Theory and Applica0on of Gas Turbine Systems

Theory and Applica0on of Gas Turbine Systems Part IV: Axial and Radial Flow Turbines Munich Summer School at University of Applied Sciences Prof. Kim A. Shollenberger

Introduc0on to Turbines Two basic types: radial flow and axial flow Majority of gas turbines are axial flow turbines because they are more efficient except for very low power and mass flowrate applica0ons Radial flow turbines are widely used in the cryogenic industry as a turbo-expander and in turbochargers for reciproca0ng engines A back-to-back radial turbine and centrifugal compressor has the benefit of a very short rigid rotor that is suitable for compact applica0ons such as auxiliary power plant units and for mobile power plants

Windmill or Air Turbine for Very Low Pressure

Schema0c Diagram of flow Through Windmill Blades

Basic Opera0on of Axial Flow Turbines Consists of a series of one or more stages where each stage contains a set of two blades : Row of sta0onary nozzle blades (or stator blades, or nozzle guide vanes) that align the flow, start the expansion process to a lower pressure and temperature which results in a higher velocity Followed by a row of rotor blades awached to the rota0ng shax that extracts power from the further expanding gas that exits at lower velocity NOTE: Opposite order for an axial flow compressor

Schema0c of Axial Flow Turbine Stage and Velocity Triangles 2 α - fluid velocity angle β - rela0ve velocity angle for fluid and blade V 1 V a,1 V θ,2 V rel, 1 Nozzle blades I V a,2 V 2 V θ,3 V a,3 V 3 V rel, 2

Axial Flow through Turbine Blades Sta0c enthalpy or temperature

Theory of Axial Flow Turbine Gas enters the row of nozzle blades (or stator blades or nozzle guide vanes) at Sec0on (1) Flow expands through the nozzle blades to a lower pressure and temperature, but a higher velocity at Sec0on (2) Through the rotor blades power is extracted from the fluid with a decrease in velocity and typically con0nued expansion un0l Sec0on (3) Blade angles are chosen to guide the flow entering at Sec0on (1) and transi0oning between stages

Analysis Assump0ons In single stage turbines the fluid velocity at Sec0on (1) is mainly axial, thus α 1 0 and V 1 V a1 For mul0-stage turbines, inlet condi0ons at Sec0on (1) approximately match outlet condi0ons at Stage (3): α 1 α 3, V 1 V 3 ; called repea0ng stage Because blade speed increases with radius, velocity triangles change from blade root to 0p, however for short blades in the radial direc0on it is reasonable to use condi0ons at the mean diameter of the annulus, r CL = D CL /2 = (D tip + D root )/4

Axial Flow Turbine Forces (V θ2 + V θ3 ) represents change in angular momentum per unit mass flow which produces torque (V a2 - V a3 ) produces an axial thrust on the rotor which may supplement or offset the pressure thrust arising from the pressure drop (p 2 p 3 ) To simplify the analysis, assume constant axial flow velocity (V a V a2 V a3 ) through rotors with a flared annulus (used to accommodate decreasing density as the gas expands)

Velocity Triangles for Flow Entering and Exi0ng Rotor Blades U V a = V a,2 = V a,3 V 3 V rel,2 β 3 α 2 V rel,3 V 2 β 2 α 3 V θ2 + V θ3 U V a = tanα 2 tan β 2 = tan β 3 tanα 3 tanα 2 + tanα 3 = tan β 2 + tan β 3

Angular Momentum Balance Angular momentum equa0on for a control volume on rotor:! r F! =! r!! V ρ dv + r V! ρ V! n! da CV t CS Assuming steady flow, negligible torque due to surface forces (fric0on) with respect to large shax torque, and mass is balanced so torques cancel out (like a top): T shaft = ( r 2 V θ 2 + r 3 V ) θ 3!m where!m = ρ V! n! da CS NOTE: Torque applied to shax equals change in angular momentum of the fluid; now the veloci0es have opposite direc0ons entering and exi0ng, so they are added together

Power Extracted from the Turbine ShaX Sha/ power or rate of work done by the fluid is:!w s, ideal = " ω! T s = ω r CL V θ 2 +V θ 3 Using velocity angle, α, get specific work:!w s, ideal!m =U V a tanα 2 + tanα 3 ( ) From velocity diagram in terms of blade angle, β:!w s, ideal!m =U V a tan β 2 + tan β 3 From the steady flow energy equa0on :!W ts!m =U V a tan β 2 + tan β 3 ( )!m =U ( V θ 2 +V ) θ 3!m ( ) ( ) = c p ( T 01 T 03s )

Turbine Efficiency Apply isentropic efficiency rela0onship to each combined nozzle and rotor blade stage: T 03 =1 η s 1 p 03 T 01 p 01 ( k 1) k where η s is the isentropic stage efficiency (or total-tototal stage efficiency). It is part of the overall isentropic turbine efficiency, η t.

Turbine Efficiency Notes Defini0on of η s does not account for exi0ng kine0c energy (KE); appropriate for the following cases: Stage is followed by others in mul0-stage turbine When KE is used for propulsion (aircrax) Last stage with diffuser or volute to recover KE Some0mes defined as total-to-sta0c isentropic efficiency for each stage and whole turbine; all KE is assumed to be wasted, thus it is generally lower

Dimensionless Parameters in Turbine Design Flow Coefficient, φ = V a /U (typically less than 1) Blade Loading Coefficient (or Temperature Drop Coefficient) work capacity of a stage; ra0o of extracted energy to blade KE ψ = c T p ( T 01 03) U 2 2 = 2 φ ( tan β 2 + tan β 3 ) Degree of ReacDon frac0on of stage expansion that occurs in the rotor; typically defined in terms of sta0c temperature drops ( ) ( ) Λ = T T 2 3 T 1 T 3

Degree of Reac0on for Ideal Repea0ng Stages For our analysis where we assume V a V a2 V a3 and for repea0ng stages where V 1 V 3 (KE cancels):!w t!m = c p ( T 1 T 3 ) = c p ( T 01 T 03 ) =U V a ( tan β 2 + tan β 3 ) For just the moving rotor blades:!w r!m = c p T 2 T 3 Combine to get: Λ = T T 2 3 T 1 T 3 ( ) = 1 2 V 2 rel,3 ( 2 V ) rel,2 = V a ( ) ( ) = φ ( 2 tan β tan β 3 2) 2 ( ) 2 tan2 β 3 tan 2 β 2

Gas Angles in Terms of φ, Λ, and ψ Combine equa0ons to get the following: tan β 2 = 1 ψ 2 φ 2 2 Λ tan β 3 = 1 " ψ $ 2 φ # 2 + 2 Λ % ' & tanα 2 = tan β 2 + 1 φ tanα 3 = tan β 3 1 φ

Degree of Reac0on Notes Λ = 0 corresponds to an impulse stage with no expansion across the blades; most efficient for extremely high pressure ra0os (such as for steam turbines) due to leakage losses For gas turbines with much lower pressure ra0os typically use Λ 50%; this corresponds to expansion being divided evenly between stator and rotor

Equa0ons for Λ = 50% Subs0tute into equa0ons for repea0ng stage to get: 1 φ = ( tan β tan β 3 2) tan β 2 = 1 ψ 2 φ 2 1 tan β 3 = 1 ψ 2 φ 2 +1 tanα 2 = tan β 2 + 1 φ = tan β 3 tanα 3 = tan β 3 1 φ = tan β 2

Velocity Diagram for Λ = 50% Results in symmetric veloci0es triangles: U V 3 V rel,2 V a = V a,2 = V a,3 Vrel,3 V2

Velocity Diagram for Λ = 50% and Constant Blade Speed For low ψ and φ: Low gas veloci0es, thus reduced fric0on losses and higher efficiencies ψ = 4 φ = 1 V 3 U V 2 Requires mores stages for a given overall turbine output ψ = 2 φ = 0.6 ψ = 1.5 φ = 0.4

Blade Loading versus Flow Coefficient for Λ = 50% 6 3 = 70 3 = 60 = c p T 0 / (U 2 /2) 4 2 3 = 80 2 = 30 2 = 20 2 = 10 0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 = V a / U

6.0 Blade Loading versus Flow Coefficient for Λ = 50% with Stage Efficiency Contours from Test Data: Ref: Horlock, J. H., Axial Flow Turbines, BuWerworth, 1966. N <l <>" C\1 II ψ = c p T 0 / (U 2 /2) E "'" u "' ;;: 0 " c. E Ll :::1 1 "' c. E 1.0-1.0 IJI=-2.0 '\ ) φ = V a / U Flow coefficient dj = C 8 /U 1],=0.94 o:3(= P2) = 1 o' (swirl)

Notes on 50% Reac0on Designs Many assump0ons were made in these calcula0ons about blade shape, thus the results are only good for qualita0ve analysis Low ψ and φ designs (which implies low gas veloci0es and hence reduced fric0on) yield the highest stage efficiencies However, low ψ means more stages for a given overall turbine output and low φ means a larger turbine annular area for a given mass flowrate

Notes on 50% Reac0on Designs, cont. For industrial gas turbines when size and weight are not very important and a low SFC is vital, typically design for low ψ and φ For aircrax propulsion (where weight and frontal area need to be minimalized) use higher values of ψ and φ (current typical values are ψ 3 to 5 and φ 0.8 to 1.0) Low swirl angle at exit (α 3 < 20 ) to decrease losses in exhaust/diffuser or jet pipe/propelling nozzle; might require lower degree of reac0on

Smith Chart from Rolls-Royce Turbine Tes0ng ψ = c p T 0 / (U 2 /2) NOTE: Results are for zero 0p clearance, thus higher than for actual engine. φ = V a / U Reference: Smith, S. F., A simple correla0on of turbine efficiency, Journal of the Royal AeronauDcal Society, 69, 1965, 467-70.

Notes on Smith Chart Useful preliminary design tool that shows contours of constant isentropic efficiency as a func0on of blade loading coefficient and flow coefficient Turbine stages have been designed for many loca0ons on Smith Chart: for example, for a typical mul0ple stage turbine high pressure stages can operate in region A and low pressure stages can operate in region B

T-s Diagram for a Reac0on Stage T 01 = T 02 Temperature 01 02 p 1 1 Nozzle 2 Rotor 3 Tip T 1 1 Root p 2 T 2 T 03 2s 2 03 p 3 T 3 3 3s 3ss Entropy

Loss Coefficients for Turbine Nozzle blades: λ N = c p ( T T 2 2s) V 2 2 2 OR Y N = p 01 p 02 p 02 p 2 Rotor blades: λ R = c T p ( T 3 3ss) 2 2 V rel, 3 OR Y N = p 02, rel p 03, rel p 03,rel p 3 NOTE: λ is easier for use in design and Y N is easier to calculated from cascade test results.

Isentropic Stage Efficiency Define for a single nozzle and rotor stage: η s = T 01 T 03 T 01 T 03s From T-s diagram can assume: Use this along with previous defini0ons to get: η s = 1+ φ 2 T03 T03s T3 T3s λ R cos 2 β 3 + ( T 3 T 2 ) λ N cos 2 α 2 tan β 3 + tanα 2 1 φ 1

Example #11 A single-stage turbine for a small, inexpensive, turbojet unit is to be designed based on the specifica0ons in Table 6. Determine the following assuming V a2 = V a3, V 1 = V 3, and α 1 = β 1 = 0 : a. blade loading coefficient, b. blade angles and degree of reac0on, c. velocity components at Sec0on 2 and Sec0on 3, d. opera0ng condi0ons (p, Τ, and ρ), e. annulus geometry, and f. rotor loss coefficient and stage efficiency.

Table 6. Single-Stage Turbine Specifica@ons Mass flowrate of working fluid,!m 20 kg/s Inlet stagna0on temperature, T 01 1100 K Stagna0on temperature drop, T 01 - T 03 145 K Inlet stagna0on pressure, p 01 400 kpa Stagna0on pressure ra0o, p 01 /p 03 1.873 Rota0onal speed, N 250 rev/s Mean blade speed, U 340 m/s Flow coefficient, φ 0.80 Loss coefficient for nozzle, λ N 0.050 Swirl angle for out flow, α 3 10

Notes on Example #11 Next steps to consider for design: Check Mach number for supersonic flow issues Three dimensional nature of flow and how it affects varia0on of gas angles with radius. Blade shapes necessary to achieve the required gas angles, and the effect of centrifugal and gas bending stresses on the design. Validate design es0mates using cascade test results to calculate λ N and λ R for comparison.

Three-Dimensional Flow Velocity triangles change shape from root to 0p of blades (increasing radius) in nozzle and rotor due to: Increasing blade velocity with radius, U = ω r Increasing sta0c pressure with radius due to centrifugal forces arising from swirl veloci0es; can result in radial veloci0es NOTE: Radial changes can be ignored if blade height, h, is small compared to mean radius, r m. (h/r m < 30%)

Vortex Blading Use of twisted blades that are designed to account for changing gas angles from blade root to 0p. Typically not used for low pressure steam turbines and some single stage gas turbines where changes in efficiency are insignificant Typically used for mul0ple stage gas turbines (and axial compressors) where even small improvements in overall efficiency are important for heavy duty use

Vortex Theory Derive basic equa0ons for conserva0on of momentum and energy to account for radial property varia0ons. r-momentum Equa0on (Differen0al Form): ρ DV r Dt V 2 θ = p r r + ρ g + r µ 2 V r For steady flow, constant V a, and V θ >> V r reduces to: V θ 2 r = 1 p ρ r radial equilibrium equadon

Vortex Theory, cont. For energy equa0on stagna0on enthalpy defined as: h 0 = h + V 2 dh 0 dr = dh dr +V a 2 = h + 1 2 V a dv a dr +V θ ( 2 2 +V ) θ for V r << V a orv θ dv θ dr From thermodynamics and momentum balance: dh = T ds + dp ρ dh dr = T ds dr + 1 dp ρ dr = T ds dr + V 2 θ r

Vortex Theory, cont. For cases where ds/dr 0 (subsonic without shocks): dh 0 dr = V 2 θ r +V a For (1) constant V a and (2) constant specific work at all radii in planes between blade rows, dh 0 /dr 0: dv θ dr = V θ r dv a dr +V θ dv θ dr integrate to get: vortex energy equadon r V θ = constant Called free vortex condidon which sa0sfies the radial equilibrium equadon (with negligible radial velocity).

Free Vortex Stage Design Work done per unit mass of gas, W = W!!m, is constant versus radius at planes 1, 2, and 3, thus calculate W at any r and account for variable density using:!w = W V a ρ da = 2 π W V a ρ r dr A For ini0al calcula0ons can approximate using the following, where subscript m denotes at mean radius:!w = π W m V a ρ ( m r 2 2 t r ) r t r r

Free Vortex Stage Design, cont. To calculate gas angle varia0ons: r V θ = r m V θ,m = constant V a = V θ tanα = V θ, m tanα m = constant Combine to get: tanα = r m tanα m r Use with earlier equa0ons to get blade angle varia0on.

Example #12 Using the results from Example #11, at both Sec0on 2 and Sec0on 3 at the root and 0p of the blades calculate the following: a. velocity angles, α, b. blade angles, β, c. reac0on, Λ, and verify it is always posi0ve, and d. Mach number, Ma, and verify subsonic flow.

Angle (degrees) Gas Angles Versus Dimensionless Radius at Sec0on 2 and Sec0on 3 80 60 40 20 0-20 -1-0.5 0 0.5 1 root r* = (r - r m )/(h/2) 0p α 1 α 2 β 1 β 2 a1 a2 b2 b3

Velocity Triangles for V rel,2 Sec0on 2 and Sec0on 3 (Drawn to Scale) V rel,3

Notes on Gas Angle and Velocity Triangle Varia0ons Need to insure Mach number is less than about 0.75 to avoid possible shocks Mach number at Sec0on 2 is highest at root for: Rela0ve velocity higher and Temperature (and speed of sound) lower Need to insure reac0on is posi0ve at all radial loca0ons Reac0on is posi0ve for V rel,3 > V rel,2

Constant Nozzle Angle Design For manufacturing simplicity, constant nozzle blade angles can be used: Not essen0al to design for free vortex flow to achieve radial equilibrium (where the flow has negligible radial veloci0es) Alterna0vely, design 0p and root radii such that the axial and swirl veloci0es will now change appropriately to achieve radial equilibrium

Constant Nozzle Angle Design, cont. To calculate velocity components with constant blade and fluid velocity angles, can use the following: tanα 2 = V θ,2 V a,2 = constant dv a,2 dr = dv θ,2 dr cotα 2 Subs0tute into vortex energy equa0on for dh 0 /dr 0: 0 = V 2 θ,2 r + dv cot2 α 2 V θ,2 θ,2 dr +V θ,2 dv θ,2 dr

Constant Nozzle Angle Design, cont. Rearrange and integrate to get: dv θ,2 V θ,2 = sin 2 α 2 dr r V θ,2 r sin2 α 2 = constant Subs0tu0ng back into original equa0on we also get: V a,2 r sin2 α 2 = constant

Blade Profile, Pitch, and Chord Design Select stator and rotor blade shapes that minimize: Profile loss due to boundary layer growth (includes separa0on loss under adverse condi0ons such as an extreme angle of incidence or high Mach number) Annulus loss due to boundary layer growth on inner and outer walls of the annulus Secondary flow loss due to secondary flows which always occur when a wall boundary layer is turned through an angle by an adjacent curved surface Tip clearance loss due to outer wall interac0ons

Typical Blade Profile

Blade Loss Coefficient Use to account for overall losses; defined as: Y! = Y! + Y P s + Y!! k overall profile annulus and tip clearance blade loss losses secondary losses coefficient losses NOTE: Typically define separately for nozzle and rotor.

Radial Flow Turbines Gas flow with a high tangen0al velocity is directed inwards and leaves the rotor with as small a whirl velocity as prac0cable near the axis of rota0on Turbine looks very similar to centrifugal compressor, but with a ring of nozzle vanes replacing the diffuser vanes Also, there normally is a diffuser at the outlet to reduce exhaust velocity to a negligible value

Radial Inflow Turbine Schema0c with Velocity Triangles Volute 1 Nozzle vanes 2 4 3 For Normal Design Condi0on: V rel, 2 V r, 2 and V 3 V a3 α 2 V rel,2 U 2 V 2 V rel,3 Diffuser r 2 r 3 β 3 U 3 V 3 = V a3

T-s Diagram for Radial Turbine Temperature T 01 = T 02 01 02 p 1 T 1 1 p 2 T 2 T 03 T 3 4s 2s 3s 2 3ss 03 3 4 p 3 Entropy p a

Normal Design Condi0on For no swirl at Sec0on 3 (V θ, 3 0), work is given by:!w!m = c p T 01 T 03 ( ) = V θ,2 U 2 =U 2 2 Including a perfect diffuser this becomes:!w d!m = c p T 01 T 4,s ( ) = C 0 2 / 2 where C 0 is called the spou0ng velocity. For ideal case U 2 /C 0 1/ 2 (actual values for good performance range from 0.68 to 0.71). Also, given by: 2 = c T 1 p a p 0 C 0 2 p 01 ( k 1) k

Example #13 A radial flow turbine with a work output of 45.9 kw axer accoun0ng for mechanical losses is to be designed based on the specifica0ons in Table 7. Determine the following: a. turbine isentropic efficiency, b. velocity components at Sec0on 2 and Sec0on 3, and c. rotor loss coefficient.

Table 7. Radial Turbine Specifica@ons Mass flowrate of working fluid,!m 0.322 kg/s Inlet stagna0on temperature, T 01 1000 K Inlet stagna0on pressure, p 01 500 kpa Stagna0on pressure ra0o, p 01 /p 03 2.0 Rota0onal speed, N 1000 rev/s Rotor inlet 0p diameter 12.7 cm Rotor exit 0p diameter 7.85 cm Hub-0p ra0o at exit 0.30 Nozzle efflux angle, α 2 70 Rotor vane outlet angle, β 3 40 Loss coefficient for nozzle, λ N 0.070