube of much smaller diameter, called a capillary tube, which is normally found on a refrigerator. The result of this restriction is an abrupt pressure drop in the fluid, as it is forced to find its way through a suddenly smaller passageway. This process is drastically different from the smoothly contoured nozzle expansion and area change, which results in a significant velocity increase.

65 When the flow through the valve or other restriction is idealized in this way, the process is called a throttling process. Throttling also plays a role in the Joule Thomson expansion. Assumptions There is typically some increase in velocity in a throttle, but both inlet and exit kinetic energies are usually small enough to be neglected. Δ KE = 0, i.e. KE i KE e There is no means for doing work, = 0 There is little or no change in potential energy.. Usually, there is neither time nor opportunity for appreciable heat transfer, = 0 So the only terms left in SSSF Eq. are the inlet and exit enthalpies. We conclude that a steady-state throttling process is approximately a pressure drop at constant enthalpy. Frequently, a throttling process involves a change in the phase of the fluid.

66 SSSF equation :

67 Turbine A turbine is a rotary steady-state machine whose purpose is to produce shaft work (power, on a rate basis) at the expense of the pressure of the working fluid. Two general classes of turbines are steam (or other working fluid) turbines, in which the steam exiting the turbine passes to a condenser, where it is condensed to liquid, and gas turbines, in which the gas usually exhausts to the atmosphere from the turbine. In either type, the turbine exit pressure is fixed by the environment into which the working fluid exhausts, and the turbine inlet pressure has been reached by previously pumping or compressing the working fluid in another process.

68 Assumptions Usually, changes in potential energy are negligible, (ΔPE 0) as is the inlet kinetic energy. Often, the exit kinetic energy is neglected, (ΔKE 0). Any heat rejection from the turbine is undesirable and is commonly small. It is normally assumed that a turbine process is adiabatic, Heat transfer from turbines is usually negligible (Q= 0) since they are typically well insulated. The work output in this case reduces to the decrease in enthalpy from the inlet to exit states. SSSF equation : The SSSF equation becomes

69 Compressor and Pump The purpose of a steady-state compressor (gas) or pump (liquid) is the same: to increase the pressure of a fluid by putting in shaft work (power, on a rate basis). The internal processes are essentially the opposite of the two processes occurring inside a turbine. The working fluid enters the compressor at low pressure, moving into a set of rotating blades, from which it exits at high velocity, a result of the shaft work input to the fluid. The fluid then passes through a diffuser section, in which it is decelerated in a manner that results in a pressure increase. The fluid then exits the compressor at high pressure.

70 Assumptions Usually, changes in potential energy are negligible, (ΔPE 0) as is the inlet kinetic energy. Often, the exit kinetic energy is neglected, (ΔKE 0). Heat rejection from the working fluid during compression would be desirable, but it is usually small in a rotary compressor, There is not sufficient time to transfer much heat from the working fluid. It is normally assumed that a rotary compressure is adiabatic, Heat transfer from compressure usually negligible (Q= 0) since they are typically well insulated. The work output in this case reduces to the change in enthalpy from the inlet to exit states. In a piston/cylinder-type compressor, the cylinder usually contains fins to promote heat rejection during compression (or the cylinder may be water-jacketed in a large compressor for even greater cooling rates). In this type of compressor, the heat transfer from the working fluid is significant and is not neglected in the first law.

71

72

73 Power Plant and Refrigerator Problem Consider the simple steam power plant, as shown in the Fig. The following data are for such a power plant. Determine the following quantities per kilogram flowing through the unit: a. Heat transfer in the line between the boiler and turbine. b. Turbine work. c. Heat transfer in the condenser. d. Heat transfer in the boiler.

74 Assumptions All processes are in Steady-state steady-flow condition. No changes in kinetic or potential energy is considered. So (ΔPE 0) and (ΔKE 0)..

75 Solution At point 1: P 1 = 2.0 MPa and T 1 = C From steam table: At P 1 = 2.0 MPa, T sat = C Since T 1 > T sat, so steam is in superheated region. From superheated steam table h 1 = kj/kg a. For the control volume for the pipeline between the boiler and the turbine, the first law and solution are At point 2: P 2 = 1.9 MPa and T 2 = C From steam table: At P 2 = 1.9 MPa, T sat = C (by interpolation) Since T 2 > T sat, so steam is in superheated region. From superheated steam table For P = 2.0 Mpa and T = C, h / = kj/kg For P = 1.8 Mpa and T = C, h // = kj/kg So for P = 1.9 Mpa and T = C, h* = kj/kg In similar way, for P = 1.9 Mpa and T = C h** = C So P 2 = 1.9 MPa and T 2 = C h 2 = kj/kg (inpolating bet n h* and h** for C)

76

77 At point 3: P 3 = 15 kpa and quality, x = 90% So steam is in saturated liquid-saturated vapor region. From steam table: h f = kj/kg and h fg = kj/kg b. A turbine is essentially an adiabatic machine. Therefore, it is reasonable to neglect heat transfer in the first law, so that h 3 = h f + x h fg = (0.9 * ) h 3 = kj/kg

78 At point 4: P 4 = 14 kpa and T 4 = 45 0 C From steam table: At T 4 = C, P sat = kpa (by interpolation) Since P 4 > P sat, so steam is in compressed liquid region. (for 14 kpa subcooled water table is not given, so we can assume the saturated liquid as the compressed liquid) h 4 = kj/kg c. There is no work for the control volume enclosing the condenser. Therefore, the first law and solution are

79 Taking a control volume around the pump: d. If we consider a control volume enclosing the boiler, the work is equal to zero, so that the first law becomes