Steam Pressure Drop Calculator

Overview

Engineering reference data for Steam Pressure Drop Calculator in fluid mechanics.

Key Formulas

Reynolds Number

$Re = \frac{\rho v D}{\mu}$

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

$A_1 v_1 = A_2 v_2$

Conservation of mass for incompressible flow.

Darcy-Weisbach

$\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}$

Pressure drop due to friction in a pipe.

Variables

Symbol	Description	Unit
$Re$	Reynolds number	—
$\rho$	Fluid density	kg/m³
$v$	Flow velocity	m/s
$D$	Characteristic dimension	m
$\mu$	Dynamic viscosity	Pa·s
$P$	Pressure	Pa
$f$	Darcy friction factor	—

Pressure Drop Formulas in LaTeX

The Unwin formulas for calculating pressure drop in steam pipes can be expressed in LaTeX notation for clarity and precision.

Metric Units:

$$dp = \frac{0.6753 \times 10^6 \cdot q^2 \cdot l \cdot \left(1 + \frac{91.4}{d}\right)}{\rho \cdot d^5}$$

where:

• $dp$ is the pressure drop in Pascals (Pa),
• $q$ is the steam flow rate in kilograms per hour (kg/h),
• $l$ is the pipe length in meters (m),
• $d$ is the pipe inside diameter in millimeters (mm),
• $\rho$ is the steam density in kilograms per cubic meter (kg/m³).

Imperial Units:

$$dp = \frac{0.0001306 \cdot q^2 \cdot l \cdot \left(1 + \frac{3.6}{d}\right)}{3600 \cdot \rho \cdot d^5}$$

where:

• $dp$ is the pressure drop in pounds per square inch (psi),
• $q$ is the steam flow rate in pounds per hour (lb/hr),
• $l$ is the pipe length in feet (ft),
• $d$ is the pipe inside diameter in inches (in),
• $\rho$ is the steam density in pounds per cubic foot (lb/ft³).

Example Calculation (Metric Units)

For a steam flow of 4000 kg/h at 10 bar (1000 kPa) with density $\rho = 5.15$ kg/m³, through a 100 m long pipe with inside diameter 102 mm:

$$dp = \frac{0.6753 \times 10^6 \cdot (4000)^2 \cdot 100 \cdot \left(1 + \frac{91.4}{102}\right)}{5.15 \cdot (102)^5} = 36030 \text{ Pa} = 36 \text{ kPa}$$

Conversion Factors

• $1 \text{ bar} = 100 \text{ kPa} = 10,197 \text{ kp/m}^2 = 10.20 \text{ m H}_2\text{O} = 0.9869 \text{ atm} = 14.50 \text{ psi}$
• $1 \text{ kg/h} = 2.778 \times 10^{-4} \text{ kg/s} = 3.67 \times 10^{-2} \text{ lb/min}$

Notes on Formula Accuracy and Usage

• Use values for specific volume corresponding to the average pressure if the pressure drop exceeds 10–15% of the initial absolute pressure.
• Note that for elevated velocities, Unwin's formula is known to give pressure drops higher than the actual values. Consider alternative methods or corrections for high-velocity scenarios.

References

• Original Source