Fluid Machines (EG 616 ME) Ram C. Poudel Department of Mechanical Engineering, Pulchowk Campus 7 December 2012 Chapter 1: Introduction 1.1 Turbo-machines 1.2 Hydraulic Machines 1.3 History of Development of Water Wheels & Water Turbines 1.4 Water Wheels Turbomachines Definition: A device in which energy is transferred either to or from a continuously flowing fluid by the dynamic action of one or more moving blades rows. Etymology: Turbo - turbinis - latin: which spins or whirls around. Categories of Turbomachines Power Absorbing:Absorb power to increase fluid pressure and head. Examples: Fans, compressors, pumps, etc. Power Producing:Produce power by expanding fluid to a lower pressure or head. Examples: Wind, hydraulic, steam, and gas turbines. Categories Based on Flow Direction Axial Flow Turbo-machine:Path of through-flow is wholly or mainly parallel to the axis of rotation. Radial Flow Turbo-machines:Path of through flow is wholly or mainly in a plane perpendicular to the rotational axis. [Figure 1.1c] Mixed flow Turbo-machines:The direction of through flow at the rotor outlet when both radial and axial velocity components are present in significant amount [Figure 1.1b] Examples of Turbomachines Single stage axial flow compressor or pump Mixed flow pump Centrifugal compressor or pump Francis turbine (mixed flow type) Kaplan turbine Pelton wheel Coordinate Systems Cylindrical polar coordinate system is used for description and analysis due to the cylindrical shape of turbomachines. Axes:Axial (x) Radial (r) Tangential (or circumferential) (r θ r\theta r θ ) Simplifications and Meridional Velocity Simplification: Flow does not vary in the tangential direction. Axi-symmetric stream surfaces: Flow moves through the machine on these surfaces. Meridional velocity (C m C_m C m ): Component of velocity along an axi-symmetric stream surface. Equation: C < e m > m = c < / e m > x 2 + c r 2 C<em>m = \sqrt{c</em>x^2 + c_r^2} C < e m > m = c < / e m > x 2 + c r 2 Purely axial-flow machines:Radius of flow path is constant, radial flow velocity is zero. C < e m > m = C < / e m > x C<em>m = C</em>x C < e m > m = C < / e m > x Purely radial flow machines:Axial flow velocity is zero. C < e m > m = c < / e m > r C<em>m = c</em>r C < e m > m = c < / e m > r Total flow velocity: c = c < e m > x 2 + c < / e m > r 2 + c < e m > θ 2 = c < / e m > m 2 + c θ 2 c = \sqrt{c<em>x^2 + c</em>r^2 + c<em>{\theta}^2} = \sqrt{c</em>m^2 + c_{\theta}^2} c = c < e m > x 2 + c < / e m > r 2 + c < e m > θ 2 = c < / e m > m 2 + c θ 2 Swirl angle: α = t a n − 1 ( c < e m > θ / C < / e m > m ) \alpha = tan^{-1}(c<em>{\theta} / C</em>m) α = t a n − 1 ( c < e m > θ / C < / e m > m ) Relative Velocities Analysis within rotating blades is performed in a frame of reference stationary relative to the blades. Flow appears steady in this frame of reference. Relative velocity is absolute velocity minus the local velocity of the blade.w < e m > θ = c < / e m > θ − U w<em>{\theta} = c</em>{\theta} - U w < e m > θ = c < / e m > θ − U w < e m > x = c < / e m > x w<em>x = c</em>x w < e m > x = c < / e m > x w < e m > r = c < / e m > r w<em>r = c</em>r w < e m > r = c < / e m > r Relative flow angle: β = t a n − 1 ( w < e m > θ / c < / e m > m ) \beta = tan^{-1}(w<em>{\theta} / c</em>m) β = t a n − 1 ( w < e m > θ / c < / e m > m ) Relationship: t a n β = t a n α − U / c m tan \beta = tan \alpha - U/c_m t an β = t an α − U / c m Fundamental Laws Continuity of Flow Equations First Law of Thermodynamics and Steady Flow Energy Equations Momentum Equation Second Law of Thermodynamics Definition of Efficiency for Turbines Overall Efficiency (η o \eta_o η o ):η o = mechanical energy available at coupling of output shaft in unit time maximum energy difference possible for the fluid in unit time \eta_o = \frac{\text{mechanical energy available at coupling of output shaft in unit time}}{\text{maximum energy difference possible for the fluid in unit time}} η o = maximum energy difference possible for the fluid in unit time mechanical energy available at coupling of output shaft in unit time Mechanical energy losses occur due to friction at bearings, glands, etc.Small machines: 5% or more. Medium and large machines: as little as 1%. Isentropic Efficiency Isentropic or hydraulic efficiency (η < e m > i \eta<em>i η < e m > i or η < / e m > h \eta</em>h η < / e m > h ):η < e m > i (or η < / e m > h ) = mechanical energy supplied to the rotor in unit time maximum energy difference possible for the fluid in unit time \eta<em>i \text{ (or } \eta</em>h) = \frac{\text{mechanical energy supplied to the rotor in unit time}}{\text{maximum energy difference possible for the fluid in unit time}} η < e m > i (or η < / e m > h ) = maximum energy difference possible for the fluid in unit time mechanical energy supplied to the rotor in unit time Mechanical efficiency (η m \eta_m η m ):η < e m > m = shaft power rotor power = η < / e m > o η < e m > i (or η < / e m > o η h ) \eta<em>m = \frac{\text{shaft power}}{\text{rotor power}} = \frac{\eta</em>o}{\eta<em>i} \text{ (or } \frac{\eta</em>o}{\eta_h}) η < e m > m = rotor power shaft power = η < e m > i η < / e m > o (or η h η < / e m > o ) Isentropic efficiency in terms of work:η < e m > t (or η < / e m > h ) = actual work ideal (maximum) work = Δ W Δ W m a x \eta<em>t \text{ (or } \eta</em>h) = \frac{\text{actual work}}{\text{ideal (maximum) work}} = \frac{\Delta W}{\Delta W_{max}} η < e m > t (or η < / e m > h ) = ideal (maximum) work actual work = Δ W ma x Δ W For an adiabatic turbine:Δ W = W < e m > x m ˙ = ( h < / e m > 01 − h < e m > 02 ) + g ( z < / e m > 1 − z 2 ) \Delta W = \frac{W<em>x}{\dot{m}} = (h</em>{01} - h<em>{02}) + g(z</em>1 - z_2) Δ W = m ˙ W < e m > x = ( h < / e m > 01 − h < e m > 02 ) + g ( z < / e m > 1 − z 2 ) Steam and Gas Turbines Figure 1.9(a) shows a Mollier diagram representing the expansion process through an adiabatic turbine. Line 1-2 represents the actual expansion and line 1-2s the ideal or reversible expansion. Actual turbine rotor specific work:Δ W < e m > x = W < / e m > x m ˙ = h < e m > 01 − h < / e m > 02 = ( h < e m > 1 − h < / e m > 2 ) + 1 2 ( c < e m > 1 2 − c < / e m > 2 2 ) \Delta W<em>x = \frac{W</em>x}{\dot{m}} = h<em>{01} - h</em>{02} = (h<em>1 - h</em>2) + \frac{1}{2}(c<em>1^2 - c</em>2^2) Δ W < e m > x = m ˙ W < / e m > x = h < e m > 01 − h < / e m > 02 = ( h < e m > 1 − h < / e m > 2 ) + 2 1 ( c < e m > 1 2 − c < / e m > 2 2 ) Ideal work output:Δ W < e m > m a x = W < / e m > m a x m ˙ = h < e m > 01 − h < / e m > 02 s = ( h < e m > 1 − h < / e m > 2 s ) + 1 2 ( c < e m > 1 2 − c < / e m > 2 s 2 ) \Delta W<em>{max} = \frac{W</em>{max}}{\dot{m}} = h<em>{01} - h</em>{02s} = (h<em>1 - h</em>{2s}) + \frac{1}{2}(c<em>1^2 - c</em>{2s}^2) Δ W < e m > ma x = m ˙ W < / e m > ma x = h < e m > 01 − h < / e m > 02 s = ( h < e m > 1 − h < / e m > 2 s ) + 2 1 ( c < e m > 1 2 − c < / e m > 2 s 2 ) Total-to-total efficiency (η t \eta_t η t ):η < e m > t = Δ W < / e m > x Δ W < e m > m a x = h < / e m > 01 − h < e m > 02 h < / e m > 01 − h 02 s \eta<em>t = \frac{\Delta W</em>x}{\Delta W<em>{max}} = \frac{h</em>{01} - h<em>{02}}{h</em>{01} - h_{02s}} η < e m > t = Δ W < e m > ma x Δ W < / e m > x = h < / e m > 01 − h 02 s h < / e m > 01 − h < e m > 02 If the difference between the inlet and outlet kinetic energies is small:η < e m > t ≈ h < / e m > 1 − h < e m > 2 h < / e m > 1 − h 2 s \eta<em>t \approx \frac{h</em>1 - h<em>2}{h</em>1 - h_{2s}} η < e m > t ≈ h < / e m > 1 − h 2 s h < / e m > 1 − h < e m > 2 Exhaust kinetic energy is not wasted in:The last stage of an aircraft gas turbine, where it contributes to jet propulsive thrust. One stage of a multistage turbine where it can be used in the following stage. Hydraulic Turbines Hydraulic efficiency (η h \eta_h η h ) is a form of the total-to-total efficiency. Steady flow energy equation in differential form for an adiabatic turbine:d W x = m ˙ [ d h + d ( c 2 2 ) + g d z ] dW_x = \dot{m}[dh + d(\frac{c^2}{2}) + gdz] d W x = m ˙ [ d h + d ( 2 c 2 ) + g d z ] For an isentropic process, T d s = 0 = d h − d p ρ Tds = 0 = dh - \frac{dp}{\rho} T d s = 0 = d h − ρ d p . Maximum work output for an expansion to the same exit static pressure, kinetic energy, and height as the actual process:W < e m > m a x = m ˙ ∫ < / e m > 1 2 d p ρ + ( c < e m > 1 2 − c < / e m > 2 2 2 ) + g ( z < e m > 1 − z < / e m > 2 ) W<em>{max} = \dot{m} \int</em>1^2 \frac{dp}{\rho} + (\frac{c<em>1^2 - c</em>2^2}{2}) + g(z<em>1 - z</em>2) W < e m > ma x = m ˙ ∫ < / e m > 1 2 ρ d p + ( 2 c < e m > 1 2 − c < / e m > 2 2 ) + g ( z < e m > 1 − z < / e m > 2 ) For an incompressible fluid, maximum work output:W < e m > m a x = m ˙ [ p < / e m > 1 − p < e m > 2 ρ + c < / e m > 1 2 − c < e m > 2 2 2 + g ( z < / e m > 1 − z < e m > 2 ) ] = m ˙ g ( H < / e m > 1 − H 2 ) W<em>{max} = \dot{m} [\frac{p</em>1 - p<em>2}{\rho} + \frac{c</em>1^2 - c<em>2^2}{2} + g(z</em>1 - z<em>2)] = \dot{m}g(H</em>1 - H_2) W < e m > ma x = m ˙ [ ρ p < / e m > 1 − p < e m > 2 + 2 c < / e m > 1 2 − c < e m > 2 2 + g ( z < / e m > 1 − z < e m > 2 )] = m ˙ g ( H < / e m > 1 − H 2 ) Where g H = p ρ + c 2 2 + g z gH = \frac{p}{\rho} + \frac{c^2}{2} + gz g H = ρ p + 2 c 2 + g z and m ˙ = ρ Q \dot{m} = \rho Q m ˙ = ρQ Turbine hydraulic efficiency:η < e m > h = W W < / e m > m a x = Δ W g [ H < e m > 1 − H < / e m > 2 ] \eta<em>h = \frac{W}{W</em>{max}} = \frac{\Delta W}{g[H<em>1 - H</em>2]} η < e m > h = W < / e m > ma x W = g [ H < e m > 1 − H < / e m > 2 ] Δ W Compressors and Pumps Isentropic efficiency of a compressor or hydraulic efficiency of a pump (η < e m > c \eta<em>c η < e m > c or η < / e m > h \eta</em>h η < / e m > h ):η < e m > c (or η < / e m > h ) = useful (hydrodynamic) energy input to fluid in unit time power input to rotor \eta<em>c \text{ (or } \eta</em>h) = \frac{\text{useful (hydrodynamic) energy input to fluid in unit time}}{\text{power input to rotor}} η < e m > c (or η < / e m > h ) = power input to rotor useful (hydrodynamic) energy input to fluid in unit time Overall efficiency:η o = useful (hydrodynamic) energy input to fluid in unit time power input to coupling of shaft \eta_o = \frac{\text{useful (hydrodynamic) energy input to fluid in unit time}}{\text{power input to coupling of shaft}} η o = power input to coupling of shaft useful (hydrodynamic) energy input to fluid in unit time Mechanical efficiency:η < e m > m = η < / e m > o η < e m > c (or η < / e m > o η h ) \eta<em>m = \frac{\eta</em>o}{\eta<em>c} \text{ (or } \frac{\eta</em>o}{\eta_h}) η < e m > m = η < e m > c η < / e m > o (or η h η < / e m > o ) Compressors and Pumps Specific Work For a complete adiabatic compression process from state 1 to state 2, the specific work input is:Δ W = ( h < e m > 02 − h < / e m > 01 ) + g ( z < e m > 2 − z < / e m > 1 ) \Delta W = (h<em>{02} - h</em>{01}) + g(z<em>2 - z</em>1) Δ W = ( h < e m > 02 − h < / e m > 01 ) + g ( z < e m > 2 − z < / e m > 1 ) Total-to-total efficiency:η < e m > o = ideal (minimum) work input actual work input = h < / e m > 02 s − h < e m > 01 h < / e m > 02 − h 01 \eta<em>o = \frac{\text{ideal (minimum) work input}}{\text{actual work input}} = \frac{h</em>{02s} - h<em>{01}}{h</em>{02} - h_{01}} η < e m > o = actual work input ideal (minimum) work input = h < / e m > 02 − h 01 h < / e m > 02 s − h < e m > 01 If the difference between inlet and outlet kinetic energies is small:η < e m > c ≈ h < / e m > 2 s − h < e m > 1 h < / e m > 2 − h 1 \eta<em>c \approx \frac{h</em>{2s} - h<em>1}{h</em>2 - h_1} η < e m > c ≈ h < / e m > 2 − h 1 h < / e m > 2 s − h < e m > 1 For incompressible flow, the minimum work input is given by:Δ W < e m > m i n = W < / e m > m i n m ˙ = [ p < e m > 2 − p < / e m > 1 ρ + c < e m > 2 2 − c < / e m > 1 2 2 + g ( z < e m > 2 − z < / e m > 1 ) ] = g [ H < e m > 2 − H < / e m > 1 ] \Delta W<em>{min} = \frac{W</em>{min}}{\dot{m}} = [\frac{p<em>2 - p</em>1}{\rho} + \frac{c<em>2^2 - c</em>1^2}{2} + g(z<em>2 - z</em>1)] = g[H<em>2 - H</em>1] Δ W < e m > min = m ˙ W < / e m > min = [ ρ p < e m > 2 − p < / e m > 1 + 2 c < e m > 2 2 − c < / e m > 1 2 + g ( z < e m > 2 − z < / e m > 1 )] = g [ H < e m > 2 − H < / e m > 1 ] For a pump the hydraulic efficiency is therefore defined as:η < e m > h = W < / e m > m i n Δ W = g [ H < e m > 2 − H < / e m > 1 ] Δ W \eta<em>h = \frac{W</em>{min}}{\Delta W} = \frac{g[H<em>2 - H</em>1]}{\Delta W} η < e m > h = Δ W W < / e m > min = Δ W g [ H < e m > 2 − H < / e m > 1 ] Introduction and Classification of Fluid Machines Positive Displacement TurbomachinesRadial-Flow (Centrifugal) Axial-Flow Mixed-Flow Machines for Doing Work on a Fluid Pumps Fans Blowers Compressors Hydraulic Turbines (Impulse and Reaction) Gas Turbines (Impulse and Reaction) Turbomachinery Analysis: The Angular Momentum Principle τ × F < e m > s → + ∫ < / e m > c v r → × g → ρ d V + T < e m > s h a f t = ∫ < / e m > c s r → × V → ρ V → ⋅ d A → \tau \times \overrightarrow{F<em>s} + \int</em>{cv} \overrightarrow{r} \times \overrightarrow{g} \rho dV + T<em>{shaft} = \int</em>{cs} \overrightarrow{r} \times \overrightarrow{V} \rho \overrightarrow{V} \cdot d \overrightarrow{A} τ × F < e m > s + ∫ < / e m > c v r × g ρ d V + T < e m > s ha f t = ∫ < / e m > cs r × V ρ V ⋅ d A Momentum Equation Newton's second law of motion relates the sum of external forces to the rate of change of momentum. Equation:∑ F < e m > x = d d t ( m c < / e m > x ) \sum F<em>x = \frac{d}{dt}(mc</em>x) ∑ F < e m > x = d t d ( m c < / e m > x ) For steady flow with uniform velocities:∑ F < e m > x = m ˙ ( c < / e m > x 2 − c x 1 ) \sum F<em>x = \dot{m}(c</em>{x2} - c_{x1}) ∑ F < e m > x = m ˙ ( c < / e m > x 2 − c x 1 ) Moment of Momentum Vector sum of moments of external forces equals the time rate of change of angular momentum. Equation:T < e m > A = d d t ( m r c < / e m > θ ) T<em>A = \frac{d}{dt} (mr c</em>\theta) T < e m > A = d t d ( m rc < / e m > θ ) where r is the distance from the axis of rotation and c θ c_\theta c θ is the velocity component perpendicular to both the axis and radius vector r. Law of Moment of Momentum For a control volume enclosing the rotor of a turbomachine with swirling fluid entering at radius r < e m > 1 r<em>1 r < e m > 1 with tangential velocity c < / e m > θ 1 c</em>{\theta 1} c < / e m > θ 1 and leaving at radius r < e m > 2 r<em>2 r < e m > 2 with tangential velocity c < / e m > θ 2 c</em>{\theta 2} c < / e m > θ 2 :T < e m > A = m ˙ ( r < / e m > 2 c < e m > θ 2 − r < / e m > 1 c θ 1 ) T<em>A = \dot{m}(r</em>2 c<em>{\theta 2} - r</em>1 c_{\theta 1}) T < e m > A = m ˙ ( r < / e m > 2 c < e m > θ 2 − r < / e m > 1 c θ 1 ) Euler Work Equation: Pump For a pump or compressor rotor running at angular velocity Ω \Omega Ω , the rate at which the rotor does work on the fluid isT < e m > A Ω = m ˙ ( U < / e m > 2 c < e m > θ 2 − U < / e m > 1 c θ 1 ) T<em>A \Omega = \dot{m}(U</em>2 c<em>{\theta 2} - U</em>1 c_{\theta 1}) T < e m > A Ω = m ˙ ( U < / e m > 2 c < e m > θ 2 − U < / e m > 1 c θ 1 ) where U = Ω r U = \Omega r U = Ω r Specific work (work done on the fluid per unit mass):\Delta W = \frac{TA \Omega}{\dot{m}} = U 2 c{\theta 2} - U 1 c_{\theta 1} > 0 This is Euler's pump equation. Euler Work Equation: Turbine For a turbine, the fluid does work on the rotor, and the sign for work is reversed. Thus, the specific work is\Delta Wt = \frac{W t}{\dot{m}} = U1 c {\theta 1} - U2 c {\theta 2} > 0 This is Euler's turbine equation. For any adiabatic turbomachine (turbine or compressor), applying the steady flow energy equation givesΔ W = ( h < e m > 01 − h < / e m > 02 ) = U < e m > 1 c < / e m > θ 1 − U < e m > 2 c < / e m > θ 2 \Delta W = (h<em>{01} - h</em>{02}) = U<em>1 c</em>{\theta 1} - U<em>2 c</em>{\theta 2} Δ W = ( h < e m > 01 − h < / e m > 02 ) = U < e m > 1 c < / e m > θ 1 − U < e m > 2 c < / e m > θ 2 This can be written asΔ h < e m > 0 = Δ ( U c < / e m > θ ) \Delta h<em>0 = \Delta(Uc</em>{\theta}) Δ h < e m > 0 = Δ ( U c < / e m > θ ) For a stationary blade row, U = 0 U = 0 U = 0 and therefore h 0 = c o n s t a n t h_0 = constant h 0 = co n s t an t . This is expected since a stationary blade cannot transfer any work to or from the fluid. Euler Turbomachine Equation < b r / > T < e m > s h a f t = ( r < / e m > 2 V < e m > t 2 − r < / e m > 1 V < e m > t 1 ) m ˙ <br />
T<em>{shaft} = (r</em>2 V<em>{t2} - r</em>1 V<em>{t1}) \dot{m} < b r / > T < e m > s ha f t = ( r < / e m > 2 V < e m > t 2 − r < / e m > 1 V < e m > t 1 ) m ˙ W < / e m > m = ( U < e m > 2 V < / e m > t 2 − U < e m > 1 V < / e m > t 1 ) m ˙ W</em>m = (U<em>2 V</em>{t2} - U<em>1 V</em>{t1}) \dot{m} W < / e m > m = ( U < e m > 2 V < / e m > t 2 − U < e m > 1 V < / e m > t 1 ) m ˙ < b r / > H = ( U < e m > 2 V < / e m > t 2 − U < e m > 1 V < / e m > t 1 ) m g <br />
H = \frac{(U<em>2V</em>{t2} - U<em>1V</em>{t1})}{mg} < b r / > H = m g ( U < e m > 2 V < / e m > t 2 − U < e m > 1 V < / e m > t 1 )
Rothalpy The Euler work equation can be rewritten asI = h < e m > 0 − U c < / e m > θ I = h<em>0 - Uc</em>\theta I = h < e m > 0 − U c < / e m > θ where I is a constant along the streamlines through a turbomachine. The function I is called rothalpy, a contraction of rotational stagnation enthalpy. Rothalpy can also be written in terms of the static enthalpyI = h + 1 2 V 2 − U c θ I=h + \frac{1}{2}V^2 - Uc_\theta I = h + 2 1 V 2 − U c θ Turbomachinery Analysis Velocity Diagrams Concepts of absolute velocity, velocity relative to blade, and rotor velocity. Turbomachinery Analysis: Idealized Centrifugal Pump Assumptions:Negligible torque due to surface forces (viscous and pressure). Inlet and exit flow tangent to blades. Uniform flow at inlet and exit. Zero inlet tangential velocity. Turbomachinery Analysis: Idealized Centrifugal Pump Head Equation Head Equation: H = C < e m > 1 − C < / e m > 2 Q H = C<em>1 - C</em>2Q H = C < e m > 1 − C < / e m > 2 Q Shutoff Head:C < e m > 1 = U < / e m > 2 2 g C<em>1 = \frac{U</em>2^2}{g} C < e m > 1 = g U < / e m > 2 2 C < e m > 2 = U < / e m > 2 c o t β < e m > 2 π D < / e m > 2 b 2 g C<em>2 = \frac{U</em>2 cot \beta<em>2}{\pi D</em>2 b_2 g} C < e m > 2 = π D < / e m > 2 b 2 g U < / e m > 2 co tβ < e m > 2 Turbomachinery Analysis: Idealized Centrifugal Pump (Continued) Relationship between head and volume flow rate for centrifugal pump with forward-curved, radial, and backward-curved impeller blades. Equation:H = ω 2 R 2 g H = \frac{\omega^2 R^2}{g} H = g ω 2 R 2 Turbomachinery Analysis: Machines for Doing Work on a Fluid < b r / > W < e m > H = ρ Q g H < / e m > a <br />
W<em>H = \rho Q g H</em>a < b r / > W < e m > H = ρQ g H < / e m > a < b r / > H < e m > a = ( P ρ g + α V 2 2 g + Z ) < / e m > d i s c h a r g e − ( P ρ g + α V 2 2 g + Z ) s u c t i o n <br />
H<em>a = (\frac{P}{\rho g} + \alpha \frac{V^2}{2g} + Z)</em>{discharge} - (\frac{P}{\rho g} + \alpha \frac{V^2}{2g} + Z)_{suction} < b r / > H < e m > a = ( ρ g P + α 2 g V 2 + Z ) < / e m > d i sc ha r g e − ( ρ g P + α 2 g V 2 + Z ) s u c t i o n
Pump Efficiency:η < e m > P = ρ Q g H W < / e m > m ˙ \eta<em>P = \frac{\rho Q g H}{\dot{W</em>m}} η < e m > P = W < / e m > m ˙ ρQ g H < b r / > W < e m > H = ρ Q g H < / e m > t <br />
W<em>H = \rho Q g H</em>t < b r / > W < e m > H = ρQ g H < / e m > t < b r / > H < e m > t = ( P ρ g + V 2 2 g + Z ) < / e m > i n l e t − ( P ρ g + V 2 2 g + Z ) o u t l e t <br />
H<em>t = (\frac{P}{\rho g} + \frac{V^2}{2g} + Z)</em>{inlet} - (\frac{P}{\rho g} + \frac{V^2}{2g} + Z)_{outlet} < b r / > H < e m > t = ( ρ g P + 2 g V 2 + Z ) < / e m > in l e t − ( ρ g P + 2 g V 2 + Z ) o u tl e t
Turbine Efficiency:η < e m > t = W < / e m > m ˙ ρ Q g H t \eta<em>t = \frac{\dot{W</em>m}}{ \rho Q g H_t} η < e m > t = ρQ g H t W < / e m > m ˙ Machines for Doing Work on a Fluid Comparison of ideal and actual head-flow curves for a centrifugal pump with backward-curved impeller blades. Typical pump performance curves from tests with three impeller diameters at constant speed. Performance curves for an impulse turbine showing ideal and actual torque and power vs. ratio of wheel speed to jet speed. Performance of typical reaction turbine as predicted by model tests and confirmed by field test. Flow Coefficient:Φ = Q A < e m > 2 U < / e m > 2 \Phi = \frac{Q}{A<em>2 U</em>2} Φ = A < e m > 2 U < / e m > 2 Q Head Coefficient:Ψ = g H U 2 2 \Psi = \frac{gH}{U_2^2} Ψ = U 2 2 g H Power Coefficient:Π = W ˙ ρ Q U < e m > 2 3 = W ˙ ρ ω 2 Q R < / e m > 2 5 \Pi = \frac{\dot{W}}{\rho Q U<em>2^3} = \frac{\dot{W}}{\rho \omega^2 Q R</em>2^5} Π = ρQ U < e m > 2 3 W ˙ = ρ ω 2 QR < / e m > 2 5 W ˙ Torque Coefficient:T = T ρ A < e m > 2 U < / e m > 2 R 2 = ψ Φ \Tau = \frac{T}{\rho A<em>2 U</em>2 R_2} = \psi \Phi T = ρ A < e m > 2 U < / e m > 2 R 2 T = ψ Φ Specific Speed:N S = ω Q 1 / 2 h 3 / 4 N_S = \frac{\omega Q^{1/2}}{h^{3/4}} N S = h 3/4 ω Q 1/2 Specific Speed (Customary Units):N S c u = N ( r p m ) [ Q ( g p m ) ] 1 / 2 [ H ( f t ) ] 3 / 4 N_{Scu} = \frac{N (rpm) [Q (gpm)]^{1/2}}{[H (ft)]^{3/4}} N S c u = [ H ( f t ) ] 3/4 N ( r p m ) [ Q ( g p m ) ] 1/2 Specific Speed (Customary Units):N S c u = N ( r p m ) [ P ( h p ) ] 1 / 2 [ H ( f t ) ] 5 / 4 N_{Scu} = \frac{N (rpm) [P (hp)]^{1/2}}{[H (ft)]^{5/4}} N S c u = [ H ( f t ) ] 5/4 N ( r p m ) [ P ( h p ) ] 1/2 Typical geometric proportions of commercial pumps as they vary with dimensionless specific speed. Typical geometric proportions of commercial hydraulic turbines as they vary with dimensionless specific speed. For Dynamic Similarity:Q < e m > 1 ω < / e m > 1 D < e m > 1 3 = Q < / e m > 2 ω < e m > 2 D < / e m > 2 3 \frac{Q<em>1}{\omega</em>1 D<em>1^3} = \frac{Q</em>2}{\omega<em>2 D</em>2^3} ω < / e m > 1 D < e m > 1 3 Q < e m > 1 = ω < e m > 2 D < / e m > 2 3 Q < / e m > 2 g h < e m > 1 ω < / e m > 1 2 D < e m > 1 2 = g h < / e m > 2 ω < e m > 2 2 D < / e m > 2 2 \frac{g h<em>1}{\omega</em>1^2 D<em>1^2} = \frac{g h</em>2}{\omega<em>2^2 D</em>2^2} ω < / e m > 1 2 D < e m > 1 2 g h < e m > 1 = ω < e m > 2 2 D < / e m > 2 2 g h < / e m > 2 W < e m > 1 ˙ ρ < / e m > 1 ω < e m > 1 3 D < / e m > 1 5 = W < e m > 2 ˙ ρ < / e m > 2 ω < e m > 2 3 D < / e m > 2 5 \frac{\dot{W<em>1}}{\rho</em>1 \omega<em>1^3 D</em>1^5} = \frac{\dot{W<em>2}}{\rho</em>2 \omega<em>2^3 D</em>2^5} ρ < / e m > 1 ω < e m > 1 3 D < / e m > 1 5 W < e m > 1 ˙ = ρ < / e m > 2 ω < e m > 2 3 D < / e m > 2 5 W < e m > 2 ˙ Applications to Fluid Systems: Machines for Doing Work on a Fluid Superimposed system head-flow and pump head-capacity curves. Equation:< b r / > P < e m > 1 ρ + α < / e m > 1 V < e m > 1 2 2 + g z < / e m > 1 − ( P < e m > 2 ρ + α < / e m > 2 V < e m > 2 2 2 + g z < / e m > 2 ) = h a l l p u m p < b r / > <br />
\frac{P<em>1}{\rho} +\alpha</em>1 \frac{V<em>1^2}{2} + g z</em>1 - (\frac{P<em>2}{\rho} + \alpha</em>2 \frac{V<em>2^2}{2} + g z</em>2 ) = h_{all pump}<br /> < b r / > ρ P < e m > 1 + α < / e m > 1 2 V < e m > 1 2 + g z < / e m > 1 − ( ρ P < e m > 2 + α < / e m > 2 2 V < e m > 2 2 + g z < / e m > 2 ) = h a llp u m p < b r / > Applications to Fluid Systems : Pump Wear Effect of pump wear on flow delivery to system. Applications to Fluid Systems: Pumps in Series Operation of two centrifugal pumps in series. Applications to Fluid Systems: Pumps in Parallel Operation of two centrifugal pumps in parallel. Applications to Fluid Systems: Fans, Blowers, and Compressors Exploded view of typical centrifugal fan. Applications to Fluid Systems: Fans, Blowers, and Compressors Typical types of blading used for centrifugal fan wheels Applications to Fluid Systems: Fans, Blowers, and Compressors Typical characteristic curves for fan with backward-curved blades Applications to Fluid Systems: Positive-Displacement Pumps Schematic of typical gear pump Applications to Fluid Systems Machines for Doing Work on a Fluid Propellers
Speed of Advance Coefficient: Thrust, Torque, Power Coefficients, and Propeller Efficiency Applications to Fluid Systems Machines for Extracting Work (Power) from a Fluid Hydraulic Turbines Wind-Power Machines