Thermodynamics by PK Nag PDF

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:~ fUf8~d :/ ~ , - ~./"\{.·· I \\ I ,:· ii I '\ '., \- ; I I.- I I. I O' C (a) (b) Fig. Ez 2.1 I ,, Ill I ' "  T,mpmzlurt -=35 Example 2.2 The e.m.f. in a ihermocouple with the test junction at t°C on gas thennometer scale and reference junction at ice point is given by e = 0.20 t-5 x 1~ r2 mV The millivoltmeter is calibrated at ice and steam points. What will this thermome&er read in a place where the gas thermometer n:ads 50°C? Solutio11 At ice point., when t = 0°C. e = 0 mV At steam point, when t = 100°C, e = 0.20 x 100 - 5 x 10-1 x (100)2 = ISmV At t= S0°C, e= 0.20 x SO- S x 10-1 (50) 2 = 8.75 mV When the gas thermometer reads S0°C, the thermocouple will read lOO X 8.75, or 58.33°C Ans. IS REVIEW QUESTIONS 2.1 What is the zeroth law of thermodynamics? 21 Define thermometric property. 2.3 What is a thermometer? 2.4 Wh.at is a fixed point? 2.5 How many fixed points were used prior to 1954? What are these? 2.6 What is the standard fixed point in thcnnomctery? Define it. 2.7 Why is a gas chosen as the standard thermometric subsiance? 2.8 What is an ideal gas? 2.9 What is the difference between the universal gas conslllllt and a characteristic gas constant? 2. IO What is a constant volume gas lhermomc:u:r? Why is it pn:fem:d to a (;()IIS!alrt pmlllwe gaa thermometer? 2.11 What do you understand by the ideal gas temperature scale? 2.12 How can the ideal gas temperature for 1hc steam point be measured·? 2.13 What is the Celsius temperature scale? 2.14 What is the advantage of a thcnnocouple in temperature measurement? 2.15 How does the resistance ihennometer measure tempemture? 2.16 What is the need of the international practical temperature scale? PROBLEMS 2.1 The limiting value of the ratio of the pressure of gas at the steam point and al the triple point of water when the gas is kepi at constant volume is found to be 1.36605. What is the ideal gas temperature of the steam point? 2.2 In a c-ODStant volume gas thermometer the following pairs of pressures read· ings were taken at the boiling poinl of water and the boiling point of sulphur, respectively: Water b.p. 50.0 JOO 200 300 Sulphur b.p. 96.4 193 387 S82 Iii I, 36=- Basic and.4ppli,d Tlinmodynamia The numbers are the gas pressures, mm Hg, each pair being taken with the same amount of gas in the thermometer, but the successive _pairs being taken with different amounts of gas in the tbennonM:tcr. Plot the ratio of Si,,p.: Hpb.p. against the reading at the water boiling point, and extrapolate the plot to zero pressure at the water boiling point.. This gives the ratio of Sb.p.: H20b.p. on a gas thennometer operating at zero gas pressure. i.e., an ideal gas thermometer. What is the boiling point of sulphur on the gas scale, from your plot? A.n.r. 445°C 23 The resistance of a platinum wire is found IO be 11,000 ohms at the ice point, 1S.247 ohms at the steam point, and 28.887 ohms at the sulphur point. Find the c:oostanta A aoo JJ in the equation R =R< 12 MJ/min =24.4 MJ/s = 24.4 MW= 24,400 kW I I !!I ii I + II  W01k and Heat Tra,ssftr -=57 Example 3,5 It is n:quired to melt 5 tonnes/h of iron from a charge at I s c to mol.ten metal at l650°C. The melting point is 1535°C, and the latent heat is 270 kJ/kg. The specific heat in solid state is 0.502 and in liquid state (29.93/atomic weight) kJ/kg K. If an electric furnace has 70% efficiency, find the kW rating needed. Ifthe density in molten state is 6900 kg/m3 and the bath volwne is three times the hourly melting rate, find the dimensions of the cylindrical furnace if the length to diameter ratio is 2. The atomic weight of iron is 56. Solution Heat required to melt I kg of iron at l 5°C to molten metal at I 650°C = Heat re(tuired to raise the temperature from 15°C to 1535°C + Latent beat + Heat required to raise the temperature from 1535°c to 1650°c = 0.502 (1535 - 15) + 270 + 29.93 (1650- 1535)/56 = 763 + 270 + 61.5 = 1094.5 kJ/kg Melting rate ~ 5 x 103 kg/h So, the rate of heat supply required =(5 X lif X 1094.5) kJ/h Since the furnace has 70% efficiency, the rating of the.furnace would be = Rate of heat supply per second Furnace etrciency = 5 X 103 X )094.5 - 217 X 103 kW Ans. 0.7x3600 103 3 Volume needed= 3 x 5 x m = 2.18 m.l 6900 If d is the diameter llnd I the length of the furnace 1E d11 = 218 m3 4 or 1E d x2d=218m3 2 4 d= 1.15 m and /=2.30m Ans. Example 3.6 If it is desired to melt aluminium with solid state sp.ecific heat 0..9 kJ/kgK, latent heat 390 kJ/kg, atomic weight 27, density in molten state 2400 kg/m 3 and final te,nperat~ 700"C, find out how much metal can be melted per hour with the above kW rating. Other data are as in the above example. Also, find the mass of aluminium that the above furnace will bold. The melting point of aluminium is 660°C. Solution Heat required per kg of aluminium Iii I, 58- Bcui, and Applild Tlu1111od]namitJ ~ 0.9 (660- 15) + 390 + 29 93 · (700- 660) 27 = 580.5 + 390 + 44.3 = 1014.8 kJ Heat to be supplied= !Ol 4.S = 1449.7 kJ/kg 0.7 With the given power, the rate at which aluminium can be melted = 2.l 1 X 103 X 3600 kg/h 1449.7 = 5.39 tonneslh Ans. Mass of aluminium that can be held in the above furnace =2.18 x 2400 kg =5.23 toMes Ans. REvlEw Q.UE.fflONS 3.1 How can a closed system and is swroundings interact? What is the effect of such interactions on ihe system? 3.2 When is work said to be done by a system? 3.3 What arc positive and negative work interactions? 3.4 What is displacement work? 2 3.S Under what conditions is !he work done equal to f pdV? I 3.6 What do you Wld.cnWl.d by path function and point ftznctioo? What &("C ex.tict Cid incuct ditfmntials? 3.7 Show that work is a path function, and not a property. 3.8 What is an indicator diagram? 3.9 What is mean effective pressure? How is it measured? 3.10 What are the indicated power and the brake power of an engine? 3.11 How does the cum:nt nowing through a resistor represent wortr. ·transfer? 3.12 What do you undersl'1Dd by now work? Is i.t different from displacement work? 3.13 Why does free expansion have zero work transfer? 3.14 What is heat trm1Sfer? What are its positive and negative directions? 3. IS What are adiabatic and diathermic walls? 3.16 What is an integrating factor? 3.17 Show that heat is a path f\lllction and not a property. 3.18 What.is lhe difference between work irnnsfer and heat transfer'] 3.19 Does heat lr.lllsfer inevitably cause a temperature rise? II 1\  Work and Htat Tran.sfer -=59 3.1 (a) A pump forces 1 m3/mln of water horizontally from an open well to a closed tank where the pressure is 0.9 MPa. Compute the work the pump must do 11pon the water i,1 ao hour just to force the water into the 1ank against the pn:ssute. Sketch the system upon which the wort is done beforl! and after the pl'0CCS$. Ans. 13.31 kJ (b) If the work done as above upon the wate.r had been used solely to raise the same amount of water vertically against gravity without change of pressure, how many meters would the water have been elevated? (c) If the work done in (a) upon the water had been used solely to accelerate the water from zero velocity without change of pressure or elevation, what velocity would the water have reached? If the work had been used to accelerate die water from an initial velocity of IO mis, what would the final velocity have been? 3.2 The pi~ton of an oil engine, of area 0.0045 m2, moves downwards 7S mm. drdwing in 0.00028 m) of fresh air from the atmosphere. The pressure in the cylinder is unifonn duri.ng the process at 80 kPa, while the atmospheric pressure is 101.325 kPa, the dilTercncc being due to the Aow resistance in the induction pipe and the inlet valve. Estimate the displacement work done by the air finally in the cylinder. Ans. 21 J 3.3 Ao engine cylinder bas a piston of area 0.12 mi and contains gas at a pressure of 1.5 M.Pa. The gas expands according ttl a process which is represented by a straight line on a pressure-volume diagr.im. The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.30 m. An.,. 29.1 U. 3.4 A mass of LS kg of air is rompn:ssed in a qu.a.si static proa9S from 0.1 MPa to 0.7 MPa for wbiehpv =constant.The initial density ofair is 1.16 kg/m 3 Find the work done by tbe piston to compress the air. Ans. 251.62kJ 3.S Am.us of gai; is compn:sst'd in a qua.si slatic procns from 80 kPa, 0.1 ml to 0..4 3 MPa. 0.03 m Assuming that the pressure and volume arc n:lated by pu" "' constant, find the work done by the gas $yStem. Ans. - 11.83 U 3.6 A si.ogle-cylinder, double-acting, reciprocating water pump has an indicator diagram which is a rectangle 0.075 m long and 0.05 m high. The indicator spring constant is 147 MPa per m. The pump runs at 50 rpm. The pump cylinder diameter is 0.1 Sm and the piston stroke is 0.20 m. Find the rate in kW at which the pii!On does work on lbe water. Ans. 43.3 kW 3.7 A single-cylinder, single-acting. 4 stroke engine of 0.15 m bore develops an indicated power of 4 kW when running at 216 rpm. Calculate the area of the indicator diagram that would be: obtained with an indicator having a spring constant or 25 x I06 N/m3 The length of the indicator diagram is 0.1 times 1be length of the stroke of the engine. Ans. 505mm2 "I' I II 60=- /Jasit and Applitd T1iermodyn.omia 3.8 A six-cylinder, 4-stroke gasoline engine is run at a speed of2520 RPM. The an:a of the indicator card of one cylinder is VIS I 03 mm2 and its lengtb is 51!.5 mm. The spring constant is 20 x 1Q6 N!m3 The bore of the cylinders is 140 mm and the piston stroke is 150 mm. Determine the indicated power, assuming that each cylinder contributes an equal power..4ns. 243.S7 kW 3.9 A closed cylinder of 0.2S m diameter is fitted with a light frictionless piston. The piston is retained in position by a catch in the cylinder wall and lhc volume on one side of the piston contains air at a pressure of750 kN/m2 The volume on the other side of the piston is evacuated. A helical spring is mounted coaxially with the cylinder in this evacuated space io give a force of 120 N on the piston in ibis position. The catch is released and the piston travels along the cylinder umil it comes to rest aller a stroke of 1.2 m. The piston is then held in its position of maximum travel by a ratchet mechanism. The spring fon:e increases linearly with ihe piston displacement to a final value of 5 kN. Calculate the work done by the compressed air on the piston. At1s. 3.07 kJ 3.10 A steam turbine drives a ship's propeller through an 8 : I reduction gear. The average resisting torqueimposed by the water on the propeller is 750 x I 03 N and the shaft power delivered by the turbine to the reduction gear is 15 MW. The turbine speed is 1450 tpm. Determine (a) the torque developed by the turbine, (b) the rower delivered to the propeller shaft, and (c) the net rate or working of the reduction gear. Ans. (a) r~ 98.84 km N. (b) 1.4.235 MW, (c) 0.765 MW 3.11 A fluid, co.ntained in a horizontal cylinder lined with a fric1ionless leakproof piston, is continuously agilated by means ofa stirrer passing through the ~'Ylinder cover. The cylinder diameter is 0.40 m. During ihe stirring process lasting 10 minutes, the piston slowly moves out a distance of 0.485m against the atmosphere. The net work done by the fluid during the process is 2 kJ. The speed of the electric motor driving the stirrer is 840 rpm. De1ermine the torque in the shaft: illld the po wet output of the motor. Ans. 0.08 mN, 6.92 W 3.12 At the beginning of the compression stroke of a two-cylinder internal combus- tion engine the air is at a pressure of IO 1.325 kPa. Compression reduces the volume to 115 of its original volume. and the law of compression is given by pv1·2 = constant. If the bore and stroke of each cylinder is 0.15 m and 0.25 m. respectively, detennine the power absorbed in kW by compression strokes when the engine speed is such that each cylinder undergoes 500 compression strokes per minute. Ant. 17.95 kW 3.13 Determine the cotal wort done by a gas sysu:m following an expansion process as shown in Fig. P. 3.13. Ans. 2952 MJ A 8 so j (\ 'I I I 0.2 i 0.4 - -+-V.m3 Fig. P. 3.13 "' "  Work and Heal TranJfor -=61 3.14 A system of volume V contains a mass m of gas at prc1;surcp aod tempcmturc T. The r.nacroscopic propenies of the system obey the following relationship: (p+ ;z )< 0.06 From these two equations a = - 60 lcN/m2 b = 1661 kN/ms Work transfer involved during the process r. r. W1_2 = 1 =f Yi pdV i ca+ bY)dY v,,2 _ "i2 =a(~-Yi)+b 2 = (;; - Yi>[n + t< Yi + Yi)] = 7 67 0.03 m 3[-6o kN/m 2 + ~ :S x 0.09 m3 ] = 8.SS kJ Work is done by the system, the magnitude being 8.55 ll. ' "  Firs.I Law of Tlurr11ody1111mit.$ ~77 :. Heat transfer involved is given by Q,_2 = U2 - u, + W,_2 "'59.5 + 8.55 = 68.05.kJ 68.05 kJ of beat flow into tbe system during the process. REVIEW QUESTIONS 4.1 State the first law for a closed system undergoing a cycle. 4.2 Whal was the con1Tibu1ion of J.P. Joule in establishing the first law'! 4.3 What is the caloric theory ofhcat? Why was it rejected'! 4.4 Which is the propeny introduced by the first law'? 4.S State the first law for a closed system undergoing a change of s1a1e. 4.6 Sbow that energy is a properly of a system. 4. 7 What arc the modes in which energy is slorcd in a system'! 4.8 Define intema.l energy. How is energy stored in molecules and atoms'/ 4.9 What is the difference between the standard symbols of E and U'! 4.10 What is the. difference between heal and internal energy? 4.11 Define enthalpy. Why does the cn1halpy of an ideal gas depe"l'ld only on tempcrn1un:'? 4.12 Define the specific heals at constant volume and cons1ant pressure. 4.13 Why should specific heal not be defined in tenns of heat transfer'~ 4.14 Which property of a system increases when heat is transferred: (-,) at constant volume, (b) at constant press11re? 4.15 Whal is a PM MI? Why is ii impossible'! PROBLEMS 4.1 An engine is tested by means of a water brake at I000 r:pm. The measured torque of the engine is I0000 mN and the water consumption of the brake is OS m3 /s, its inlet temperature being 20°C. Calculate the water temperature at exit, assuming that the whole of the engine power is ultimately transfonned into heat which is absorbed by the cooling water. An.,. 20.5°C 4.2 lb a cyclic process, heat minsfers are + 14.7 kJ, - 25.2 kJ. - 3.S6 kJ and + 31.5 kJ. What is tile net work for lhis cycle process? Ans. 17.34 kJ 4.3 A slow chemical reaction takes place in a fluid at the constant pressure of 0.1 MPa_ The fluid is surrounded by a perfect heat in~,i.lator during lhe reaction which begins at st.ate I aod ends at state 2. The insulntio11 is then removed and I05 kJ of beat flow to the surroundings as the fluid goes to stnte 3. The following wua are observed for the fluid at states 1. 2 and 3. Stale l( C) I 20.ii' 78=- 2 0.3 370 3 0.06 20 For the fluid sySlem, calculate E1 and E3, if E1 = 0 Ans. £ 1 a - 29.7 kl, £ 3 - J 10.7 kJ 4.4 During one qcle 1he working flaid in an engine engages in rwo worir. interactions: IS kJ to the fluid and 44 kJ from the fluid, and three heat interaetions, two of which an: known: 75 le.I to lhe fluid and 40 kJ from the fluid. Evaluate the magnitude and direction of the third beat transfer. Ans. -6kJ 4.5 A domestic refrigerator is loaded with food and the door closed. Owing a certain period the machine consumes I kW h of energy and the internal energy of the system drops by 5000 kJ. Find the net heal transfer for the system. Ans. - 8.6 M1 4.6 1.S kg ofliqwd having a c.an&Wrl ,pecmc beat of2.S kJ/kg K is stined in a wcll- imulatcd chamber cau.,ing lhe tempm1twe co rise by 1S°C. Find 4£ and W fur the process. Ans. /J.E = 56.25 kJ, W =- 56.25 kJ 4.7 The same liquid as in Problem 4.6 is stirred in a conducting chamber. During the process 1.7 kJ of beat arc transferred from the liquid to the surroundings, while the temperature of the liquid is rising by 15°C. Find /J..E and Wfor ihe process. A.ns. tJ.E =56.25 kJ, W =S1.9S kJ 4.8 The properties of a certain fluid an: related as follows u = 196 +0.718 t pV z 0.287 (I+ 273) where u is the ~i>ecific internal energy (kJ/kg), t is in °C, p is pressure (kN/m2), and vis specific volume (m3/kg). For this fluid, find c. and cP Ans. 0.718, 1.005 kJ/kg K 4.9 A system composed of2 kg of the above.fluid expands in a frictionless piston and cylinder machine from an inil.ial state of I MPa, IOO"C to a final temperature of 30°C. If then: is no heat transfer, ftnd lhe net work for the process. An.,. IOO.S2 kJ 4.10 If all lhe work in the expansion of Problem 4.9 is done on the moving piston, show that the equation representing the path of the expansion in the pv-plane is given by pvu = constant. 4.11 A stationary system consistingof2 kg ofthenuid of Problem 4.8 expands in ao · adiabatic process according to pv 1 2 "'constant. The initial conditions are I MPa and 200°C, and the final pressure is 0.1 MPa. Find W and tJ. U for the process. J Why is the wort 11'8.Mfer not equal to pd r? Ans. ff's 216.83, l!U O - 216.83 kJ, j pdV ; 434.4 kJ 4.12 A mixture of gases expands at constant pressure from I MPa, 0.03 m3 to 0.06 m1 with 84 kJ positive heat transfer. There is no work other than that done on a piston. Find.1. E for the gaseous mixture. Ans. S4 kJ The same mixture expands through the same siaie path while a stirring device does 21 kJ of work on the system. Find t.E, W, and Q for the process. Ans. 54 kJ. - 21 kJ, 33 kJ d I I II I I II  Fust Law ofTAnmodynamics -=79 4.13 A mass of 8 kg gas expands within a flexible container so that the p-v relationship is of the fonn pvi.2 = const. The initial pressun: is I000 kPa and the initial volume is I m3 The final pressure is 5 kPa. Ifspecific internal energy of the gas decreases by 40 kl/kg, find the heat transfer in magnitude and direction. Ans.+ 2615 lcJ 4.14 A gas of mass I.S kg undergoes a quasi-static expansion which follows n relationship, p = o + bV, where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 ml and 1.20 ml. The specific internal energy of the gas is given by the relation 11 ; 15 pv - 8HJ/lcg where p is in kPa and v is in m 3/lcg. Calculate the net heat transfer and the maximum illtemal energy oflhe gas anaincd during expansion. Ans. 660 kJ, 503.3 kJ 4.15 The heat capacity at constan1 pressure of a unain system is a function of temperature only and may be expressed as C = 2.093 + 41·87 J/K P I+ 100 where t is the temperature of the system in °C. The system is heated while it is mainillined at a pressure of l atmosphere until its volume increases from 2000 cm 3 to 2400 cm1 and its tempcratun: increases from 0°C to 100°C. {a) Find the magnitude of the heat interaction. (b) How much docs the int.emal energy of the system increase? Ans. (a) 238.32J (b) 197.79 J 4.16 An imaginary engine receives heat and docs work on a slowly moving piston at such rotes that the cycle of operation of I kg of working fluid can be represented as a circle 10 cm in diameter on a p-v diagram on which I cm = 300 k.Pa and I cm = 0.1 m3/kg. (a) How much work is done by each kg of working lluid for each cycle of operation? (b) The thennal efficiency of an engine is defmed as the ratio of work done and heat input in a cycle. If the heat rejected by the engine in a cycle is I000 kJ per kg of working fluid, what would be its thermal efficiency'? Ans, (a) 2356.19 kJ/kg, (b) 0.702 4.17 A gas undergoes a thennodynamic cycle con~isting of three processes beginning at an i.niiial state where p, = l bar, V1 = I. 5 m3 and U1 = 512 kJ. The processes arc as follows: (i) Process 1-2: Compression withpV = constant top2 = 2 bar, U2 = 690 kJ (ii) Process 2--3: W2l ·= 0. Q23 = - 150 kJ, and (iii) Process 3-1: W31 = + 50kJ. Neglecting KE and PE changes, dcten:ni11e the heat interactions Q 12 and Q31 Attr. 74 kJ. 22 kJ 4.18 A gas undergoes a thermoy1u:mic cycle cons.isting of the follO\ving processes: (i) Process 1- 2: Constant pressure p = 1.4 bar. 111 = 0.028 m\ W12 = 10.5 kJ. (ii) Process 2- 3: Compression with pV = constant.. U3 = U1 , (iii) Process 3- 1: Constani volume, U1 - U3 = - 26.4 lcJ. There are no significani changes in KE and PE. (a) Sketch the cycle on ap-V diagram. (b) Calculate the net work for the 80~ Basic and Appl~d T1tnmodynamie1 cycle in kJ. (c) Calculate the heat transfer for process 1-2 (d) Show thai IQ IW. -,.1< c,d Ans. (b) - 8.28 kJ, (c) 36.9 kJ 4. 19 A certain gas of mass 4 kg is contained within a piston cylinder assembly. The gas undergoes a process for which pVI.> = constant. The initial state is given by 3 bar, 0.1 nl. The change in internal energy of the gas in the process is 112 - u 1 = - 4.6 kJ/kg. Find the net heat transfer for the process when. the final volume is 0.2 m 3 Neglect the changes in KE and PE. Ans. - 0.8 lLI 4.20 An electric gener,llur coupled to a windmill produces an aver.ige electrical power output of5 kW. The power is used to charge a storage battery. Heat.transfer from the battery to the surroundings occurs at a constant rate of 0.6 kW. Determine the total amount of energy stored i.n the battery in 81, of operation. Ans. 1.27 x lOs kJ 4.21 A gas in a piston-cylinder assembly undergoes two processes in series. From state I to state 2 then: is energy transfer by heat to the gas of 500 kJ. and the gas does work on the piston amounting 800 kJ. The second process. from state 2 to state 3. is a i:onstant pressure compression at 400 kPa, during which there is a heat iransfer from the gas amounting 450 kJ. The following data are also known: U1 - 2000 kJ and U3 ~ 3500 kJ. Neglecting changes in KE and PE, calculate the change in volume oft.he gas during process 2- 3. Ans, - 5.625 nl 4.22 Air is contained in a rigid well-insulated rank with a volume of 0.2 m3 The tank is fitted with a paddle wheel which transfers energy to the air at a constant rate of 4 W for20 min. The initial density of the air is 1.2 kg/m3. If no changes in KE or PE occur, determine (a) the specific volume at ihe final state. (b) the change in specific internal energy of the air. A,u. (a) 0.833 m3Jkg. (b) 20 ltJ/kg fll I  First Law Applied to Flow Processes 5.1 Control Volume For any system and in any process, the first law can be written as Q=tt.E+ W where£ represents all fonns of energy stored in the system. For a pure.substance £=EK+ Ep+ U where EK is the K.E., Ep the P.E., and U the residual energy stored in the molecular structure of the substance. Q=tt.Er..+tt.EP+tt.U+ W (5.1) When there is mass transfer across the sytcm boundary, the system is called an open system. Most of the engineering devices are open systems involving the now of fluids through them. Equation (5.1) refers to a system having a particular mass ofsubstance, and is free to move from place to place. Consider a steam turbine (Fig. 5.1) in which steam enters at a hlgb pressure, does work upon the turbine rotor, and then leaves the turbine at low pressure through the exhaust pipe. If a certain mass ofsteam is e-0nsidered as the thenn.odynamie system, then the energy equation becomes Q =tt.EK +tt.Ep+ tt.U+ w and in order to analyze !he expansion process in turbine ihe moving system is to be followed as it travels throngh the tnrbine, taking into account the work and heat interactions all the way through. This method of analysis is similor to that of Langrange in fluid mechanics. I I I 11 ' ' 82=- Bask and A.pplitd 11imnodynamia i MOYtng syetem Control autfaca _.\ I~ / ----------- / :> / :.... w I ~ :o Shall Q - Exhaust pipe Fig. 5.1 Flow ptt>ttu invokiin,g work and luat intnaction.s Although the system approach is quite valid, there is another approach which is found to be highly convenient. Instead of concentrating attention upon a certain quantity of fluid, which constitutes a moving system in flow process, attention is focussed upon a cenain fixed region in space called a control volume through which the moving substance flows. This is simi.lar to th.e analysis of Euler in fluid mechanics. To distinguish the two concepts, it may be noted U1at while the system (closed) bonndary usually changes shape, position and orientation relative to the observer, the control volnme boundary remains fixed and unaltered. Again, while matter usually crosses the control volume boundary no such flow occurs across the system boundary. The broken line in Fig. 5.1 represents the surface of the control volume which is known as the control surfuce. This is the same as the system boundary of the open system. The method of analysis is to inspect the control surface and account for all energy quantities transferred through this surface. Since there is mass transfer across the control sw:face, a mass balance also has to be made. Sections l and 2 allow mass transfer to take place, and Q and Ware the heat and work interactions respectively. 5.2 Steady Flow Process As a fluid ·flows through a certain control volume, its thermodynanic properties may vary along the space coordinates as well as with time. If the rates of flow of mass and energy through the control surface change with time, the mass and energy within the control volume also would change with time. 'Steady flow' means that the rates of flow of mass and energy across the control surface are constant. In most engineering devices, there is a coostant rate of flow of mass and energy through the control surface, and the control volume in course of time attains a steady state. At the steady srare of a system. any thermodynamic properry will I! Ji.\ II  Finl Law Applied to Flow Procnm have a fued value at a particular location, and will not alter with time. Thennodynamic properties may vary along space coordinates, but do not vary with time. 'Steady state' means that the state is steady or invariant with time. 5.3 Mass Balance and Energy Balance in a Simple Steady Flow Process In Fig. 5.2, a steady flow system has been shown in which one stream of fluid enters and one siream leaves the control volume. There is no accumulation of mass or energy within lhe control volume, and the properties at any location within the control. volume are steady with time. Sections 1.1 and 2.2 indicate, respectively, the entrance and exit of the fluid across the control surface. The following quantities are defined with reference to Fig. 5.2. dQ ~ : : 1-2-~~ftT Flowoul ~~ T j· ~ ~ y !.' Steady now devic:8 ------- ~~~=- -;:'._ _ 1 l : / 777 777 / 7 77 /7777 ,77 7 77 7 77// / / / / 7/777 : di'"" c... z I : Fig. 5.2 Sttady ffew Process A1, A2-(:ross-section ofstream, m1 w1, w2-mass flow rate, kg/s P 1, p 2-pressure, absolute, N/m2 v 1, v2-specific volwne, m3/kg u1, ur-specific internal energy, J/kg V 1, V 2- velocity, mis 2 1, Z2-elevation above an arbitrary datum, m d'Q -net rate of heat transfer th.rough the control surface, J/s df dW. d,r" -net rate ofworlt t:ran.sfer through the control surface, J/s exclusive of work done at sections I and 2 in transferring the Ouid through the control surface. t-time,s. Subscripts I and 2 refer to the inlet and exit sections. I I !I !! I 84=- Basic an4 Applied Thmt!odyaamics 5.3.1 Mass Balanu By the conservation of mass, if there is no accumulation of mass within the control volume., the mass flow rate entering must equal the mass flow rate leaving, or WI "'W2 A1V1 = A2V2 or (5.2) VI V2 This e.la,Cria 108=- /JaJic a,u/ Applitd Tlurmotlynamies current was measured to be 0.5 amp and the electrostatic potenti3l at {I) was 0.8 volt above tbat at (2). Energy transfer as heat to the hot side of the generator was taking place at a r.itc of S.S wans. Determine the rate ol'energy transfer as heal from the cold side and the energy conversion efficiency. At!$. Q1 S. l watts, 11 - 0.073 S.6 A turbocompressor deliver., 2.33 m3/s of air at 0.276 MPa. 43°C which is heated at th.is pressure 10 430°C and finally eitpanded in a turbine whicb delivers 1860 kW. During the expansion, there is a beat transfer of 0.09 MJls to the surroundings. Calculate the turbine exhaust temperature if changes in kinetic and potential energy are negligible. Take cp a 1.005 kjlkgK A11s. IS7°C S.7 A reciprocating air compressor takes in 2 m1/min at 0.11 MPa, 20°C which it delivers at 1.5 M.Pa. I. I 1°C to an aflercooler whc.ro the air is cooled at constant pressure to 2S°C. The power absorbed by the compressor is 4. l5 kW. Determine the heat transfer in (a) the compressor, and (b) the cooler. State your assumptions. Ans. - 0.17 kJ/s, - 3.76 ld/s. S.8 l.n a water cooling tower air enters at a height of I m above the ground level and leaves at a height of 7 m. The inlet and outlet velocities are 20 mis and 30 mis respectively. Water entei:s al a height of8 m and leaves at a height ofO.S m. The velocity of water al entry and ex.it are 3 mis and I mis n:spcctively. Water temperatures arc 80°C and 50°C at the entry and exit respectively. Air temperatures are 30°C and 70°C at ihe entry and exit respectively. Tiie colling tower is well insulated and a fan of 2.25 kW drives the air through the cooler. Find the amount of air per second required for I kg/s of water flow. The values of er of air and water are 1.005 and 4.1871kg K respectively. ,411.r. 3.16 kg 5. 9 Air at 101.325 kPa, 20°C is iaken into a gas iurblne power plant al a velocity of 140 mis through an opening of 0.1 S m 2 cross-sectional area. The air is compressed heated. expanded through a rurbine, and ex.baustcd at 0.18 MPa, 150-C through an opening of 0.10 mi cross-sectional area. T11e power output is 375 kW. Calculate the net amount of heat added to the air in kJ/lcg. Assume that air obeys the law pt ~ 0.287 (t + 273) whe1e p is the pressure in kPa v is the spccilk volume in m3/kg, and, is the temperature in c. Take er= l.005 kJ/kg K..4ns. 150.23 kj/kg S.10 A gas llows steadily through a rotary compressor. The gas enters the compressor at a temperature of I69 C, a pressure of I00 kPa. and an enthalpy of 39 l.2 kJ1kg. The gas leaves the compressor at a temperature of245 9 C, a pressure of 0.6 MPa. and an entha.lpy of 534.5 kJ/Jcg. There is tJO heat transfer to or from the gas as it flows through the compressor. (a) Evaluate the eKtemal work done per unit mass of gas assuming the gas velocities at entry and exit to be negligible. (b) Evaluate the cxtemal work done per uo.it mass of gas when the gas velocity at entry is 80 m/ s and that at ex.it is 160 mis. A11s. 143.3 kJ/kg, 152.9 kJ/kg S.11 The steam supply to an engine comprises two streams which mix before entering the engine. One stream is supplied at the rate of 0.01 kg/s with an enthalpy of 2952 kJ/kg and a velocity of 20 mis. The other stream is supplied at ihe rate of 0.1 kg/s with an enthalpy of2569 kJ/kg and a velocity of 120 mis. At the c:x.it "I'  Finl l..llw Applid to Flow Proussts -=109 from the engine the nuid leaves as two streams, one of water at the rate of0.001 kg/s with an enthalpy of 420 kJ/kg and the other of steam; the Huid velocities at the exit are negligible. The engine develops a shaft power of 25 kW. The heat transfer is negligible. Evaluate the enthalpy of the secoud cxii stream. A11s. 2401 kJ/kg S.12 The stream of air and gasoline vapour. in the ratio of 14:1 by mass, enters a gasoline engine at a temper.1ture of 30°C and leaves as combustion products ai a tempetllture of 790 c. The engine has a specific fuel consumption of 0.3 kg/kWh. The net heat transfer rate from the fuel-air steam to the jacket cooling water and to the surroundings is 35 kW. The shaft power delivered by the engine is 26 kW. Compute the increase in the specific enthalpy of the fuel-air s~am. assuming the changes in kinetic energy and in elevation to be negligible. A11s. - 1877 kJ/kg mixture 5.13 An air turbine forms part of an aircraft refrigerating plant. Air at a pre.ssure of 295 kPa and a tcmperdture of58°C nows steadily into the turbine with a velocity of 45 mis. The air leaves the turbine at a pressure of 115 kPa, a temperature of 2°C, and a velocity of 150 mis. The shaft work delivered by the turbine is 54 kJ/kg_ of air. Neglecting changes in elevation. determine the magnitude and sign of the he.at uansfer per unit mass of air Oowing. For air, take cP = 1.005 kJ/lcg X. and the enlhalpy h "'cP t. Ans.+ 7.96 kJ/kg 5.14 In a turbomachiochandliug an iocomprcssible Ouid with a density of 1000 kghn3 the conditions of the Oniil at the rotor entry and ex.it are as given below Inlet Exit Pressure 1.15 MPa 0.05 MPa Velocity 30 mis IS.S mis Height above datum IO ro 2m If !ht volume now raie of lhe nuid is 40 m3/s, eslimate the net energy transfer lrom the fluid as work. A11s. 60.3 MW 5.15 A room for four persons has two fans, each consuming0.18 kW power, and three 100 W lamps. Vent.ilation air at the rate of 80 kg/h enters with an enthalpy of 84 kJ/kg and leaves with an enthalpy of 59 kJ/kg. If each person puts out heat al the rate of 630 kg/h determine the rate at whicb heat is to be removed by a room cooler, so that a steady state is maintained in the room. AIIS. 1.92 kW S.16 Air nows steadily at the rate of 0.4 kgfs through an air com~ressor. entering at 6 mis wid1 a pressure of I bar and a specific volume of0.8S m /kg. and leaving at 4.S mis with a pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The internal energy of the air leaving is 88 kJ/kg greater ihan that of the air entering. Cooling water in a jacket, surrounding the cylinder absorbs heat from the air al the.rate of 59 W. Calculate the power required to drive the compressor and the inlet and outlet cross-sectional areas. (.411s. 4S.4 kW, 0.057 m2, 0.0142 m2) 5.17 Steam flowing in a pipel.ine is at a steady state represented bypP' ' P' up, up, hp and Vr· A small amount of the total flow is led througb a small tube to an evacuated chamber which is allowed to fill slowly until the pressure is equal to the pipeline II I I 110=- pressure. lftbcrc is no heat transfer, derive an eKpression for the final specific internal energy in the chamber, in terms of the propelties in the pipeline. 5.18 The internal energy of air is given, at ordinary temperatures, by u:u0 +0.718r where u is in Ju/1cg, u0 is any arbitrary value of u at 0°C, kJ/lcg, and t is temperature in c. Also for air, pv = 0.287 (I+ 273) wbere pis in kPa and vis in ml/kg. (a) An cvacuat.cd bottle is fitted with a va]ve through which air from the atmosphere, at 760 mm Hg and 25°C, is allowed to flow slowly to fill the bottle. If no beat is transferred to OT from the air in the boitle, what will its lempent~ be when lhe pres.sure in w bottle reaches 760 nw Hg? Ans. 144.2°C (b) lfthe bottle initially contains 0.03 m1 of air at 400 mm Hg and 2S°C, what will Ille tempm1ture be when the pn!s.sute ill Ille bottle reaches 760 mm Hg? Ans. 7 l.6°C 5.19 A pressure cylinder of volume /1 contains air at pressurep 0 and temperature 70 It is to be filled from a compressed air line maintained at cons.tan! pressure p1 and temperature T1 Show that the temperature of the air in the cylinder after it has been charged to the pressure of the line is given by T= 71j I+ P1 k(1.!i._-1) To 5.20 A smal.l reciprocating vacuum pump having the rate of volume displacement Vd is used to evacuate a large vessel of volume V. The air in the vessel is maintained at a constant temperature T by energy transfer as heat. lf the initial and final pressures are p 1 and Pi respectively, find the time taken for the pressure drop and the necessary energy transfer as heat during evacuation. Assume that for air, pV= mR.T, where mis the mass and R. is a constant, and u is a function of Tonly. [ Ans.t = L In 12-; Q=(p1 -.Pi) v.. Pl v] [Hint: dm - - p(.Vd·dt)l(R1) = V dpl(R1)]. 5.21 A tank containing 45 kg of water initially at 45°C has one inlet and one exit with equal mass flow r.ues. Liquid water enters at 45°C and a mass flow rate of 270 kg/h. A cooling coil immersed in the warcr removes energy at a rate of 7.6 kW. The water is well mixed by a paddle wheel with a power input of 0.6 kW. The pres.~ures at inlet and eKit are equal. Ignoring changes in KE and PE. find the variation of water temperature with time. A11.r. T 0, i.e., there has always to be a beat rejection. To produce net work in a thermodynamic cycle, a heat engine has thus to exch.ange heat with two reservoirs, the source and the sink. The Kelvin-Planck statement of the second law states: It is impossible for a heat engine Jo produce 11et work ill o complete cycle if it exchanges lteai 011/y with bodies at a single fixed temperature. IfQ2 = 0 (i.e., w = Q1, or 11 = 1.00), the heat engine will produce net work in a completecycle by exchanging heat with only one reservoir, thus violating the Kelvin-Planck statemeni (Fig. 6.5). Such a heat engine is called a perpetual motion mac/tine of the second killd; abbreviated to PMM2. A PMM2 is impossible. A heat engine has, therefore, to exchange heat with two thennal energy reservoirs at two different temperatures to produce net work in a complete cycle (Fig. 6.6). So long as there is a diffe.rence in temperature, motive power (i.e. work) can be produced. Ifthe bodies with which the heat engine exchanges heat are offinite heat capacities, work will be produc-ed by the heat engine till the temperatures of the two bodies are equalized. Iii I, 116=- Basic a,u/ Applied Tlttrmodynami '111 I I +! It 1111 130=- Basit and Applitd 11imnoJynamia L... _-,-__s_ou_r_ce_;_',_...,...._ _J 1... l~. ~ w, ITp-% r~ r~ Fig. 6.25 Two cyclic luat engines E_. and E8 optTating bttwun the same soum and sink, of wliid1 E8 is mmihle WA> We Now, lett.'8 be reversed. Since £,8 is a reversible heat engine, the magnitudes of heat and work transfer quantities will remain the same, but their directions will be reversed, as shown in Fig. 6.26. Since WA > W8 , some part of W" (equal to We) may be fed to drive the reversed heat engine 313 Soutce, t, r- - Sink.ft Flg. 6.26 E8 is rru,md Since Q1" = Q18 = Q1, the heat discharged by 38 may be supplied toE. The source may, therefore, be eliminated (Fig. 6.27). The net result is that EA and 3 0 together con.stitute a heat engine which, operating in a cycl.e, produces net work WA - W0 , while exchanging heat with a single reservoir at t2. This violates the Kelvin-Planck statement of the second law. Hence the assumption that 'IA > 118 is wrong. Therefore fie~ '7A.. ,.  -=131 Sink, 12 Fig. 6.27 E., aru! 38 togetlur oilate llte K·P flatmfflt 6.14 Corollary of Carnot's Theorem The efficiency of all reversible heat engines operating between the same temperature levels is the same. Let both the heat engines EA and £ 8 (Fig. 6.25) be reversible. C.et us assume rt;. > 718. Similar to the procedure outlined in the preceding article, if £ 8 is reversed to run, say, as a heat pump using some part of the work output (WA) of engine E", we sec that the combined system of heat pump £ 0 and engine E,,, becomes a PMM2. So 71A cannot 'be greater than 71 8. Similarly, if we assume 170 > 1/;. and reverse the engine£.,., we observe that71 8 cannot be greatcrthan IJA· Therefore rtA : 71s Si.nee the efficiencie.s of all reversible heat engines operating between the same heat reservoirs are the same, the efficiency ofa reversible engi11e is independent ofthe nah1re or amount ofthe working substance undergoing the cycle. 6.15 Absolute Thermodynamic Temperature Scale The efficiency of any beat engine cycle receiving heatQ 1 and rejecting heat Q2 is given by (6.18) By the second law, it is necessary to have a temperature difference (t1 - t2) to obtain work of any cycle. We know that the efficiency of all heat engines operating between the same temperature level.s is the same, and it is independent of the working substance. Therefore, for a reversible cycle (Carnot cycle), the efficiency will depend solely upon the temperatures /1 and 12, at which heat is transferred, or (6.19) where/signifies some function of the temperatures. From Eqs (6.18) and (6.19) Ill" ii.  Btuic and. Applied TMrm.odynamics 1- ~ = f (t1, tz) In terms of a new function F t =F(t1, t 2) (6.20) If some functional relationship is assigned between t 1, 12 e.nd Q,tQ2, the equation becomes the definition of a temperature scale. Let us consider two reversible heat engines,£1 receiving heat from the source at t 1, and rejecting heat at t2 to £ 2 which, in tum, rejects heat to the sink at t3 (Fig. 6.28). H&at raserwlr, 11 _J O; Heat reservoir, 13 Fig. 6.28 Time Oamol mgiMi Now EI and £ 2 together constitute W1othcr heat engine £ 1 operating between t I and t3 ~ \ I I) {b =Fft 3 Now ~ ; Q1l The energy rejected from ihis engine is received by a second reversible engine at the same temperature T. The second engine rejec.ts energy al temperature T2 (T2 0 It is thus proved that the entropy ofan isolated system can never decrease. It always increases and remains constant only when the process is reversible. This is known as the principle of increase of entropy, or simply the entropy principle. It is the quantitative general statement of second law from the macroscopic viewpoint. An isolated system can always be formed by including any system and its surroundings within a single boundary (Fig. 7.12). Sometimes ihe origiual system which is then only a part of the isolated system is called a 'subsyste,n'. - System ~ ,-~ ~ ~.... 0 - Svrroundings isolated (oomposite) system Flg. 7.12 isolated :rystnn The system and the surroundings together (the univcn;e or the isolated system) include everything which is affected by the process. For all possible processes that a system in the given surroundings can undergo d.~univ 0. s,. The magnitude of Sgen q11:3nti fies th~ irreversibility of the process. If systems A and B operate so that (Sgc.JA > (S8 0 ) 8 it can be ~aid that the system A operates more irreversibly than system B. The unit of Sg is WIK. At steady state, the continuity equation gives (7.36) " the energy equation becomes o= Q- w.,, +~m;(1r+ !:+gz)_ - I,m.(1a + V +gz) 2 (7.37) I 2 I e 2 e and the entropy equation reduces to o= Q + I,m,s, - I,mc-1., + sscn (7.38) T i These equations often must be solved simultaneously, together with appropriate propeny relations. Mass and energy are conserved quantities, but entropy is not generally conserved. The rate al which entropy is transferred out must exc.eed the rate at which entropy enters the CV, the difference being the rate of entropy generated within the CV owing to irreversibilities. For one-inlet and one-exit control volumes, the entropy equation becomes O= %+ m(s1 -s2) + s,.., (7.39) 7.13 First and Second Laws Combined By the second law 4Qn:v= TdS and by the first law, for a closed non-flow system, dQ=dU+ pdV "I' ' II  Entropy -=179 TdS=dU+pdV (740) Again, the enthalpy H=U+pV dH=dU+pdV+ Vdp = TdS+ Vdp TdS =dH - Vdp (7.41) Equations (7.40) and {7.41) are the thermodynamic equations relating the properties of the system. Let us now examine the following equations as obtained from lhe first and second laws: (a) (t Q = dE + d: W- This equation holds good for any process, reversible or irreversible, and for any system. (b) (t Q= dU + pd· W- This equation holds good for any process undergone by a closed stationary system. (c) dQ = dU + pdV- This equation holds good for a closed system when only pdV-work is present. This is true only for a reversible (quasi-static) process. (d) ct Q = TdS-This equation is true only for a reversible process. (e) TdS = dU + pdV- This equation holds good for any process reversible or irreversible, undergone by a closed system, since ii is a relation among properties which arc independent of the path. (f) TdS = dH - Vdp-This equation also relates only the propenies of a system. There is no path function tem1 in the equation. Hence the equation holds good for any process. The use of the term 'irreversible process' is doubtful, since no irreversible path or process can be plotted on thennodynamic coordinates. It is more logical to state tbat 'the change of state is irreversible, rather than say 'it is an irreversible process'. A natural process whic.h is inherently irreversible is indicated by a dotted line connecting the initial and !iDal states, both of which arc in equilibrium. The dotted line has no other meaning, since it can be drawn in any way. To determine lhe entropy change for a real process, a known reversible path is made to connect the two end states, and integration is performed on this path using either equation (e) or equation (f), as gi.ven above. Therefore, the entropy change of a system between two identifiable equilibrium states is the same whether the intervening process is reversible or the change of state is irreversible. 7.14 Reversible Adiabatic Work in a Steady Flow System ln. the differential form, the steady now energy equation per unit mass is given by Eq. (5.11), 11Q = dh + VdV + gdZ + d: Wx ,, i,I I I II 180~ &uic and Applitd 11lmnody1111mia For a reversible process, 2, the problem is indetenninate. If one set of p'; s is chosen arbitrarily, some outcomes are certainly overemphasized. The problem is then how to avoid bias in selecting a certain set ofp', s. Tbe answer according 10 E.T. Jaynes is to assign that set of values of p'; s which is consistent with the given infonnation and which maximizes the uncertainty. This is the principle of minimum prejudice, enunciated by Jaynes in the following words: "The le"st prejudiced or most unbiased estimate of probabilities is rhut assignment which maximizes the entropy S, subject to the given i11formation ", The entropy S represents the uncertainty of an evenl, which is to be maximized: S "" - K !. p, In p, (7.54) subject to the two constraints of Eq. {7.55). Lagrange's method of undeter· mined multipliet'! (see Chap. 21) will be used for the solution. Differentiating Eqs {7.54) and (7.55) Idp, =:0 (7.56) Ip; dpi =O (7.57) dS = - K [I. lnp, dp, + L:lp,) = 0 or, I. In P; dp; = 0 (7.58) where V,s are held constant and dS = 0 for S to be maximum. Multiplying Eq. (7.56) by ,land Eq. (7.57) by {J and adding to Eq. (7.58). I(ln P; + A + PY;] dp1=0 (7.59) where ,land fJ are the Lagrange's multipliers. Since dp;'s are non-zero, In p, + A + /W; = 0 or P; = e-i. e·flV, (7.60) The Eq. (7.60) represents a set of c equations: P1 =e·>. e-llV1 P2=e->.e-l1V2 (7.61) Pc =e·). e·P v. The above equations are the desired unbiased set ofp'; s. iii I  EntrfJPJ -=187 Again. _..~, C LP; =I,e-.le-/ill', I = I ~ = :,Ee-llv, or.t = In I,e-pv, (7.62) From Eq. (7.60}, e-pv, Pi= I.e-/JV, (7.63) Thus.t has been estimated. Using Eq. (7.SS) LY;e-Pv, _ I.Pi ; f'; = L -fJV e ' =V (7.64) Since P is known, {j can be detennined. The entropy can be expressed in temts of l and Pas: S = - K 1: P; [-.t - /jV,) = KA.l:pi + K fJE p; Y; S = K.t + KP (7.65} The above procedure is referred to as the Jaynes· formalism. 7.18.2 Information Theory Applud to a System of Particles Let us consider a system having a large number of particles. According to the quantum theory (see Chap. 19). the energy that the system can have is discretely distributed. The system cannot have any energy, bui only certain values of energy levels. Let us denote by P; the _pcobability or the energy level E:;. A high probability signifies that the corresponding energy level is more frequently attained by the system, i.e,, the system can be found for longer durations or time in that energy level. The problem is to determine the most probable state or the system subje.ct to the constraints imposed by the nature of the system. Usually, the average energy of the system, -which is also the expectation energy is known by physical measurements. Therefore, = !p;E; = E and l:p, = I In stati.stical thermodynamics, one proceeds to determ.ine the numbe.r of microstates corresponding to the most probable rnacrostaie of the system (see Chap. 19). corresponding to the thermodynamic equilibrium state when W is maximized to yield S = K In W. The entropy signifies the uncertainty inherent in lhe system. , Ill I, 188=- Basic and Applied 1111rmodynamia Jaynes' fonnalism can be applied here. The outcomes are the various possible energy levels corresponding to the various macrostates of the system. The probabilities P; of the energy levels are to be determined according to the principle of minimum prejudice, snbject to the constraints, l:P; = I Ipi£; = Maximizing the uncenainty or the entropy: S=-Kip; lnpi we get the probability distribution, as obtained earlier in Eq. (7.60) Pi= e-,.. e-Pti (7.66) where i =te-11-c.m"' -,- J mcl' -T = I x 4.187 In 293 273 273 ""0.296 kJ/K Total entropy change of ice as it melts into water (~101A1 =6Si +.:\Sn + Mm = 0.0389 + 1.22 + 0.296 = 1.5549 kJ/K The en11opy-temperature diagram for the system as ice at -5°C converts to water at 20°C is shown in Fig. Ex. 7.3.2 :. Entropy increase of the universe (~univ =(.:\S),,...,an + (~ IJ!I = 1.5549 - 1.46 =0.0949 kJ/K Ans. (a) ' "  Entropy -=195 293K 4 -·-·-·-·················-·······-···-···-·············-.... t ,n K..... rr.7777lii':,'7'7nf!:S 268K - ---s Fig. Ex. 7.3.2 (b) To convert 1 kg of water at 20°C Atmosphere to ice at-5°C, 427.S kJ of heat have to at293 K be removed from it, and the system has to be brought from state 4 to state I (Fig. Ex. 7.3.2). A refrigerator cycle, as w shown in Fig. Ex. 7.3.3, is assumed to accomplish this. The entropy change of the system would be the same, i.e. S4 - S1, with the only difference that its sign will be 1 kg Water at 20°C System negative, because heat is removed to Ice at - S"C from the system (Fig. Ex. 7.3.2). Fig. Ex. 7.3.3 (t\S) Ytl= =S1 - S4 (qegative) The entropy change of i.he working fluid in the refrigerator would be zero, since it is operating in a cycle, i.e., (~r=O The entropy change of the atmosphere (positive) (t\S)- = Q;W :. Entropy change of the universe (M>wuv = (LlS),ystcm + (L\S),cr + (t\S)aun -= (S1 - S ) + Q + W T By the principle of increase of entropy (L\S)wuv or i1olated system~ 0 [+ Q;w] ~ o f IJ II 196==- Basic and Applied Tllnmodynamia Q+W ~(SrS1) T w~ T(S4 - s,) -Q ffclNII, = 7lS4 - S1) - Q Here Q= 421.S k1 T=293K Sr S 1 = 1.5549 kIIK W(min) - 293 X l.5549 - 427.5 = 28.5 kJ Aiu. (b) Example 7., Two identical bodies of co.nstant heat capacity are at the same initial temperature T;. A refrigerator operates between these two bodies until one body is cooled to temperature T2 If the bodies.remain at constant pressure and undergo no change of phase, show that the minimum amount of work needed to do this is w,mm> = cl'( ;:.. 1i - 27;) Solution Both the finite bodies A andB are initially at the same temperature~. Body A is to be cooled to temperature T2 by operating the refrigerator cycle, as shown in fig. Ex. 7.4. Let r{ be the final temperature of body B. w Fig. Elt. 7.4 Heat removed from body A to cool it from T; to T2 Q = Cp(T;- T2) where CP is the constant preasure heal capacity of the identical bodies A and B. Heat dischaIBed to body B = Q + W = CP (7i - T.) Work input, W = Cp(Ti. - T;) - Cl' (T; - 12) = Cp (T2+ T2 - 27;) (7.4.1) 1, "' !! '  Entropy -=197 Now, th.e entropy change of body A liS,. = f T, CP.!!I= C11 In 72 (negative) T T; The entropy change of body.B t' u.,B = fc T, 11 -dT T =CP In -r; 1j (pos1hve).. Entropy c:hangc (cycle) of refrigerant = 0 :. Entropy change of the universe (6S>..,,,v =Ll.S,. + liSs r. C In ---1... + C In _1... T,' = p r. p T; By the entropy principle (~$)univ ~ 0 C In T, + C. In T{ ) 2: 0 ( p T; p 1i CP!n 7;;7i 2 ~ O (7.4.2) In equation (7.4.1) with Cp, T2, and T, being given, W will be a minimum when T2 is a minimum. From Eq. (7.4.2), the minimum value of r 2 would correspond to T. T-' C.p In -Ll.. T; =0 =In I r,2 T;=-·- 1i From equation (7.4.1) ·l=Cp W.iysi.rn = - 0.047 kJ kg K. Since the duct is insulated (&S)SWT = 0 (&S>uruv = -0.047 lc.J/ltg K This is impossible. So the flow must be from B to A. Example 7.9 A hypothetical devic.e is supplied with 2 kg/s of air at 4 bar, 300 K. Two separate streams of air leave the device, as shown in figure below. Each stream is at an ambien.t pressure of I bar, and the mass tlow rate is the same for both streams. One of the exit streams is said to be al 330 K while the other is at 270 K. The ambient temperature is at 300 K. Determine whether such a device is possible. Sol11tion The entropy generation rate for the conlrOI volume (Fig. Ex. 7.9) is ssm = I,m.s. - I,m;si = m~2 + m~3 - m1s1 = '"1!2 + '">13 - ( '"1 + iit3).f, = ntz{S1 - s1) + m3(s3 - s,) Now, 1i -Rln -P2 s2 -s1 =cJ>ln- 7j P1 = 1.005 ln 330 - 0.287 ln _!_ 300 4 = 0,494 kJ/kgK 7; s3 -s1 =cpln- -Rln -p3 7i P1 = LOOS ln 270 - 0.287 In.!.. 300 4 = 0.292 kJ/kgK S..., = I X 0.494 + l X 0.292 =0.786kW/K Since Sgen > 0, the device is possible. Such devices actually exist and are called vortex tllbes. Although they have low efficiencies, they are suitable for certain applications like rapid c-0oling of soldered parts, electronic compo11ent cooli.ng, cooling of machining operations and so on. The vortex tube is essentially a passive device with no moving parts. It is relatively maintenance free and durable. f I! ii I  Entropy -=203 Air in 2kg/$ -f-] 111[aln i + b(~ - 7j)] (b) 23.9 kJ/k.mol K 7.35 An iron block of unknown mass at ss c is dropped into an insulated tank that contains O, Im3 of water at 20°C. At the same time a paddle-wheel driven by a 200 W motor is activated to stir the water. Thennal equilibrium is established after 20 min when the final temperature is 24°C. Detennine the ma.u of the iron block and the entropy generated during the process. Ans. 52.2 kg, 1.285 kJ/K 7.36 A piston-cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27°C. The gas is now compressed slowly in a polytropic process during which pVIJ = constant.The process ends when the volume is reduced by one-half. Dctennine the entTopy ch.Inge of nitrogen during this process. Ans. - 0.0615 kJ/K. Iii I ' II 212=- Basic and Applied 17amnodynamics 7.37 Air enters a compressor at ambient conditions of96 kPa and 17°C with a low velocity and exits at I MPa, 327°C. and 120 mis. The compressor is cooled by the ambient air at 17°C at- a rate of 1.500 kJ/min. The power input to the compres.sor is 300 kW. Oct.ermine (a) the mass now rate of air and (b) the raie of entropy generation. Ans. (a) 0.85 l kg.ts, (b) 0.144 kW/K 7.38 A gearbox operating at steady slate receives 0. I kW along the input shaft and delivers 0.09.5 kW along the output shaft. The outer surface of the gearbox is at so c. For the gearbox, determine (a) the rate. of heat transfer, (b) the rate at which entropy is produced. Atts. (a)-0.005 kW.(b) l.54x 10-skW/K 7.39 At steady state, an electric inotor develops power along its output shaft at the r.ite of 2 kW while dr.awing 20 amperes at 120 volts. The outer surface of the motor is at S0°C. For the motor. determine the rate of heat transfer and the rate of entropy generation. Ans. - 0.4 kW, 1.2.4 x io-·3 kW/K 7.40 Show that the minimum theoretical work input required by a refrigeration cycle to bring two.finite bodies from the same initial temperature to the final temperatures of T1 1111d T2 (T2 < T1) is given by Wmil =111c (2(T1Ti)'12 - T1 - T1 ) 7.41 A rigid tank coniains an ideal gas at 40°C that is being ~1irred by a paddle wheel. The paddle wheel docs 200 kJ of work on the ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as l rewlt of heat transfer between the syslem and the surroundings at 25°C. Delermine (a) Ille eno:opy change of the ideal gas and (b) the total entropy generation. Ans. (a) 0, (b) 0.671 kJ/K 7.42 A cylindrical rod of length L insulated on its lateral surface is initially in contact at one end with a wall at temperature T1 and at the other end with a wall at a lower temperature T2 The temperature within the rod initially varies linearly with position x according to: 1'(.r) ,. f1 - 1j - Ji X l The rod is insulated on its ends and eventually comes to a final equilibrium state where the temperature is T,. Evaluate Tr and in terms of T1 and Ti, and show that the amount of entropy generated is: S 1'!(;[1 + In Tr+ _!i._ In Tz - __li_ In 7j] ~ 1j-~ lj-~ where c is the specific heat of the rod. AIIS. r,.. er,+ r1112 7.43 Air nowing throll8h a horizontal, insulalcd duc:t was studied by stvdenu in a Laboratory. One mident group measured the pressun:, icmperatun:, and velocity at a location in the duct as 0.95 bar, 67°C, 7S mis. At another location the respective values were found to be 0.8 bar, 22°C, 310 mis. The group I I ,, " '  Entropy -=213 neglected to note the direction of flow, however. Using the kno\1/1\ dat11. deter mine lhe dim:1ion. Ans. Flow is from right to left 7.44 Nitrogen gas at 6 bar, 21 °C entcrn an in.sulated c-0ntrol volume operating at steady slate for which Wc~v. ~ 0. Half of tbe nitrogen exits the device at I bar, 82°C and the other half exits at I bar, - 40°C. The effects of KE and PE are negligible. Employing ihe ideal gas model, decide whether the device can operate as described. Ans. Yes, the device can opcr.il.t' as described Iii I ,  Available Energy, Exergy and Irreversibility 8.1 Available Energy The sources ofenergy can be divided into two groups, viz. high grade energy and low grade energy. The conversion of high grade energy to shaft work is exempted from the limitations of the second Jaw, while conversion oflow grade energy is subject to them. The examples of two kinds of energy are: High grade energy Low grade energy (a) Mechanical work (a) Heat or thermal energy (b) Electrical energy (b) Heat derived from nuclear fission or fusion (c) Water power (c) Heat derived from combustion of fossi I fuels (d) Wind power (e) Kineticenergyofajet (t) Tidal power The bulk oflhe high grade energy in the fonn of mechanical work or electrical energy is obtained from sources of low grade energy, such as fuels, through the medium of the cyclic heat engine. The complete conversion oflow grade energy, heat, into high grade energy, shaft-work, is impossible by virtue of the second.law of thermodynamics. That part of the low grade energy which is available for conversion is referred to as available energy, while the part which, according to the second law, must be rejected, is known as 11na11ailable energy. Josiah Willard Gibbs is accredited with being the originator of the availability concept. He indicated that environment plays an important part in evaluating the available energy. , I , I I I,.  -=215 8.2 Available Energy Referred to a Cycle The maximum work output obtainable from a certain beat input in a ~yclic heat engine (Fig. 8.1) is called the availab/eenerg)'(A..E.), or the available part of the energy supplied. The minimum energy that has to be rejected to t~e sink by the second law is called the U11avai/able energy (U.E), or the unavailable part of the energy supplied. Therefore, Q1 = A.E. + U.E. or w"""' = A.E. =QI - U.E. For the given T1 and T2, Tim-= 1-r.1i For a given T1, Tfrcv will increase with the decrease of T2 The lowest practicable temperature ofheat rejection is the temperature ofthe surroundiTtgs, To and Let us co.nsider a finite process x-y, in which heat is supplied reversibly to a heat engine (Fig. 8.2). Taking an elementary cycle, ifd Q1 is !he heat received by the engine reversibly at T1, then 1i-To To - - ctQ1 =4Q1 - -4Q dWmu = 1.. A.E. 1j 7j T1 I' Y/ ro, ,.... I O S2-S1- ~ >O To Qi Q1 (8.16) Therefore, from Eqs (8.13) and (8.16), WR>W1 (8.17) Therefore, the work done by a closed system by interacting only with the swroundings stpo, T0 in a reversible process is always more than that done by it in an irreversible process between the same end stat1ls. 8.4.1 Work done in all Rnersible Processes is the Same Let us assume two reversible processes R 1 and R 2 between the same end siates I and 2 undergone by a closed system by exchanging energy only with the surroundings (Fig. 8.10). Let one of the processes be reversed. Surroundings Po,To Syslem ~ 1·>-2... t w - $ Fig. 8.10 £pal r.tJOrk dou in all rt'llirm6lt ro,usts bttwmt t!u sarn, n,d slates hI h I I t 224=- BaJie and Applitd Thermodynamics Then the system would execute a cycle l-2-1 and produce network represented by the area enclosed by exchanging energy with only one reservoir, i.e. the surroundin.gs. This violates the Kelvin-Planck statement. Therefore, tire two reversible processes must coincide and produce equal amounts of work. 8.5 Reversible Work by an Open System Exchanging Heat only with the Surroundings Let us consider an open system exchanging energy only with the surroundings at constant temperature T0 and at constant _pressure Po (Fig. 8.11). A mass dm 1 enters the system at state l, a mass dn1 2 leaves the system at state 2, an amount of heat 2 - To,1"2) (8.43) 8.6.1 Muimum Useful Work Obtainal>k when the System &c.hanges Heat u,ith a T1Jmnal Rum,oir in, Addition to the Atmosphere If the open system discussed in Sec. 8.5 exchanges heat with a thennal energy reservoir at temperature TR in addition to the atmosphere, the maximum useful work will be increased by d QR ( l - ~ ) , where i! QR is the beat m:eived by the system. For a steady !low process, (WJmu'"' w_ =( H1 - ToS1 + m:l + mgz1) mV} - ( H2-To.s':i+-2-+111gz2 10 ) +Q11. ( l-Ta. ) = \V1 - ¥'2 + QR( 1 - ~) (8.44) For a closed system {Wu)mu = W"'..,. -po (Vi - Y1)+ QR ( 1- ~) or (W11)mu =£ 1 - E2 + Po (V1 - Y2) - T0(S1 - S2) + QR ( 1- ~) (8.45) lfK.E. and P.E. changes are neglected, then for a steady/low process: (W11),.... ~ (H 1- H2 ) - T0 (S1 - S2 ) + Q11. ( 1- ~) (8.46) and for a closed system: (W11)ow"' U, - U2 + Pof = (U1 - U1)r- TtS1 - S2 >,, (8.56) "I'  AllflilaMt Enn-gy, E.xtTg'/ and /1m'el'si6ility ~233 Fro.m Eqs (8.54) and (8.56), (Wr)- = (F, - F2tr (8.57} or W1 S (F1 - F1h (8.58) The wo.r.k done by a sysiem in any process between two equilibrium states at the same temperature during which the system exchanges heat only with the environment is equal to or Jess 1han the decrease in the.Helmholtz functio.n of the system during the process. The maximum work is done when the process is reversible and the equality sign holds. If the process is irreversible, the work is less than the maximum. (b) Let us now consider a system which is in both pressure and temperature equilibrium with the surroundings before and after the process. When the volume of the system increases some work is done by the system against lhe surroundings (pdV work), and this is not available for doing useful work. The availability of the system, as defined by Eq. (8.51), neglecting the K.E. and P.E. changes, can be expressed in the form A= (W.Jmu =(U + PoY- To,S)-(Uo + Po'Yo -ToSo) = ~-'° The maximum work obtainable during a change of slate is the decrease in availability of the system, as given by Eq.(8.SJ) for unit mass. (Wu)mu =A, -A2 = ~1- ~2 = (U, - U2) + PoW1 - Y2) - To(S1 - S2) lfthe initial and final equilibrium states of the system are at the same pressure and temperature of the surroundings, say p 1 = p 2 ; Po = p, and T1 = T2 =T0 =T. Then, cw.. >.... =(U1 - u~p. 1 + P< r'i - Y2 lp. r - ns1 - s,.>p. r (8.59) The Gibbs junctions G is defined as G=H-TS =U+pV-TS (8.60) Then for two equilibrium states at the same pressure p and temperature T (G1 - G1)p. r =(U1 - Uz)p. r + p(Y1 - Yi\i. r -T{S, - S1\. r (8.61) From Eqs (8.59) and (8.61) (W.. ),.... = (G, -G2)p,T {8.62} p. T ( W)p.T S ( G1 - G2)p, r (8.63) T1le decrease in the Gibbs function of a system sets an upper limit to the work that can be performed, exclusive of pdV work, in any process between two equilibrium states at the same temperature and pressure, provided the system exchanges heat only with the environment which is at the same temperature and p.ressure as the end states of the system. If the 'Process is irreversible, the useful work is less than the maximum. ,, Iii I ' "  Basic and.Applied 17imnodynamia 8.10 Irreversibility and Gouy-Stodola Theorem The actual work d.one by a system is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process. (8.64) This is also sometimes refened to as 'degradation' or 'dissipation'. For a non-flow process between the equilibrium states, when the system exchanges beat only with the environment I= [(U1 - U,)- TJS1 - S2)J-((U1 - U2) + Q] = T0 (S2-S1)-Q = To(M>sy,tcm + To(AS)surr = To{(M),yns of Gouy-Stodola EtJUation (a) Heat Transfer through a Finite Temperature Difference If heat transfer Qocc\llS from the hot reservoir at temperature T1 lo the cold reservoir at tcmperatw:e T2 (Fig. 8.14a) II '  ~235 T2 (bl Fig. 8.14 DulTudion ofaoailabk work or e:cn1r1 f,y luat ITa,ufrr througlr a finite t1111ptratart diffirttl(t ;, i>.T,-T. slffl =.lL :,; -.Ji. 7j =Q-'--" 1j 7; and. ·( 7i) · 7i-7i W.m=Q 1 - - =Q-- 1j 1j lfrmc = T2Spa ff the heat transfer Q from T1 to T2 takes place through a reversible engine£, the entire work output Jf'is dissipated in the brake, from which an equal amowtt of heat is rejected to the reservoir at T2 (Fig. 8.14b). Heat transfer through a finite temperature difference is equivalent to the destruction of its exergy. (b) Flow with Friction Let us consider the steady and adiabatic now of an Ideal gr:u through Ille segment ofa pipe (Fig. 8. lSa). By the fint law, and by the second law, Tds =dh vdp 2 2 dlr 2 2 I J J--J ds.. IT IT.E..dp =-fIT.E..dp f Insulation f , v ad/f(////( //C(! s, "7".7"7"7"7"'77--,..,...,,...,....,.-,..._ Bz --j-... -fl'- (mb) out ,')77777?7777777777~/ '. P, containing air at I00 kPa, 400K. An internally reversible Carnot heat pump is then thermally connected between them so that it heats one up and cools the other down. In order to transfer heat at a reasonable rate, the teroper,ttore di.f'fere11ce between the working fluid inside the beat pump and the air in the cont.ii.ners is sci to 20°c. The p1·occss stops when the air in the coldest tank reaches 300K. Find the final temperature of the air that is heated up, ihc work input to the hc:at pump, and the overall second law efficiency. An.r. 550 K, 31.2 k.J, 0.816 "';  Properties of Pure Substances A pure substance is a substance of constant chemical composition throughout its mass. It is a one-component system. It may exist in one or more phases. 9.1 Jrv Diagram for a Pure Substance Assume a unit mass of ice (solid water) at -I 0°C and I atm contained in a cylinder and piston machine (Fig. 9.1 ). Let the ice be heated slowly so that its temperature i:; always uniform. The changes which occur in the mass of water would be traced as the temperature is increased while the pressure is held constant. Let the state changes of water be plotted on p-v coordinates. The distinct regimes of heating, as shown in Fig. 9.2, are: 1 aim Fig. 9.1 Healing ofH20 at a co,utanl pms,m of 1 aim 1-2 The temperature of ice increases from - I 0°C to 0°C. The volume of ice would increase, as would be the case for any solid upon heating. At state 2, i.e. 0°C, the ice would start melting. 2-J lee melts into water at a constant temperature of 0°C. At state 3, the melting process ends. There is a decrease in volume, which is a peculiarity of water. I I ,., !,I ' II 280=- 81JJ1t and Ap;lied Tlttrmodynamiu 3-4 The temperature of water iacrcases, upon heating, from 0°C to 100°C. The volume of water increases bcc:ause of thermal expansion. 4-5 The water starts boiling at state 4 and boiling ends at state 5. This ,phase change from liquid to vapour occurs at a constant temperature of 100°C (the pressure being constant at I atm). There is a large increase in volume. 5-6 The vapour is heated to, say, 250°C (state 6). The volume of vapour increases from v5 to v6 Water existed in the solid phase between 1 and 2, in the liquid phase between 3 and 4, and io the gas phase beyond 5. Between 2 and 3, the solid changed intO' the liquid ·phase by absorbing the latent heat of fusion and between 4 and 5, the liquid changed into the vapour phase by absorbing the latent heat of vaporization, both at constant temperature and pressure. The states 2, 3, 4 and 5 are known as saturation states. A saturation state is a state from which a change of phase may occur without a change of pressure or temperature. State 2 is a saturated ,folid state because a solid can change into liquid at constant.Pressure and temperature from state 2. States 3 and 4 are both saturated liquid states. In state 3, the liquid is saturated with respect to solidification, whereas in state 4, the liquid is saturated with respect to vaporizaiion. State 5 is a saturated vapour state, because from state 5, the vapour can condense into Iiquid without a change of.pressure or temperature. If the heating of ice at -10°C to steam at 250°C were done at a constant pressure of 2 atm, similar regimes of heating would have been obtained with similar saturation states 2. 3, 4 and 5, as shown in Fig. 9.2. All the state changes of the system can similarly be plotted oo thep-vcoordinates, when it is heated at ditfe.rcnt constant pressures. All the saturated solid states 2 at various pressures are joined by a line., as shown in Fig. 9.3. t :.:.. ;~ ; ;-i ir-7~-~- -~,~-~ 3 4 1 2 ~ 6 -···- - - -... v Fig. 9.2 Cliangu in tlu aolumt a/ wattT during lualing at co,u~nt pressure Similarly, all the saturated liquid states 3 with respect to solidification, all the saturated liquid states 4 with respect to vaporization, and all the saturated vapour states 5, are joined together. Figure 9.4 shows state changes of a pure substance other ihan water whose volume increases on melting. The line passing through all the saturated solid states 2 (Figs 9.3 and 9.4) is called the saturated solid line. The lines passing through all the saturated liquid states 3 and 4 with respect to solidification and vaporization respectively are ,, ' II  Propertfa of Pim Sul»tanas -=281 known as the saturated liquid lines. and the line passing through all the snturated vapour states 5, is the saturaJed vapo1ir line. The saturated liquid line with respect to vaporization and the saturated vapour line incline towards each other and form what is known as the saruratio,i or i apm1r dome. The two lines meet al the critical state. Q. 4 Saturated vapQUr lin.e I Saturated_ Hquid Jnes -v Flg. 9.3 p-u diagrom ofWflitr, whose 'DOlamt decuasts OIi mt/ling -----v Fig. 9.4 ~r, duz,,tr,n of P " sw/lwrlrtr otNr tflon woltr, w,\ow wt.w itu1tc1a 4lt ""l.tutt To the left of the saturated solid line is the solid (S) region (Fig. 9.4). Between the satunued solid line and saturated liquid line with respect to solidification there exists the solid-liquid mixture (S + L) region. Between the two s:itunlled liquid lines is the compre.ued liquid region. The liquid-vapour mixture region (L + V) exists with.in the v11pour dome between the saturated liquid and saturated vapour line.~. To the right of the saturated vapour line is the vapour region. The triple point is a line on the p-v diagram. where aU the three phases, solid, liquid.. ,. 282~ Basit and Applied Tltermodynamies and gas, exit in equilibrium. At a pressure below the triple point line, the substance cannot exist in the liqujcJ phase, and the substance, whe.n heated, transforms from solid to vaponr (knov,11 as sublimation) by absorbing the latent heat of sublimation from the surroundings. The region below the triple point line is, therefore, the solid-vapour (S + JI) mixture region. Table 9.1 gives the triple point data for a number ofsubstances. Table 9.1 Tripl,-Poinl /)ata S11bstance Temperature. K Pressure, mm Hg Acetylene, C2 H2 192.4 962 Ammonia. NH 1 195.42 45.58 Argon, A