ME6301- ENGINEERING THERMODYNAMICS

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1 ME6301- ENGINEERING THERMODYNAMICS UNIT-I PART-B 1. During a flow process 5kw paddle wheel work is supplied while the internal energy of the system increase in one minute is 200kJ.Find the heat transfer when there is no other form of energy transfer. Given data: To Find: Heat transfer (Q) Solution: Work done=5kw (since work is supplied to the system) Internal energy, u=200kj/min=200kj/sec=3.33kj/sec By first law of thermodynamics, Q=w+ u = Q=-1.67kw 2. In an internal combustion engine, during the compression stroke the heat rejected to the cooling water is 50kj/kg and the work input is 100kJ/kg. Calculate the change in internal energy of the working fluid starting whether it is a gain or loss. Given data: To Find: Q=-50kj/kg(heat is rejected) Work input, W=-100kJ/kg E Solution:

2 Q= E+W -50= E-100 E=50kj/kg Gain in internal energy=50kj/kg. 3. A Fluid system, contained in a piston and cylinder machine, oases through a complete cycle of four processes The sum of all heat transferred, during a cycle is -340kJ. The system completes 200 cycles per minute. Complete the following table showing the method for each item and complete the net rate of work output in KW. PROCESS Q(KJ/min) W(KJ/min) E(KJ/min) Solution: Using First law of thermodynamics, PROCESS 1-2 Q= E+w O= E PROCESS 2-3 E=-4340KJ/min Q= E+w 42000= E+0 E=42000KJ/min PROCESS 3-4 Q= E+w -4200= w

3 =w W=69000KJ/min Q=-340KJ The system completes 200 cycles/min Qcycle 200=Q1-2+Q2-3+Q3-4+Q = Q4-1 Q4-1= KJ/min Since cycle integral of any property is zero E1-2+ E2-3+ E3-4+ E4-1= ( )+ E4-1=0 E4-1=35540KJ/min Q4-1= E4-1+W = W4-1 W4-1= KJ/min Since Q cycle = W cycle Q= Qcycle-68000KJ/min Rate of work output=-68000kj/min =-68000/60KJ/sec = Kw The complete table is given below, PROCESS Q(KJ/min) W(KJ/min) E(KJ/min)

4 A Piston and cylinder machine contains a fluid system which process through a complete cycle of four processes. During a cycle, the sum of all heat transfer is -170KJ. The system completes 100 cycles/min.complete the following table showing the method for item and completes the net rate of work output in KW PROCESS Q (KJ/min) W(KJ/min) E(KJ/min) a-b b-c c-d d-a [Similar as above problem] 5. Mass of 15 kg of air in a piston cylinder device is heated from 25 to90 by passing current through a resistance heater inside the cylinder.the pressure inside the cylinder is held constant at 300Kpa during the process and a heat loss of 60KJ occurs. Determine the electrical energy supplied in kw-hr and changing in internal energy. Given data: M=15kg; T1=25 =25+273=298k T2=90 =90+273=363k P1=P2=300kpa=300KN/m² To find: Q=-60Kj a) Electrical energy supplied b) Changing in internal energy Solution: Work done, W=mR(T2-T1)

5 = ( ) = KJ Work done, W in KW-hr=workdone 3600 = Electrical energy supplied = ³Kw-hr Change in internal energy= U=Q-W = U= KJ 6. 5Kg of air at 40 and 1 bar is heated in a reversible non-flow constant pressure until the volume is doubled find (a) Change in volume(b)work done(c) change in internal energy and (d)change in enthalpy. Given data: M=5Kg T1=40 c P1=1bar=100kN/m² V2=2V1 P=constant To find: a) V2 -V1=? b) W=? c) U=? d) H=? Solution From ideal gas equation, P1 V1=mR T1 V1=

6 100 V1=4049m³ The final volume V2= V1 = V2 =8.98 m³ 1) Change in volume V2 -V1= =4.49m² 2) Work done transfer W=P(V2 -V1) =100(4.49) =449KJ 3) Change in internal energy U=mCv(T2-T1) V1/ V2= T2/ T1 T2= T1( V1/ V2) =313 ( 2 V1/ V2) =626k 4) Change in enthalpy H= mcp(t2-t1) = ( ) = KJ 7. An ideal gas of molecular weight 30 and specific heat ratio 1.4 is compressed according to the law pv1.25=c from 1 bar absolute and 27 c to a pressure of 16 bar calculated the temperature

7 at the end of compression the heat received or rejected work done on the gas during the process and changes in enthalpy, assume mass of the gas as 1kg Given data: Molecular weight (M) =30 Cp/Cv=ɤ=1.4 M=1kg P1=1bar=100KN/m² P2=16 bar=1600kn/m² T1=27 c+273=300k Pv1.25=C To find; T2, Q and w Solution: For polytrophic process the P, V and T relation. T2/T2= (P1/ P2) n-1 T2= T1 (P1/ P2) n-1/n T2=300(1600/100) /1.25 T2=522.33k Work done W=mR (T2-T1) n-1 Gas constant R=Ru/M =8.314/30 R=0.277KJ/kgk W= ( )

8 W= KJ Heat transfer Q= W (r-n/r-1) = ( /1.4-1) Q= KJ Change in enthalpy H= mcp(t2-t1) = ( ) H=233.49KJ 8. 50kg/min of air enters the control volume in a steady flow system at 2 bar and 100 c and at an envelope elevation of 100m above the datum. The same mass leaves the control volume at 150m elevation with a pressure of 10 bar and temperature of 300 c. The entrance velocity as 2400m/min and the exit velocity is 1200m/min. During the process 50000kj/hr of heat is transferred to the control volume and the rice in enthalpy is 8kj/kg. Calculate the power developed. Given data: 1. M=50kj/kg=50/60=0.83kj/sec 2. P1=2bar=200kN/m 2 3. T 1 =100 c=373k 4. Z 1 =100m 5. Z 2 =150m 6. P 2 =10bar=1000KN/m 2 7. T 2 =300 c=573k 8. C 1 =2400m/min=2400/60=40m/sec 9. C 2 =1200m/min=1200/60=20m/sec 10. Q=50000KJ/hr=50000/3600=13.89KJ/sec

9 11. H 2 -h 1 =8KJ/kg To find: Solution: Power developed P=? SFEE is Power developed Gz 1 +c 1 2 /2+u 1 +p 1 v 1 +Q=gz 2 +c 2 2 /2+u 2 +p 2 v 52 +W W=g (z 1 -z 2 ) +c 1 2 -c 2 2 /2+ (h 1 -h 2 ) +Q W=9.81( ) / W=5999.5J/kg W=6Kj/kg P=W mass = P=4.98KJ/sec P=4.98KN 9. In a steady flow of air through a No 22 the enthalpy decreases by 50kj between two sections. Assuming that there are no other energy changes than the kinetic energy determine the increase in velocity is 90m/s. Given data: Enthalpy decrease (h 1 -h 2 ) = 50kj = J Velocity at section 1 C 1 =90m/sec To find: Solution: We know that Increases in velocity (C 1 -C 2 ) =?

10 Steady flow energy equation for a nozzle is H 1 -h 2 =C 2 2 -C 1 2 /2 Velocity at exit C 2 = 2(H 1 -h 2 )+ C 1 2 Increase velocity = C 2 =328.78m/s C 1 -C 2 = =238.78m/s. 10. A room for four persons has two for each consuming 0.18KW power and three lamps. Ventilation air at the rate of 80kg/hr enters with an enthalpy of 84kj/kg. In each person put out heat of the rate of 630Kj/hr. Determine the rate at which heat is to be removed by a room cooler so that a steady state is maintained in the room. Given data: To find: Solution: N p =4(person); n f =2 W f =0.18kw W l =100w Mass of air m=80 kg/hr=80/3600=0.022kg/s h 1 =84kj/kg h 2 =59kj/kg Q p =630kj/hr Rate of heat is to be removed =? Assuming that, E=m (h 1 -c 1 2 /2+gz 1 +Q)-m (h 2 +C 2 2 /2+Z 2 g+w) C 1 2 -C 2 2 /2=0. (z 1 -z 2 )g=0.

11 Q=E-m (h 1 -h 2 )-w E=-npQp =-4 630/3600 W=-0.7Kw m (h 1 -h 2 )=80/3600(84-59) =0.55kw W=electrical energy in put W=n f w f +n e w e = /1000 W=0.66kw Q= E- m (h 1 -h 2 )-w = Q=-1.916kw.

12 UNIT - II PART-B 1. A heat engine is used to drive a heat pump. The heat transfer from the heat engine and from the heat pump is used to heat the water circulating through the radiators of building. The efficiency of the heat engine is 27% and COP of the heat pump is 4. (i) Draw the heat diagram of the arrangement and (ii) evaluate the ratio of heat transfer to the circulating water to the heat transfer to the heat engine. [Oct 95] Given data: H.E = 27% COP H.E = 4 TO FIND: SOLUTION: =? H.E = = 0.27 = 1- = 0.73 QR 1 = 0.73 QS 1 (1)

13 COP H.P = = 4 = (2) Substituting (1) (2) 4 = = QR 2 = 1.08 QS 1 (3) Total heat supplied to the water, Q = QR 1 + QR 2 = 0.73 QS QS 1 = 1.81 QS 1 = 1.81 RESULT: The ratio of heat transfer to the circulating water to the heat transfer to the heat engine, = A Carnot heat engine takes heat from an infinite reservoir at 550 c and rejects it to a sink a 275 c. Half of the work delivered by the engine is used to run generator and the other half is used to run heat pump which takes heat at 275 c and rejects it at 440 c. Express the heat rejected at 440 c by the heat pump as % of heat supplied to the engine at 550 c. If the operation of the generator is 500KW, find the heat rejected per hour by the heat pump at 440 c. GIVEN DATA: T 1 = 550 c T 2 = 275 c T 4 = 440 c Wg = 500 KW TO FIND: 1. =? 2. Heat rejected, QR 2 =? SOLUTION: For Carnot heat engine

14 = = Q R1 = Q R1 (1) Work done by the heat engine, W = Q S1 Q R1 = Q R1 Q R1 = Q R1 Generator input, Wg = = Q R1 Work input to the heat pump, W HP = Q R1 Heat rejected by the heat pump, Q R2 = Q S2 + W HP = Q S Q R1 (2) For reverse heat pump, Q S2 = Q R2 = Q R2 Q S2 = 0.77 Q R2 Substituting Q S2 (2) Q R2 = 0.77 Q R Q R1 = 0.77 Q R Q S1 From characteristic gas equation, PV = mrt V 1 = = = 2.17 m 3 From = V 2 = = = m 3

15 Similarly, = T2 = 303 = K = c Change in entropy during compression, S 2 S 1 = mc V Ln = mc P Ln S 2 S 1 = Ln = KJ/K Process 2-3 is a constant volume process Change in entropy,s 3 S 1 = mc V ln = ln = KJ/K RESULT: 1. Final volume at the end of compression, V 2 = m 3 2. The corresponding temperature, T 2 = c 3. Change of entropy during compression S 2 S 1 = KJ/K 4. Change of entropy during constant volume, S 3 S 2 = KJ/K 3. One Kg of Ice at -5 c is exposed to the atmosphere which is at 20 c. The ice melts and comes into thermal equilibrium with the atmosphere (i) Determine the entropy increase of the turbine (ii) what is the minimum amount of work necessary to convert the water back to ice at -5 c? Assume CP for Ice as KJ/Kg K and the latent heat of fusion of ice as KJ/Kg. Given data: Ti = -5 c = = 268 K Ta = 20 c = = 293 K CPi = KJ/Kg K L = KJ/Kg To Find: 1. Entropy increase of universe ( S) univ =? 2. Minimum amount of work W/min =? Solution:

16 Heat absorbed by air from atmosphere ( ) = Heat absorbed in solid phase + Latent heat + Head absorbed in liquid phase = mc pi (To-Ti) + ml + mc pw (Tu-To) Assuming m =1 Kg and cpw = KJ/Kg K Q = (0-1-5) (20-0) = KJ Entropy change of atm ( S) atm = - = = KJ/K Entropy change of system, ( S) system = ( S) ice + ( S) fusion + ( S) liquid = + = ln ln = KJ/K Entropy of universe, ( S) univ = ( S) sys + ( S) atm = ( S)univ = KJ/K If water is to be converted back to ice using a reversible refrigerator heat to be removed from water. Q = KJ Now, ( S) sys = KJ/K But ( S) atm = ( S) ref = 0 (as the refrigerant operates in a cycle) ( S) univ = ( S) sys + ( S) ref +( S) atm Q + W KJ W , so, W min = KJ

17 4. Two reversible heat engines A and B are arranged in series. A rejecting heat directly to B. engine receive 200 KJ at a temperature of 421 c from a hot source while engine B is in communication with a cold sink at a temperature of 4.4 c if the work output of A is twice that of B, find (i) The intermediate temperature b/w (A) & (B) (ii) The efficiency of each engine, and (iii) The heat rejected to the cold sink Given data: 1. T H = 421 c = = 694 K 2. T L = 4.4 c = = 277.4K 3. Q S1 = 200 KJ, W A = W B To find: 1. The intermediate temperature b/w (A) & (B), T =? 2. The efficiency of each engine A & B =? 3. The heat rejected to the cold sink iq R2 =? Solution: Work output from engine A, W A = Q S1 Q R1 = 200- Q R1 For reversible heat engine, (1) o, = QR1 = (2) o, W A = But W B = (3) ( ) and also WB = Q S2 Q R2 = Q R (4) Equating equations (4) and (5) = Q R2 Q R2 = (5) Similarly, for reversible engine B,

18 so, T = K (or) T = c So, Q R1 = = KJ and Q R2 = Q R2 = KJ Efficiency of engine A, A = 1- Efficiency of engine A, B = 1- = = = 40.04% = 33.39% Result: 1. The intermediate temperature between A and B, T = c 2. The efficiency of each engine A = 40.04% and B = 33.39% 3. The heat rejected to the cold sink Q R2 = KJ.

19 UNIT-III PART-B 1. A vessel of volume 0.04m 3 contains a mixture of saturated water and steam at a temperature of 25 0 C. The mass of the liquid present is a 9 Kg find the pressure, mass, specific volume, enthalpy, entropy and Internal energy. Given data V=0.04m 3 T=250 o C M 1 =9kg To find: P, m, v, h, s and U Solution: From steam tables corresponding o C read V f = V 1 = m 3 /Kg V g = V s = m 3 /Kg P = bar Total volume occupied by the liquid V 1 = m 1 v 1 = 9 x V 1 = m 3 Total volum of the vessel V = Volume of liquid + volume of steam V = V 1 + V s 0.04 = Vs V s = m 3 Mass of steam, Ms = =0.574 kg Mass of mixture of liquid and steam, m = m 1 + m s M = =9.574 kg Total Specific volume of mixture

20 V= We know that V = V f + xv fg = x ( ) X = 0.06 From steam tables corresponding to 250 o C, h f = KJ/Kg h fg = KJ/kg sf = KJ/KgK sfg = KJ/Kg K. Enthalpy of mixture, h = hf + xhfg = x H = KJ/kg Entropy of mixture, S = sf + xsfg = x S = 2.99 KJ/Kg K Internal energy u = h-pv = x 10 2 x ) u = 1172 KJ/kg 2. In steam generator compressed water at 10 MPa, 30 o C enters a 30mm diameter tube at the rate of 3 litres1sec. Steam at 9 MPa, 400 o C exist the tube. Find the rate of heat transfer. (November -2003) Given data P 1 = 10 bar T W = 30 o C D = 30mm = 0.03m N 1 = 3 litres / sec = m 3 /sec P 2 = 9 bar T 2 = 400 o C To find : Q Solution: From steam tables corresponding to 30 o C V f1 = m 3 /kg H f1 = h 1 = KJ/Kg

21 V 1 = mvf1 Mass flow rate of steam m = M = Kg/s Area of the tube A = A = 7.068x10-4 m 2 Inlet Velocity C 1 C 1 From steam tables corresponding to 9MPa and 400 o C V 2 = m 3 /kg H 2 = KJ/kg Final velocity, C 2 = C 2 = m/s. From SFEE (steady How energy equation) Q = KW Result: The rate of heat Transfer to the water Q = KW 3. Steam at 0.8MPa, 250 o C and flowing the rate of 1 Kg / s passes into a pipe carrying wet steam at 0.8MPa, 0.95 dry, after adiabatic mixing the flow rate is 2.3 kgs /s. Determine the properties of the steam after mixing. Given data P 1 = 0.8 MPa = 8 bar T 1 = 250 o C M 1 = 1 Kg/s P 2 = 0.8 MPa = 8 bar X 2 = 0.95 M 3 = 2.3 kg/s

22 To find Properties Solution The sum of mass of the steam before mixing = The sum of mass of the steam after mixing. M 1 + m 2 = m 3 M 2 = m 3 m 1 = = 1.3 Kg/sec The energy balance equation for adiabatic mixing. M 1 h 1 + m 2 h 2 = m 3 h 3 1 Corresponding to 8 and 250 o C h 1 = KJ/Kg Corresponding to8 bar, read hf and hfg h f2 = KJ/kg. h fg2 =2046.5KJ/kg h 2 = h f2 + x 2 h fq2 = x h 2 = KJ/Kg Sub all the values to the equation 1 1x x = 2.3 x h 3 H 3 = KJ/kg Corresponding to 8 bar, read, hg = KJ/Kg Since h37hg the steam is in super heated condition from the molier chart, corresponding to 8 bar and h 3 = KJ/kg Super heated temperature T 3 = 180 o C Entropy, S 3 = KJ/Kg K Specific volume, V 3 = 2.5m 3 / Kg The conditions of steam, after mixing is 0.8 MPa and180 o C Result: The conditions of steam after mixing is 0.8 MPa and 180 o C Enthalpy, h 3 = KJ/kg Entropy, S 3 = 6.645KJ/Kg K Specific volume V 3 = 2.5 m 3 /kg 4. In a steam power plant operating on an ideal reheat Rankine cycle, the steam enters the High-pressure turbine at 3 MPa and 400 o C after expansion to 0.6MPa. The steam is reheated to 400 o C and then expanded the logo pressure Turbine to the condenser

23 pressure of 10kPa. Determine the thermal efficiency of the cycle and the quality of the steam at the outlet of the pressure turbine Given data: P 1 = 3 MPa and T 1 = 400 o C P 2 = 0.6 MPa P 3 = 10 kpa To find Reheat x 4. Solution: From Super heated steam table. A+ P 1 = 30bar and 400 o C H 1 = KJ/kg S 1 = KJ/Kg K From saturated steam table. Sq 2 = KJ/KgK; Sf 2 = KJ/Kgk Sfg 2 = KJ/Kg K: hfg 2 = KJ / Kg Hf 2 = KJ/Kg Since S 1 >Sg 2 the steam is in super heated conditions, So equating the superheated entropy at 6 bar you will get superheated temperature. By interpolation superheated temp is192 o C from super heated table at 6 bar and 192 o C. h 2 = KJ/kg From super heated steam table at 6 bar and 400 o C h 3 = KJ/kg: S 3 = KJ/KgK From Saturated steam table at 0.15 bar. Hf4 = KJ/Kg hfg4= KJ/kg Sf4 = KJ/Kgk Sfg4 = KJ/KgK Vf4 = m 3 /Kg We know that S 3 = S 4 = Sf 4 + X 4 x Sf g = X 4 x 7.502

24 X 4 =0.941 Dryness fraction of stream at the end of the Turbine. X 4 = h 4 = hf 4 + X 4 x hf g4 = x h 4 = KJ/kg Enthalpy of steam the end of the Turbine. h 4 = KJ/kg h 5 = h f4 = KJ/kg Pump work Wp = V f4 (P 1 -P 4 ) = ( ) WP = KJ/kg Efficiency of reheat Rankine Cycle Reheat = Reheat = % = Result Efficiency of reheat Rankine cycle % Quality of steam at outlet of L.P. Turbine X 4 = = 94.1% dry) 5. A reheat cycle operating between 30 and 0.04 bar has a superheat and reheat temperature of 450 o C. The first expansion takes place till the steam is dry saturated and then reheat is given. Neglecting feed pump work, determine the ideal cycle efficiency.

25 Given data: P 1 = 30 bar, P 4 = 0.04 bar, T 1 = 450 o C, T 3 = 450 o C, X 2 = 1 To find: Efficiency of the Cycle Solution From steam tables at 30 bar and 450 o C H 1 = KJ/Kg S 1 = 7.08 KJ/Kgk A = 0.04 bar H f4 = KJ/Kg s f4 = KJ/Kgk hf g4 = KJ/Kg Sf g4 = KJ/Kgk 1-2 isentropic process S 1 = S 2 = KJ / KgK S 2 = Sg = dry saturated steam P 2 = Psat at sg (From steam table) P 2 = 2.3 bar At 2.3 bar h 2 = KJ/Kg At 2.3 bar and450 o C h 3 = KJ/Kg, S 3 = KJ/Kgk 3-4 isentropic process, S 3 = S 4 = KJ/Kg k S4 = Sf 4 + x 4 x sf g4

26 = X 4 x S4 = X 4 x X 4 = 0.98 h 4 = h f4 + x 4 x hf g4 = x h 4 = KJ/Kg The cycle efficiency M = M = Result M = = 38.73% The cycle efficiency = 38.73% 6. In a generative cycle, the steam pressure at Turbine inlet is 30 bar and the exhaust is at 0.04 bar. The steam is initially saturated. Enough steam is bled off at the optimum pressure of 3 bar to heat the feed water. Determine the cycle efficiency, neglect pump work. Given data P 1 = 30 bar P2 = 3 bar P3 = 0.04 bar To find:

27 cycle Solution: From Steam tables A + 30 bar and dry h 1 = KJ/Kg S 1 = KJ/Kg K At 0.04 bar h f3 = KJ/kg s f3 = KJ/Kg K h fg3 = KJ/kg S fg3 = KJ/ KgK V f3 = m 3 /kg A+3 bar H f2 = KJ/Kg ; h fg2 = KJ/kg S f2 = KJ/Kgk ; S fg2 = KJ/Kg k S g2 = KJ/Kgk ; F fb = m 3 / kg 1-2 = isentropic process S 1 = S 2 = KJ/KgK Since S g2 >S 1 : the condition of steam is wet S 1 = S f2 + X 2 x S fg = X 2 x X 2 = 0.85 h 2 = hf 2 + x 2 x h fg2 = x h 2 = KJ/Kg K Similarly process 1-3 insentropic process S 1 = S 3 = KJ/Kg K

28 S 3 = S f3 + X 3 x Sf g = X 3 x X 3 = 0.72 h 3 = h f3 + X 3 x h fg3 h 3 = x h 3 = KJ/Kg h 4 = h f3 = KJ/Kg Pump work during 4-5 Wp = (1-m)(h 5 -h 4 ) = (1-m) V f3 (P 5 -P 4 ) h 5 -h 4 = x (300-4) = h 5 = (h 4 h f3 ) h 5 = KJ/Kg Similarly for pump work during 6-7 h 7 -h 0 = V 6 (P 7 -P 6 ) = V f2 (P 1 -P 2 ) = ( ) h 7 -h 6 = (h 6 = h f2 = KJ/Kg) h 7 = h 7 = KJ/kg The amount of steam bled (m) M = M = Thermal efficiency of the Regenerative cycle = = regenerative = = 36.83% 07. A Steam Power plant uses steam at boiler pressure of 150 bar and temperature 550 o with reheat at 40 bar and 500 o C at condenser pressure of 0.1 bar. Find the quality of steam at Turbine exhaust, cycle efficiency and steam rate. (Apr 2004) Given data:

29 P 1 = 150 ba P 2 = 40 bar P 3 = 0.1 bar T 1 = 550 o C T 3 = 550 o C To find: X 4 =? : SSC =? Solution: Properties of steam from steam table at 150 bar and 550 o C h 1 = KJ/kg, S 1 = KJ/ Kg K A + 40 bar and 550 o C h 3 = KJ/Kg, S 3 = KJ/Kg K A + 40 bar T sat = o C = 523. K h f = KJ/Kg : hfg = KJ/Kg K A+0.1bar : : s f = KJ / Kg : Sfg = KJ / Kg K 1-2 isentropic S 1 = S = KJ / Kg K S 2 > S g at 40 bar Exist of H p turbine is super heat Trup 322 o C h 2 = KJ/kg

30 S 3 < S g at 0.1 bar Steam is at wet conditions S 4 = S 3 = KJ / Kg K S 4 = S f4 + X 4 x S fg4 X 4 = X 4 = h 4 = hf + x 4 x h fg4 h 4 = x h 4 = KJ / Kg K Cycle efficiency = = = x 100 = 44.26% Steam state = = = 2.16 Kg / Kw hr = 2.16 kg / kw - hr

31 UNIT IV PART-B 1. Derive MAXWELL S equations:- The Maxwell s equation relates entropy to the Three directly measurable properties P, V and T for pure simple compressible substances. From first Law of thermodynamics Q= W+ U Rearranging the Parameters Q - U + W Tds = du + pdv [ ds = Q/T by second Law of thermodynamics W = Pdv by first Law of thermodynamics] du=tds Pdv -1 We know that h = u + pv dh = du + d (pv) = du +Vdp + pdv -2 Substituting the value du in equation(2), dh = Tds-Pdv +Vdp+Pdv dh = Tds + Vdp -3 By Helmotz s function a = U-Ts da = du d (Ts) = du Tds SdT -4 (By differentiation rule, d(uv) udv+vd) Substituting the Value fo du in equation (4), Da = Tds Pdv Tds sdt = - Pdv SdT -5 By Gibbs functions G = h Ts dg = dh d (Ts) dg = dh Tds SdT -6 Substituting the value of dh in equation (6), So, dg becomes dg = Tds _ Vdp Tds - SdT ( dh = Tds + Vd) dg = Vdp SdT -7 By inverse exact differential we can write equation (1) as du = Tds pdv

32 Similarly, equation (3) can be written as dh = Tds + Vdp -8 Similarly, equation (5) can be written as -9 Similarly,equation (7) can be written as dg = Vdp sdt -10 These equations 8, 9, 10 and 11 are maxwell s equation Derive the ENTROPY relations (Tds Equation) The entropy (s) of a pure substance canbe Expressed as a function of temperature (T) and pressure (P) S = f(t, P) We know that and From Maxwell equation, we know that Substituting in ds equation, Multiplying by T on both Sides of the equation, _ 1 _ 2 This is known as the first form of entropy equation (or) the first Tds equation

33 By considering The entropy of a pure substance as a function of temperature and specific volume. i.e. s = f(t,v) We know that From Maxwell Equations, We know that Substituting in Ds Equation, ds = Multiplying by T, Tds = _ 3 This is known as the second form of Entropy equation (or) the Second Tds equation 3. Derive CLAPEYRON equation Clapeyron equation which involves the relationship between the saturation pressure, Saturation, Temperature, the enthalpy of evaporation and the Specific volume of the two phases involved. This equation provides a basis for calculation of properties in a two phase region. It gives the slope of a curve Separating the two phases in the P-T diagram. Let, Entropy (s) is a function of Temperature (T) and volume (V) i.e. S = f (T, V) ds = _ 5 When the phase is changing from saturated liquid to saturated Vapour temperature remains constant. So, ds equation reduces to ds = _ 6 [ Temperature remains constant dt = 0] From Maxwell equations, We know that, Substituting in Equation (6)

34 S = 7 The term is the slope of the saturation curve. Integrating the above equation between saturated liquid (f) and saturated vapour (g), _ 8 ( S fg = S g S f, V fg V g V f ) From Second Law of Tharmodynomics, We knowthat ds = For constant pressure process dq = dh d3 = S fg = Substituting in Equation _ 8 _ 9 This equation is known as clapeyron Euation. 4. Derive JOULE THOMSON co-efficient Joule Thomson Coefficient is defined as the change in temperature with change in pressure keeping the enthalpy remains constant. It is denoted by Throttling process:- Throttling process is defined as the fluid expansion through a minute Orifice or slightly opened Value. During the throttling process, Pressure and velocity are reduced. But there is no heat transfer and no work done by the system. In this process, enthalpy remains constant.

35 Joule Thomson Experiment:- In this experiment, a stream of gas at a pressure P 1 and Temperature T 1 is allowed to flow continuously through a porous plug. The gas comes out from the other side of the porous plug at a pressure and Temperature. The whole apparatus is completely insulates. Therefore, no heat transfer takes place i.e. Q = 0 The system does not exchange work with the surrounding. So, W = OF from steady flow energy equation, we know that, Since There is no considerable change velocity V 1 = V 2 and Z 1 = Z 2 Q = 0 W= 0 V 1 = V 2 and Z 1 = Z 2 are applied in steady flow energy equation so the equation (1) becomes h 1 = h 2 Enthalpy at inlet, h 1 = Enthalpy at outlet h z It indicates that the Enthalpy in constant for throttling process It is assumed that a series of Experiment performed on a real gas keeping the initial pressure P 1 and Temperature T 1 constant with various reduced down steam pressures. It is found that the down steam temperature also changes. The results from these experiments can be plotted as constant enthalpy curve on T-p plane. The slope of a constant Enthalpy is known as Joule Thomson Co-efficient-it it denoted by Case (i) There is always pressure drop in throttling process - & 0 P and temperature change are negative. Therefore is positive. This throttling process produces cooling effect Case (ii) There is always pressure drop in throttling process So p in negative. When the temperature change in positive u is negative

36 This throttling process produces heating effect Case (iii) When is zero the temperature of the gas remains constant with throttling. The temperature at which = 0 is called inversion temperature for a given pressure Inversion Curve: The maximum point on each curve is called inversion point and the lows of the inversion point is called inversion curve. A Generalized equation of the Joule Thomson Co-efficient can be derived by using change of enthalpy equation. WKT dh = = = Dividing by C p on both sides _ 3 Differentiating this equation with respect to pressure at Constant enthalpy. _ 4 From eqn 4, we can determine the Joule Thomson Co-efficient Joule Thomson Coefficient for ideal gas: The Joule Thomson. Coefficient is defined as the change in temperature with change in pressure, keeping the enthalpy remains constant. It is denoted by W.K.T PV = RT Differentiating the above equation of state with respect to T by keeping pressure p constant

37 =0 It implies that the Joule Thomson Coefficient is Zero. Constant temperature coefficient, Lets Enthalpy is a function of pressure and temperature i.e., h = f (p, T) For throttling process, enthalpy remains constant. h = C dh = 0 Sub-dk value in eqn Diving by dt. The property C p = is known as constant temperature to efficient. 5. A mixture of 2 kg Oxygen and 2 kg Argon is in an insulated piston cylinder arrangement at 100Kpo. 300k. the piston now compresses the mixture to half its initial volume. Molecular weight of oxygen is 32 and for organ is 40. Ratio of specific heats for oxygen is1.39 and for argon is i-667. Given data: Mo z = 2 kg M AR = 2 kg

38 P 1 = 100 kpa T 1 = 300k V 2 = ½ V 1 MO 2 = 32 M ar = 40 r oz = P ar = Solution Mars fraction XO z = PO 2 = XO 2 x p = 0.56 x100 = 556kpa Similarly, X Ar = P a = x 100 = 44.4 kpa From equation of state V oz = = = 2.804m 3 = 2.81 M 3 Volume after compression Ratio of Specified heat of mixture V = X O2 V O2 + X AR V AR = 0.55 x x = Jnoulot process refers reersibel adiabatic process P 1 V 1 r = P 2 V 2 r P 2 = = 285 Kpa Similarly

39 T z = k Piston Work W = = KJ = -IMJ (-Ve sign indicates the work input to the piston) kg of Co and 1 Kg of air is contained in a vessel of volume 0.4m 3 at 15 o C. Air has 23.3% of O 2 and 76.7% of N 2 by mass. Calculate the partial press are of each constitutent and total pressure in the vessel Motor masses of Co. Co 2 and N 2 are and 28kg/km Given Date M co = 0.49kg M ar = 1 kg V = 0.4m 3 T=15 0 C X O2 = X n2 = M co = 28kg / kmol M CO2 = 32 kg kmol M N2 = 28kg / kmol Solution R a = X co = R = X co R co + X a R a = x = KJ/KgK P CO = x = kpa P a = x = kpa WKT, Air contains 23.3% O 2 and 76.7% N 2 partial pressure of 02 = x = kpa Partial pressure of N 2 = x = 130.3kpa

40 UNIT-V PART B 1. Air at 20 o C, 40% RH is mixed adiabatically with air at 40 o C, 40% RH in the ratio of 1 Kg of the forces with 2kg of the Latter (on dry basic) find the final condition of air (Nov 03) Given data:- Dry bulb temperature td 1 = 20 o C Relative humidity Q 1 = 40% Dry bulb temperature td 2 = 40 o C Relative humidity Q 2 = 40% Solution:- KlKT By mass balance M1 +m2 = m3 M1w1 + m2w2 = m3w3 By energy balance M1h1 + m2h2 = m3h3 Sub the value of m3 ch eq m1w1+m2h2 = (m1+m2)w3 m1w1 m1w3 = m2w3 m2w2 m1(w1-w3) m2 (w3-w2) From Psychrometriy chart w1 = kg/kg of dry w2 = kg/kg of dry W3 = kg/kg of air Similarly m1h1 + m2h2 = (m1+m2) h3 m1h1 m1h3 = m2h3 m2h2 m1(h1-h3) = m2 (h3-h2) From psychrometry h1 = 35 KJ/kg H2 = 90 KJ/Kg

41 ½ = h3 = KJ/kg 2. An air water vapour mixure at 0.1 mpa, 30 o C 80% RH has a volume of 50m 3, calculate the specific humidity, dew point, wet bulb temp, mass of dry air andmass of water vapour Given data P1 = 0.1 mpa Td =30 o C Q =80% Va = 50 m 3 Solution:- 1. Relative humidity Q = Pv/PS From steam table. K/e find that fro 30 o C dry bulb temperature corresponding pressure is bar. Ps = bar Q = PV/PS 0.8 = PV / Vapour Pressure Pv = bar Specific humidity w = PV / Pb-Pv = x W = Kj/Kg of dry air From ST corresponding to vapour Pr. Pv = bar temp is o C So the (DBT) Dew point temp tdp = o C From gas Law Pava = marat Ma = When Volume Va = 50m 3 Gas constant Ra = KJ/kg-K Temperature T = =303K Pb = Pa + Pv Pa = Pb Pv = Pa = bar Pa = x 100 kpa = 96.6 kpa (1 bar = 100 kpa)

42 Ma = Specific humidity W= = Mv = kg From psychometry chart, corresponding to dry bulb temperature 30 o C and relative humidity 80% the wet bulb temp is 27 0 C Tw = 27 o C 3. Air at 16 o C and 25% relative humidity passes through a heater and then through a humidify to reach final dry bulb temperature of 30 o C and 50% relative humidity. Calculate the heat and moisture added to the air what is the sensible heat factor? Given: Td1 = 16 o C Q1 = 25% Td2 = 30 o C Q2 = 50% Solution Step 1 The dry of ai at 16 o C- dry bulb temp and25% (RH) is marked on the psychometric chart at point 1. The dry of air 30 o C dry bulb temp and 50% (RH) is marked on the psychometric chart at point 2. Step 2 Draw ahorizontal line from point 1 and draw a vertical line. From paint 2. There named as point 3.

43 Step 3 Draw a inclined line from pant 1 to 2 read enthalpies and specific humidty values at point 1, 2 and 3 from psychonotric chart At point 1 enthalpy h1 = 24.5 KJ/kg Specific humidity w1 = kg/kg of dry air At point 2, enthalpy h2 = 67.5 KJ/kg Specific humidity w2 = kg/kg of dry air At point3, enthalpyh3 = 38KJ/kg Heat added Q = h2-h1 = = 43 KJ/kg Moisture added w = w2 w1 = = kg / kg of dry air Sensible heat SH =- h3-h1 = = 13.5 KJ/Kg Latnet heat LH = h2-h3 = = 29.5 KJ/Kg SHF = = = Saturated air at 20 o C at s rate of 1.16 m 3 /sec is mixed adiabatically with the outside at ari 35 o C and 50% relative humidity at a rate of 0.5 m 3 /sec. Assuming a adiabatic mixing condition at 1ctm, determine specific humidity, relative humidity, dry bulb temperature and volume flow rate of the mixture. Given:- First Stream of air DBT td1 = 20 o C Flow ratev1 = m 3 /s Second stream of ai: DBT + d2 = 35 o C

44 RH Q2 = 50% Flow rate V2 = 0.5 m 3 /S Solution:- Step 1 : The first steam of air i.e. 20 o C dry bulb temp upto the saturation curve I marked on the psychometric short at point 1. Step 2 The second stream of air i.e. 35 o C, dbt & 50o of (RH) is marked in the psychometric chart at point 2. Step 3

45 Joint the points 1 and 2 from the psychrometric chart WKT Specific humidity of the first stream of air W1 = kg / kg of dry air. Speciify humidity of the second stream of air W2 = kg/kg of dry air Step 4: WKT First stream flow rate v1 = 0.3 m 3 /s (given) Mass m1 = From psychrometric chart WKT (Specific volun) (vs) passing through point 1is V 1 = m3 / kg M 1 = Second stream flwo rate v2 = 0.5m 3 /3 From psychrometric chart MKT (S.V) (V) Passing through point 2 is V2 = m 3 /kg M2 = Sub m1 & m2 Value W3 = w w3 = W3 = kg/kg of dry air Specific humidity after mixing w3 = kg/kg of dry air Step 5

46 From psychrometric chart at point 3 is m 3 /kg V 3 = m 3 /kg Q = 82% Td 3 = 23.5 o C