[...]... "State 2" p2=60 [kPa] s2=sl sf2=ENTR0PY(Steam,P=p2,X=0) sg2=ENTR0PY(Steam,P=p2,X=l) x2=(s2-sf2)/(sg2-sf2) hf2=ENTHALPY(Steam,P=p2,X=0) hg2=ENTHALPY(Steam,P=p2,X=l) h2=(l-x2)*hf2+x2*hg2 "Performance C a l c u l a t i o n s " wt=hl-h2 The results are: hl=3177 [ k J / k g ] hf2=359.8 [kJ/kg] pl=6000 [kPa] s l = 6 5 4 [kJ/kg-K] Tl=400 [C] h2 =22 97 [ k J / k g ] hg2 =26 53 [ k J / k g ] p2=60 [kPa] s2=6.54... • 10 02 V\ = —,— = = 1 12. 3 m/s 1 A1 14 .25 ' At the exit the flow area is A2 = b(s cos a2 -t2)= 3[5cos(70°) - 0 .20 ] = 4.53cm 2 hence the velocity is rhv2 0 .2 ■ 1.00 • 10 02 _ = 441.5 m/s V2 = —r~ = A2 4.53 ' 2. 2 FIRST LAW OF THERMODYNAMICS For a uniform steady flow in a channel, the first law of thermodynamics has the form m (m + pwi +-V? + gzA + Q = fa (u2 + p2v2 + -V 22 + gz2) +W (2. 2) The sum of specific... CONTENTS Hydraulic Transmission of Power 385 11.1 385 386 388 390 391 3 92 394 398 11 .2 12 Fluid couplings 11.1.1 Fundamental relations 11.1 .2 Flow rate and hydrodynamic losses 11.1.3 Partially filled coupling Torque converters 11 .2. 1 Fundamental relations 11 .2. 2 Performance Exercises Wind turbines 401 12. 1 12. 2 4 02 403 403 407 409 4 12 415 415 419 424 425 429 430 12. 3 12. 4 Horizontal-axis wind turbine... 12. 3 12. 4 Horizontal-axis wind turbine Momentum and blade element theory of wind turbines 12. 2.1 Momentum Theory 12. 2 .2 Ducted wind turbine 12. 2.3 Blade element theory and wake rotation 12. 2.4 Irrotational wake Blade Forces 12. 3.1 Nonrotating wake 12. 3 .2 Wake with rotation 12. 3.3 Ideal wind turbine 12. 3.4 Prandtl's tip correction Turbomachinery and future prospects for energy Exercises Appendix A: Streamline... diameter is D2 = 2. 5 cm, with specific volume at the outlet v2 = 3.80m 3 /kg Find the change in enthalpy neglecting any change in the potential energy Figure 2. 2 Row through a diffuser Solution: The areas at the inlet and outlet are TTD2 ^O.Ol2 „oc 1A_5 5 Ai = — - 1 = — - — = 7.85 -10 irD2 TTO. 025 2 2 m2 4.91 - 1 0 " 4 m 2 SECOND LAW OF THERMODYNAMICS 19 The velocity at the inlet is rhui 0.01 -2. 4 V\ =... forms V ■ n A = VA cos 9 = VnA = VAn Principles of Turbomachinery By Seppo A Korpela Copyright © 20 11 John Wiley & Sons, Inc 15 16 PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW in which An is the area normal to the flow The principle of conservation of mass for a uniform steady flow through a control volume with one inlet and one exit takes the form PiViAnl = p2V2An2 Turbomachinery flows are steady only... s2=6.54 [kJ/kg-K] x2=0.8448 wt=879.9 [kJ/kg] Calculation of enthalpy and steam quality at state 2 could have been shortened by simply writing [kPa] P 2= 60 h2=ENTHALPY(Steam, P=p2, S=sl) x2=QUALITY(Steam, P=p2, S=sl) The Ts diagram is a convenient representation of the properties of steam, for lines of constant temperature on this chart are horizontal in the two-phase region, as are the lines of ... used It is defined as the mass of vapor divided by the mass of the mixture In terms of quality, thermodynamic properties of a two-phase mixture 22 PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW are calculated as a weighted average of the saturation properties Thus, for example h = (1 — a:)/if + xhg or h = h{ + xhfg in which h{ denotes the enthalpy of saturated liquid, hg that of saturated vapor, and their... 7.5314 - 1.1451 in which s f = 1.1451 k J / ( k g • K) and s g = 7.5314kJ/(kg • K) are the values of entropy for saturated liquid and saturated vapor at p2 = 60 kPa Exit enthalpy is then obtained from h2 = h(+ x2h{g = 359.79 + 0.8448 • 22 93.1 = 22 97.0 k J / k g (b) Work delivered is Ws = hi-h2= 3177.0 - 22 97.0 = 8 8 0 k J / k g The calculations have been carried out using the EES script shown below "... t\ = 2. 5 mm, and at the outlet it is t2 = 2. 0 mm Blade height is b — 3.0 cm Nozzle angle is a2 = 70° Find the steam velocity at the inlet and at the outlet Figure 2. 1 Turning offlowby steam nozzles Solution: The area at the inlet is Ax =b(s-h) = 3 ( 5 - 0 2 5 ) = 14 .25 cm 2 Velocity at the inlet is solved from the mass balance m = piVxAx = Vi FIRST LAW OF THERMODYNAMICS which gives 17 mVl 0 .20 • 0.80 . analysis 22 2 7.1.1 Stage temperature and pressure rise 22 3 7.1 .2 Analysis of a repeating stage 22 5 7 .2 Design deflection 23 0 7 .2. 1 Compressor performance map 23 4 7.3 Radial equilibrium 23 5 7.3.1. 2. 4 Equations of state 20 2. 4.1 Properties of steam 21 2. 4 .2 Ideal gases 27 2. 4.3 Air tables and isentropic relations 29 2. 4.4 Ideal gas mixtures 31 2. 4.5 Incompressibility 35 2. 4.6 Stagnation. Theory 403 12. 2 .2 Ducted wind turbine 407 12. 2.3 Blade element theory and wake rotation 409 12. 2.4 Irrotational wake 4 12 12. 3 Blade Forces 415 12. 3.1 Nonrotating wake 415 12. 3 .2 Wake with