Expansion Tank Sizing for Closed Tanks with Air/Water Interface

System Water Volume (\( V_s \)):

Specific Volume at Lower Temperature (\( V_1 \)):

Specific Volume at Higher Temperature (\( V_2 \)):

Linear Coefficient of Thermal Expansion (\( \alpha \)):

Temperature Difference (\( \Delta T \)):

Atmospheric Pressure (\( P_a \)):

Pressure at Lower Temperature (\( P_1 \)):

Pressure at Higher Temperature (\( P_2 \)):

Expansion Tank Volume (\( V_t \)):

1. What is an Expansion Tank Sizing Calculator?

Definition: This calculator computes the volume (\( V_t \)) of an expansion tank required for a closed hydronic system with an air/water interface, accounting for water volume changes due to temperature variations.

Purpose: It is used in hydronic system design to size expansion tanks, ensuring they can accommodate thermal expansion of water, preventing system overpressure and damage.

2. How Does the Calculator Work?

The calculator uses the following formula for expansion tank volume:

Expansion Tank Volume: \[ V_t = V_s \frac{\left( \left[ \left( \frac{V_2}{V_1} \right) - 1 \right] - 3 \alpha \Delta T \right)}{\left( \frac{P_a}{P_1} - \frac{P_a}{P_2} \right)} \]

Where:

\( V_t \): Volume of expansion tank (gal, m³)
\( V_s \): Volume of water in system (gal, m³)
\( V_1 \): Specific volume of water at lower temperature (ft³/lb, m³/kg)
\( V_2 \): Specific volume of water at higher temperature (ft³/lb, m³/kg)
\( \alpha \): Linear coefficient of thermal expansion (in/in·°F, m/m·°C)
\( \Delta T \): Temperature difference (°F, °C)
\( P_a \): Atmospheric pressure (psia, Pa)
\( P_1 \): Pressure at lower temperature (psia, Pa)
\( P_2 \): Pressure at higher temperature (psia, Pa)

Unit Conversions:

System Water Volume (\( V_s \)) and Tank Volume (\( V_t \)): gal, m³ (1 m³ = 264.172 gal)
Specific Volumes (\( V_1 \), \( V_2 \)): ft³/lb, m³/kg (1 m³/kg = 16.0185 ft³/lb)
Linear Coefficient (\( \alpha \)): in/in·°F, m/m·°C (1 m/m·°C = 0.555556 in/in·°F)
Temperature Difference (\( \Delta T \)): °F, °C (1 °C = 1.8 °F)
Pressures (\( P_a \), \( P_1 \), \( P_2 \)): psia, Pa (1 Pa = 0.000145038 psia)

Steps:

Enter the system water volume (\( V_s \)), specific volumes (\( V_1 \), \( V_2 \)), linear coefficient of thermal expansion (\( \alpha \)), temperature difference (\( \Delta T \)), atmospheric pressure (\( P_a \)), and pressures at lower and higher temperatures (\( P_1 \), \( P_2 \)), and select their units.
Convert all inputs to consistent units: \( V_s \) to gal, \( V_1 \) and \( V_2 \) to ft³/lb, \( \alpha \) to in/in·°F, \( \Delta T \) to °F, \( P_a \), \( P_1 \), and \( P_2 \) to psia.
Calculate the expansion tank volume using the formula.
Convert the result to the selected unit (gal or m³).
Display the result with 5 decimal places, or in scientific notation if the value is greater than 10,000 or less than 0.00001.

3. Importance of Expansion Tank Sizing

Calculating the correct expansion tank size is crucial for:

System Safety: Prevents overpressure and potential damage by accommodating water expansion due to temperature increases.
System Design: Ensures proper tank sizing to maintain system pressure within safe limits.
System Reliability: Supports stable operation of hydronic systems under varying temperature conditions.

4. Using the Calculator

Examples:

Example 1: For \( V_s = 2500 \, \text{gal} \), \( V_1 = 0.01602 \, \text{ft³/lb} \), \( V_2 = 0.01663 \, \text{ft³/lb} \), \( \alpha = 6.5e-6 \, \text{in/in·°F} \), \( \Delta T = 160 \, \text{°F} \), \( P_a = 14.7 \, \text{psia} \), \( P_1 = 62.3 \, \text{psia} \), \( P_2 = 117.3 \, \text{psia} \), tank volume in gal:
- Numerator: \( \left( \frac{V_2}{V_1} - 1 \right) - 3 \alpha \Delta T = \left( \frac{0.01663}{0.01602} - 1 \right) - 3 \times 6.5e-6 \times 160 \approx (1.03808 - 1) - 0.00312 \approx 0.03808 - 0.00312 \approx 0.03496 \)
- Denominator: \( \frac{P_a}{P_1} - \frac{P_a}{P_2} = \frac{14.7}{62.3} - \frac{14.7}{117.3} \approx 0.23596 - 0.12532 \approx 0.11064 \)
- \( V_t = 2500 \times \frac{0.03496}{0.11064} \approx 2500 \times 0.31598 \approx 789.95 \)
- Since 789.95 < 10000 and > 0.00001, display with 5 decimal places: \( 789.95000 \)
- Note: The calculated value (789.95 gal) is very close to the example output of 787 gal, with minor differences likely due to rounding in the original example.
Example 2: For \( V_s = 9.4635 \, \text{m³} \), \( V_1 = 0.001 \, \text{m³/kg} \), \( V_2 = 0.001038 \, \text{m³/kg} \), \( \alpha = 1.17e-5 \, \text{m/m·°C} \), \( \Delta T = 88.89 \, \text{°C} \), \( P_a = 101325 \, \text{Pa} \), \( P_1 = 429159 \, \text{Pa} \), \( P_2 = 808115 \, \text{Pa} \), tank volume in m³:
- Convert: \( V_s = 9.4635 \times 264.172 \approx 2500 \, \text{gal} \)
- \( V_1 = 0.001 \times 16.0185 \approx 0.0160185 \, \text{ft³/lb} \)
- \( V_2 = 0.001038 \times 16.0185 \approx 0.0166272 \, \text{ft³/lb} \)
- \( \alpha = 1.17e-5 \times 0.555556 \approx 6.5e-6 \, \text{in/in·°F} \)
- \( \Delta T = 88.89 \times 1.8 \approx 160 \, \text{°F} \)
- \( P_a = 101325 \times 0.000145038 \approx 14.7 \, \text{psia} \)
- \( P_1 = 429159 \times 0.000145038 \approx 62.3 \, \text{psia} \)
- \( P_2 = 808115 \times 0.000145038 \approx 117.3 \, \text{psia} \)
- Numerator: \( \left( \frac{0.0166272}{0.0160185} - 1 \right) - 3 \times 6.5e-6 \times 160 \approx (1.03796 - 1) - 0.00312 \approx 0.03484 \)
- Denominator: \( \frac{14.7}{62.3} - \frac{14.7}{117.3} \approx 0.11064 \)
- \( V_t = 2500 \times \frac{0.03484}{0.11064} \approx 787.45 \, \text{gal} \)
- Convert to m³: \( 787.45 \times 0.00378541 \approx 2.981 \)
- Since 2.981 < 10000 and > 0.00001, display with 5 decimal places: \( 2.98100 \)
Example 3: For \( V_s = 1000 \, \text{gal} \), \( V_1 = 0.0161 \, \text{ft³/lb} \), \( V_2 = 0.0165 \, \text{ft³/lb} \), \( \alpha = 7e-6 \, \text{in/in·°F} \), \( \Delta T = 100 \, \text{°F} \), \( P_a = 14.7 \, \text{psia} \), \( P_1 = 50 \, \text{psia} \), \( P_2 = 100 \, \text{psia} \), tank volume in gal:
- Numerator: \( \left( \frac{0.0165}{0.0161} - 1 \right) - 3 \times 7e-6 \times 100 \approx (1.02484 - 1) - 0.0021 \approx 0.02274 \)
- Denominator: \( \frac{14.7}{50} - \frac{14.7}{100} \approx 0.294 - 0.147 \approx 0.147 \)
- \( V_t = 1000 \times \frac{0.02274}{0.147} \approx 154.69 \)
- Since 154.69 < 10000 and > 0.00001, display with 5 decimal places: \( 154.69000 \)

5. Frequently Asked Questions (FAQ)

Q: What does expansion tank sizing represent?
A: Expansion tank sizing (\( V_t \)) determines the volume of a tank needed to accommodate the thermal expansion of water in a closed hydronic system, preventing overpressure and ensuring system safety.

Q: How can I determine the input parameters?
A: System water volume (\( V_s \)) is estimated from the total water content in the system. Specific volumes (\( V_1 \), \( V_2 \)) are obtained from water property tables at the lower and higher temperatures (e.g., 0.01602 ft³/lb at 40°F, 0.01663 ft³/lb at 200°F). Linear coefficient of thermal expansion (\( \alpha \)) depends on the pipe material (e.g., 6.5e-6 in/in·°F for steel). Temperature difference (\( \Delta T \)) is the expected temperature range. Atmospheric pressure (\( P_a \)) is typically 14.7 psia. Pressures (\( P_1 \), \( P_2 \)) are system pressures at the lower and higher temperatures.

Q: Why is expansion tank sizing important in hydronic system design?
A: It ensures the system can handle water expansion due to temperature changes, preventing pressure-related damage and maintaining safe, reliable operation.