3.4 Pumps, Compressors, and NPSH Calculations

3.4 Pumps, Compressors, and NPSH Calculations

Pumps and compressors add mechanical energy to fluids to overcome friction, increase elevation, or raise pressure. On the NCEES PE Chemical exam, fluid machinery questions emphasize centrifugal pump hydraulics, affinity laws, motor sizing, and above all, Net Positive Suction Head (NPSH) calculations to prevent cavitation.

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1. Machinery Classification: Centrifugal vs. Positive Displacement

Process pumps fall into two major thermodynamic categories:

Kinetic (Centrifugal) Pumps

Centrifugal pumps impart kinetic energy to the fluid using a high-speed rotating impeller. As fluid exits the impeller vanes radially or axially, it enters a diffusing volute casing where velocity converts to static pressure. Centrifugal pumps are the workhorses of the chemical industry, favored for continuous, non-pulsating delivery of low-viscosity fluids.

Positive Displacement (PD) Pumps

PD pumps trap a fixed volume of fluid inside an enclosed chamber and physically displace it against discharge pressure using reciprocating pistons/plungers or rotating gears, lobes, or screws.

• Key Distinction: A centrifugal pump produces a variable flow rate depending on discharge pressure ($H$ vs. $Q$ curve). A PD pump delivers a virtually constant volumetric flow rate regardless of discharge pressure, limited only by drive motor torque and casing burst strength. For this reason, PD pumps must always be protected by a dedicated pressure relief valve (PRV) installed upstream of the first discharge block valve.

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2. Total Dynamic Head (TDH) and Pump Power

Applying the Mechanical Energy Balance across a pump from suction flange ($s$) to discharge flange ($d$) yields the Total Dynamic Head (TDH) imparted to the fluid:

$TDH = H = \left(\frac{P_d - P_s}{\rho g}\right) + (z_d - z_s) + \left(\frac{v_d^2 - v_s^2}{2g}\right)$

If expressed between the free surface of an upstream supply vessel ($1$) and a downstream destination vessel ($2$):

$TDH = \left(\frac{P_{2,abs} - P_{1,abs}}{\rho g}\right) + (z_2 - z_1) + h_{f,total}$

where $h_{f,total} = h_{f,suct} + h_{f,disch}$ is the sum of all piping and minor losses in the suction and discharge lines.

Power and Efficiency Formulations

1. Water (Hydraulic) Horsepower ($WHP$): The theoretical mechanical power transferred to the fluid: $WHP = \dot{m} g (TDH) = \rho g Q (TDH) = Q \Delta P$ In US Customary units ($Q$ in gpm, $TDH$ in ft, $SG$ relative to water at $60^\circ\text{F}$): $WHP = \frac{Q \times TDH \times SG}{3,960} = \frac{Q \times \Delta P\text{ [psi]}}{1,714}$

2. Brake Horsepower ($BHP$): The actual shaft power required from the electric motor or turbine, accounting for pump hydraulic, mechanical, and volumetric efficiency $\eta_p$: $BHP = \frac{WHP}{\eta_p} = \frac{Q \times TDH \times SG}{3,960 \times \eta_p}$

3. Electrical Motor Power ($P_{elec}$): Factoring in driver/motor electrical efficiency $\eta_m$: $P_{elec} = \frac{BHP}{\eta_m} = \frac{WHP}{\eta_p \eta_m} = \frac{WHP}{\eta_{overall}}$

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3. System Head Curves and Operating Duty Points

A piping system presents a system head curve $H_{sys}(Q)$ composed of static head and friction:

$H_{sys}(Q) = H_{static} + C_{sys} Q^2 = \left[ \frac{\Delta P_{static}}{\rho g} + \Delta z \right] + \left[ \sum \left(f \frac{L}{D} + K\right) \frac{1}{2 g A_c^2} \right] Q^2$

The pump's manufacturer-tested performance curve $H_{pump}(Q)$ slopes downward as flow increases. The intersection of $H_{pump}(Q)$ and $H_{sys}(Q)$ establishes the unique operating duty point $(Q_{op}, H_{op})$.

  Head (H) ^                     Pump Head Curve H_pump(Q)
           |  \ 
           |   \                   System Curve H_sys = H_stat + C*Q^2
           |    \               /
           |     \             /
       H_op|......*-----------/  <-- Operating Point (Duty Point)
           |     / \         /
           |    /   \       /
     H_stat|---+     \     /
           |          \   /
           |           \ /
           +-----------------------------> Flow Rate (Q)
                       Q_op

Pumps Operating in Series vs. Parallel

• Series Operation: Pumps arranged in series handle the same flow rate while their heads add: $H_{total}(Q) = H_A(Q) + H_B(Q)$. Used for high-pressure boiler feed systems.
• Parallel Operation: Pumps discharging into a common header experience identical head while their capacities add: $Q_{total}(H) = Q_A(H) + Q_B(H)$. Used for high-capacity cooling water circulation.

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4. The Affinity Laws for Centrifugal Pumps

The affinity laws scale pump performance when modifying rotational shaft speed ($N$, in rpm) or trimming impeller outer diameter ($D$):

Variable Speed ($N$) with Constant Impeller Diameter ($D$)

$\frac{Q_2}{Q_1} = \frac{N_2}{N_1}, \quad \frac{H_2}{H_1} = \left(\frac{N_2}{N_1}\right)^2, \quad \frac{BHP_2}{BHP_1} = \left(\frac{N_2}{N_1}\right)^3$

Impeller Trimming ($D$) at Constant Speed ($N$)

(Valid for small trims within $10% - 20%$ of original diameter): $\frac{Q_2}{Q_1} = \frac{D_2}{D_1}, \quad \frac{H_2}{H_1} = \left(\frac{D_2}{D_1}\right)^2, \quad \frac{BHP_2}{BHP_1} = \left(\frac{D_2}{D_1}\right)^3$

[!TIP] Affinity Exponents Memory Aid: Flow is linear ($1$), Head is squared ($2$), Power is cubed ($3$). A $10%$ increase in pump speed yields a $10%$ increase in flow, a $21%$ increase in head ($(1.10)^2 = 1.21$), and a $33.1%$ surge in motor power ($(1.10)^3 = 1.331$)!

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5. Cavitation Physics and Net Positive Suction Head (NPSH)

The Mechanism of Cavitation

As fluid accelerates into the eye of a centrifugal impeller, local static pressure drops to its lowest point in the system. If this localized pressure falls to or below the fluid's vapor pressure at the operating temperature ($P_{local} \le P_{vp}$), the liquid spontaneously boils, generating vapor bubbles. As these bubbles sweep outward into high-pressure regions along the impeller vanes, they collapse violently in microseconds. The resulting micro-jets generate localized shock pressures exceeding $1,000\text{ MPa}$ ($145,000\text{ psi}$), pitting metal, eroding impellers, generating intense gravel-like noise, and destroying mechanical seals.

NPSHR vs. NPSHA

• NPSHR (Net Positive Suction Head Required): A function of pump design determined by the manufacturer via water testing (standardized at a $3%$ head drop criterion, $NPSH_3$). It is the minimum suction head over vapor pressure required at the suction nozzle to suppress cavitation.
• NPSHA (Net Positive Suction Head Available): A property of the plant process piping system. It represents the absolute suction head over vapor pressure actually delivered to the pump suction nozzle.

The Governing NPSHA Equation

$NPSHA = \frac{P_{suct,abs} - P_{vp}}{\rho g} + \frac{v_s^2}{2g}$

Referenced back to the upstream supply vessel free liquid surface:

$NPSHA = \frac{P_{source,abs}}{\rho g} + z_s - h_{f,suct} - \frac{P_{vp}}{\rho g}$

where:

• $P_{source,abs}$ is the absolute pressure on the liquid surface in the supply tank ($\text{Pa}$ or $\text{lb}_f/\text{ft}^2$),
• $P_{vp}$ is the true vapor pressure of the liquid at pumping temperature ($\text{Pa}$ or $\text{lb}_f/\text{ft}^2$),
• $z_s$ is the static elevation of the supply liquid level relative to the pump suction centerline ($+z_s$ for flooded suction head; $-z_s$ for suction lift),
• $h_{f,suct}$ is the cumulative frictional head loss (pipe skin friction + valves/fittings + entrance) in the suction line.

The Saturated / Boiling Liquid Case (Reboilers, Deaerators, Reflux Accumulators)

In chemical processing, liquids often leave separators or bottoms sumps at their bubble point (saturation pressure): $P_{source,abs} = P_{vp}$.

When $P_{source,abs} = P_{vp}$, the pressure terms cancel completely:

$NPSHA = z_s - h_{f,suct}$

[!WARNING] Critical PE Exam Rule: For a boiling liquid, the tank pressure cannot help you! The only available suction head comes from the physical liquid elevation $z_s$ above the pump centerline minus suction piping friction $h_{f,suct}$. If a boiling liquid pump is placed at grade with zero elevation head ($z_s = 0$), $NPSHA$ is negative ($-h_{f,suct}$), and the pump will cavitate instantly!

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6. Gas Compressors and Blowers

When compressing compressible gases where density varies substantially, fluid compression work is evaluated using thermodynamic path functions:

Isentropic (Reversible Adiabatic) Compression Power

For an ideal gas with heat capacity ratio $k = C_p / C_v$:

$W_{isen} = \dot{m} \left(\frac{k}{k-1}\right) R T_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} - 1 \right] = \left(\frac{k}{k-1}\right) P_1 Q_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} - 1 \right]$

The isentropic discharge temperature is:

$T_{2,isen} = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}$

Polytropic Compression

Real industrial compressors deviate from isentropic paths due to internal friction and heat dissipation, described by the polytropic exponent $n$:

$\frac{n}{n-1} = \left(\frac{k}{k-1}\right) \eta_{poly}$

To limit discharge temperatures below lubricant breakdown limits ($T_2 < 150^\circ\text{C}$ / $300^\circ\text{F}$) and minimize overall compression work, large compression ratios are split into multistage compression with intercooling between stages. For equal work distribution across $N_{stages}$ with perfect intercooling to $T_1$, the stage pressure ratio is constant:

$r_p = \left(\frac{P_{final}}{P_{initial}}\right)^{1 / N_{stages}}$

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7. Comparison Table of Industrial Process Pumps

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8. Step-by-Step Worked Numerical Example

Problem Statement

A centrifugal pump transfers liquid n-hexane ($SG = 0.655$, $\rho = 655\text{ kg/m}^3$) from an enclosed reflux accumulator to an overhead distillation column. The accumulator operates at a blanket pressure of $P_1 = 180\text{ kPa(a)}$ (absolute). At the operating temperature of $55^\circ\text{C}$, the vapor pressure of n-hexane is $P_{vp} = 65.0\text{ kPa(a)}$.

System parameters:

• Liquid level in accumulator: $z_1 = +3.50\text{ m}$ above the pump centerline.
• Suction line frictional head loss at design flow: $h_{f,suct} = 0.65\text{ m}$.
• Destination column nozzle elevation: $z_2 = +24.0\text{ m}$ above the pump centerline.
• Destination column operating pressure: $P_2 = 320\text{ kPa(a)}$.
• Discharge line frictional head loss: $h_{f,disch} = 4.80\text{ m}$.
• Design flow rate: $Q = 45.0\text{ m}^3/\text{h} = 0.0125\text{ m}^3/\text{s}$.
• Pump efficiency: $\eta_p = 0.68$.
• Manufacturer's required suction head: $NPSHR = 2.20\text{ m}$.
• Gravitational acceleration: $g = 9.807\text{ m/s}^2$.

Calculate:

1. The Net Positive Suction Head Available ($NPSHA$) in meters of hexane.
2. The available NPSH safety margin over $NPSHR$.
3. The Total Dynamic Head ($TDH$) developed by the pump in meters.
4. The Brake Horsepower ($BHP$) required from the electric motor in kilowatts and horsepower.

Solution

Step 1: Calculate NPSHA.

$NPSHA = \frac{P_{1,abs} - P_{vp}}{\rho g} + z_1 - h_{f,suct}$

Calculate the net static pressure head over vapor pressure:

$\Delta P_{suct} = P_{1,abs} - P_{vp} = 180.0\text{ kPa} - 65.0\text{ kPa} = 115.0\text{ kPa} = 115,000\text{ Pa}$ $\text{Pressure Head} = \frac{115,000\text{ Pa}}{655\text{ kg/m}^3 \times 9.807\text{ m/s}^2} = \frac{115,000}{6,423.6} = 17.903\text{ m}$

Insert into the NPSHA equation:

$NPSHA = 17.903\text{ m} + 3.50\text{ m} - 0.65\text{ m} = 20.753\text{ m} \approx 20.75\text{ m}$

Step 2: Check NPSH safety margin.

$\text{Margin} = NPSHA - NPSHR = 20.75\text{ m} - 2.20\text{ m} = 18.55\text{ m}$ $\frac{NPSHA}{NPSHR} = \frac{20.75}{2.20} = 9.43$

Because the margin far exceeds the API 610 minimum of $1.0\text{ m}$ ($3.3\text{ ft}$) and the $NPSHA/NPSHR$ ratio exceeds $1.2$, the pump is fully protected against cavitation.

Step 3: Calculate Total Dynamic Head (TDH). Applying the MEB between liquid surfaces $1$ and $2$:

$TDH = \frac{P_{2,abs} - P_{1,abs}}{\rho g} + (z_2 - z_1) + h_{f,suct} + h_{f,disch}$

Evaluate pressure head difference:

$\Delta P_{vessels} = P_{2,abs} - P_{1,abs} = 320.0\text{ kPa} - 180.0\text{ kPa} = 140.0\text{ kPa} = 140,000\text{ Pa}$ $\text{Pressure Head} = \frac{140,000\text{ Pa}}{655\text{ kg/m}^3 \times 9.807\text{ m/s}^2} = 21.795\text{ m}$

Evaluate elevation and friction:

• Static elevation difference: $z_2 - z_1 = 24.0\text{ m} - 3.50\text{ m} = 20.50\text{ m}$
• Total friction loss: $h_{f,total} = 0.65\text{ m} + 4.80\text{ m} = 5.45\text{ m}$

Sum all components:

$TDH = 21.795\text{ m} + 20.50\text{ m} + 5.45\text{ m} = 47.745\text{ m} \approx 47.75\text{ m}$

Step 4: Calculate pump power requirements. Hydraulic power delivered to the liquid ($WHP$):

$WHP = \rho g Q (TDH) = 655\text{ kg/m}^3 \times 9.807\text{ m/s}^2 \times 0.0125\text{ m}^3/\text{s} \times 47.745\text{ m}$ $WHP = 6,423.6 \times 0.0125 \times 47.745 = 3,833.7\text{ W} = 3.834\text{ kW}$

Brake Horsepower ($BHP$) required at pump shaft:

$BHP = \frac{WHP}{\eta_p} = \frac{3.834\text{ kW}}{0.68} = 5.638\text{ kW}$

Convert to mechanical horsepower:

$BHP = 5.638\text{ kW} \times \left(\frac{1\text{ hp}}{0.7457\text{ kW}}\right) = 7.56\text{ hp}$

For industrial reliability, select a standard $10\text{ hp}$ ($7.5\text{ kW}$) motor to provide adequate margin against end-of-curve motor overload.

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9. Common PE Exam Traps in Pumps, Compressors, and NPSH

1. Using Gauge Pressure in NPSHA: The single most catastrophic error on the exam. Atmospheric pressure must be included in absolute pressure terms. Using gauge pressure causes calculated NPSHA to be negative or off by $10.33\text{ m of water}$ ($33.9\text{ ft}$). Always convert to absolute pressure before calculating NPSH.
2. Overlooking the Saturated Fluid Simplification: For boiling liquids in reboilers, flash drums, and deaerators, $P_{source,abs} = P_{vp}$. Never attempt to look up atmospheric pressure for closed saturation vessels; the pressure terms cancel identically, leaving $NPSHA = z_s - h_f$.
3. Mixing Up Affinity Law Exponents: Remember that head scales with speed squared ($N^2$) and power scales with speed cubed ($N^3$). Applying linear scaling to head or power is a fatal mistake.
4. Forgetting Specific Gravity in Pump Power: Centrifugal pump head ($TDH$, in feet or meters) is independent of fluid density. However, Brake Horsepower is directly proportional to density ($SG$). Pumping brine ($SG = 1.2$) requires $20%$ more motor power than pumping pure water ($SG = 1.0$) at the identical head and flow!