3.1 Fluid Statics, Pressure Measurement, and Manometry

3.1 Fluid Statics, Pressure Measurement, and Manometry

Fluid statics is the branch of chemical engineering fluid mechanics that examines fluids at rest, where shear stresses are identically zero ($\tau = 0$) and the stress state is purely isotropic. In stationary fluids, pressure acts perpendicularly on any boundary surface, regardless of surface orientation. Mastering fluid statics is essential for the NCEES PE Chemical exam, where questions routinely test hydrostatic tank loadings, level measurement transmitter calibrations, purge gas blanket pressures, and multi-fluid manometer calculations.

---

1. Fundamental Hydrostatic Equations

Consider a differential fluid element of dimensions $dx$, $dy$, and $dz$ with density $\rho$ subjected to the local gravitational field $g$. Setting the vertical forces into equilibrium yields the basic equation of fluid statics:

$$\frac{dP}{dz} = -\rho g = -\gamma$$

where:

• $P$ is local static pressure ($\text{Pa}$ or $\text{lb}_f/\text{ft}^2$),
• $z$ is the vertical coordinate oriented positively upwards ($\text{m}$ or $\text{ft}$),
• $\rho$ is fluid mass density ($\text{kg/m}^3$ or $\text{lb}_m/\text{ft}^3$),
• $g$ is gravitational acceleration ($9.80665\text{ m/s}^2$ or $32.174\text{ ft/s}^2$),
• $\gamma = \rho g$ is specific weight ($\text{N/m}^3$ or $\text{lb}_f/\text{ft}^3$ in standard gravity).

For an incompressible fluid (where density $\rho$ is constant throughout the depth $h = z_1 - z_2$):

$$P_2 - P_1 = \rho g (z_1 - z_2) = \rho g h = \gamma h$$

In US Customary units, remember to apply Newton's second law proportionality constant $g_c = 32.174\text{ lb}_m\cdot\text{ft}/(\text{lb}_f\cdot\text{s}^2)$:

$$P_2 - P_1 = \frac{\rho g h}{g_c}$$

If $g = g_c$ (standard terrestrial gravity), a fluid with density $\rho\text{ [lb}_m/\text{ft}^3\text{]}$ exerts a pressure gradient of $\rho / 144\text{ psi per vertical foot}$ of depth.

---

2. Pressure Datums and Terminology

Pressure measurement requires an explicit reference datum. Misinterpreting gauge versus absolute pressure is among the most frequent error modes on the PE exam.

Critical Unit Conversions at 4 °C / 60 °F

• $1\text{ atm} = 101.325\text{ kPa} = 1.01325\text{ bar} = 14.696\text{ psia} = 760\text{ mmHg} = 29.92\text{ inHg} = 33.91\text{ ft H}_2\text{O} = 407.0\text{ inH}_2\text{O}$
• $1\text{ psi} = 6.89476\text{ kPa} = 2.307\text{ ft H}_2\text{O} = 27.68\text{ inH}_2\text{O} = 51.715\text{ mmHg}$
• $1\text{ bar} = 100\text{ kPa} = 14.5038\text{ psi} = 0.98692\text{ atm}$

---

3. Multi-Fluid Manometers and Differential Pressure Balances

Manometers quantify pressure differences by balancing the weight of one or more fluid columns. The governing operational rule is Pascal's principle combined with hydrostatic continuity: points at the same elevation in a continuous, homogeneous, stationary fluid possess identical pressures.

The Systematic "Manometer Walk" Algorithm

To write the governing hydrostatic equation for any complex, multi-leg manometer without algebraic error, follow this four-step procedure:

1. Select a starting boundary: Write down the known pressure at one terminus (e.g., $P_A$).
2. Descend fluid columns: As you move downward through a fluid leg of vertical height $h_i$ with density $\rho_i$, pressure increases. Add $+\rho_i g h_i$.
3. Ascend fluid columns: As you move upward through a fluid leg of vertical height $h_j$ with density $\rho_j$, pressure decreases. Subtract $-\rho_j g h_j$.
4. Equate to destination boundary: Upon reaching the terminal tap or open interface, set the entire summation equal to the destination pressure (e.g., $= P_B$ or $= P_{atm}$).

The Differential U-Tube Manometer with Process Leads

In chemical process piping, a differential U-tube manometer measures pressure drop $\Delta P = P_1 - P_2$ across an orifice plate, venturi, or packed bed. The process fluid (density $\rho_f$) fills the instrument tap leads above the manometer sealing fluid (density $\rho_m$, where $\rho_m > \rho_f$):

$$P_1 + \rho_f g z_1 - \rho_f g (z_1 - z_2) - \rho_m g \Delta h + \rho_f g (z_1 - z_2 - \Delta h) = P_2$$

Simplifying directly yields the standard differential manometry equation:

$$P_1 - P_2 = (\rho_m - \rho_f) g \Delta h$$

---

4. Inclined Manometers and Sensitivity Magnification

When measuring minute pressure differences—such as draft pressure inside a furnace firebox (typically $-0.1\text{ to }-0.5\text{ inH}_2\text{O}$) or differential pressure across a HEPA filter—a standard vertical manometer produces meniscus displacements too tiny to read accurately. An inclined manometer magnifies this reading by canting one leg at an angle $\theta$ relative to the horizontal.

The vertical hydrostatic elevation difference $h$ is related to the linear scale displacement $L$ along the tube bore by:

$$h = L \sin\theta$$

Substituting this into the hydrostatic balance gives:

$$\Delta P = \rho_m g L \sin\theta$$

The scale magnification factor is defined as:

$$\text{Magnification Factor} = \frac{L}{h} = \frac{1}{\sin\theta}$$

For an incline of $\theta = 5.74^\circ$, $\sin\theta = 0.100$, yielding a $10\times$ magnification: a mere $0.10\text{ in}$ vertical water head produces a full $1.0\text{ in}$ readable travel along the inclined scale.

---

5. Hydrostatic Forces on Submerged Surfaces & Displacer Level Transmitters

Forces on Submerged Planar Surfaces

The total resultant hydrostatic force $F_R$ acting on a planar submerged surface of area $A$ immersed in a liquid of density $\rho$ is governed by the pressure at its area centroid ($h_c$):

$$F_R = P_c A = (\rho g h_c) A$$

The resultant force acts not through the centroid, but through the center of pressure ($y_{cp}$), which always lies deeper than the centroid due to the trapezoidal pressure distribution:

$$y_{cp} = y_c + \frac{I_{xx,c}}{y_c A}$$

where $I_{xx,c}$ is the area moment of inertia about the horizontal centroidal axis, and $y_c$ is the inclined distance from the liquid surface to the centroid.

Buoyancy and Displacer Level Measurement

Archimedes' principle states that any body completely or partially submerged in a fluid experiences an upward buoyant force $F_B$ equal to the weight of the displaced fluid:

$$F_B = \rho_{fluid} g V_{displaced}$$

In chemical reactors and distillation accumulators, liquid level is commonly measured using a torque tube displacer. A solid cylindrical rod of cross-sectional area $A_d$ and mass $m_d$ hangs inside a stilling well. As liquid level rises to submerge height $h_d$ of the displacer, the net apparent weight $W_{app}$ registered by the torque spring decreases linearly:

$$W_{app} = m_d g - F_B = m_d g - \rho_{liq} g A_d h_d$$

For dual-liquid interfaces (e.g., water-hydrocarbon phase separation in an overhead decanter), two liquids of densities $\rho_1$ and $\rho_2$ submerge fractions $h_1$ and $h_2$ of the displacer:

$$F_B = g A_d [\rho_1 h_1 + \rho_2 h_2]$$

---

6. Industrial Pressure Measurement Technologies

---

7. Step-by-Step Worked Numerical Example

Problem Statement

A differential mercury manometer ($\rho_m = 13,550\text{ kg/m}^3$) is connected across an orifice meter installed in a vertical pipe carrying liquid kerosene ($SG = 0.820$, $\rho_k = 820\text{ kg/m}^3$) flowing downwards. The pressure taps $A$ (upstream) and $B$ (throat) are separated by a vertical distance of $z_A - z_B = 0.450\text{ m}$. The instrument impulse leads are completely flooded with kerosene. The mercury manometer indicates a deflection of $\Delta h = 260\text{ mm} = 0.260\text{ m}$, with the lower mercury meniscus residing on the side connected to tap $A$.

Calculate the differential static pressure $\Delta P = P_A - P_B$ in units of kilopascals ($\text{kPa}$) and pounds per square inch ($\text{psi}$). ($g = 9.807\text{ m/s}^2$).

              Tap A (z = +0.450 m)
                |
                | [Kerosene-filled lead]
                |
              Tap B (z = 0.000 m)
                |
       =========+=========
       |                 |
       |                 | (z2 = elevation of higher Hg meniscus)
       |                 v
       |               [Hg]  <-- meniscus 2 (connected to B)
       |                 |
       v                 |    delta_h = 0.260 m
     [Hg]  <-------------+   <-- meniscus 1 (connected to A)
       |_________________|
          Mercury Basin

Solution

Step 1: Formulate the continuous hydrostatic balance. Start at Tap $A$ at elevation $z_A$ and write the manometer walk downwards to the lower mercury interface (Meniscus 1, at arbitrary elevation $z_1 = 0$):

$$P_{\text{meniscus 1}} = P_A + \rho_k g (z_A - z_1)$$

Cross through the continuous mercury leg to Meniscus 2 at elevation $z_2 = z_1 + \Delta h = 0.260\text{ m}$. Since we are moving upward through mercury, subtract the mercury column:

$$P_{\text{meniscus 2}} = P_{\text{meniscus 1}} - \rho_m g \Delta h = P_A + \rho_k g (z_A - z_1) - \rho_m g \Delta h$$

Now ascend from Meniscus 2 through the kerosene-filled lead to Tap $B$ at elevation $z_B = 0.450\text{ m}$. Subtract the kerosene column between $z_2$ and $z_B$:

$$P_B = P_{\text{meniscus 2}} - \rho_k g (z_B - z_2)$$

Step 2: Consolidate and isolate $P_A - P_B$. Substitute $P_{\text{meniscus 2}}$ into the expression for $P_B$:

$$P_B = P_A + \rho_k g (z_A - z_1) - \rho_m g \Delta h - \rho_k g (z_B - (z_1 + \Delta h))$$

Expand the terms:

$$P_B = P_A + \rho_k g z_A - \rho_k g z_1 - \rho_m g \Delta h - \rho_k g z_B + \rho_k g z_1 + \rho_k g \Delta h$$

Notice that the reference datum $\rho_k g z_1$ cancels completely:

$$P_B = P_A + \rho_k g (z_A - z_B) - (\rho_m - \rho_k) g \Delta h$$

Rearranging for the pressure difference $P_A - P_B$:

$$P_A - P_B = (\rho_m - \rho_k) g \Delta h - \rho_k g (z_A - z_B)$$

Step 3: Insert numerical parameters.

• Density difference: $(\rho_m - \rho_k) = 13,550 - 820 = 12,730\text{ kg/m}^3$
• Mercury contribution: $12,730\text{ kg/m}^3 \times 9.807\text{ m/s}^2 \times 0.260\text{ m} = 32,459.7\text{ Pa} = 32.46\text{ kPa}$
• Elevation head of kerosene: $\rho_k g (z_A - z_B) = 820\text{ kg/m}^3 \times 9.807\text{ m/s}^2 \times 0.450\text{ m} = 3,618.8\text{ Pa} = 3.62\text{ kPa}$

$$P_A - P_B = 32,459.7\text{ Pa} - 3,618.8\text{ Pa} = 28,840.9\text{ Pa} = 28.84\text{ kPa}$$

Convert to psi:

$$P_A - P_B = 28,840.9\text{ Pa} \times \left(\frac{1\text{ psi}}{6,894.76\text{ Pa}}\right) = 4.183\text{ psi}$$

---

8. Common PE Exam Traps in Fluid Statics

1. Neglecting Seal-Leg Fluid Density: Assuming the fluid above the manometer liquid is air when it is process liquid. Always write $(\rho_m - \rho_f) g \Delta h$, not simply $\rho_m g \Delta h$.
2. Sign Errors Across Vertical Pipe Runs: When taps are at differing elevations, failing to add or subtract the fluid column $\rho_f g \Delta z$ between the taps based on the directional orientation of the pipe leads.
3. Using Specific Gravity without Density Units: Multiplying directly by $SG$ without multiplying by the base density of water ($\rho_{w} = 1,000\text{ kg/m}^3$ or $62.4\text{ lb}_m/\text{ft}^3$). Remember: $SG$ is dimensionless.
4. Confusing Head with Pressure: Reporting head $h$ in feet or meters when the problem asks for pressure in psi or kPa. Remember: $\Delta P = \rho g h$, so $10\text{ ft of kerosene}$ ($SG = 0.82$) exerts only $3.55\text{ psi}$, whereas $10\text{ ft of water}$ exerts $4.33\text{ psi}$.