工程系热力学6

• The efficiency of a turbine:
& & & & WShaft WFluid − Loss WS / mg = = ηT = & & WFluid WFluid V12 p1 V2 2 p2 − 2 + ρg + z1 2 g + ρg + z 2
T = ρQ(r2Vt 2 − r1Vt1 )
• The power = ωT and also = γQH t, where Ht is the theoretical head developed by the pump.
Ht =
ω 2 r22
g
• β2>90o -- forward curving blades; − not desirable for pumps due to the potential for instability − becomes a turbine for sufficient -ve Q. • β2<90o -- backward curving blades; − typical arrangement for pumps − becomes a turbine for sufficiently large +ve Q.
Vr1 cosα1 = V1 cos β1 − VB
VB
• Choose exit blade angle • Reaction force in terms of exit blade angle
• Hence
& Rx = m(V1 cos β1 − V2 cos β 2 ) & Rx = m(V1 cos β1 − (VB − Vr 2 cosα 2 )
& & W = NRxVB = NVB (m(V1 cos β1 − (VB − Vr 2 cosα 2 ))
Example E4.17
Radial Flow Pumps
• Use a control volume which encloses the impeller and is fixed in the frame of reference of the casing.
• Assuming steady incompressible flow, the moment of momentum equation is: d ~ ∑ (M ) = dt ∫ r × VρdV + ∫ r × V(n .V) ρdA c .v . c. s . This becomes;
& V12 p1 − WS V2 2 p2 mg − hlpump = 2 g + ρg + z 2 − 2 + ρg + z1 ≡ H P &
• Ie “Pump head” is total head increase of fluid and is less than shaft power because of losses
ωcotβ − Q 2πb2 g
Real pump performance
• As Q → ∞ losses increase due to viscous shear forces in passages. • As Q → 0 losses increase due to viscous shear forces around circumference of impeller. • Max efficiency occurs for smallest value of Htheoretical-Hp.
Figure 4.15 – Fluid striking a series of vanes.
& − Rx = m(V2 x − V1x )
& = m(V2 cos β 2 − V1 cos β1 )
& W = NRxVB
VB
• Design analysis in blade frame of reference • Jet angle ( β1 ) is fixed, inlet blade angle should be tangential to relative velocity
B
Figure 12.21 – Reaction turbine (Francis type): (a) schematic; (b) Francis spiral turbine for the Victoria Falls power station, Australia; (c) runner for one of the turbines for the Itaipu power station, Brazil. (Courtesy of Voith Siemens Hydro Power Generation, Inc.) C
2 & V12 p1 − WS V2 p2 fL V 2 = + + z2 − + + z1 + hlturb + ∑ K + 2 ρg & mg 2 g ρ g D 2g
• Analyse the pump / turbine with no pipe. • For a pump :
• Analyse with no pipe:
& V2 2 p2 WS V12 p1 − mg + hlturb = 2 + ρg + z1 2 g + ρg + z 2 ≡ H T &
• Ie “Turbine head” is total head decrease of fluid, shaft power is less because of losses
Forces on axial flow blades
• A stationary blade, no losses, P const, V1 = V2
& ∑ F = m(V
2
− V1 )
& & − Rx = m(V2 cosα − V1 ) = mV1 (cosα − 1)
& & − Ry = mV2 sin α = mV1 sin α
• Converting terms to “heads”:
2 & V12 p1 − WS V2 p2 = 2 g + ρg + z 2 − 2 + ρg + z1 + hl & mg
• Head loss is in the pipework and the turbine / pump itself so:Βιβλιοθήκη • Rearranging:
& Loss = (1 − ηT )WFluid
Different types
Figure 12.7 – Idealized axial-flow impeller: (a) impeller control volume; (b) velocity diagrams at control surface.
Figure 12.27 – Impulse turbine (Pelton type): (a) typical arrangement of twin-jet machine. (Courtesy of Gilbert Gilkes and Gordon, Ltd.); jet striking a bucket; (c) Pelton runner in the New Colglate power station, USA. (Courtesy Voith Siemens Hydro Power Generation, Inc.)
• Consider steady state energy eqn:
& & Q − WS = ∫
c . s ..
VΙ 2 p ~ 2 + gz + u + ρ ρn .VdA
• For a pump / turbine situation it becomes:
V2 2 p2 V12 p1 & − WS = ρ 2V2 A2 2 + ρ + gz 2 − ρ1V1 A1 2 + ρ + gz 2 1 + Losses
Turbo Machines
• Pumps or turbines • Fluid flowing through a rotating impeller/runner with some change in angular momentum results in a torque (either +ve or -ve) on the impeller/runner. • Flow leaves the impeller in either a radial or axial direction, or in between (mixed). • Pure radial and axial cases can be analysed as a one dimensional problem assuming ideal conditions.
• Rearranging:
1 & − 1WFluid Loss = ηP
• For a turbine:
2 & V12 p1 − WS V2 p2 fL V 2 = 2 g + ρg + z 2 − 2 + ρg + z1 + hlturb + ∑ K + D 2 g & mg