UNIT 5: Turbomachinery

I. Fundamental Principles and Governing Equations

Definition and Classification

• Turbomachinery: Devices that transfer energy between a rotor and a fluid via dynamic interaction.

• Classification:

  • By energy transfer: Turbines (extract energy, produce work) and Compressors/Pumps (add energy, increase pressure).

  • By flow path: Axial (flow parallel to axis), Radial/Centrifugal (flow perpendicular to axis), Mixed-flow (combination).

Thermodynamic Foundations

• Steady Flow Energy Equation (SFEE):

$$ \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{C_2^2 - C_1^2}{2} + g(z_2 - z_1) \right) $$

For turbomachines, often $ \dot{Q} \approx 0 $, $ \Delta z \approx 0 $.

• Second Law: Entropy generation $$\displaystyle \dot{S}_{gen} = \dot{m}(s_2 - s_1) - \frac{\dot{Q}}{T_{boundary}} \ge 0 $$.

  Exergy destruction: $$\displaystyle \dot{X}_{dest} = T_0 \dot{S}_{gen} $$.

Euler’s Turbomachine Equation

• Derivation from moment of momentum principle for a rotor:

$$ \dot{W} = \dot{m} (U_2 C_{w2} - U_1 C_{w1}) $$

where $ U $ = blade speed, $$\displaystyle C_w $$ = whirl (tangential) component of absolute velocity.

• For turbines: Work output $$\displaystyle \dot{W} > 0 $$; for compressors/pumps: work input $$\displaystyle \dot{W} < 0 $$.

• Significance: Relates energy transfer to blade speed and change in whirl velocity.

• Limitations: Assumes ideal (lossless), 1D flow, no radial variation.

Velocity Triangles

• Velocities:

  • Absolute $ \vec{C} $: fluid velocity relative to stationary frame.

  • Relative $ \vec{W} $: fluid velocity relative to moving blade.

  • Blade speed $ \vec{U} $: tangential velocity of blade.

  • Relation: $$\displaystyle \vec{C} = \vec{U} + \vec{W} $$.

• Angles:

  • $ \alpha $: angle between $ \vec{C} $ and axial direction (for axial machines) or tangential (for radial).

  • $ \beta $: angle between $ \vec{W} $ and axial/tangential direction.

• Construction:

  • Impulse turbine: Pressure constant in rotor; $$\displaystyle C_1 \approx C_2 $$ (with friction, $$\displaystyle C_2 < C_1 $$). Blade angles $$\displaystyle \beta_1, \beta_2 $$ set for smooth entry/exit.

  • Reaction turbine: Pressure drop in rotor; $$\displaystyle C_1 \neq C_2 $$. For Parsons (50% reaction), $$\displaystyle \alpha_1 = \beta_2 $$, $$\displaystyle \beta_1 = \alpha_2 $$, and often constant axial velocity $$\displaystyle C_{a1} = C_{a2} $$.

Stagnation Properties and Efficiencies

• Stagnation (total) enthalpy: $$\displaystyle h_0 = h + \frac{C^2}{2} $$.

• Total-to-total efficiency (turbine):

$$ \eta_{tt} = \frac{h_{01} - h_{02}}{h_{01} - h_{02s}} $$

where subscript $ s $ denotes isentropic process.

II. Steam Turbines

A. Types and Configurations

• Impulse (De Laval):

  • Single-stage: One nozzle row and one blade row.

  • Multi-stage: Multiple impulse stages in series.

  • Used for high head, low mass flow.

• Reaction:

  • Parsons: 50% reaction, symmetrical blades, axial flow.

  • Rateau: Pressure compounding (multiple nozzle/blade stages).

• Mixed-flow: Radial-inflow or axial-outflow combinations.

B. Velocity Diagrams and Calculations

Single-Stage Impulse Turbine

• Blade angles (symmetric blades, $$\displaystyle \beta_1 = \beta_2 $$):

$$ \tan\beta_1 = \frac{C_{a1}}{U - C_{w1}}, \quad \tan\beta_2 = \frac{C_{a2}}{U - C_{w2}} $$

For shockless entry, $$\displaystyle W_1 $$ must be relative to blade inlet angle.

• Whirl velocity: $$\displaystyle C_w = C \cos\alpha $$ (if $ \alpha $ measured from tangential) or $ C \sin\alpha $ (if from axial). Clarify convention in problem.

• Axial thrust (on blades):

$$ F_a = \dot{m} (C_{a1} - C_{a2}) $$

• Work done per kg: $$\displaystyle w = U(C_{w1} - C_{w2}) $$.

• Diagram (blade) efficiency:

$$ \eta_b = \frac{2U}{C_1} \left( \cos\alpha_1 - \frac{C_2}{C_1} \cos\alpha_2 \right) \quad \text{(if } \alpha \text{ from tangential)} $$

For friction, use velocity coefficient $$\displaystyle K = C_2/C_1 $$.

Single-Stage Reaction Turbine (Parsons)

• Symmetrical conditions: $$\displaystyle \alpha_1 = \beta_2 $$, $$\displaystyle \beta_1 = \alpha_2 $$, and often $$\displaystyle C_{a1} = C_{a2} $$ (constant axial velocity).

• Given: Mean diameter $ D $, RPM $ N $, steam velocity at exit from blades $$\displaystyle C_2 $$, blade outlet angle $$\displaystyle \beta_2 $$, mass flow $ \dot{m} $.

• Steps:

  1. Blade speed: $$\displaystyle U = \frac{\pi D N}{60} $$.

  2. Since Parsons (50% reaction), $$\displaystyle \alpha_1 = \beta_2 $$.

  3. Assume constant axial velocity: $$\displaystyle C_{a1} = C_{a2} = C_a $$.

  4. Condition for 50% reaction with constant $$\displaystyle C_a $$: $$\displaystyle C_{w1} + C_{w2} = U $$.

  5. From geometry:

$$ C_{w1} = C_a \tan\alpha_1 = C_a \tan\beta_2, \quad C_{w2} = \sqrt{C_2^2 - C_a^2} $$

6. Solve for $$\displaystyle C_a $$ from:

$$ C_a \tan\beta_2 + \sqrt{C_2^2 - C_a^2} = U $$

7. Then $$\displaystyle \alpha_2 = \cos^{-1}(C_a / C_2) $$, and blade inlet angle $$\displaystyle \beta_1 = \alpha_2 $$.

8. $$\displaystyle C_1 = C_a / \cos\alpha_1 $$.

9. Tangential force on moving blades:

$$ F_t = \dot{m} (C_{w1} - C_{w2}) $$

10. Power developed:

$$ P = \dot{m} U (C_{w1} - C_{w2}) $$

11. Axial thrust (momentum change only):

$$ F_a = \dot{m} (C_{a1} - C_{a2}) = 0 \quad (\text{since } C_{a1}=C_{a2}) $$

  *Note: In practice, pressure difference causes axial thrust, but not captured in simple momentum analysis.*

C. Degree of Reaction

• Definition: Ratio of enthalpy drop in rotor to total enthalpy drop per stage.

$$ R = \frac{\text{Enthalpy drop in rotor}}{\text{Total enthalpy drop per stage}} = \frac{h_1 - h_2}{h_0 - h_3} $$

where $$\displaystyle h_0 $$ = stagnation enthalpy at nozzle inlet, $$\displaystyle h_3 $$ = stagnation enthalpy at stage exit.

• For 50% reaction turbine: $$\displaystyle R = 0.5 $$, meaning equal enthalpy drop in nozzle and rotor.

• Relation to velocity triangles (with constant $$\displaystyle C_a $$):

$$ R = \frac{1}{2} + \frac{C_{w2} - C_{w1}}{2U} $$

For $$\displaystyle R=0.5 $$, $$\displaystyle C_{w1} + C_{w2} = U $$.

• Calculation from given parameters: Use Euler’s equation and energy balance.

D. Compounding

• Velocity compounding (Curtis turbine):

  • Multiple blade rows in one stage, separated by guide vanes.

  • Steam velocity is compounded (reduced) in steps.

  • Used for high pressure ratio in single stage.

  • DiagramSEARCH: Curtis turbine velocity diagram

• Pressure compounding (Rateau turbine):

  • Multiple stages, each with its own nozzle and blade rows.

  • Pressure drop divided among stages.

  • More common for large turbines.

• Comparison:

  | Feature | Velocity Compounding | Pressure Compounding | |------------------|----------------------|----------------------| | Stages per stage | Multiple blade rows | Multiple full stages | | Pressure drop | All in first nozzle | Divided among nozzles| | Efficiency | Lower (more losses) | Higher | | Application | High head, small flow| Large turbines |

E. Losses and Efficiencies

• Sources of losses:

  • Nozzle friction (nozzle efficiency $$\displaystyle \eta_n $$).

  • Blade friction (surface roughness, profile losses).

  • Leaving losses (kinetic energy of exit velocity not utilized).

  • Disc friction (bearing losses).

  • Leakage and clearance losses.

• Efficiencies:

  • Nozzle efficiency: $$\displaystyle \eta_n = \frac{C_1^2/2}{h_0 - h_1} $$.

  • Blade efficiency (diagram efficiency):

$$ \eta_b = \frac{U(C_{w1} - C_{w2})}{C_1^2/2} $$

• Stage efficiency: $$\displaystyle \eta_s = \eta_n \cdot \eta_b $$ (for simple stage).

• Overall efficiency: $$\displaystyle \eta_o = \frac{\text{Actual work output}}{\text{Heat input}} $$.

• Condition for maximum blade efficiency in impulse turbine:

  For given $ U $ and $$\displaystyle C_1 $$, optimize $$\displaystyle \alpha_1 $$:

$$ \cos\alpha_1 = \sqrt{\frac{r}{2}} \quad \text{where } r = \frac{C_2}{C_1} \text{ (velocity coefficient)} $$

For frictionless case ($$\displaystyle r=1 $$), $$\displaystyle \alpha_1 \approx 35.26^\circ $$.

F. Multi-Stage Turbines

• Reheat factor $$\displaystyle R_f $$:

  • Definition: Ratio of total isentropic enthalpy drop to the sum of stage isentropic drops.

$$ R_f = \frac{h_0 - h_{exit,s}}{ \sum (h_{in,i} - h_{out,i,s}) } $$

• Cause: Variation of stage efficiency with pressure and temperature (stages at lower pressure have lower efficiency due to higher moisture or lower density).

• Effect: Overall efficiency $$\displaystyle \eta_o > \eta_{avg} $$ (average stage efficiency) due to reheat factor.

• Efficiency improvement: Multi-staging allows better utilization of steam expansion, reduces moisture content in later stages, and improves overall efficiency compared to single-stage with same average stage efficiency.

G. Governing of Steam Turbines

• Throttle governing: Control steam flow by throttling nozzle inlet pressure. Simple, but wasteful at part load.

• Nozzle governing: Control by opening/closing groups of nozzles (e.g., in reaction turbines). More efficient at part load.

• Bypass governing: Steam bypassed to condenser or later stages. Used for overload conditions.

• Applications: Throttle for small turbines; nozzle for large reaction turbines.

H. Nozzle Flow

• Convergent-Divergent Nozzle: Used for supersonic flow. throat at minimum area.

• Critical pressure ratio: For steam, $$\displaystyle \left( \frac{p^*}{p_0} \right)_{\text{crit}} \approx 0.546 $$ for superheated steam, 0.577 for saturated vapor.

• Supersaturation: When steam expands rapidly, it may remain supercooled beyond saturation line; effective pressure ratio reduced.

• Nozzle efficiency: $$\displaystyle \eta_n = \frac{\text{Actual kinetic energy}}{\text{Isentropic enthalpy drop}} = \frac{C_1^2/2}{h_0 - h_1} $$.

• Friction loss: Reduces exit velocity, increases entropy.

III. Hydraulic Turbines

A. Classification

• Based on head $ H $ and flow $ Q $:

  • Pelton: High head ($$\displaystyle >300 $$ m), low flow.

  • Francis: Medium head ($ 30–300 $ m), medium flow.

  • Kaplan: Low head ($$\displaystyle <30 $$ m), high flow.

B. Pelton Wheel

• Construction: Bucket (split to reduce thrust), nozzle, runner, casing.

• Velocity Diagram:

  • Jet velocity: $$\displaystyle V_j = \sqrt{2gH} $$ (theoretical), actual $$\displaystyle V_j = C_v \sqrt{2gH} $$, $$\displaystyle C_v $$ = coefficient of velocity.

  • Bucket speed: $ U $.

  • Relative velocity: $ W $, deflection angle $$\displaystyle \phi \approx 165^\circ $$ for split bucket.

  • For ideal (frictionless), $$\displaystyle W_1 = V_j - U $$, $$\displaystyle W_2 = V_j - U $$ (magnitude), direction change $ \phi $.

• Design Parameters:

  • Jet ratio $$\displaystyle m = D/d $$ (runner diameter to jet diameter), typical $ m \approx 10–20 $.

  • Number of buckets $ z \approx 15 + 0.5D/d $ (to prevent jet impact on adjacent bucket).

  • Side clearance angle $$\displaystyle \approx 1^\circ–2^\circ $$.

• Performance Calculations:

  • Power available at nozzle: $$\displaystyle P_{\text{avail}} = \rho g Q H $$.

  • Hydraulic efficiency:

$$ \eta_h = \frac{2U V_j \cos(\phi/2)}{V_j^2} \quad (\text{for ideal bucket}) $$

For split bucket, $$\displaystyle \phi = 165^\circ $$, $$\displaystyle \cos(\phi/2) = \cos 82.5^\circ \approx 0.13 $$? Actually, $$\displaystyle \cos(82.5^\circ) \approx 0.13 $$, but typical formula: $$\displaystyle \eta_h = \frac{2U}{V_j} \cos\alpha_2 $$, where $$\displaystyle \alpha_2 $$ is angle of $$\displaystyle W_2 $$ with $ U $. For ideal, $$\displaystyle \alpha_2 = \phi/2 $$? In many texts, $$\displaystyle \eta_h = \frac{2U}{V_j} \cos\alpha_2 $$, and for Pelton, $$\displaystyle \alpha_2 \approx 10^\circ $$? I need to recall standard formula.

Actually, from velocity diagram: Work per kg = $$\displaystyle U(V_w1 + V_w2) $$. For ideal, $$\displaystyle V_w1 = V_j $$ (if jet tangential), $$\displaystyle V_w2 = -V_j \cos\phi $$? If deflection $ \phi $, then $$\displaystyle V_w2 = -V_j \cos\phi $$? For $$\displaystyle \phi = 180^\circ $$, $$\displaystyle V_w2 = -V_j $$, so work = $$\displaystyle U(V_j - (-V_j)) = 2U V_j $$. But for $$\displaystyle \phi < 180^\circ $$, $$\displaystyle V_w2 = -V_j \cos\phi $$? Actually, if the bucket reverses the jet by $ \phi $, the change in whirl velocity is $$\displaystyle V_j - (-V_j \cos\phi) = V_j(1 + \cos\phi) $$. But for split bucket, $$\displaystyle \phi \approx 165^\circ $$, $ \cos\phi \approx -0.9659 $, so $ 1 + \cos\phi \approx 0.0341 $, which is small. That can't be right because Pelton efficiency is high. I think I have the sign wrong.

Standard: For Pelton, the jet strikes the bucket and is deflected by nearly $$\displaystyle 180^\circ $$. The relative velocity magnitude remains nearly constant ($$\displaystyle W_1 \approx W_2 $$). The whirl component of absolute velocity: $$\displaystyle V_{w1} = V_j $$ (if jet tangential), $$\displaystyle V_{w2} = -V_j \cos\phi $$? Actually, if the bucket moves with speed $ U $, and the jet velocity is $$\displaystyle V_j $$, then relative velocity $$\displaystyle W_1 = V_j - U $$ (if same direction). After deflection, $$\displaystyle W_2 $$ has same magnitude but direction changed by $ \phi $. Then $$\displaystyle V_{w2} = U + W_2 \cos(\pi - \phi) $$? This is messy.

Better to use formula: Work done per kg = $$\displaystyle U(V_{w1} + V_{w2}) $$. For ideal Pelton with frictionless bucket and $$\displaystyle \phi = 180^\circ $$, $$\displaystyle V_{w1} = V_j $$, $$\displaystyle V_{w2} = -V_j $$, so work = $$\displaystyle U( V_j - (-V_j) ) = 2U V_j $$. But then maximum efficiency when $$\displaystyle U = V_j/2 $$, giving $$\displaystyle \eta_h = 2U/V_j = 1 $$? That's not right because there is also the axial component? Actually, for Pelton, the jet is tangential, so $$\displaystyle V_{a1}=0 $$? No, the jet is usually directed to hit the bucket at an angle? In Pelton, the nozzle is shaped to give a circular jet that strikes the bucket at the splitter. The bucket is designed so that the jet enters axially relative to the bucket? I think for Pelton, the absolute velocity is tangential, so $$\displaystyle V_{a1}=0 $$, $$\displaystyle V_{w1}=V_j $$. After deflection, the absolute velocity has a tangential component $$\displaystyle V_{w2} $$ and an axial component $$\displaystyle V_{a2} $$. But the work is only from tangential component. So $$\displaystyle V_{w2} = -V_j \cos\phi $$? For $$\displaystyle \phi=180^\circ $$, $$\displaystyle V_{w2} = -V_j $$, so work = $$\displaystyle U(V_j - (-V_j)) = 2U V_j $$. But then the kinetic energy loss is $$\displaystyle (V_{w2}^2 + V_{a2}^2)/2 $$. For $$\displaystyle \phi=180^\circ $$, $$\displaystyle V_{a2}=0 $$, so loss = $$\displaystyle V_j^2/2 $$. Then efficiency = $$\displaystyle 2U V_j / V_j^2 = 2U/V_j $$. Max when $$\displaystyle U = V_j/2 $$, efficiency = 1. But actual efficiency is less due to friction and $$\displaystyle \phi < 180^\circ $$. So the formula $$\displaystyle \eta_h = \frac{2U}{V_j} \cos\alpha_2 $$ is common, where $$\displaystyle \alpha_2 $$ is the angle of $$\displaystyle V_2 $$ with the tangential direction? Actually, $$\displaystyle \cos\alpha_2 = -V_{w2}/V_2 $$? I'll use the standard formula from textbooks:

Hydraulic efficiency:

$$ \eta_h = \frac{U(V_{w1} + V_{w2})}{V_j^2} $$

For ideal bucket with $$\displaystyle \phi = 180^\circ $$ and no friction, $$\displaystyle V_{w1}=V_j $$, $$\displaystyle V_{w2}=-V_j $$, so $$\displaystyle \eta_h = \frac{2U V_j}{V_j^2} = \frac{2U}{V_j} $$.

With friction, $$\displaystyle V_2 < V_1 $$, and $$\displaystyle \phi < 180^\circ $$, so $$\displaystyle V_{w2} = -V_2 \cos\phi $$? Actually, from vector diagram, if $$\displaystyle W_1 $$ and $$\displaystyle W_2 $$ have same magnitude $ W $, and $ U $ is horizontal, then $$\displaystyle V_{w1} = U + W \cos\beta_1 $$? This is confusing.

To avoid confusion, I'll state: For Pelton, the work done per kg is $$\displaystyle U(V_{w1} + V_{w2}) $$, and hydraulic efficiency $$\displaystyle \eta_h = \frac{U(V_{w1} + V_{w2})}{gH} $$, where $ gH $ is the available head. Since $$\displaystyle V_j = \sqrt{2gH} $$ ideally, $$\displaystyle \eta_h = \frac{U(V_{w1} + V_{w2})}{V_j^2} $$.

In many problems, they use $$\displaystyle \eta_h = \frac{2U}{V_j} \cos\alpha_2 $$, where $$\displaystyle \alpha_2 $$ is the angle of $$\displaystyle V_2 $$ with the direction of $ U $. For a well-designed bucket, $$\displaystyle \alpha_2 \approx 10^\circ–15^\circ $$.

• Force on bucket: $$\displaystyle F = \dot{m} (V_{w1} - V_{w2}) $$ (since $$\displaystyle V_{w2} $$ negative).

• Work done per kg: $$\displaystyle w = U(V_{w1} + V_{w2}) $$.

C. Draft Tubes

• Function:

  • Regain kinetic energy from exit of runner.

  • Increase net head by reducing exit pressure (pressure recovery).

  • Provide a passage for water to exit to tailrace.

• Types:

  • Conical: Simple, straight taper.

  • Elbow: Saves space, changes direction from vertical to horizontal.

  • Moody (spreading): Inlet is cylindrical, then spreads to reduce velocity.

• Theory: Based on Bernoulli’s equation. Efficiency:

$$ \eta_d = \frac{H_{\text{actual}}}{H_{\text{theoretical}}} = \frac{(p_2 - p_3)/\rho g + (V_2^2 - V_3^2)/(2g)}{V_2^2/(2g)} $$

where 2 = runner exit, 3 = draft tube exit.

D. Specific Speed of Turbines ($$\displaystyle N_s $$)

• Definition:

$$ N_s = \frac{N \sqrt{P}}{H^{5/4}} $$

where $ N $ = speed (rpm), $ P $ = power (kW), $ H $ = head (m).

• Derivation from similarity laws: $$\displaystyle \phi = Q/(ND^3) $$, $$\displaystyle \psi = gH/(N^2 D^2) $$, $$\displaystyle \lambda = P/(\rho N^3 D^5) $$. Eliminate $ D $ to get $$\displaystyle N_s $$.

• Significance: Characterizes turbine type and geometry. Higher $$\displaystyle N_s $$ → lower head, higher flow (e.g., Kaplan $$\displaystyle N_s > 300 $$, Francis $ 60–300 $, Pelton $$\displaystyle <60 $$).

E. Efficiencies

• Hydraulic efficiency $$\displaystyle \eta_h $$: $$\displaystyle \frac{\text{Power delivered to runner}}{\rho g Q H} $$.

• Mechanical efficiency $$\displaystyle \eta_m $$: $$\displaystyle \frac{\text{Shaft power}}{\text{Runner power}} $$ (accounts for bearing losses).

• Volumetric efficiency $$\displaystyle \eta_v $$: $$\displaystyle \frac{\text{Water used}}{\text{Water supplied}} $$ (accounts for leakage).

• Overall efficiency $$\displaystyle \eta_o = \eta_h \eta_m \eta_v $$.

• Manometric head: Head measured by pressure gauge; includes losses in delivery pipe.

F. Cavitation

• Causes: Local pressure drops below vapor pressure due to high velocity on suction side of blades.

• Effects: Pitting, noise, vibration, efficiency drop, damage to blades.

• Prevention:

  • Ensure sufficient Net Positive Suction Head (NPSH): $$\displaystyle \text{NPSH}_a > \text{NPSH}_r $$.

  • Proper design to avoid low-pressure zones.

  • Use cavitation-resistant materials.

IV. Pumps

A. Centrifugal Pumps

• Main parts: Impeller, casing (volute or diffuser), suction/delivery pipes, stuffing box, bearings.

• Working principle: Impeller rotates, fluid gains kinetic energy and pressure via centrifugal force. Volute converts kinetic energy to pressure.

• Velocity diagram at inlet (shockless inflow): $$\displaystyle C_{u1} = 0 $$ (if no pre-swirl), $$\displaystyle \alpha_1 = 90^\circ $$, $$\displaystyle W_1 $$ relative to blade angle $$\displaystyle \beta_1 $$.

  At outlet: $$\displaystyle C_{u2} = U_2 - W_{u2} $$, $$\displaystyle C_{a2} = W_{a2} $$.

• Head-capacity curve: $ H $ decreases as $ Q $ increases. Best Efficiency Point (BEP) at design flow.

• Slip factor $ \sigma $: Accounts for deviation from ideal due to fluid slip. $$\displaystyle \sigma = \frac{C_{u2,\text{actual}}}{C_{u2,\text{ideal}}} $$.

• Specific speed of pumps $$\displaystyle N_s $$:

$$ N_s = \frac{N \sqrt{Q}}{H^{3/4}} $$

(Note: Different exponent than turbines.) Used for selection: radial impeller ($$\displaystyle N_s < 50 $$), mixed-flow ($ 50–500 $), axial ($$\displaystyle >500 $$).

B. Positive Displacement Pumps

• Types: Reciprocating (piston/plunger), rotary (gear, vane, screw).

• Characteristics:

  • Constant flow (independent of pressure, within limits).

  • High pressure generation.

  • Pulsating flow (reciprocating).

  • Self-priming.

  • Used for metering, high-viscosity fluids.

C. Comparison: Centrifugal vs Reciprocating Pumps

D. Cavitation in Pumps

• NPSH:

  • Required $$\displaystyle \text{NPSH}_r $$: Minimum NPSH to avoid cavitation (given by manufacturer).

  • Available $$\displaystyle \text{NPSH}_a = \frac{p_{\text{suction}}}{\rho g} + \frac{V_{\text{suction}}^2}{2g} - \frac{p_v}{\rho g} $$.

• Prevention: Ensure $$\displaystyle \text{NPSH}_a > \text{NPSH}_r $$, optimize suction pipe design (large diameter, short length, minimal bends).

V. Compressors

A. Centrifugal Compressors

• Construction: Impeller (radial or backward-curved blades), diffuser (vaned or vaneless), volute.

• Vector diagram: Inlet: $$\displaystyle C_{u1} \approx 0 $$ (often with pre-swirl), $$\displaystyle C_{a1} $$. Outlet: $$\displaystyle C_{u2} = U_2 - W_{u2} $$, $$\displaystyle C_{a2} = W_{a2} $$.

• Work input (ideal, adiabatic):

$$ w = U_2 C_{u2} - U_1 C_{u1} \approx U_2 C_{u2} $$

• Isentropic efficiency:

$$ \eta_s = \frac{\text{Isentropic work input}}{\text{Actual work input}} = \frac{h_{02s} - h_{01}}{h_{02} - h_{01}} $$

• Pressure ratio: $$\displaystyle \frac{p_{02}}{p_{01}} = \left(1 + \frac{\eta_s (T_{02s} - T_{01})}{T_{01}}\right)^{\gamma/(\gamma-1)} $$.

• Multi-staging: To achieve high pressure ratio; intercooling between stages improves efficiency.

B. Axial Flow Compressors

• Construction: Rotor and stator blades alternating, multi-stage, often with variable stator vanes.

• Vector diagram for a stage:

  • Inlet to rotor: $$\displaystyle C_1 $$, $$\displaystyle \alpha_1 $$; relative $$\displaystyle W_1 $$, $$\displaystyle \beta_1 $$.

  • Outlet from rotor = inlet to stator: $$\displaystyle C_2 $$, $$\displaystyle \alpha_2 $$; relative $$\displaystyle W_2 $$, $$\displaystyle \beta_2 $$.

  • Outlet from stator: $$\displaystyle C_3 $$, $$\displaystyle \alpha_3 $$, etc.

• Degree of reaction $ R $: Typically 50% for axial compressors (equal pressure drop in rotor and stator).

$$ R = \frac{\text{Static enthalpy drop in rotor}}{\text{Stage static enthalpy drop}} = \frac{h_1 - h_2}{h_1 - h_3} $$

For 50% reaction, $$\displaystyle \alpha_1 = \beta_2 $$, $$\displaystyle \beta_1 = \alpha_2 $$, and $$\displaystyle C_1 = C_3 $$, $$\displaystyle C_2 = C_4 $$, etc.

• Stage loading:

$$ \Delta h_0 = U (C_{w2} - C_{w1}) $$

Often expressed as $$\displaystyle \psi = \frac{\Delta h_0}{U^2} $$ (head coefficient).

• Polytropic efficiency $$\displaystyle \eta_p $$:

  • Definition: Efficiency of an infinitesimal stage; constant for multi-stage compressor.

  • Relation to isentropic efficiency:

$$ \eta_s = \frac{r^{(\gamma-1)/\gamma} - 1}{r^{\eta_p (\gamma-1)/\gamma} - 1} $$

where $$\displaystyle r = p_{0,\text{out}}/p_{0,\text{in}} $$.

• Importance: More accurate for multi-stage compression with intercooling.

C. Performance and Operational Issues

• Surging:

  • Definition: Complete flow reversal due to stall on entire compressor blade row.

  • Cause: Flow separation at high incidence angles (low flow, high pressure ratio).

  • Characteristics: On compressor map, surge line is the left boundary; violent oscillations, possible damage.

  • Prevention: Bleed valves, variable stator vanes, control systems.

• Choking:

  • Definition: Mach number reaches 1 at blade tips or throats; flow becomes sonic and constant.

  • Cause: Sonic velocity limit at minimum area.

  • Effect: Maximum mass flow rate; further increase in pressure ratio does not increase flow.

  • On map: Choke line is the right boundary.

D. Comparison: Axial vs Centrifugal Compressors

VI. Power Transmission Devices

A. Fluid Coupling

• Construction: Impeller (pump) on driving shaft, runner (turbine) on driven shaft, casing filled with fluid.

• Working principle: Momentum transfer: driving impeller imparts kinetic energy to fluid, which drives runner. Slip occurs (runner speed < impeller speed).

• Efficiency:

$$ \eta = \frac{2n}{1+n} \quad \text{where } n = \frac{N_t}{N_p} \text{ (speed ratio)} $$

• Applications: Soft start, overload protection, torque dampening in conveyors, crushers, mills.

B. Torque Converter

• Construction: Pump (impeller), turbine, stator (with one-way clutch).

• Working principle: Pump throws fluid onto turbine blades; stator redirects fluid to improve torque multiplication.

• Vector diagram: Shows velocities at pump, turbine, stator.

• Torque ratio:

$$ \frac{T_t}{T_p} = \frac{U_t (C_{w2} - C_{w1})}{U_p (C_{w2} - C_{w1})} \quad \text{? Actually, standard: } \frac{T_t}{T_p} = \frac{\dot{m}(U_t C_{w2t} - U_t C_{w1t})}{\dot{m}(U_p C_{w2p} - U_p C_{w1p})} \text{ but usually pump and turbine have different radii.} $$

More common: $$\displaystyle \frac{T_t}{T_p} = \frac{U_t}{U_p} \cdot \frac{C_{w2t} - C_{w1t}}{C_{w2p} - C_{w1p}} $$, but with stator, the torque multiplication occurs at low speed ratios.

• Applications: Automotive automatic transmissions, industrial drives.

C. Industrial Applications of Power Transmitting Devices

• Fluid coupling: Conveyors, crushers, mills, pumps, fans, compressors.

• Torque converter: Automobiles, forklifts, agricultural equipment, marine propulsion.

VII. Analysis and Design Methods

A. Dimensional Analysis and Similarity

• Buckingham Pi Theorem: For $ n $ variables and $ k $ fundamental dimensions, there are $ n-k $ independent dimensionless groups.

• Example: Fan efficiency $$\displaystyle \eta = f(\rho, \mu, \omega, D, Q) $$.

  • Variables: 5; dimensions: $ M, L, T $ → 3; so 2 Pi groups.

  • Choose repeating variables: $ \rho, \omega, D $.

  • Pi groups: $$\displaystyle \Pi_1 = \eta $$, $$\displaystyle \Pi_2 = \frac{\rho \omega D^2}{\mu} = \text{Re} $$, $$\displaystyle \Pi_3 = \frac{Q}{\omega D^3} = \phi $$? Actually, with 5 variables and 3 dimensions, we have 2 Pi groups. But efficiency is dimensionless, so one Pi is $ \eta $. The other Pi must combine the remaining variables. Typically, we get $$\displaystyle \eta = f(\text{Re}, \phi) $$, where $$\displaystyle \phi = Q/(\omega D^3) $$.

• Similarity Laws for Turbomachines:

  • Flow coefficient: $$\displaystyle \phi = \frac{Q}{N D^3} $$

  • Head coefficient: $$\displaystyle \psi = \frac{gH}{N^2 D^2} $$

  • Power coefficient: $$\displaystyle \lambda = \frac{P}{\rho N^3 D^5} $$

• Scaling: For model and prototype:

$$ \frac{Q_p}{Q_m} = \frac{N_p D_p^3}{N_m D_m^3}, \quad \frac{H_p}{H_m} = \frac{N_p^2 D_p^2}{N_m^2 D_m^2}, \quad \frac{P_p}{P_m} = \frac{\rho_p N_p^3 D_p^5}{\rho_m N_m^3 D_m^5} $$

B. Performance Maps

• Compressor map:

  • Axes: Mass flow vs pressure ratio.

  • Lines: Surge line (left), choke line (right), efficiency contours.

  • Shows operating range.

• Turbine map: Flow function vs pressure ratio, efficiency contours.

C. Specific Speed

• Unified concept: For pumps: $$\displaystyle N_s = \frac{N \sqrt{Q}}{H^{3/4}} $$; for turbines: $$\displaystyle N_s = \frac{N \sqrt{P}}{H^{5/4}} $$.

• Use: Preliminary design and selection of machine type (radial, mixed, axial).

VIII. Ancillary and Short Note Topics

A. Centrifugal Blowers

• Similar to centrifugal compressors but for low-pressure gas movement (pressure rise typically < 0.5 bar).

• Applications: Ventilation, combustion air supply, drying, dust collection.

B. Hydraulic Intensifier

• Principle: Uses area ratio to increase pressure: $$\displaystyle P_2 = P_1 \times \frac{A_1}{A_2} $$.

• Construction: Two cylinders (high-pressure and low-pressure), ram, reservoirs.

• Applications: High-pressure hydraulic systems for testing, presses, where pump cannot generate required pressure directly.

C. Draft Tube

• (See III.C)

  Short note: Draft tube is a diffuser at turbine exit to recover kinetic energy and increase net head. Types: conical, elbow, Moody. Efficiency defined as ratio of actual head gain to theoretical velocity head recovery.

D. Other Short Note Topics (from Exams)

• Cavitation in Turbines and Pumps: See III.F and IV.D.

• Governing of Steam Turbines: See II.G.

• Polytropic Efficiency: See V.B.3.

• Slip in Fluid Coupling: Slip $$\displaystyle s = 1 - n $$, where $$\displaystyle n = N_t/N_p $$. Efficiency $$\displaystyle \eta = \frac{2n}{1+n} $$.

[!TIP] Exam Tips

• Velocity diagrams: Always draw clearly; label $ U, C, W $ and angles $ \alpha, \beta $. Use consistent convention (axial or tangential).

• Parsons turbine: Remember key relations: $$\displaystyle \alpha_1 = \beta_2 $$, $$\displaystyle \beta_1 = \alpha_2 $$, and for 50% reaction with constant $$\displaystyle C_a $$: $$\displaystyle C_{w1} + C_{w2} = U $$.

• Euler equation: $$\displaystyle \Delta h = U(C_{w1} - C_{w2}) $$ for turbine; sign matters.

• Specific speed: Know formulas for pumps ($$\displaystyle N_s \propto N\sqrt{Q}/H^{3/4} $$) and turbines ($$\displaystyle N_s \propto N\sqrt{P}/H^{5/4} $$).

• Surging vs Choking: Surging = flow reversal at low flow; choking = sonic limit at high flow.

• Dimensional analysis: Practice forming Pi groups; common groups: Re, $ \phi $, $ \psi $, $ \lambda $.

• Numericals: For Parsons, first compute $ U $, then solve for $$\displaystyle C_a $$ using $$\displaystyle C_a \tan\beta_2 + \sqrt{C_2^2 - C_a^2} = U $$. Then find $$\displaystyle \beta_1 = \cos^{-1}(C_a/C_2) $$.