C H A P T E R ,1752'8&7,21

Transcription

1 CHAPTER The first law of thermodynamics is a secial case of the fundamental and general law of conseration of energy, which is alied to thermal henomena in thermodynamic system. The basic law of energy conseration which was discoered by M.V. Lomonso states that energy can neither be created nor destroyed, if mass is consered. Howeer, it can be conerted from one form to another form. Where, the sum of all kinds of energy in the isolated system (e.g., unierse) will remain constant. Whereas, the first law includes systems in which there is a flow of work and heat energy only. The conersion of mechanical work into heat has been known to man since long, but conersion of heat into mechanical work was realized only when in 766 a recirocating engine was created by Russian inentor I.I. Polzuno. Although, the first attemt to conert heat into mechanical work had been undertaken in first century B.C., when Heron of Alexandria inented a ball that rotated under the action of reactie forces created by water aours (steam) that escaed from the nozzle laced on the closed cylindrical shell, when water was heated in the shell. Regarding flow of heat there were two rial theories of heat caloric theory and molecular theory. According to caloric theory, heat is an indestructible fluid which sreads through the matter and flows from hot body to cold body whereas molecular theory beliees that the raid ibrations of the molecules of a matter are resonsible for the flow of heat energy. The caloric theory found greater suort until the middle of the nineteenth century. Howeer, most significant exeriments based on the caloric theory were done much earlier. Such one obseration was made by Black in 76. According to his obserations two ails each containing ice cold water and ice, were laced in a warm room, the temerature of a ail of ice cold water rose quite quickly, whereas, the temerature of the ail containing ice remained constant for longer duration, till ice melted. On the basis of his obseration he argued that ice-cold water contains more heat than water. Such concets and theories continued to reail until 840s, when Joule in his work ut molecular theory on sound basis by demonstrating quantitatie equialence of work and heat. Further, in 84 J.R. Mayer established direct relationshi between work and heat.

2 discussed that Q and W are ath deendent terms. Therefore, a quantity (Q W) is a differential term and in the first law of thermodynamics it is reresented by a new term de Therefore, Q W = de (.4) The aboe exression reresents the first law of thermodynamics for a close system. Where Q and W reresent the transfer of amount of heat and work energy, resectiely and E reresents the total energy of the system during the rocess. The symbol d before the term E in Eq. (.4), has exact differential and hence it is a oint function which means the term E, total energy, can be calculated between two end states. For any gien finite rocess integrate the term in Eq. (.4) Q W = de Q = W = E E The term E is a stored energy within the system. To understand this new term, let us consider a iston cylinder assembly with stoers and filled with a comressible fluid (Fig..). Initially, fluid is at temerature T and on adding heat, it attains temerature T. Mathematically, exression for this is Q W = E E (.5) Comressible fluid T, V, P V, T, P Q heat added (a) (b) In thermodynamics work and heat are two forms of transient energy, therefore according to the first law heat and work are mutually conertible; since energy can neither be created nor destroyed, the energy of the system disaears in one form, simultaneously it aears in other form but the total energy of the system associated with energy conersion will remain same. The total energy change E of the system is the summation of seeral forms of energy like kinetic energy KE, otential energy, PE, internal energy, U, magnetic energy, ME, surface tension, SE, etc. E = KE + PE + U + ME + SE +... (.6)

3 In closed thermodynamic system the energies (heat and work) are transferred across the boundary and the system as a whole is considered to be stationary. In such cases the KE and PE of the system do not change and other forms of energies like ME and SE did not hae considerable imact and neglected. Thus, aboe Eq. (.6) for the first law can be written as Q W = U (.7) The aboe Eq. (.7) is known as the non-flow energy equation as it does not consider the flow of mass. It is an imaginary deice that roduces a work without absorbing energy from the surrounding or the system. Such deice iolates the first law of thermodynamics by roducing work from nothing is called eretual motion machine of first kind (PMMFK). This different form of energy contained in a body or system of bodies is called internal energy. This energy can be reresented as the sum of seeral forms of energy like the kinetic energy of molecules, comrising the energy of their translational and rotational motion and also of the oscillatory motion of atoms. Other forms of energy include the the energy stored in a battery, i.e., electron energy, nuclear energy, the otential energy or the energy of molecules laced in some external fields. The internal energy of a body can be exressed as, U = U kin + U ot + U o (.8) where U kin and U ot reresent the kinetic and otential energies of molecules. U o is integration constant and reresents internal energy at the absolute zero temerature. At T = 0 the internal motion of molecules and atoms ceases, but not the motion of articles (electron, roton) inside the atoms. The motion of electrons can takes lace at any temerature, including absolute zero is therefore cannot be considered thermal energy. Therefore, in thermodynamic analysis U o is usually assumed zero. For ideal gas the internal energy is summation of kinetic energy and otential energy and it is a function of state only, indeendent to ath of rocess. Also the internal energy of an ideal gas, in which there are no forces of molecular interaction, is indeendent of olume or ressure, and deends only on temerature, exressed as, U = f (T ) f (T ) (.9) Thus, according to the first law of thermodynamics, which is always associated with the system in definite amount with a gien change of state is brought about by the erformance of work alone. Therefore, the work done on the systems is simly the change in the internal energy in going from the initial to the final state, U = W (.0) where, U is a function of state because W is indeendent of ath.

4 It is eident from Eq. (.0) that in a thermally isolated system it is ossible to change the state of a system with doing work on it. But when a system is not thermally isolated Eq. (.0) is no longer alid, it must be modified to, U = Q + W (.) where Q reresents heat, which can be measured to the extent that the change is not adiabatic. In case of adiabatic system Q is considered to be zero. Therefore, in the statement of the first law it is considered that energy is consered if heat is taken into account. As stated earlier, heat is a form of energy which is entirely equialent in its effect on the total energy of the system as communicated by the erformance of some kind of work on the same system. Therefore, sometimes it is difficult to distinguish whether a contribution of articular energy on the system should be categorised as heat or work. Suose an electric heater and fluid which is required to be heated were laced in a close box, in this case, we should say certain amount of electric work is done. On the other hand, instead of utting fluid and electric heated in a close box, we attach heater from outside and choose not to make the heater art of the system. In such case, we should instinctiely hae considered the energy to be sulied as heat. In both the cases energy was communicated but in former case, it is considered to be work in the latter by heat. Thus, the most conenient way to make the distinction between the kind of energy enters into the system is a macroscoic or microscoic way. Suose, a iston moes in a cylinder to comress a gas, the iston moement is considered macroscoic in the sense, since the elocity of iston is suerimosed on gas molecules, and iston does work on the gas. On the other hand, if iston is hot, then energy is communicated to gas as heat, without the actual moement of iston, and can be ordered on microscoic scale. It is clear from the aboe discussion that it is always imossible to make a shar distinction between heat and work, therefore, it is stated in the first law that they are equialent, in certain ways. According to first law, it is necessary to define U, wheneer a system undergoes a change. Since U is a function of state, and so to define the sum of Q and W, that how the system asses from initial to final state, U is need to know or define. Thus, for a gien infinitesimal change we write du = dq + dw (.) Equation (.) is known as the differential form of the first law. Here symbol d indicates that the infinitesimal quantities dq and dw are not exact differentials: they cannot be ealuated from a knowledge of the initial and final states alone: du is of course always exacts since U is a function of state. If we consider a number of systems, interacts with each other but isolated from their surroundings, then total internal energy among the grou of systems must be consered, i.e., U = 0. Such a grou of system together form a comosite system, in which neither heat nor work can enters. But in the secial case when the system interact by exchange of heat alone, heat is also said to be consered, we hae U total = Ui ( Q Wi) = 0 (.)

5 While using the first law of thermodynamics in the form of Eq. (.) or (.), it is imortant to be clear about the signs of the terms. If U is the change in internal energy of the system in going from its initial to its final state, then W must be the work done on the system and taken as e work, while Q is the heat transfer to the system and assigned +e sign. Haing adoted these sign conention, it is ossible to define whether body is getting hotter or colder. Consider a comressible fluid (gas) contained in an air tight, frictionless cylinder iston assembly (Fig..a). Let the surface area of the iston be A and ressure exerted by the gas on it is. Then the force, exerted on the iston by the gas is gien F =. A F = A F = A dl dl Solid Gas, Water (a) (b) Now suose the iston is moed to a small distance dl. The work done on the gas is dw = F. dl =. A. dl =. d (.4) where d is the change in olume of the fluid. The sign should be noted. While increasing its olume a gas does work on its surrounding and ositie (+e) work is said to be done. We shall use this result a great deal, since it is customary to follow the sign conention and generally use a fluid system as a conenient model through the deeloment of thermodynamics. In the case of work done on a solid by hydrostatic ressure we immerse the solid in an incomressible fluid (water) and again ut the whole system in a cylinder iston assembly (Fig..b). Since the fluid is incomressible, no work can be done on it by changing the ressure and all the work done by the iston must be communicated to the solid. Thus, for work done on the system by hydrostatic ressure is always gien by Eq..4.

6 But from the definition of the work done on the surface of liquid due to surface tension is dw = da = 8r dr (.7) These two works done will be equal and oosite under equilibrium conditions, i.e.,. 4r dr = 8r dr = r (.8) This is the ressure difference across a single sherical surface and ressure intensity inside the drolet increases with decrease of diameter of the drolet. If we require to use T and V as indeendent ariables. Then we must exress du in terms dt and dv as U du = U dt T V dv (.9) V T We know, dq = U + dv (.0) Substituting, U dq = U dt T dv (.) V V T If we assume that V is constant then secific heat at constant olume in its simle form can be written as, U C V = T (.) V We may obtain similar exression if we choose and T as our indeendent ariables to substitute in Eq..0 to obtain du and dv in terms of d and dt: and Then dq = U U du = V dv = V T T U d T V d T U d T dt dt V T T T dt

7 In these cases two heat caacities are obtained U C T = T V T (.) U V C P = (.4) T T Equation.4 can be written in a further simlified way by constructing an energy function H, and we hae H C P = T (.5) Equating Eqs. (.4) and (.5) H T = U V ( T T dt U V ) Therefore, from the aboe equations H = U + V (.6) H is called enthaly. As all terms U, and V on right hand side of Eq. (.6), defining enthaly, are functions of state, hence, enthaly is also a function or roerty of state. Its differential form gies another mathematical statement of the first law of thermodynamics when ressure and temerature T are chosen as indeendent arameters: dh = du + dv + Vd = dq + Vd (.7) We see from Eqs. (.5) and (.6) that when the system undergoes a reersible isobaric change, all the heat inoled in the rocess is exended to change the enthaly. Hence, Eq. (.7) will reduce to From Eq. (.7) enthaly H, is dh = d Q (.8) dh = H H = Q Vd (.9) Sometimes enthaly H has been gien misleading name of heat content. But it is obious from Eq. (.9) that the enthaly is greater than external heat added to the system by the amount of work, Vd, which is reresented on a -V diagram by the elementary area abcd as shown in Fig..5. A change in enthaly is fully determined by initial and final states of working medium and is indeendent of the intermediate states. In a cyclic rocess the change in enthaly is zero. dh = 0

8 Therefore, like internal energy, enthaly of an ideal gas is also a function of temerature. As described in Section.5 wheneer one or more roerties (ressure, olume, temerature, etc.) of a system change, we may say that a change in state has occurred. That successie ath through which the system asses from one state to another is called the rocess. Any change in the roerty or roerties of the system during the rocess results in the flow of energy across the boundary of the system. On the basis of these facts the thermodynamic rocess may be classified as Constant olume (isochoric) Constant ressure (isobaric) Constant temerature (isothermal) Reersible adiabatic (isentroic) Polytroic Throttling or restricted flow A constant olume rocess for a erfect gas is shown in a P -V diagram in Fig..7. In this case when a gas is heated in a closed essel, it does not undergo change in olume but the rocess results in an increase of ressure. We know that the general exression for first law of thermodynamics is gien in Eq. (.). Ealuating each term of the equation, we get: Work. Since there is no change in olume, i.e., d = 0 and area under cure is zero, therefore, P P W =. d = 0 (.) V= V V Internal Energy. For a erfect gas in constant olume rocess the internal energy of the system is U = U U = mc (T T ) (.) Heat Transferred. From Eqs. (.), (.) and (.), we get Q = W + U = 0 + mc (T T ) (.) It is, therefore, concluded that in this rocess all the heat sulied during the rocess is stored in the system to raise its internal energy. Relation between P, V and T:

9 From characteristic gas Eq. (.) P V T = P V T Since the rocess is isochoric, i.e., V = V, then Eq. (.4) can be modified to (.4) P P = T T (.5) A gas enclosed inside the cylinder beneath a sliding iston is heated. The heating of the gas will take lace at constant ressure, because the force exerted by the exansion of gases due to heat, make the iston to moe through a certain distance. Then work is said to be done by the gas on its surroundings. The rocess is reresented on the diagram by a horizontal line as shown in Fig..8. Work. If the state oint moes from state to, it reresents exansion rocess and if end state oint moes in the reerse direction, from to, it is said to be comression rocess. From Eq..4 w = d for reersible rocess Since is constant, therefore w = d ( ) (.6) From characteristic gas equation V = mrt for constant ressure Eq. (.6) can be written as w = mr (T T ) (.7) Internal Energy. For an ideal gas in the constant ressure rocess the equation for internal energy can be written as U = mc (T T ) (.8) Heat Transfer. From the non-flow energy Eq. (.), we can write Q = (u u ) + w Q = mc (T T ) + mr (T T ) Q = m (T T ) (C + R) Q = mc (T T ) [(C + R) = C ] (.9) = W.D

10 i.e., w = log e (.4) Also, since = then Hence, substituting in Eq. (.4) = Using Eq. (.), we can write w = log e = RT (.4) Hence, substituting in Eq. (.4) or for mass m, of the gas w = RT log w = mrt log (.4) Internal Energy. Since the temerature is constant during the rocess, therefore the internal energy remain constant, hence, u = u u = 0 Heat Transfer. Heat transfer during the rocess can be deried from the non-flow energy Eq. (.). Q = u u + w Q = w since u = u Therefore, it is concluded that in an isothermal rocess the amount of heat added to the system is conerted into work and ice ersa. An adiabatic rocess is one in which no heat is transferred to or from the system (gas) during the rocess. Such a rocess may be reersible or irreersible. In this section it is considered that the rocess is reersible. From non-flow energy Eq. (.), for an adiabatic rocess, we can write w = u u Since Q = 0 The aboe equation is true for an adiabatic rocess irresectie of the fact whether the rocess is reersible or irreersible. During adiabatic exansion, the work done by the gas is at the exense of reduction of internal energy of the gas. On the other hand,

11 during adiabatic comression all the work done on the system (gas) results into increase of internal energy. Since there is no transfer of heat energy, hence whole of the work contributed to raise the alue of internal energy. Goerning equation for a reersible adiabatic rocess: From non-flow energy Eq. (.), Q = du + w For a reersible rocess, dw =. d, hence, dq = du +. d = 0 (.44) Since du = C dt Eq. (.44) can be exressed as, du +. d = 0 C dt +. d = 0 Since = RT or = RT C dt + RT d = 0 Diiding aboe equation by T, we get C dt R d = 0 (.45) T Integrated the Eq. (.45), we get C log e T + R log e = const Using characteristic gas equation we hae T = /R C log e R + R log e = constant Diiding through by C, we get Also from Eq. (.) log e R + R C log e = constant (.46) R C = R = C Hence, substituting the alue of R C in Eq. (.46),

12 W = (.49) Since = mrt, the exression of W (work done) in Eq. (.49) can be written as W = mrt ( T ) (.50) Internal Energy. In this rocess change in internal energy is equal to the work done. Since rocess is adiabatic, the amount of work done on the system contributes to raise its internal energy or work is done by the system at the exense of internal energy. Therefore, gain in internal energy is, u u = mc (T T ) (.5) Heat Transfer. For an adiabatic rocess, Q = 0 Relation between, and T Using Eq. = RT, relationshi between T & and T &, may be deried as = RT = RT Substituting alue of in Eq. (.47), we get RT = constant T = constant (.5) Also, = (RT)/; hence substituting in Eq. (.47), we get RT = constant T T ( ) / = constant = constant (.5) Therefore, for a erfect gas for a reersible adiabatic rocess, we can write From Eq. (.48) = or (.54) From Eq. (.5) T = T or T T (.55)

13 From Eq. (.5) T ( )/ = T ( )/ or T T = / (.56) From Eqs. (.55) and (.56), we hae T T = / The olytroic rocess is a rocess that can be described by the general equation n = constant, where n is a constant haing any alue between 0 and. Many actual rocesses hae exansion and comression cure defined by the relation n = constant. From the definition of olytroic rocess it is concluded that the basic thermodynamic rocesses, the isochoric, isobaric, isothermal and adiabatic are secial cases of the olytroic rocess. As examle, for a chosen alue of n, the general equation n = C will change as, alue of index n Equation Process when n = 0 0 = C or = C Isobaric when n = = C or / = C i.e., = C Isochoric when n = = C Isothermal when n = = C Adiabatic For the gien alue of n the aboe exression is illustrated in Fig... C n = D D n = n = A n = 0 n = A n = D D C Comression Exansion Work. The olytroic rocess of an ideal gas is mathematically similar to the adiabatic rocess excet that the index n is used in lace of index. Therefore, Eq. (.49) can be used to calculate the work done by substituting n for. Thus,

14 Note that If n = then n 0, so Q = 0, i.e., this is the adiabatic case. If n = then n, so Q = work done this is the isothermal case. We know that Q = ( n ) mrt ( T ) ( ) n Since R = C ( ) Q = n T mc T ( ) ( ) n Q = mc n n (T T ) Q = mc n (T T ) (.6) n where C n = C n C n is called the olytroic secific heat caacity. EXAMPLE. A system undergoes the cyclic rocess 4. The alue of Q, W and U for indiidual rocess are gien below: Process U (kj) Q (kj) W (kj) Comlete the aboe table. The system undergoing a cyclic rocess 4 is shown in Fig. Ex.. For each rocess alying the first law of thermodynamic: u = dq dw Process: 60 = dq 5 dq = 5 kj Process: u = du = 6 kj 4

15 Process: 4 65 = 55 dw dw = 0 kj Process: 4 Since it is a cyclic rocess, therefore, sum of internal energy for all the rocesses will be zero. U + U + U 4 + U 4 = U 4 = 0 U 4 = kj Alying first law of thermodynamics U 4 = dq dw = dq 88 dq = 99 kj EXAMPLE. One kg of gas at 50 C is heated at constant olume until the ressure is doubled. Determine (a) the final temerature, (b) the change in internal energy, and (c) the change in enthaly. Take C =.005 kj/kg-k and =.69 for the gas. Gien data: m = kg T = 50 C, K = = =.69, C =.005 kj/kg-k For constant olume rocess, = = T T Final temerature, T = T = K = 646 K Ans. Internal energy, u = mc (T T ) =.005 kj/kg K (646 ) K ( kg) Enthaly, = 4.6 kj Ans. C H = mc (T T ) =.69 C =.698 kj/kg K (646 ) K ( kg) C =.698 = kj Ans. EXAMPLE. One kg of air at 50 kpa and 0 C undergoes a constant ressure rocess until olume is triled. Determine the change in internal energy and enthaly. Take C =.005 kj/kg K and C = kj/kg K for air.

16 Change in internal energy, U U = mc (T T ) U = (56 04) = 06.7 kj Work done by the gas, W = d = P ( ) or Heat transfer, W = ( ) W = J.96 0 J Since the rocess is isobaric, hence, Q = H Q = mc (T T ) Q =.5.64 (56 04) Q = 4. kj EXAMPLE.5 Four kg of nitrogen is exanded from initial state = 76 kpa and T = 50 C to the final ressure = 96 kpa. Find, the work done, the change in internal energy and the heat transfer if the exansion is isothermal, olytroic (n =.), and isentroic. Gien data: m = 4 kg = 76 kpa T = 50 C and = 96 kpa For Isothermal Process: T = T = 50 C = = 5 K Alying characteristic equation for initial state of gas, = mrt = m 76 Isothermal Polytroic Isentroic For isothermal rocess, = Work done by gas, = =.6 m W = log e = 96 = log e = 0.9 kj

17 Since there is no change in temerature, change in internal energy is zero. Here the amount of work done by the gas has been done at the cost of heat. Heat gien by gas Q = W Polytroic Process: Considering olytroic rocess n n Q = 0.9 kj. T = T = 5 5 = 78.6 K 76 0 Work done by gas, W = = mrt ( T ) n n ( 78. 6) W = (. ) W = kj Change in internal energy, U U = mc (T T ) U = (78.6 5) U = 48.0 kj Heat transfer from the gas, Q = W. D n (. 4. ) = kj (. 4 ) Isentroic Process: Considering isentroic rocess, where C = (C R) C = (.08.97) C = 0.74 kj/kg K T = T =.4 K 76 0 Work done by gas, W = mrt ( T ) (. 4) = = 6.5 kj (. 4) Change in internal energy, U U = mc (T T ) = (.4 5) = 6.5 kj Since the rocess is an isentroic (reersible adiabatic) there is no transfer of heat.

18 T = 00 K = 4 Initial olume of the air is determined by characteristic gas equation, = mrt ( 0kg) ( kj/kg K) ( 00 K) =.45 m Nm / Final olume, = 4 = 4.45 m = 5.74 m Final ressure, Work done, = = (600 kpa) (. 45 m ) ( 574. m ) = 50 kpa W = log e = (600 0 N/m ) (.45 m ) log e (4) = J = 9.6 kj Ans. In isothermal rocess, amount of work done is equal to the heat added. Q = 9.6 kj Ans. EXAMPLE.8 The ressure of 0.5 m of air increases from 500 kpa to.5 MPa while undergoing isothermal comression. Determine the alue of the transferred heat. Gien data: = 0.5 m Final olume, = 500 kpa, N/m =.5 MPa, N/m = = 0.05 m = 500 kpa 0.5 m 500 kpa In case of isothermal rocess work done is equal to heat transferred, WD = Q = log e ( / ) Q = (500 0 N/m ) (0.5 m 005. m ) loge 0. 5 m Q = J = 8.4 kj Ans. EXAMPLE.9 One kg of an ideal gas at 0 C and 00 kpa is allowed to exand isothermally till the ressure reduces to one half of the original ressure followed by a constant olume cooling till its ressure reduces to one fourth of the original alue. Then it is resorted to the initial state by a reersible adiabatic rocess. Calculate the net work done by the gas. Take R = 0.94 kj/kg K.

19 Gien data: = 00 kpa T = = 0 K = / = 50 kpa = /4 = 5 kpa Isothermal rocess, W = RT log e / W = log 00 e 50 W = 6.74 kj Constant olume rocess During this rocess W = 0, because change in olume does not takes lace. Temeratures at state = T T T = T T 0 = 5.5 K Adiabatic ressure / /4 = C = C = const W = mr ( T T ) = (. 4) (. ) W =.5 kj Net WD = W + W W = = 49.6 kj EXAMPLE.0 Half kg of an ideal gas (C = 0.74 kj/kg K) at 00 K and 00 kpa is contained in an insulated and rigid cylinder. A addle wheel is inserted into the gas sace and rotated by an electric motor of 50 W caacity for minutes. Estimate the final temerature and ressure of the gas. Gien data: m = 0.5 kg, T = 00 K, = 00 kpa Time = minutes Work done by a addle wheel in minutes, W addle = 50 W = 50 J/s Time W addle = 50 0 = 6000 J = 6 kj Final temerature Since the cylinder is rigid and insulated, whole of the work done by addle wheel is conerted into raising the internal energy of the gas.

20 Heat rejected Q = mc n T T n. 4. ( ) ( kg) ( 0. 7 kj/ kg K). ( ) K Q = 9. kj Ans. EXAMPLE. Two kg of oxygen is comressed olytroically from a ressure of 48 kpa and 7 C to 740 kpa. If the comression is according to the law,. = constant, find (a) the final temerature and olume of the gas, (b) the amount of work done, (c) the change in internal energy, and (d) the heat transferred. Take R = 0.6 kj/kg K and C = 0.9 kj/kg K for oxygen. Gien data: m = kg = 48 kpa = 740 kpa T = 7 C, 90 K n =. C = (C R) = ( ) = 0.65 kj/kg K Initial olume = mrt ( kg) ( 0. 6 kj/ kg K) ( 90 K) =.08 m ( 48 kpa) Final olume = Final temerature n 48 kpa 740 kpa. (.08 m ) = 0.95 m n. n. T = T 740 kpa 90 ( K) 48 kpa = 40.4 K Work done w = ( 48 0 Nm / ) (. 08 m) ( Nm / ) ( m ) n (. ) = 0 J =. kj Change in internal energy U = mc (T T ) = kg (0.65 kj/kg K) (40.4 K 90 K) = 70. kj Heat transferred Q = n WD Q = 4... kj = 58. kj 4.

21 EXAMPLE. A gas of mass kg haing a ressure and temerature, 4 MPa and 75 C is exanded to four times the original olume according to the law V. = C. Determine: (a) the initial and final olume of the gas; (b) the final ressure of the gas; and (c) the final temerature of the gas. Take R = 0.87 kj/kg K. Gien data: m = kg = 4 MPa T = 75 C or 548 K R = 0.87 kj/kg K, = 4 V = mrt = mrt 4 mpa = = m 4 0 ( kpa) = 4 = = 0.0 m n n = n = = 4 0 kpa 4. =. 0 kpa = T T T =. 0 4 T = 548 K = 60. K 4 0 kpa T = = 87. C EXAMPLE.4 A gas of 0.7 kg occuies olume 0.5 m at a ressure of 40 kpa and comressed to a ressure.4 MPa according to the law, V.5 = C. Determine: (a) the change of internal energy of the air; (b) the work done on or by the air; and (c) the heat receied or rejected by the air. Take C =.005 kj/kg K, C = 0.78 kj/kg K Gien data: m = 0.7 kg = 0.5 m = 40 kpa, = 400 kpa (.4 MPa) R = C C R = = 0.87 kj/kg K

22 During isothermal rocess = =, = 400 kpa 0 kpa = 6.6 For adiabatic rocess = P P = Since = 400 kpa 0 kpa = (6.6) = =.4 C Also, = C or C = C =. 005 kj/ kg K = kj/kg K 4. Temerature at end state. For constant olume rocess, = T T T = T = 0 kpa 60 K = 89. K 0 kpa T = 89. K or 6. C Change in internal energy, U = mc (T T ) = 0. kg kj/kg K (89. K 60 K) = 55.4 kj (loss of internal energy) Isothermal adiabatic EXAMPLE.6 A gas cylinder has a caacity of 90 litres and contains gas at a ressure of.0 MPa at temerature 90 K. The sto ale is oen and some gas is used. If the temerature and ressure of the gas falls to.6 MPa and 87 K, resectiely after the release of gas. Determine the mass of gas used. If after the sto ale closed the oxygen remaining in the cylinder gradually attains its initial temerature of 90 K, determine the amount of heat transferred through the cylinder wall from the atmoshere. The density of gas at 0 C and 0. kpa may be taken as.4 kg/m and =.4.

23 Gien data: Cylinder caacity, = 00 liters or 00 0 m =.0 MPa or 0 kpa T = 90 K After release of sto ale, =.6 MPa or.6 0 kpa T = 87 K at ambient condition, = 0. kpa T = 7 K density, =.4 kg/m At ambient condition alue of R will be, PV = mrt R = mt = 0. kpa = 0.6 kj/kg K 4. kg/ m 7 K Initially, mass of gas inside the cylinder, m = Nm / 00 0 m RT Jkg / K 90 K m =.9 kg Final mass of gas after release of sto ale, m = Nm / 00 0 m RT Jkg / K 87 K m = 6.4 kg Mass of gas used = m m = = 5.5 kg Heat transfer during the rocess: Aboe is the case of non-flow rocess, therefore, Q = U + W W = 0, because olume of gas remains same, hence no work is done. Q = U = m C (T T ) Q = m R ( ) (T T ) Since C = Q = 6.4 kg 06. kj/ kg K (90 K 87 K) (. 4) Q =.5 kj R ( ) Ans.

24 (a) the temerature and ressure of the air at the end of the comression; (b) the temerature and ressure of the air at the end of the constant olume rocess; and (c) the heat transfer required to carry out constant olume rocess. For the air, take C =.005 kj/kg K, R = 0.4 kj/kg K. Gien data: s = 5 liter or 5 0 m (comression ratio) r = 4 : = = 95 kn/m T = 0 C or 0 K =.6 : PV.4 = C n =.4.4 =.4 = = 95 kn/m 4 = 6. kn/m Temerature at the end of comression, T = T T = T = 0 (4) (.4 ) = 74. K or 470. C Pressure and temerature at the end of constant olume rocess from state, = : 6 : =.6 = 6. kn/m.6 = kn/m T = T T = T = K = 89. K or 96. C Mass of air entered into the cylinder, = mrt m = RT 95kN/ m 50 m 04. kj/ kg K 0 K

25 m = kg Also C = C R = = kj/kg K Heat transfer carrying the constant olume rocess Q = mc (T T ) = kg kj/kg K (89. K 74. K) Q = 6.65 kj EXAMPLE.9 The cylinder of an engine has a stroke of 00 mm and a bore of 50 mm. The olume ratio of comression is 4 :. Air in the cylinder at the beginning of comression has a ressure of 96 kn/m and temerature of 9 C. The air is comressed for the full stroke according to the law PV. = C. Determine: (a) the mass of air, (b) the work transfer, and (c) the heat transfer. For air, take =.4, C =.006 kj/kg K. Gien Data: Stroke, l = 00 mm, bore, d = 50 mm comression ratio r = = 96 kn/m, T = 9 C or 66 K = 4 : Gas law, PV. = C = d l (. ) (. ) 4 m = m Volume at the and of comression, = =.05 0 m r 4 Pressure at the end of comression, =. = c 96 kn/m = 96 kn/m (4) (. ) =.88 kn/m Temerature at the end of comression, T n = T n T = T n = 66 K (4) 0. = K

26 Mass of air inside the cylinder, Since = mrt m = RT 96 kn/ m m 087. kj/ kg K 66 K m = 0.04 kg Work transfer during the rocess, W = mrt ( T ) n 004. kg 087. kj/ kg K (66 K K) W = (. ) W = 5.66 kj (work is done on the system) Heat transfer during the rocess, Q = n n W.D R = C C = C C = C 4. = R = 0.87 kj/kg K 4.. Q = ( 5.66) kj. Q =.88 kj (heat rejected by the system) EXAMPLE.0 An air main is connected to a cylinder through a ale. A iston slides in the cylinder. The air in the main is maintained at a constant ressure and temerature of MN/m and 400 C resectiely. The initial ressure and olume of air in the cylinder are 40 kn/m and litres, resectiely. The ale is oened, 0. kg of air enters the cylinder then the ale is closed. As a result of this mass transfer, the ressure in the cylinder becomes 700 kn/m and the olume becomes 5 litres. Assuming the rocess to be adiabatic, determine the work done on the iston. For air, take c =.006 kj/kg K, c = 0.77 kj/kg K. Assume M 0 = initial mass of air in the cylinder M f = final mass of air in the cylinder P = Pressure of air in main line = s. olume of air in main line u = s. internal energy of air in main line Q = 0 Main air line Cylinder

27 T f T f K or C At 700 kn/m, 9 C s. olume = 0.99 m /kg Work done, W = (m f m 0 ) = = 5.4 kj EXAMPLE. A ressure essel contains a gas at an initial ressure of.5 MN/m and at a temerature of 60 C. It is connected through a ale to a ertical cylinder in which there is a iston. Initially, there is no gas under the iston. The ale is oened, gas enters the ertical cylinder, and work is done in moing the iston. The ale is closed and the ressure and temerature of the remaining gas in the essel are.7 MN/ m and 5 C, resectiely. Determine the temerature of the gas in the ertical cylinder if the rocess is assumed to be adiabatic. Take =.4. Pressure essel =.5 MN/m T = 60 C Volume = V Mass = m Piston Cylinder Initial condition =.7 MN/m T = 5 C Volume = Mass = m m Volume = Mass = m Press. = Tem. = T Cylinder Piston Final condition

28 At initial condition in ressure essel, P V = mrt R = P V mt At final condition in ressure essel, Equating Eqs. (i) and (ii), 6 P V = (m m ) RT P V mt. 50 N/m V m K Since V = V = cylinder olume R = = PV ( m m ) T PV ( m m ) T 6 = 7. 0 N/m V ( m m ) 98 K.5 98 (m m ) = (.7 ) m (i) (ii) [(.5 98) (.7 )] m = (.5 98) m m = 04 m m m =.87 Assuming the rocess is non-flow, aly the equation of the first law, Q = U + W Q = 0, since the rocess is adiabatic W + U = 0 (iii) W = U = (U U ) = (Internal energy before the ale is oened Internal energy after ale is oened) Internal energy before ale is oened, U = mc T (i) Internal energy after ale is oened, U = Internal energy of gas in ressure essel + Internal energy of gas in cylinder U = (m m ) C T + m C T () Work required to moe the iston, W = = m RT = m (C C ) T = mc ( )T (i) From Eqs. (iii), (i), (), and (i), we hae, m C ( ) T = mc T [(m m )C T + m C T ]

29 during comression the ressure of the gas at the gien olume, V will not be equal to that of what it was during exansion rocess. Only reersible rocesses can be shown on P-V diagrams, since on these diagrams each oint reresents an equilibrium state of the gas. Steady state flow rocesses are of rimary imortance in engineering because mass roduction of materials and energy flow demand continuous oeration of rocesses. A flow rocess in thermodynamics constitutes an oen system through which working medium (gas) is allowed to flow in and out, also the interaction of energy (heat, work, internal energy, KE and PE) between system and surroundings. Where steady state imlies that inflow and outflow of mass and energies at any instant of time must be equal, in other words there should be no accumulation of mass and energy in the system. Consider the schematic diagram of Fig.., which shows a region/sace bounded by a dotted line is called control surface and the region thus bounded is also called control olume. During a small time interal, dt the fluid of mass, m enters the control olume at section with aerage elocity. V, ressure P, s. olume s, and internal energy U, whereas at section mass leaes the control olume and the roerties are mentioned by using subscrit instead of. heat added Q h in m,,, U s, Control olume mechanical or shaft work w out m,,,, U s h datum The kinetic energy of mass, m flowing in the control olume is gien by, KE = mv and otential energy can be exressed as, PE = mgh where, h is the height aboe some arbitrary chosen datum. Also, when the fluid flow across the boundary of any chosen system, heat Q and mechanical or shaft work W, may also cross the boundary.

30 Now to write the steady state flow energy equation we may aly the first law in a form already deeloed (Eq..) roided that we add the terms to account for change in kinetic energy, otential energy and flow work which require to ush the fluid in and out of the control olume considering the uniersal definition of the law of conseration of energy that energy can neither be created nor destroyed, it can be transformed from one form to another, the steady flow energy equation, aear as: m u V gh + Q = m u V gh + W s (.6) m H V gh Q = m h V gh W s (.64) Equation.64, for steady flow rocess, for unit mass of gas flowing in and out of system can be written as H + + gh + Q = H + + gh + W s (.65) or Q W = (H H ) + + (h h ) g (.66) In differential form Q W = dh + d( /) + d(gh) (.67) Now, throughout the discussion mechanical work done during steady flow rocess and non-flow rocess can be differentiated by utting suffix sf for steady flow and nf for non-flow rocesses. Therefore, Eq. (.67) can be written as, Q W sf = dh + d( /) + d(gh) (.68) Equation (.) reresents non-flow rocess, considering the effect of motion and graity it can be written as, Q W nf = du + d / + d(gh) (.69) or, Q = du + W nf + d / + d(gh) (.70) Substituting the alue of Q from Eq. (.70) in Eq. (.67), we hae, du + W nf + d / + d(gh) W sf = dh + d / + d(gh) (.7) i.e., du + W nf W sf = dh (.7) W sf = dh du W nf W sf = du + du W nf dh = du + W sf = W nf

31 W sf = W nf ( ) W sf = + W nf (.7) For non-flow rocess, W nf = P d Substituting the alue of W nf in Eq. (.7), we hae W sf = + P d = d (.74) Note that the quantity of work required during the steady flow rocess is gien by the shaded area in Fig..4(a). This area can be obtained by using Eq. (.74). wsf (a) Steady flow work d (b) Non-flow work wnf Pd In Eq. (.74), the term reresents the flow work brought into the control olume and the term shows the work leaing the control olume by irtue of flowing fluid. Where, the term P d reresents the work done within the control olume. The sum of three terms therefore, reresents the work done during steady flow rocess and can be exressed as W sf = d (.75) The steady flow work W sf gien in the aboe equation if alied to a rocess follows the law, P n = C, can be calculated as = C P / n / n / n / n dp = C dp = C P / n / n / n P ( / n ) Using C /n /n /n = P = P W sf = dp n P P n (.76) Place the alue W sf in Eq. (.68) we get, Q n P P dh d dgh n

32 Power outut = W.D/sec = kj/kg 4 kg/s = 0.75 kj/s = 0.75 kw Examle. Lead is extruded slowly through a horizontal die. The ressure difference across the die is MN/m. Assuming there is no cooling through the die, determine the temerature rise of the lead. Assume the lead is incomressible and to hae a density of 60 kg/m and secific heat caacity of 0 J/kg K. Consider the steady flow energy equation (.6) alied to the gien extrusion rocess: u gh + Q = u gh + W (i) Since the die is horizontal, there will be no change in the otential energy, therefore, gh = gh The elocity of lead is also low, and hence kinetic energy can be neglected. There is no cooling, and no external work is done, hence term Q and W can be neglected. Now the energy equation becomes, + u = + u u u = (ii) Rise in temerature of any substance = mass s. heat tem. rise, = mct Here in this articular case lead is incomressible, so energy due to rise in temerature of lead is associated with change in internal energy instead of work of exansion or comression mct = u u Also, lead is incomressible, there is no change in olume, = = Therefore, Eq. (ii) can be written as, mct = ( ) t = 6 m N/ m 60 kg/ m 0 J/ kg K t = K Examle.4 Air enters a gas turbine system with a elocity of 05 m/s and has a secific olume of 0.8 m /kg. The inlet area of the gas turbine system is 0.05 m. At the exit the air has a elocity of 5 m/s and secific olume of.5 m /kg. In its assage through the turbine system, the secific enthaly of the air is reduced by 45 kj/kg and the air also has a heat transfer loss of 7 kj/kg. Determine:

33 (a) the mass flow rate of the air through the turbine system in kg/s; (b) the exit area of the turbine system in m ; and (c) the ower deeloed by the turbine system in kw. Gien data: = 05 m/s = 5 m/s s = 0.8 m /kg s =.5 m /kg A = 0.05 m H H = 45 kj/kg Q = 7 kj/kg Mass of air at inlet, m = A 005. m 05 m/ s s 08. m / kg = 6.56 kg/s Area at outlet of the turbine, Air in m = A s A = m A = 0.07 m Steady flow energy equation, H + + Q = H + + W s kg/ s 5. m / kg 5 m/ s W = (H H ) + Q Turbine Air out 05 5 W = 45 kj/kg + 0 m /s 7 kj/kg W = 4.4 kj/kg Power deeloed = 4.4 kj/kg 6.56 kg/s = kw EXAMPLE.5 Steam at 50 kpa and 47 K enters a nozzle with a elocity of m/s at the rate of kg/s and leaes as dry saturated steam at 0 kpa. The energy loss as heat from the nozzle is 0 kj/s. Calculate the exit elocity of steam. Gien data: = 50 kpa, T = 47 K, = m/s, m = kg/s = 0 kpa, Q = 0 kj/s

34 A comressor is a deice, which takes in ambient air and comresses it to higher ressure. Comressor may or may not be insulated. Here, work is taken negatie, since it is done on the gas to comress it. In this case if comressor is insulated and change in kinetic and otential energies are zero then SFEE can be written as, m (h h ) = W s Here gases are comressed adiabatically, therefore, it can be calculated as, W s = C (T T ) W s = mr (T T ) W s = ( ) (. 4) W s = 96. kw EXAMPLE.7 A comressor is required to roide 500 kpa and 48 K air for a stationary ower lant. The air enters into the comressor at 0 kpa and 9 K. The outlet elocity is 5 m/s. Ealuate the work er unit mass need for an adiabatic comressor. Solution Gien data: = 500 kpa T = 48 K m = unity = 0 kpa T = 95 K = 5 m/s In this roblem initial kinetic energy ( ) is taken zero because in ractice inlet area to the comressor is often larger than the outlet area. Also the comressor is adiabatic, so Q = 0. Thus, Eq. (.6) reduces to m (H + gh ) = m (H + + gh ) + ( W s ) Comressor work is done on the system, therefore, W s is taken negatie. Assuming, h = h, we get mh = mh + W s W s = m(h H ) + W s = mc (T T ) + since (H = C dt) Considering er unit mass flow of air, i.e., m = kg/s

35 W s = [(.005 kj/kg K) (48 95) K] + W s = 89.5 kj/kg 5 kj/ kg 000 EXAMPLE.8 A comressor comresses air of an initial secific olume of 0.89 m / kg at the rate of 5 m /min. During comression the enthaly of air increases by 85.4 kj/kg. The comressor is considered to be water jacketed. While in oeration heat is transferred to the cooling water at the rate of 50 kj/min. If change in kinetic and otential energies are considered negligible, determine the ower required to derie the comressor. Gien data: s. olume of air, s = 0.89 m /kg Cold water in Rate of comression of air = 5 m /min Mass of air, m = 5 = 5.6 kg/min 089. Increase in ethaly of air = 85.4 kj/kg Rate of heat transfer to cooling water, Q = 50 kj/min In this case KE = 0, PE = 0 For this case SFEE (Eq..6) can be written as mh + Q = mh + ( W s ) ( W s ) = m(h H ) + Q = mh + Q W s = (5.6 kg/min) (85.4 kj/kg) + 50 kj/min W s = kj/min or.5 kj/s W s =.5 kw work mass of air in mass of air out Hot water out Ans. EXAMPLE.9 A centrifugal air comressor comresses 0 kg/min of air from 0 kpa, C to a final ressure of 45 kpa. The comression is olytroic with a olytroic index of.5. The olume of air changes from m /kg to 0.85 m /kg. If the suction line diameter is 5 cm and discharge line diameter is 0 cm, and deliery connection is 0 m aboe the inlet, calculate ower outut. Polytroic index, n =.5

36 EXAMPLE.0 A steam turbine, designed to oerate under steady flow condition, receies steam at the rate of.5 kg/s. The steam enters the turbine at a elocity of 50 m/s at an eleation of 6 m and has secific enthaly of 780 kj/kg. The steam leaes the turbine at an eleation of m with 00 m/s and enthaly of outgoing steam is 60 kj/kg. The rate of heat transfer to the surrounding from the turbine is 5 kj/s. Determine the ower outut of the turbine. Gien data: m =.5 kg/s Q = 5 kj/s H = 780 kj/kg = 50 m/s h =6 m Turbine W =? Steady flow energy equation H = 60 kj/kg =00 m/s h = m Q W = m(h H ) + m ( ) + g(h h ). 5kg/ s ( 00m/ s) ( 50m/ s) 5 kj/s W = (.5 kg/s) [(60 kj/kg) (780 kj/kg)] ( m 6m) 9. 8 m/ s +. 5 kg/ s kj/s W =.5 kg/s [ (50 kj/kg) + (.7 kj/kg) (0.049 kj/kg)] 5 kj/s W = ( 56.6 kj/kg).5 kg/s W = 5 kj/s kj/s W = 96.7 kj/s or 96.7 kw Ans. EXAMPLE. A steam turbine is designed to oerate under steady flow condition and receies steam at the rate of.5 kg/s. The steam entering at the rate of 60 m/s has ressure and temerature equal to 000 kpa and 400 C, resectiely. The steam leaes the turbine with a elocity of 50 m/s and is 98% dry, the ressure of the steam at the exit of the turbine is 00 kpa. The difference in eleation between the inlet to the outlet

37 is m and rate of heat loss is 000 kj/min. Calculate: (i) ower outut of the turbine, and (ii) Diameter of inlet and exit ie. Gien data: m =.5 kg/s = 000 kpa = 00 kpa, = 50 m/s T = 400 C h h = m = 60 m/s Dryness fraction, x = 0.98 Q = 000 kj/min, 50 kj/s The steam turbine is a deice in which steam exands, work is roduced on the shaft of the turbine. Unlike the comressor the work done by the turbine is considered to be ositie. In steam turbine, most often eleation (height of inlet and outlet steam ie) is neglected, hence PE = 0. Also the exansion of steam is considered to be adiabatic, thus Q = 0. But in this roblem both eleation and heat transfer are considered. Therefore, in the steady flow energy equation. Equation.6 will be written as Q W = m (H H ) + m ( ) + mg(h h ) Proerties from Steam Table State P kpa T C x H kj/kg, m /kg Steam in Steam out Enthaly H for steam leaing turbine will be Steam in Turbine Steam out Q = heat loss Work outut H = h f + xh fg = (5.0) = kj/s W = (.5 kg/s) (60. kj/kg) 47.6 kj/kg} + ( 50m/ s) ( 60m/ s) m/ s ( m) kj/s W = (.5 kg/s) [( 67. kj/kg) + (9.45 kj/kg) ( kj/kg)] W = 9.8 kj/s or 9.8 kw Ans.

38 Q W = m (H H ) + m (V V ) + mg(h h ) ( 9 m/ s) ( 7. 5m/ s) W = (8.9 kg/s) [{(86 kj/kg) (50 kj/kg)} ( 0 0 m) W = (8.9 kg/s) [( 5 kj/kg) + (7.6 kj/kg) (0.098 kj/kg)] W = 47.5 kj/s W = 47.5 kw Ans. If heat loss, Q = 879 kj/min From aboe equation, Q W = 47.5 kj/s 879 kj/s W = 47.5 kj/s 60 W = kw Ans. EXAMPLE. A gas turbine receies air at 05 kpa and 555 K and discharges to ressure of 0.4 kpa. The actual temerature at discharge is K. If s. heat of the gas, C, can taken to be constant oer this temerature range and is equal to.86 kj/kg K, determine the work outut of the turbine er kg of working fluid. At inlet condition the secific olume is 0.5 m /kg, and at outlet it is 0.9 m /kg. In this case Q = 0 and inlet and outlet ies are laced at same datum leel. Therefore, steady flow energy equation is reduced to W = m(h H ) P = 05 kpa T = 555 K V = 0.5 m /kg h Q = 0 h P = 0.4 kpa T = K V = 0.9 m /kg Change in enthalies H H = C (T T ) = (.86 kj/kg K) [(555 K) ( K)] H H = 405. kj/kg Work out er kg, W = ( kg/s) (405.) kj/kg (Taking mass flow rate as unit mass er unit time) W = 405. kj/s or kw Ans.

39 EXAMPLE.4 Steam is generated in a boiler. The water at 7000 kpa as saturated liquid sulied to the boiler and leaes as suerheated steam at 6 MPa and 500 C temerature. How much heat is added to the steam to generate kg of steam under these conditions. Neglect kinetic and otential energy, also assume that there is no heat loss. Boiler, condenser and heat exchanger are the deices. Their objectie is to transfer heat energy therefore heat transfer rate cannot be taken zero. Water in Boundary Steam out System Boiler h = h If change in kinetic energy and otential energies are negligible, as in resent case, then V = V = 0 and h = h There is no shaft work, therefore, W S = 0 Therefore, steady flow energy equation can be exressed as, Q = m(h H ) Proerties from Steam Table State Pressure kpa Temerature C Enthaly kj/kg Entry Exit Q = ( kg) [(4. kj/kg) (66.9 kj/kg)] Q = 55 kj Ans. EXAMPLE.5 A heat exchanger is used to cool 8 kg/min of lubricating oil. Hot oil enters at 0 C and leaes at 5 C. The s. heat of oil is.59 kj/kg K. Cooling water enters the unit at 5.5 C and leaes a. C. If s. heat of the water is.87 kj/kg K, determine the required water flow rate if heat losses are negligible.

40 For this case steady flow energy equation is u + + Q = u + Q = (u u ) + ( ) Q = C (T T ) + ( ) Q = (0.599 kj/kg K) [577 K 78 K] + [(54 kpa 0.4 m /kg) (689 kpa 0.5 m /kg)] Q =.9 kj/kg kj/kg Q = kj/kg Ans. EXAMPLE.7 If the ie referred to in Examle.6 is a ertical run of ie such that section - is 0 m aboe section - determine the direction and magnitude of the heat transfer. Gien data: h = 0 h = 0 m Other arameters are same as in the reious examle. Steady flow energy equation u + + Q = u + + h Q = (u u ) + ( ) + h From reious examle u u = C (T T ) =.9 kj/kg = kj/kg For gien alue of h, PE is PE = h = gh 00 kj/kg = ( 98. m/s ) ( 0 m ) 000 PE = h = kj/kg Q =.9 kj/kg kj/kg kj/kg Q = kj/kg 0 m Ans. EXAMPLE.8 Determine the ower required to drie a um which extract the water from the dee well at the rate of 0.06 m /s. Water is extracted from 65 m dee well and raises 4 m aboe the ground. The diameter of the ie is 8 cm at the inlet and cm at exit. Neglect heat and frictional losses. About System The centrifugal um is a deice that transfers the mechanical energy of a motor or an engine into otential energy of a liquid. These deices um the liquid from lower leel to higher leel and therefore ower consuming deices.

41 Here work done is taken as negatie because it is done on the system, W = e Also Q = 0 and U = 0 as there is no change in temerature and s. olume of water. = = 0.06 m /s 4 m d = cm Motor 60 m d = 8 cm Therefore, steady flow energy equation reduces to, V W = m V ( h h ) g Water flow rate, q = 0.06 m /s Mass flow rate of water m = (0.06 m /s) (000 kg/m ) = 6 kg/s Steady flow energy equation, V V Q W = m ( H H) gh ( h ) Velocity at inlet, V = q 006. m / s A ( m) = 0.68 m/s Velocity at exit, V = q 006. m / s =.4 m/s A ( 4 0. m)

42 Velocity at exit, V = q 0075 (. m / kg) = 9.54 m/s A ( 4 0. m). Change in internal energy is zero, since temerature remains constant, u = 0. Change in kinetic energy, KE = V V ( m/s) ( 4. 4 m/s) Change in otential energy, = 0.06 kj/kg PE = gh ( ) ( 98. m/s) ( 9 m) = kj/kg Change in flow work, = = ( ) = (0.005 m /kg) (40 0 N/m 80 0 N/m ) = 0.5 kj/kg Substituting the alue of each term in steady flow energy equation we get, Q W = 50 [ ] =.7 kj/s or W =.7 kw Ans. In many engineering roblems the rate of mass flow in and out of the system is not same. Similarly, amount of heat transfer to or from the system and or the rate at which work is done on or by the fluid is not necessarily constant with time. Therefore, in such cases the total energy of the system within the boundary is no longer constant, such rocesses are called non-steady flow rocesses. For such a rocess energy equation can be written as: Energy entening Energy leaing Inc system system rease of energy stored in the system While writing steady flow energy Eq. (.65) if we consider that all the energy and mass of flowing fluid are time deendent, then it becomes non-steady flow equation and can be exressed as: