Flow Rate Relation to Pressure Calculator

Flow Rate Formulas:

\[ V = \sqrt{\frac{2 \times (P_1 - P_2) \times 144}{\rho}} \] \[ Q = V \times A \times 448.83 \]

Inlet Pressure (P₁):

psi

Outlet Pressure (P₂):

psi

Fluid Density (ρ):

lb/ft³

Cross-sectional Area (A):

ft²

Velocity (V): ft/s

Flow Rate (Q): GPM

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1. What is Flow Rate Relation to Pressure?

Definition: This calculator determines the fluid velocity and volumetric flow rate based on pressure difference, fluid density, and pipe cross-sectional area.

Purpose: It helps engineers and technicians analyze fluid systems and design piping networks with proper flow characteristics.

2. How Does the Calculator Work?

The calculator uses these formulas:

\[ V = \sqrt{\frac{2 \times (P_1 - P_2) \times 144}{\rho}} \] \[ Q = V \times A \times 448.83 \]

Where:

\( V \) — Fluid velocity (ft/s)
\( P_1 \) — Inlet pressure (psi)
\( P_2 \) — Outlet pressure (psi)
\( \rho \) — Fluid density (lb/ft³)
\( Q \) — Volumetric flow rate (GPM)
\( A \) — Cross-sectional area (ft²)

Explanation: The first equation calculates velocity from pressure difference using Bernoulli's principle. The second converts velocity to flow rate.

3. Importance of Flow Rate Calculation

Details: Proper flow rate estimation ensures efficient system design, prevents pipe erosion, and maintains desired process conditions.

4. Using the Calculator

Tips: Enter pressure values (psi), fluid density (default 62.4 lb/ft³ for water), and cross-sectional area (ft²). All values must be positive.

5. Frequently Asked Questions (FAQ)

Q1: What's the 144 factor in the equation?
A: It converts psi to lb/ft² (since 1 psi = 144 lb/ft²).

Q2: What's the 448.83 conversion factor?
A: This converts ft³/s to gallons per minute (GPM).

Q3: When would I change the fluid density?
A: For fluids other than water (e.g., oil ρ ≈ 55 lb/ft³, seawater ρ ≈ 64 lb/ft³).

Q4: How do I find cross-sectional area?
A: For circular pipes, A = π × (diameter/2)². Convert diameter from inches to feet.

Q5: Does this account for friction losses?
A: No, this is ideal flow. For real systems, consider additional pressure drop calculations.