12.6 Turbomachinery: Pumps, Turbines, Fans, and Compressors

12.6 Turbomachinery: Pumps, Turbines, Fans, and Compressors

Core FE Exam Principle: Turbomachines transfer energy between a fluid continuous medium and a rotating shaft. Pumps add mechanical energy to fluids, whereas turbines extract hydraulic energy. Gas dynamics analyzes compressible fluid flows where density varies significantly with pressure and temperature.

Pump Performance and System Operating Points

A pump operates at the intersection of its Pump Head-Capacity Curve ((H_{\text{pump}}) vs. (Q)) and the system's System Head Curve ((H_{\text{sys}}) vs. (Q)).

System Head Curve Equation

$H_{\text{sys}}(Q) = (z_2 - z_1) + \frac{P_2 - P_1}{\gamma} + \left[ \sum f \left(\frac{L}{D}\right) + \sum K \right] \frac{Q^2}{2g A^2} = h_{\text{static}} + K_{\text{sys}} Q^2$

• Operating Point: Found by equating (H_{\text{pump}}(Q) = H_{\text{sys}}(Q)).

  Head [m] ^ 
           |   Pump Performance Curve H_pump(Q)
           |   \\ 
           |    \\           Operating Point
           |-----\\---------> (Q_op, H_op)
           |      \\         /
           |       \\       /
           |        \\     /  System Head Curve H_sys = h_static + K*Q^2
           |________(o)__/________________________>
           0                                        Flow Rate Q [m^3/s]

Efficiencies

• Pump Hydraulic Efficiency ((\eta_{\text{pump}})): $\eta_{\text{pump}} = \frac{W_{\text{hyd}}}{P_{\text{shaft}}} = \frac{\gamma Q h_p}{P_{\text{shaft}}}$
• Overall Wire-to-Water Efficiency ((\eta_{\text{overall}})): $\eta_{\text{overall}} = \eta_{\text{pump}} \cdot \eta_{\text{motor}} = \frac{\gamma Q h_p}{P_{\text{electrical}}}$

Net Positive Suction Head (NPSH) and Cavitation

To prevent cavitation inside a pump impeller, the static pressure at the pump inlet flange must remain safely above the fluid vapor pressure (P_v).

Net Positive Suction Head Available (NPSHA)

Calculated based on suction piping system hydraulics:

$NPSHA = \frac{P_{s,\text{abs}}}{\gamma} + \frac{V_s^2}{2g} - \frac{P_v}{\gamma} = \frac{P_{\text{atm}}}{\gamma} + \frac{P_{s,\text{gage}}}{\gamma} + \frac{V_s^2}{2g} - \frac{P_v}{\gamma}$

If the suction supply reservoir is an open atmospheric tank located distance (h_z) relative to the pump inlet centerline:

$NPSHA = \frac{P_{\text{atm}} - P_v}{\gamma} \pm h_z - h_{L,\text{suction}}$

• Use (+h_z) if the liquid level is above the pump inlet (static suction head).
• Use (-h_z) if the liquid level is below the pump inlet (static suction lift).

Cavitation Prevention Criterion

$NPSHA \ge NPSHR$

where (NPSHR) is the Net Positive Suction Head Required published by the pump manufacturer based on experimental testing.

Homologous Turbomachinery and Pump Affinity Laws

The Affinity Laws govern performance changes for geometrically similar centrifugal pumps operating at different rotational speeds ((N)) or impeller diameters ((D)).

Varying Rotational Speed (N) (Fixed Impeller Diameter (D))

1. Volumetric Flow Rate: Proportional to rotational speed ((Q \propto N)): $\frac{Q_1}{Q_2} = \frac{N_1}{N_2}$
2. Pump Head: Proportional to the square of rotational speed ((h_p \propto N^2)): $\frac{h_{p1}}{h_{p2}} = \left(\frac{N_1}{N_2}\right)^2$
3. Shaft Power: Proportional to the cube of rotational speed ((P \propto N^3)): $\frac{P_1}{P_2} = \left(\frac{N_1}{N_2}\right)^3$

Combined Scaling Laws (Varying Speed (N) and Diameter (D))

$\frac{Q_1}{Q_2} = \left(\frac{N_1}{N_2}\right) \left(\frac{D_1}{D_2}\right), \quad \frac{h_{p1}}{h_{p2}} = \left(\frac{N_1}{N_2}\right)^2 \left(\frac{D_1}{D_2}\right)^2, \quad \frac{P_1}{P_2} = \left(\frac{N_1}{N_2}\right)^3 \left(\frac{D_1}{D_2}\right)^5$

Compressible Flow Basics and Gas Dynamics

When flow velocities approach the speed of sound, fluid density variations cannot be neglected (typically when (Ma > 0.3)).

Speed of Sound and Mach Number

• Speed of Sound ((c)): Propagation speed of infinitesimal pressure waves through an ideal gas: $c = \sqrt{k R T} = \sqrt{\gamma R T}$ where:

  • (k = c_p/c_v) = Specific heat ratio ((k = 1.4) for diatomic air)
  • (R) = Specific gas constant ((R_{\text{air}} = 287 \text{ J/(kg}\cdot\text{K)} = 1716 \text{ ft}\cdot\text{lbf/(slug}\cdot^\circ\text{R)}))
  • (T) = Absolute static temperature (Kelvin or Rankine)

• Mach Number ((Ma)): Ratio of local flow velocity to local speed of sound: $Ma = \frac{V}{c} = \frac{V}{\sqrt{k R T}}$

Isentropic Stagnation Relations for Ideal Gases

$\frac{T_0}{T} = 1 + \frac{k-1}{2} Ma^2$ $\frac{P_0}{P} = \left( 1 + \frac{k-1}{2} Ma^2 \right)^{\frac{k}{k-1}}$ $\frac{\rho_0}{\rho} = \left( 1 + \frac{k-1}{2} Ma^2 \right)^{\frac{1}{k-1}}$

Isentropic Converging-Diverging Nozzles and Choked Flow

For flow accelerated from a stagnant reservoir ((P_0, T_0)) through a nozzle, sonic flow ((Ma=1.0)) can occur only at the minimum cross-sectional area (throat, (A^*)).

• Critical Pressure Ratio ((P^*/P_0)): Minimum backpressure ratio required to achieve choked flow at the throat: $\frac{P^*}{P_0} = \left( \frac{2}{k+1} \right)^{\frac{k}{k-1}}$ For standard air ((k = 1.4)): $\frac{P^*}{P_0} = \left( \frac{2}{1.4+1} \right)^{\frac{1.4}{0.4}} = (0.8333)^{3.5} = 0.5283$

Key Principle: Once a nozzle is choked ((Ma_{\text{throat}} = 1.0)), decreasing downstream backpressure further cannot increase mass flow rate through the nozzle.

Comprehensive Worked Engineering Example

Problem Statement

Part A: A water pumping system draws water ((\gamma = 9.81 \text{ kN/m}^3), vapor pressure (P_v = 2.34 \text{ kPa abs})) from an open sump through a suction line. Atmospheric pressure is (P_{\text{atm}} = 101.3 \text{ kPa}). The pump centerline is located (h_z = 2.50 \text{ m}) above the open water surface (suction lift). The suction pipe losses total (h_{L,\text{suction}} = 1.20 \text{ m}) at a flow rate (Q = 0.050 \text{ m}^3/\text{s}). The pump delivers (h_p = 42.0 \text{ m}) of head with an efficiency (\eta_{\text{pump}} = 76%).

Calculate:

1. The Net Positive Suction Head Available ((NPSHA)).
2. The required brake horsepower (shaft input power (P_{\text{shaft}})).
3. The new shaft power (P_{\text{shaft},2}) if the pump speed is increased by (15%).

Part B: Air ((k = 1.4), (R = 287 \text{ J/(kg}\cdot\text{K)})) expands from a large supply reservoir at (P_0 = 400 \text{ kPa abs}) and (T_0 = 350 \text{ K}) through an isentropic converging-diverging nozzle. At a specific nozzle cross-section, the measured Mach number is (Ma = 1.60).

Calculate: 4. The local static temperature (T), static pressure (P), and flow velocity (V) at this nozzle section.

Step-by-Step Solution

Step 1: Compute NPSHA for the Pump System

• Equivalent atmospheric head: (\frac{P_{\text{atm}}}{\gamma} = \frac{101.3 \text{ kPa}}{9.81 \text{ kN/m}^3} = 10.326 \text{ m}).
• Vapor pressure head: (\frac{P_v}{\gamma} = \frac{2.34 \text{ kPa}}{9.81 \text{ kN/m}^3} = 0.239 \text{ m}).
• NPSHA formula for suction lift: $NPSHA = \frac{P_{\text{atm}} - P_v}{\gamma} - h_z - h_{L,\text{suction}}$ $NPSHA = 10.326 - 0.239 - 2.50 - 1.20 = 6.387 \text{ m} \approx 6.39 \text{ m}$

Step 2: Compute Initial Pump Shaft Power (P_{\text{shaft}})

• Hydraulic power delivered to fluid: $W_{\text{hyd}} = \gamma Q h_p = 9810 \text{ N/m}^3 \times 0.050 \text{ m}^3/\text{s} \times 42.0 \text{ m} = 20,601 \text{ W} = 20.60 \text{ kW}$
• Shaft input power: $P_{\text{shaft}} = \frac{W_{\text{hyd}}}{\eta_{\text{pump}}} = \frac{20.601 \text{ kW}}{0.76} = 27.11 \text{ kW}$

Step 3: Compute New Shaft Power After 15% Speed Increase

• Speed ratio: (\frac{N_2}{N_1} = 1.15).
• By the affinity power law: (\frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3 = (1.15)^3 = 1.520875).
• New shaft power: $P_{\text{shaft},2} = 27.11 \text{ kW} \times 1.520875 = 41.23 \text{ kW}$

Step 4: Compute Compressible Flow Properties at (Ma = 1.60)

• Temperature ratio: $\frac{T_0}{T} = 1 + \frac{k-1}{2} Ma^2 = 1 + \frac{1.4-1}{2} (1.60)^2 = 1 + 0.2(2.56) = 1 + 0.512 = 1.512$ $T = \frac{T_0}{1.512} = \frac{350 \text{ K}}{1.512} = 231.48 \text{ K}$
• Static pressure ratio: $\frac{P_0}{P} = \left(\frac{T_0}{T}\right)^{\frac{k}{k-1}} = (1.512)^{\frac{1.4}{0.4}} = (1.512)^{3.5} = 4.2504$ $P = \frac{P_0}{4.2504} = \frac{400 \text{ kPa}}{4.2504} = 94.11 \text{ kPa}$
• Local speed of sound: $c = \sqrt{k R T} = \sqrt{1.4 \times 287 \times 231.48} = \sqrt{92,999} = 304.96 \text{ m/s}$
• Flow velocity: $V = Ma \cdot c = 1.60 \times 304.96 \text{ m/s} = 487.94 \text{ m/s}$

Final Answer: (NPSHA = 6.39 \text{ m}), initial shaft power (P_{\text{shaft}} = 27.11 \text{ kW}), speed-scaled shaft power (P_{\text{shaft},2} = 41.23 \text{ kW}), and compressible flow parameters (T = 231.5 \text{ K}), (P = 94.11 \text{ kPa}), (V = 487.9 \text{ m/s}).