Unit-I, Steam Nozzles - thermal engineering

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1. Thermal Engineering-II Unit-I Steam Nozzles

2. Thermal Engineering Thermal Engineering  is a specialized sub-discipline of  mechanical engineering  and  chemical engineering  that deals with the movement of heat energy and transfer. The  energy  can be transformed between two mediums or transferred into other forms of energy. A thermal engineer will have knowledge of  thermodynamics  and the process to convert generated energy from thermal sources into chemical, mechanical, or electrical energy.  Thermal Engineering-I Thermal Engineering-II 1. Gas and Steam Power Cycles 2. Reciprocating air Compressor 3. I.C. Engines and Combustion 4. I.C. Engine Performance and systems 5. GAS TURBINES 1. Steam Nozzle 2. Boilers 3. Steam Turbines 4. Cogeneration and Residual Heat Recovery 5. Refrigeration and Air Conditioning

3. Thermal Engineering-II UNIT I STEAM NOZZLE Types and Shapes of nozzles, Flow of steam through nozzles, Critical pressure ratio, Variation of mass flow rate with pressure ratio. Effect of friction. Metastable flow. UNIT II BOILERS Types and comparison. Mountings and Accessories. Fuels - Solid, Liquid and Gas. Performance calculations, Boiler trial. UNIT III STEAM TURBINES Types, Impulse and reaction principles, Velocity diagrams, Work done and efficiency – optimal operating conditions. Multi-staging, compounding and governing. UNIT IV COGENERATION AND RESIDUAL HEAT RECOVERY Cogeneration Principles, Cycle Analysis, Applications, Source and utilization of residual heat. Heat pipes, Heat pumps, Recuperative and Regenerative heat exchangers. Economic Aspects. UNIT V REFRIGERATION AND AIR – CONDITIONING Vapour compression refrigeration cycle, Effect of Superheat and Sub-cooling, Performance calculations, Working principle of air cycle, vapour absorption system, and Thermoelectric refrigeration. Air conditioning systems, concept of RSHF, GSHF and ESHF, Cooling load calculations. Cooling towers – concept and types.

4. Contents Introduction Types of Nozzles Flow of steam through nozzles Critical pressure ratio Variation of mass flow rate with pressure ratio Effect of friction Metastable flow. Sample Problems

5. Introduction A steam nozzle may be defined as a passage of varying cross section through which heat energy of steam is converted to kinetic energy Major function is to produce steam jet with high velocity to drive steam turbines Turbine nozzles performs two functions It transforms a portion of energy of steam into kinetic energy In the impulse turbine it direct the steam jet of high velocity against blades which are free to move in order to convert K.E into shaft work Application : Steam turbines , Gas Turbines ,In jet engines, Rocket motors , Flow measurement, Spray painting

6. Types of Nozzles Convergent Nozzle Divergent Nozzle Convergent-Divergent Nozzle Convergent Nozzle: In a convergent nozzle, the cross-sectional area decreases continuously from its entrance to exit. It is used in a case where the backpressure is equal to or greater than the critical pressure ratio. Divergent Nozzle The cross-sectional area of the divergent nozzle increases continuously from its entrance to exit. It is used in a case, where the back pressure is less than the critical pressure ratio.

7. Types of Nozzles Convergent-Divergent Nozzle: In this case, the cross-sectional area first decreases from its entrance to the throat and then increases from throat to exit.it is widely used in many types of steam turbines.

8. Flow of steam through nozzles The steam enters the nozzle with a high pressure, but with a negligible velocity. In the converging portion, there is a drop in the steam pressure with a rise in its velocity. There is also a drop in the enthalpy or total heat of the steam. This drop of heat is not utilized in doing some external work, but is converted into kinetic Energy. In the divergent portion there is further drop of steam pressure with a further rise in its velocity-drop in enthalpy-Converted into Kinetic Energy Entry (High pressure, Low Velocity) - Exit (Low Pressure , High Velocity) Back pressure-The steam which leaves the nozzle Isentropic Flow No Heat Transfer (Q=0),No work Transfer (W=0)

9. Velocity of steam Flowing through the nozzle Steam enters the nozzle with high pressure and low velocity and leaves with high velocity and low pressure The outlet velocity (V2) of steam can be found as follows. Consider a unit mass flow of steam through a nozzle V 1 =Velocity of steam at entrance V 2 = Velocity of steam at any section h 1 = Enthalpy of steam entering, Kj/Kg h 2 = Enthalpy of steam at any section, Kj/Kg The steady flow process (Application: Boiler, Condenser, Nozzles, Turbines etc.,) gZ 1 +C 1 2 /2+u 1 +p 1 V 1 +Q=gZ 2 +C 2 2 /2+u 2 +p 2 V 2 +W The SFEE can be written as C 1 2 /2+ h 1 =C 2 2 /2+ h 2 +Losses WKT for unit mass flow rate (Q=0,W=0)

10. Velocity of steam Flowing through the nozzle m=1 neglect the losses C 1 2 /2+ h 1 =C 2 2 /2+ h 2 , h 1 - h 2 = C 2 2 /2 - C 1 2 /2 1000(h 1 - h 2 )= C 2 2 /2 - C 1 2 /2 (h 1 - h 2 )= C 2 2 -C 1 2 /2x1000 2000(h 1 - h 2 ) +C 1 2 = C 2 2 C 2 2 = C 1 2 +2000(h 1 - h 2 ) C 2 = √C 1 2 +2000(h 1 - h 2 ) Inlet velocity C 1 is negligible as compared to outlet C 2 = √2000(h 1 - h 2 ) C 2 = 44.72√(h 1 - h 2 ) In Actual practice there is always a certain amount of friction present between the steam and nozzle surfaces. This reduces the heat drop by 10 to 15 percent and thus the exit velocity of steam is also reduced correspondingly C 2 = 44.72√K(h 1 - h 2 )

11. Steam and its Properties Steam may be defined as water vapors suspended in air. It can be classified in wet, saturated or superheated. Vapors may be considered as a fluid in gaseous state as it may contain liquid particles and it’s temperature is not far from it’s boiling point. It is produced by heating of water and carries large quantities of heat within itself. Hence, it could be used as a working substance for heat engines and steam turbines.

12. Steam and its Properties Constant Pressure formation of steam Steam is generated in boilers at constant pressure. Consider a unit mass of ice at -20 degree Celsius and 1 atm contained in a cylinder-piston system. Let the ice be heated slowly so that it’s

13. Steam and its Properties Let the state changes be plotted on temperature – heat added condiments, as shown in figure below.

14. Steam and its Properties Enthalpy or Total heat of water ( hf ) It is defined as the quantity of heat required to raise the temperature of one kg of water from 0 C to it’s boiling point or saturation temperature corresponding to the pressure applied. It is also called enthalpy of saturated water or liquid heat and is represented by hf . hf = specific heat of water( cpw ) x rise in temperature, hf = 4.187 x Δt , kJ/kg Enthalpy of wet steam , h = hf + x hfg Enthalpy of dry steam , h =hg= hf + x hfg, (then x = 1) Enthalpy of dry steam , h = hg + cps( tsup + ts)

15. Steam and its Properties Types of Steam Enthalpy Entropy Specific Volume Wet steam h = h f  + x. h fg     s ws  = s f  + x. s fg      v ws = x.v g Dry steam h g  = h f  + h fg S= S g V=V g Superheated steam h sup  =   h g  + C s ( t sup   -t s  )   S sup = S g +Cp s log e ( T sup / T s )  V= Vsup Entropy  is defined as the quantitative measure of disorder or randomness in a system. The concept comes out of  thermodynamics , which deals with the transfer of heat energy within a system.

16. Mass of steam discharge through nozzle The isentropic process in nozzle may be approximately represented by an equation pV n =c , n= 1.135 for saturated steam, n=1.3 for superheated steam P 1 =Initial pressure of steam V 1 = specific volume of steam at entry P 2 =pressure of steam at throat V 2 =Specific volume of steam at pressure P 2 C 1 = velocity of steam at entry C 2 = Velocity of steam at exit Steam passes through the nozzle its pressure is dropped so the h is also reduced Reduction of enthalpy must be equal to increase in kinetic energy Here the work done by the steam upon itself is equal to enthalpy drop pV n =c W.D=n/n-1 (P 1 V 1 -P 2 V 2 )

17. Mass of steam discharge through nozzle Gain in Kinetic energy = Work done during isentropic C 2 2 /2 -C 1 2 /2 = n/n-1 (P 1 V 1 -P 2 V 2 ) C 1 is very less compared to C 2 it can be negligible C 2 2 /2 = n/n-1 P 1 V 1 (1-P 2 V 2 /P 1 V 1 ) ---------------(1) WKT pV n =c P 1 V 1 n =P 2 V 2 n (P 1 /P 2 ) = (V 2 /V 1 ) n (V 2 /V 1 )= (P 1 /P 2 ) 1/n Substitute V 2 /V 1 in equation 1 C 2 2 /2 = n/n-1 P 1 V 1 (1-P 2 /P 1 (P 1 /P 2 ) 1/n ) C 2 2 /2 = n/n-1 P 1 V 1 (1-P 2 /P 1 (P 2 /P 1 ) -1/n ) C 2 2 /2 = n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) C 2 =√2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n )

18. Mass of steam discharge through nozzle Mass of steam discharge through nozzle /sec Density=Mass/Volume, Mass=Density x volume ,(V=Volume/Mass , V=1/ρ) m= Volume of the steam flow/Specific volume of the steam m=Area x Velocity/Specific volume, m= A x C 2 /V 2 Substitute C 2 m= A /V 2 x √2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) Specific volume (V 2 /V 1 )= (P 1 /P 2 ) 1/n , V 2 = V 1 (P 1 /P 2 ) 1/n m= A / V 1 (P 1 /P 2 ) 1/n x √2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) m= A / V 1 (P 2 /P 1 ) 1/n x √2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) m= A / V 1 x √2 n/n-1 P 1 V 1 (P 2 /P 1 ) 2/n (1- (P 2 /P 1 ) n -1/n ) m= A / V 1 x √2 n/n-1 P 1 V 1 [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) 2/n+ n -1/n )] m= A / V 1 x √2 n/n-1 P 1 V 1 [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) n +1/n )] m= A x √2 n/n-1 P 1 V 1 / V 1 2 [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) n +1/n )] m= A x √2 n/n-1 P 1 / V 1 [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) n +1/n )]

19. Condition for maximum discharge WKT the mass of steam discharge through the nozzle m= A x √2 n/n-1 P 1 / V 1 [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) n +1/n )] There is only one value of the ratio critical pressure ratio P 2 /P 1 which will produce the maximum discharge .this can be obtained by differentiate m with respect to P 2 /P 1 and equating it to zero other quantities except the ratio P 2 /P 1 are constant d/d (P 2 /P 1 )= [(P 2 /P 1 ) 2/n - (P 2 /P 1 ) n +1/n )]=0 2/n (P 2 /P 1 ) 2/n-1 - (n+1/n) (P 2 /P 1 ) (n +1/n)-1 =0 (P 2 /P 1 ) 2-n/n = n /2 (n+1/ n ) (P 2 /P 1 ) 1/n For cancelling n multiplying n on both sides power (P 2 /P 1 ) 2-n/n x n =[ (n+1/2) (P 2 /P 1 ) 1/n ] n (P 2 /P 1 ) 2-n =(n+1/2) n (P 2 /P 1 ) 1/ nxn (P 2 /P 1 ) 1-n =(n+1/2) n

20. Condition for maximum discharge Simplifying divide by 1-n on both sides power (P 2 /P 1 ) 1-n/1-n =(n+1/2) n/1-n (P 2 /P 1 ) =(2/n+1) -n/1-n (P 2 /P 1 ) =(2/n+1) n/n-1 Substitute in equation m max = A x √2 n/n-1 P 1 / V 1 (2/n+1) n/n-1x2/n - (2/n+1) n/n-1xn +1/n )] m max = A x √2 n/n-1 P 1 / V 1 [ (2/n+1) 2/n-1 - (2/n+1) n+1/n-1 )] Sub n=1.135 and n=1.3 for saturated steam and superheated steam m max = 0.6356√2 P 1 / V 1 for saturated steam m max = 0.667√2 P 1 / V 1 for superheated steam When the flow is isentropic γ =n=1.4 sub this value in above equation m max = 0.685√2 P 1 / V 1

21. Nozzle Efficiency When the steam flow through a nozzle the final velocity of steam for a given pressure drop is reduced due to following reasons Due to friction between the nozzle surface and steam Due to internal fluid friction in the steam Due to shock losses Most of these frictional losses occur between the throat and exit in convergent divergent nozzle. Nozzle efficiency = Actual Enthalpy drop/Isentropic enthalpy drop η N =h 1 -h 3 /h 1 -h 2

22. Critical Pressure Ratio There is only one value of the ratio P 2 /P 1 which produces maximum discharge from the nozzle This ratio is called Critical pressure ratio P 1 = Inlet Pressure P 2 = Outlet pressure For Saturated steam n=1.135 (P 2 /P 1 ) =(2/n+1) n/n-1 (P 2 /P 1 ) =0.577 For Superheated steam n=1.3 (P 2 /P 1 ) =(2/n+1) n/n-1 (P 2 /P 1 ) =0.546 For gases n=1.4 (P 2 /P 1 ) =(2/n+1) n/n-1 (P 2 /P 1 ) =0.5282

23. Effect of friction & Metastable expansion of steam The expansion is no more isentropic enthalpy drop is reduced The final dryness fraction of steam is increased as the kinetic energy gets converted into heat due to friction and is absorbed by steam The specific volume of steam is increased as the steam becomes more dry due to this frictional reheating Metastable expansion of steam\ When steam flow through a nozzle it would normally be expected that the discharge of steam through a nozzle would be slightly less than the theoretical value .but it has been observed during experiments on flow of wet steam that the discharge is slightly greater that calculated by the formula

24. Metastable expansion of steam The convergent part of nozzle is so short and the steam velocity is so high that the molecules of steam have insufficient time to collect and form droplets so that normal condensation does not take place such rapid expansion is said to be metastable C 2 =√2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) V 2 = V 1 (P 2 /P 1 ) 1/n A 2 = m x V 2 / C 2

25. Introduction of steam Steam  is water in the gas phase. It is commonly  formed   by boiling or evaporating water.  Steam  that is saturated or superheated is invisible; however, " steam " often refers to wet  steam , the visible mist or aerosol of water droplets  formed  as water vapour condenses. Types of Steam Enthalpy Entropy Specific Volume Wet steam h = h f  + x. h fg     s ws  = s f  + x. s fg      v ws = x.v g Dry steam h g  = h f  + h fg S= S g V=V g Superheated steam h sup  =   h g  + C s ( t sup   -t s  )   S sup = S g +Cp s log e ( T sup / T s )  V= Vsup

26. Sample problems Problem: 1 Dry saturated steam at 3.5bar is supplied to a convergent divergent nozzle whose that area is 4.4 cm 2 .the exit pressure is 1.1bar.determine maximum possible discharge through nozzle/minute and area of the nozzle at exit when the flow is maximum assume the flow is frictionless adiabatic. Solution: The critical pressure ratio when the steam is initially dry saturated Pt/P1= 0.577 , Pt = 0.577 x P1 = 2 bar From Steam Table P 1 = 3.5 bar

27. Sample problems P t = 2 bar P2 = 1.1 bar From steam table at 3.5bar and dry saturated steam h 1 = 2731.6 Kj/Kg , S 1 =6.936 kj / kgK From 1-t = isentropic expansion b/w inlet and throat S 1 =S t =6.936 Kj/ KgK S t = S ft + x t . S fgt , 6.936 = 1.530 + x t x 5.597 , x t = 0.966 Enthalpy of steam at throat h t = h ft + x t .h fgt, h t =504.7+ 0.966 x 2201.6 , h t = 2631.5 Kj/kg

28. Sample problems Specific volume of steam at throat V t = x t .V gt, V t = 0.966 x 0.88544 , V t = 0.85566 m 3 /kg S t = S 2 = 6.939 Kj/ KgK S 2 =S f2 + x 2 .S fg2, 6.939=1.3330 + x 2 x 5.9947 , x 2 = 0.935 h 2 =h f2 + x 2 .h fg2, h 2 =428.84 + 0.935 x 2250.8 , h 2 = 2533.3 kj /kg V 2 = x 2 .V g2, V 2 = 0.935 x 1.5492 , V 2 = 1.448 m 3 /kg Velocity at throat portion C t = √2000(h 1 - h t ) ,C t = √2000(2731.6 – 2631.5) ,C t = 447.55 m/s C 2 = √2000(h 1 - h 2 ) ,C 2 = √2000(2731.6 – 2533.3) ,C 2 = 629.72 m/s M max = A t x C t / V t , , M max = 4.4 x 10 -4 x 447.55/0.85566 , M max = 0.23 kg/sec,13.81 kg/min Area of Nozzle at exit: A 2 = m max x V 2 /C 2 , A 2 = 0.23 x 1.4485/629.72, A 2 = 5.3 x 10 -4 m 2

29. Sample problems Problem: 2 Steam at a pressure of 15bar saturated is discharged through a convergent divergent nozzle to a back pressure of 0.2 bar .the mass flow rate 9kg/kwhr. if the power developed is 220kw determine number of nozzles required .if each nozzle has a throat of rectangular cross section of 4mm x 8mm .if 12% of overall isentropic enthalpy drop occurs in the divergent portion due to friction. Find the cross section of the exit rectangle Solution: The critical pressure ratio when the steam is initially dry saturated P t /P 1 = 0.577, P t = 0.577 x P 1 = 8.7 bar Properties of steam P 1 = 15 bar

30. Sample problems P t = 8.7 bar P 2 = 0.2 bar From steam table at 15bar and dry saturated steam h 1 = 2789.9 Kj/Kg S 1 =6.441 kj / kgK From 1-t = isentropic expansion b/w inlet and throat S 1 =S t =6.441 Kj/ KgK S t = S ft + x t . S fgt , 6.441 = 2.08 + x t x 4.55 , x t = 0.958 Enthalpy of steam at throat h t = h ft + x t .h fgt, h t =736.65+ 0.958 x 2034.5 , h t = 2685.4 Kj/kg

31. Sample problems Specific volume of steam at throat V t = x t .V gt, V t = 0.958 x 0.221885 , V t = 0.212565 m 3 /kg S t = S 2 = 6.441 Kj/Kg K S 2 =S f2 + x 2 .S fg2, 6.441=0.832 + x 2 x 7.077 , x 2 = 0.793 h 2 =h f2 + x 2 .h fg2, h 2 =251.5 + 0.793 x 2358.4 , h 2 = 2121.712 kj /kg V 2 = x 2 .V g2, V 2 = 0.793 x 7.6492 , V 2 = 6.058 m 3 /kg Velocity at throat portion C t = √2000(h 1 - h t ) ,C t = √2000(2789.9 – 2685.4) ,C t = 457.17 m/s C 2 = √2000(h 1 - h t )+ η( h t - h 2 ) , ( η= 1- Loss = 1-0.12 = 0.88 ) C 2 = √2000(2789.9 – 2685.4)+ 0.88 (2685.4-2121.72) , C 2 = 1095.94 m/s Total Mass flow rate , m= 9 x 220/3600 ,m=0.55kg/s Mass flow rate of steam per nozzle , M f = A t x C t / V t , , M max = 0.068823 kg/sec Number of Nozzle: N=m/mf = 0.55/0.068823=7.99=8 Area of Nozzle at exit: A 2 = m f x V 2 /C 2 , A 2 = 3.81 x 10 -4 m 2

32. Sample problems Problem:3 It is proposed to design steam nozzles for the following data Initial Pressure = 30bar Initial Temperature =450 c Back Pressure=6 bar Nozzle Efficiency= 90 % Initial steam velocity=60m/s Mass flow rate = 2kg/s Assuming circular cross section, Calculate the Inlet, Throat and Exit diameter of nozzle. Solution: P 1 =30bar, P 3 =6bar,C 1 =60m/s , m=2kg/s Inlet: From steam tableP 1 =30bar, T 1 =450 c From steam table at 30 bar and 450 c Superheated steam h 1 = 3343 Kj/Kg, V 1, = 0.1078 m 3 /kg (From mollier chart) Mass flow rate of steam per nozzle M f = A 1 x C 1 /V 1, , A 1 = m x V 1, / C 1 = 2 x 0.1078/60, A 1 = 0.003593 m 2

33. Sample problems A 1 =π / 4 X D 1 2 = 0.003593 D 1 = 0.0676 m Throat assume the frictional losses P t /P 1 = 0.546, P t = 0.546x P 1 = 16.38 bar From mollier chart Enthalpy of steam at throat h t = 3155 Kj/kg Specific volume of steam at throat V t = 0.19 m 3 /kg Velocity at throat portion C t = √2000(h 1 - h t ) ,C t = √2000(3343 – 3155) ,C t = 616 m/s A t = m f x V t /C t , A t = 0.000617 m 2 , A t =π / 4 X D t 2 = 0.000617, D t = 0.028m Exit portion: Mollier chart Enthalpy of steam at Exit h 2 = 2905 Kj/kg Specific volume of steam at Exit V 2 = 0.4 m 3 /kg C 2 = √2000(h 1 - h t )+ η( h t - h 2 ) , η= 1- Loss = 1-0.1 = 0.9 C 2 = √2000(3343 – 3155)+ 0.9 (3155-2905) , C 2 = 890 m/s A 2 = m 2 x V 2 /C 2 , A 2 = 0.000899 m 2 A 2 =π / 4 X D 2 2 = 0.000899, D 2 = 0.0338m

34. Sample problems Problem:4 Determine the throat and exit height of Delaval nozzle to discharge 27kg of perfect gas/min .The inlet and exit pressure are 480kpa and 138kpa respectively. Initial temperature of the gas is 535 C nozzle efficiency is 90 % friction losses occur only after the throat the molecular weight of the gas is 29 and its adiabatic index n =1.4 Assume Square cross section of the nozzle. Solution: ( P t /P 1 ) = (2/n+1) n/n-1 ( P t /P 1 ) = 0.5282 , P t = 0.5282 x 480 = 253.44 kpa T t /T 1 = 2/n+1 ( P t /P 1 ) = ( T t /T 1 ) n/n-1 T t /T 1 = 2/1.4+1=0.833 T t = 673.3K R= 8.314 /29 = 0.2867kj/kg C Cp = Rγ /γ-1 = 0.2867 x 1.4/1.4-1 = 1.0034 Kj/Kg C

35. Sample problems C t 2 /2 = Cp (T 1 -T t ) C t 2 = 2Cp (T 1 -T t ) C t = √2Cp (T 1 -T t ) C t = √2 x 1.0034x1000(808-673.3) C t = 519.9 m/s m= ρ t . A t .C t m= P t / RT t .A t . C t A t = mRT t / P t . C t A t = (27/60)(0.2867x1000)x673.3/ (253.44x1000)x519.9 A t = 6.59 x 10 -4 m 2 Height of the nozzle = √6.59 x 10 -4 Height of the nozzle = 0.0257 m

36. Sample problems (T 2 /T 1 ) = (P 2 /P 1 ) n-1/n ,T 2 = T 1 x (P 2 /P 1 ) n-1/n T 2 = 808 x (138/480) 1.4-1/1.4 , T 2 = 565.9 K η N =h 1 -h 2 /h 1 -h t , η N =T 1 -T 2 1 /T 1 -T 2 0.9 = 808-T 2 /808-565.9 T 2 1 = 808 – 0.9(808-565.9), T 2 1 = 590.1 K C 2 2 /2 = Cp (T 1 - T 2 1 ),C 2 2 = 2Cp (T 1 - T 2 1 ) C 2 = √2Cp (T 1 - T 2 1 ),C 2 = √2 x 1.0034x1000(808-590.1) C 2 = 661.27 m/s m= ρ 2 .A 2 .C 2 ,m= P 2 /RT 2 .A 2 .C 2 A 2 = mRT 2 / P 2 . C 2 A 2 = (27/60)(0.2867x1000)x590.1/ (138x1000)x661.27 A 2 = 8.343 x 10 -4 m 2 Height of the nozzle = √8.343 x 10 -4 , Height of the nozzle = 0.02888 m

37. Sample problems Problem:5 A convergent divergent nozzle is to be designed in which steam initially at 14 bar and 80 C of superheat is to be expanded down to a back pressure of 1.05 bar .Determine the necessary throat and exit diameter of the nozzles of a steam discharge of 500 kg/hr. assuming that the expansion is in thermal equilibrium throughout and friction reheat amounting to 12% of the total isentropic enthalpy drop to be effective in the divergent part of the nozzle. Solution: P 1 =14bar, T sup - T sat = 80 C, T sup = 80 C + T sat = 80+ 195 =275 C P 2 = 1.05 bar We know that P t / P 1 = (2/n+1) n/n-1 ,P t / P 1 = (2/1.3+1) 1.3/1.3-1 , P t / P 1 = 0.546 P t = 0.546 x14 =7.64 bar For P 1 =14bar and 450 c Superheated steam h 1 = 2980 Kj/Kg, h t = 2850 Kj/kg (From mollier chart)

38. Sample problems h 2 = 2490 Kj/Kg, x 2 = 0.921 V t = 0.287 m 3 /kg For throat C t = √2000(h 1 - h t ) ,C t = √2000(2980– 2850) ,C t = 509.8 m/s m = A t x C t / V t , A t = m f x V t /C t , A t = 500 x 0.287/3600x509.8, A t = 7.82 x 10 -5 m 2 , A t =π / 4 X D t 2 = 7.82 x 10 -5 , D t = 0.009978 m C 2 = √2000η(h 1 - h 2 ) ,η= 1- Loss = 1-0.12 = 0.88 C 2 = √2000 x 0.88 (2980-2490) ,C 2 = 928.6 m/s V 2 = x 2 .V g2, V 2 = 0.921 x 1.69 , V 2 = 1.556 m 3 /kg A 2 = m 2 x V 2 /C 2 , A 2 = 0.0002327 m 2 A 2 =π / 4 X D 2 2 = 0.0002327, D 2 = 17.2 x 10 -3 m

39. Sample problems Problem:6 Determine the throat area ,exit area and exit velocity for a steam nozzle to pass a mass flow of 0.2 kg/s when inlet conditions are 10bar and 250 C and the final pressure is 2bar .assume expansion is isentropic and the inlet velocity is negligible use pV 1.3 = constant.do not calculate from h-S chart Solution: From superheated table p 1 =10bar,T 1 = 250 C P t / P 1 = (2/n+1) n/n-1 ,P t / P 1 = (2/1.3+1) 1.3/1.3-1 ,P t / P 1 = 0.546 P t = 0.546 x10 =5.45 bar C t =√2 n/n-1 P 1 V 1 (1- (P t /P 1 ) n -1/n ) C t =√2 x 1.3/1.3-1 (10x10 5 ) (0.233) (1- (5.45/10) 1.3 -1/1.3 ) C t =√2019333.3 (1-0.8696) C t = 513.15 m/s

40. Sample problems ( V t /V 1 ) = (P 1 /P t ) 1/n , ( V t /0.233) = (10/5.457) 1/1.3 V t = 0.3712 m 3 /kg m = A t x C t / V t , A t = m f x V t /C t , A t = 0.2 x 0.3712/513.15 , A t = 1.446 x 10 -4 m 2 C 2 =√2 n/n-1 P 1 V 1 (1- (P 2 /P 1 ) n -1/n ) C 2 =√2 x 1.3/1.3-1 (10x10 5 ) (0.233) (1- (2/10) 1.3 -1/1.3 ) C 2 = 791.45 m/s (V 2 /V 1 ) = (P 1 /P 2 ) 1/n (V 2 /0.233) = (10/2) 1/1.3 ,V 2 = 0.8035 m 3 /kg m = A 2 x C 2 /V 2 A 2 = m f x V 2 /C 2 , A 2 = 0.2 x 0.8035/791.45 A 2 = 2.03 x 10 -4 m 2

41. Sample problems Problem:7 The steam at 3 bar with 10 C superheat is passed through a convergent nozzle the velocity of steam entering the nozzle is 91.5 m/s the back pressure is 1.5 bar the assuming nozzle efficiency 90% determine the area of nozzle at throat maximum discharge through the nozzle is limited to 0.45 kg/sec. Take Cp =2.2kj/kg C use steam table only Solution: P 1 = 3 bar P 2 = 1.5 bar

42. Sample problems Isentropic Expansion S 1 =S 2 S 1 = Sg + Cps. Loge (Tsup/Ts) S 1 = 6.991 + 2.2 Loge (133.5+10+273/133.5+273) S 1 = 7.015 kj / kg C S 2 = S f 2 + x 2 s fg2 , S 2 = 1.434 + x 2 . 5.79, x 2 = 0.964 h 1 = hg + Cps(Tsup-Ts), h 1 = 2724.7 + 2.2 (10) h 1 = 2746.7 kj /kg h 2 = h f 2 + x 2 h fg2 , h 2 = 467.1+ x 2 .2226.2 h 2 = 2613.2 kj /kg C 2 2 /2 – C 1 2 /2 = η(h 1 - h 2 ) C 2 = √ 91.5 2 +2000 x 0.9 (2746.7-2613.2) C 2 = √8372+240.3x10 3 , C 2 = 1038 m/s m = A 2 x C 2 /x 2 . V g2 , A 2 = 0.45x0.964x1.159/1038, A 2 = 4.84 x 10 -4 m 2

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