6.3 Flow Measurement and Fluid Machinery

Flow Measurement Devices: Orifice, Venturi, and Rotameters

Accurate flow rate measurement is essential for process control and mass balance verification. The three most common flow meters in chemical processes are orifice meters, Venturi meters, and rotameters.

Orifice and Venturi Meters: Both are obstruction-type meters that create a local constriction in the flow channel. According to the Bernoulli and continuity equations, restricting the flow increases fluid velocity, which results in a localized pressure drop. The volumetric flow rate ($q$) is calculated using the standard flow meter formula:

$q = C_d A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho (1 - \beta^4)}}$

where:

• $C_d$ is the dimensionless discharge coefficient.
• $A_2$ is the cross-sectional area of the throat or orifice opening.
• $P_1 - P_2$ is the differential pressure measured across the meter.
• $\rho$ is the fluid density.
• $\beta = D_2 / D_1$ is the ratio of throat/orifice diameter to the pipe diameter. The term $\frac{1}{\sqrt{1 - \beta^4}}$ is known as the velocity of approach factor.

The primary difference between the two meters lies in their geometry, which dictates their performance:

1. Orifice Meter: Consists of a thin plate with a sharp-edged hole. It is inexpensive, simple to install, and occupies little space, but it creates high unrecoverable pressure loss due to intense eddy formation downstream. Its discharge coefficient $C_d$ typically ranges from $0.60$ to $0.62$ because the fluid jet contracts to a minimum area (the vena contracta) that is smaller than the physical orifice opening.
2. Venturi Meter: Consists of a converging section, a throat, and a diverging recovery cone. It is streamlined to minimize flow separation, resulting in excellent pressure recovery (low unrecoverable loss). However, it is expensive, bulky, and difficult to manufacture. Its discharge coefficient $C_d$ is much higher, typically between $0.97$ and $0.99$.

Rotameter: A rotameter is a variable-area meter. It consists of a vertical, tapered glass tube containing a float. As fluid flows upward, the float rises until the upward drag force and buoyancy force equal the downward gravitational force:

$F_{\text{drag}} + F_{\text{buoyancy}} = F_{\text{gravity}}$

The equilibrium height of the float is directly proportional to the volumetric flow rate, read from a calibrated scale on the tube.

Pump Performance, Operating Point, and Affinity Laws

Pumps transfer mechanical energy to liquids to increase pressure or elevation. Centrifugal pumps are the workhorses of chemical plants.

The Operating Point: A centrifugal pump operates at the intersection of the pump performance curve (head vs. flow rate, provided by the manufacturer) and the system curve (required head vs. flow rate, determined by elevation change and pipe friction):

$h_{\text{pump}}(Q) = h_{\text{system}}(Q) = \Delta z + \frac{\Delta P}{\rho g} + \left( \sum f_D \frac{L}{D_H} + \sum K_L \right) \frac{Q^2}{2g A_c^2}$

Changing a valve setting alters the system curve, moving the operating point.

Pump Affinity Laws: For geometrically similar (homologous) pumps, performance changes with shaft speed ($N$, rpm) and impeller diameter ($D$) according to the affinity laws. When impeller diameter is held constant ($D_1 = D_2$):

1. Flow Rate: $Q_2 / Q_1 = N_2 / N_1$
2. Head: $H_2 / H_1 = (N_2 / N_1)^2$
3. Power: $P_2 / P_1 = (N_2 / N_1)^3$

Cavitation and Net Positive Suction Head (NPSH)

Cavitation is a destructive phenomenon that occurs when the local static pressure inside a pump drops below the liquid's vapor pressure ($P_{vp}$). This causes the liquid to flash into vapor bubbles. As these bubbles move into higher-pressure regions near the impeller, they collapse violently, generating high-velocity micro-jets and shockwaves that pit metal surfaces, cause severe vibration, and reduce pump efficiency.

To prevent cavitation, the Net Positive Suction Head Available ($NPSH_A$) at the pump suction nozzle must exceed the Net Positive Suction Head Required ($NPSH_R$) specified by the pump manufacturer:

$NPSH_A > NPSH_R$

$NPSH_A$ represents the total head at the pump suction nozzle above the vapor pressure head:

$NPSH_A = \frac{P_s}{\rho g} + \frac{v_s^2}{2g} - \frac{P_{vp}}{\rho g}$

In terms of a source reservoir open to the atmosphere:

$NPSH_A = H_{\text{atm}} \pm z_s - h_{f,\text{suction}} - H_{\text{vp}}$

where $H_{\text{atm}} = P_{\text{atm}} / \rho g$, $z_s$ is the elevation of the suction liquid level relative to the pump inlet (positive if liquid level is above the pump, negative if below), $h_{f,\text{suction}}$ is the frictional head loss in the suction piping, and $H_{\text{vp}} = P_{vp} / \rho g$.

Compressors, Turbines, and Vacuum Systems

Compressors and Turbines: Compressors add energy to gases (increasing pressure), while turbines extract energy from expanding fluids to perform shaft work. For gas compressors and turbines, thermodynamic paths (isentropic vs. polytropic) are analyzed. The isentropic efficiency ($\eta$) relates actual work to ideal work:

$\eta_{\text{compressor}} = \frac{w_{\text{isentropic}}}{w_{\text{actual}}} \quad \text{and} \quad \eta_{\text{turbine}} = \frac{w_{\text{actual}}}{w_{\text{isentropic}}}$

Vacuum Systems: Steam-jet ejectors and liquid-ring vacuum pumps are used to maintain sub-atmospheric pressures in process vessels (e.g., vacuum distillation columns). Ejectors utilize high-velocity steam jets through a Venturi nozzle to entrain process gases, converting kinetic energy into pressure rise.

Worked Example 1: Venturi Meter Flow Rate

Problem: A Venturi meter with a throat diameter of $5.0\text{ cm}$ is installed in a $10.0\text{ cm}$ diameter pipe carrying water ($\rho = 1000\text{ kg/m}^3$). A differential pressure transducer reads $40\text{ kPa}$ across the meter. Assuming a discharge coefficient $C_d = 0.98$, calculate the volumetric flow rate.

Solution:

Calculate the diameter ratio ($\beta$) and throat area ($A_2$):

$\beta = \frac{D_2}{D_1} = \frac{5.0\text{ cm}}{10.0\text{ cm}} = 0.50$

$A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.05\text{ m})^2 \approx 1.963 \times 10^{-3}\text{ m}^2$

Substitute the values into the flow meter formula:

$q = C_d A_2 \sqrt{\frac{2 \Delta P}{\rho (1 - \beta^4)}} = 0.98 \cdot (1.963 \times 10^{-3}) \cdot \sqrt{\frac{2 \cdot 40,000\text{ Pa}}{1000\text{ kg/m}^3 \cdot (1 - 0.5^4)}}$

$q = 1.924 \times 10^{-3} \cdot \sqrt{\frac{80,000}{1000 \cdot (1 - 0.0625)}} = 1.924 \times 10^{-3} \cdot \sqrt{\frac{80,000}{937.5}}$

$q = 1.924 \times 10^{-3} \cdot \sqrt{85.33} \approx 1.924 \times 10^{-3} \cdot 9.238 \approx 0.0178\text{ m}^3\text{/s} = 17.8\text{ L/s}$

Worked Example 2: NPSHA and Cavitation Verification

Problem: A centrifugal pump draws water at $50^\circ\text{C}$ ($\rho = 988\text{ kg/m}^3$, vapor pressure $P_{vp} = 12.35\text{ kPa}$) from an open tank. The water level in the tank is $3.0 ext{ m}$ below the centerline of the pump inlet. The suction line has a total frictional head loss of $1.2\text{ m}$. If the atmospheric pressure is $101.3\text{ kPa}$, calculate $NPSH_A$. If the pump requires $NPSH_R = 4.5\text{ m}$, will the pump cavitate?

Solution:

Convert pressures to head:

$H_{\text{atm}} = \frac{P_{\text{atm}}}{\rho g} = \frac{101,300\text{ Pa}}{988\text{ kg/m}^3 \cdot 9.807\text{ m/s}^2} = \frac{101,300}{9689.3} \approx 10.45\text{ m}$

$H_{\text{vp}} = \frac{P_{\text{vp}}}{\rho g} = \frac{12,350\text{ Pa}}{988\text{ kg/m}^3 \cdot 9.807\text{ m/s}^2} = \frac{12,350}{9689.3} \approx 1.27\text{ m}$

Since the water level is below the pump, $z_s = -3.0\text{ m}$ (suction lift).

Calculate $NPSH_A$:

$NPSH_A = H_{\text{atm}} + z_s - h_{f,\text{suction}} - H_{\text{vp}} = 10.45\text{ m} - 3.0\text{ m} - 1.2\text{ m} - 1.27\text{ m} = 4.98\text{ m}$

Compare $NPSH_A$ with $NPSH_R$:

$NPSH_A = 4.98\text{ m} > NPSH_R = 4.5\text{ m}$

The available head exceeds the required head, so the pump will not cavitate.