Unit Conversions & Physical Process Principles

Pressure Units and Pressure Scale Conversions

Pressure ($P$) is defined as force ($F$) exerted per unit area ($A$):

$$P = \frac{F}{A}$$

Technicians must routinely convert between imperial, metric, and liquid column units.

Pressure Unit Equivalencies

Standard atmospheric pressure at sea level serves as the primary baseline for unit conversion:

$$1.000\text{ atm} = 14.696\text{ psi} = 101.325\text{ kPa} = 1.01325\text{ bar} = 760.0\text{ mmHg (Torr)} = 29.92\text{ inHg} = 407.2\text{ inH}_2\text{O}$$

Gauge, Absolute, Vacuum, and Differential Pressure

  Absolute Scale (psia)            Gauge Scale (psig)          Vacuum Scale (inHg Vac)
  ====================            ===================          =======================
  P_abs = 29.7 psia ------------> P_gauge = 15.0 psig

  P_atm = 14.7 psia ------------> P_gauge = 0.0 psig ---------> 0.0 inHg Vac

  P_abs = 9.7 psia -------------> P_gauge = -5.0 psig --------> 10.18 inHg Vac

  P_abs = 0.0 psia (Absolute Zero)                               29.92 inHg Vac (Full Vac)

1. Absolute Pressure ($P_{\text{abs}}$ / psia): Measured relative to a perfect vacuum ($0.0\text{ psia}$). Absolute pressure is nonnegative, with zero representing a perfect vacuum.
2. Gauge Pressure ($P_{\text{gauge}}$ / psig): Measured relative to local ambient atmospheric pressure ($P_{\text{atm}}$). Standard sea level $P_{\text{atm}} = 14.7\text{ psi}$. $$P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}$$ $$P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}}$$
3. Vacuum Pressure ($P_{\text{vac}}$): Pressure below atmospheric pressure, often measured in inches of mercury vacuum (inHg Vac) or mmHg. $$P_{\text{vac\_inHg}} = (P_{\text{atm\_psia}} - P_{\text{abs\_psia}}) \times 2.0360\text{ inHg/psi}$$
4. Differential Pressure ($\Delta P$ / psid / $\text{inH}_2\text{O}$): The algebraic difference between two pressure points ($P_{\text{High}} - P_{\text{Low}}$).

Worked Example: Pressure Scale Conversion

Scenario: A vacuum distillation column operates at an absolute pressure of $3.20\text{ psia}$. Calculate the equivalent pressure in gauge pressure ($\text{psig}$) and inches of mercury vacuum ($\text{inHg Vac}$) assuming standard $P_{\text{atm}} = 14.70\text{ psia}$.

$$\text{Gauge Pressure: } P_{\text{gauge}} = 3.20\text{ psia} - 14.70\text{ psia} = -11.50\text{ psig}$$ $$\text{Vacuum Pressure: } P_{\text{vac}} = (14.70\text{ psia} - 3.20\text{ psia}) \times 2.0360\text{ inHg/psi} = 11.50 \times 2.0360 = 23.414\text{ inHg Vac}$$

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Temperature Scales and Exact Conversion Formulas

Process measurement utilizes four temperature scales: two relative scales (Fahrenheit and Celsius) and two absolute scales (Rankine and Kelvin).

+-----------------------------------------------------------------------------------------+
|                         TEMPERATURE SCALE RELATIONSHIPS                                 |
|                                                                                         |
| Water Boiling Point   212 °F --------- 100 °C --------- 373.15 K --------- 671.67 °R    |
| Water Freezing Point   32 °F ---------   0 °C --------- 273.15 K --------- 491.67 °R    |
| Absolute Zero       -459.67 °F ------- -273.15 °C -------   0 K -----------   0 °R      |
+-----------------------------------------------------------------------------------------+

Exact Conversion Formulas

$$\text{Celsius to Fahrenheit: } T_{^\circ\text{F}} = (T_{^\circ\text{C}} \times 1.8) + 32$$ $$\text{Fahrenheit to Celsius: } T_{^\circ\text{C}} = \frac{T_{^\circ\text{F}} - 32}{1.8}$$ $$\text{Celsius to Kelvin (Absolute): } T_{\text{K}} = T_{^\circ\text{C}} + 273.15$$ $$\text{Fahrenheit to Rankine (Absolute): } T_{^\circ\text{R}} = T_{^\circ\text{F}} + 459.67$$ $$\text{Kelvin to Rankine: } T_{^\circ\text{R}} = T_{\text{K}} \times 1.8$$

Worked Example: Temperature Conversion

Scenario: A heat exchanger outlet temperature transmitter reads $356.0\text{ }^\circ\text{F}$. Convert this reading to Celsius ($^\circ\text{C}$) and Kelvin ($\text{K}$).

$$T_{^\circ\text{C}} = \frac{356.0 - 32}{1.8} = \frac{324.0}{1.8} = 180.0\text{ }^\circ\text{C}$$ $$T_{\text{K}} = 180.0 + 273.15 = 453.15\text{ K}$$

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Volumetric Flow, Mass Flow, and Standard Volumetric Flow

Flow rate measurement is categorized into three fundamental physical types:

1. Volumetric Flow ($Q_v$)

Measures physical fluid volume passing a point per unit time (e.g., Gallons Per Minute [GPM], Actual Cubic Feet Per Minute [ACFM], Cubic Meters Per Hour [$\text{m}^3/\text{hr}$]).

• Limitation: Gas and liquid volume expands or contracts with changing temperature and pressure, making raw volumetric flow inaccurate for mass balance custody transfer.

2. Mass Flow ($Q_m$)

Measures actual mass of fluid passing a point per unit time (e.g., Pounds Per Hour [lb/hr], Kilograms Per Second [kg/s]).

$$Q_m = \rho \times Q_v$$

Where $\rho$ is fluid density.

• Advantage: Mass flow expresses quantity directly as mass per time and does not require choosing standard volumetric base conditions; the actual process and meter performance can still depend on temperature, pressure, composition, and phase.

3. Standard Volumetric Flow ($Q_s$ - SCFM vs. ACFM)

Gas flow is frequently expressed in Standard Cubic Feet Per Minute (SCFM). A standard-volume result is meaningful only when its reference pressure and temperature are stated. The worked examples here use $T_{\text{std}} = 60.0\text{ }^\circ\text{F} = 519.67\text{ }^\circ\text{R}$ and $P_{\text{std}} = 14.696\text{ psia}$; contracts, industries, and meter configurations may use different base conditions.

For the simplified case in which ideal-gas behavior is assumed and compressibility effects cancel, the Ideal Gas Law ($P V = n R T$) converts Actual Cubic Feet Per Minute (ACFM) measured at actual process pressure ($P_{\text{act}}$) and actual process temperature ($T_{\text{act}}$) is converted to SCFM:

$$\text{SCFM} = \text{ACFM} \times \left( \frac{P_{\text{actual\_psia}}}{P_{\text{std\_psia}}} \right) \times \left( \frac{T_{\text{std\_Rankine}}}{T_{\text{actual\_Rankine}}} \right)$$

CRITICAL EXAM RULE: Temperatures MUST be converted to Absolute Rankine ($^\circ\text{R} = ^\circ\text{F} + 459.67$) and pressures MUST be converted to Absolute Pressure ($\text{psia} = \text{psig} + 14.7$).

Worked Example: ACFM to SCFM Conversion

Scenario: An air compressor delivers $400.0\text{ ACFM}$ through a discharge line operating at $100.0\text{ psig}$ pressure and $140.0\text{ }^\circ\text{F}$ temperature. Calculate the mass-equivalent flow rate in $\text{SCFM}$.

Step 1: Convert pressure to Absolute Pressure (psia)

$$P_{\text{actual}} = 100.0\text{ psig} + 14.7 = 114.7\text{ psia}$$ $$P_{\text{standard}} = 14.7\text{ psia}$$

Step 2: Convert temperatures to Absolute Temperature (Rankine)

$$T_{\text{actual}} = 140.0\text{ }^\circ\text{F} + 459.67 = 599.67\text{ }^\circ\text{R}$$ $$T_{\text{standard}} = 60.0\text{ }^\circ\text{F} + 459.67 = 519.67\text{ }^\circ\text{R}$$

Step 3: Calculate SCFM

$$\text{SCFM} = 400.0 \times \left( \frac{114.7}{14.7} \right) \times \left( \frac{519.67}{599.67} \right)$$ $$\text{SCFM} = 400.0 \times (7.8027) \times (0.86659) = 2{,}704.7\text{ SCFM}$$

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Hydrostatic Head Pressure and Liquid Level Calculations

A column of liquid exerts hydrostatic pressure at its base due to gravity:

$$P = h \times \rho \times g$$

In industrial engineering units, hydrostatic pressure is governed by Specific Gravity (SG), the dimensionless ratio of fluid density to a stated reference-fluid density at stated conditions. The following water-column examples use the stated SG value directly.

$$P\text{ (inH}_2\text{O)} = h\text{ (inches)} \times \text{SG}$$ $$P\text{ (psi)} = h\text{ (inches)} \times \text{SG} \times 0.036127\text{ psi/inH}_2\text{O}$$ $$P\text{ (psi)} = \frac{h\text{ (inches)} \times \text{SG}}{27.680\text{ inH}_2\text{O/psi}}$$

Worked Example: Tank Level Hydrostatic Pressure

Scenario: An open storage tank contains sulfuric acid (Specific Gravity $\text{SG} = 1.84$) filled to a liquid height of $150.0\text{ inches}$. Calculate the hydrostatic pressure at the bottom tap in $\text{inH}_2\text{O}$ and $\text{psig}$.

$$P\text{ (inH}_2\text{O)} = 150.0\text{ in} \times 1.84 = 276.0\text{ inH}_2\text{O}$$ $$P\text{ (psig)} = \frac{276.0\text{ inH}_2\text{O}}{27.680\text{ inH}_2\text{O/psi}} = 9.971\text{ psig}$$

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Differential Pressure Level Connections: Dry Leg vs. Wet Leg

When measuring liquid level in enclosed pressure vessels, static vapor space pressure must be canceled out using a Differential Pressure (DP) transmitter.

      DRY LEG INSTALLATION                         WET LEG INSTALLATION
   (Non-Condensable Vapor)                      (Condensable Vapor / Steam)

       +--------------+                            +--------------+
       | Vapor Space  | (Dry Leg)                  | Vapor Space  | (Condensate Leg)
       |              |----[LP]                    |              |====[LP]
       |              |      |                     |              |    |  (Pre-filled
       | Liquid (SG)  |      |                     | Liquid (SG)  |    |   SG_leg)
       |              |----[HP]                    |              |====[HP]
       +--------------+                            +--------------+
         DP Transmitter                              DP Transmitter
      P_HP = h*SG; P_LP = 0                       P_HP = h*SG; P_LP = H*SG_leg

Dry Leg Installations

Used when the gas or vapor above the liquid does not condense at ambient room temperature.

• High Pressure (HP) Port: Connected to bottom liquid tap.
• Low Pressure (LP) Port: Connected to top vapor space tap via a dry pipe.
• Pressure Equations:
  • At 0% Level ($h = 0$): $P_{\text{HP}} = P_{\text{vapor}}$, $P_{\text{LP}} = P_{\text{vapor}}$. $$\Delta P_{\text{LRV}} = P_{\text{HP}} - P_{\text{LP}} = 0.0\text{ inH}_2\text{O}$$
  • At 100% Level ($h = h_{\text{max}}$): $$\Delta P_{\text{URV}} = h_{\text{max}} \times \text{SG}$$
• Calibration Range: $0.0\text{ to } (h_{\text{max}} \times \text{SG})\text{ inH}_2\text{O}$.

Wet Leg Installations (Zero Elevation)

Used when vapor above the liquid condenses into liquid (e.g., steam boilers, distillation columns). The Low Pressure (LP) reference pipe is intentionally pre-filled with a reference liquid of known density ($\text{SG}_{\text{leg}}$) to maintain a constant, stable liquid column height ($H$).

The Zero Elevation Effect:

In the illustrated geometry, the full wet-leg head on the LP port exceeds the HP-side static head at empty, so the 0% differential is negative. Calculate the actual LRV from tap, transmitter, and reference-leg elevations rather than assuming every wet-leg installation has the same sign.

Calibration Equations for Wet Leg DP Transmitter:

$$\Delta P = P_{\text{HP}} - P_{\text{LP}}$$ $$\Delta P_{\text{LRV (0\% Level)}} = (0 \times \text{SG}) - (H \times \text{SG}_{\text{leg}}) = -(H \times \text{SG}_{\text{leg}})$$ $$\Delta P_{\text{URV (100\% Level)}} = (h_{\text{max}} \times \text{SG}) - (H \times \text{SG}_{\text{leg}})$$ $$\text{Span} = \Delta P_{\text{URV}} - \Delta P_{\text{LRV}} = h_{\text{max}} \times \text{SG}$$

Comprehensive Worked Example: Wet Leg Transmitter Calibration

Scenario: A closed boiler steam drum uses a wet leg DP level transmitter. The vertical distance between vessel taps is $H = 100.0\text{ inches}$. Maximum liquid measuring height is $h_{\text{max}} = 80.0\text{ inches}$. The boiler water specific gravity is $\text{SG} = 0.90$. The reference wet leg is filled with water/glycol condensate having specific gravity $\text{SG}_{\text{leg}} = 1.00$. The transmitter is mounted exactly level with the bottom tap.

  Top Tap (LP) =============================================+ (H = 100.0 in, Wet Leg SG_leg = 1.00)
  |                                                         |
  |--- 100% Level (h_max = 80.0 in, Process SG = 0.90)      |
  |                                                         |
  |                                                         |
  |--- 0% Level (h = 0.0 in)                                |
  Bottom Tap (HP) ==========================================+ (Transmitter mounted here)

Step 1: Calculate $\Delta P$ at 0% Level (LRV)

At $0\%$ level ($h = 0.0\text{ in}$):

• $P_{\text{HP}} = 0.0 \times 0.90 = 0.0\text{ inH}_2\text{O}$
• $P_{\text{LP}} = H \times \text{SG}_{\text{leg}} = 100.0\text{ in} \times 1.00 = 100.0\text{ inH}_2\text{O}$

$$\Delta P_{\text{LRV}} = P_{\text{HP}} - P_{\text{LP}} = 0.0 - 100.0 = -100.0\text{ inH}_2\text{O}$$

Step 2: Calculate $\Delta P$ at 100% Level (URV)

At $100\%$ level ($h = 80.0\text{ in}$):

• $P_{\text{HP}} = h_{\text{max}} \times \text{SG} = 80.0\text{ in} \times 0.90 = 72.0\text{ inH}_2\text{O}$
• $P_{\text{LP}} = 100.0\text{ inH}_2\text{O}$

$$\Delta P_{\text{URV}} = P_{\text{HP}} - P_{\text{LP}} = 72.0 - 100.0 = -28.0\text{ inH}_2\text{O}$$

Step 3: Calculate Calibrated Span

$$\text{Span} = \Delta P_{\text{URV}} - \Delta P_{\text{LRV}} = -28.0 - (-100.0) = +72.0\text{ inH}_2\text{O}$$

Summary Calibration Table:

• LRV (4.00 mA / 0% Level): $-100.0\text{ inH}_2\text{O}$
• URV (20.00 mA / 100% Level): $-28.0\text{ inH}_2\text{O}$
• Calibrated Span: $72.0\text{ inH}_2\text{O}$

Validation: Notice that Span equals $h_{\text{max}} \times \text{SG} = 80.0 \times 0.90 = 72.0\text{ inH}_2\text{O}$. The wet leg creates a constant $-100.0\text{ inH}_2\text{O}$ zero elevation offset, but does not alter the span.