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Heat Loss Open Water Tanks

Reference data and engineering information about heat loss open water tanks for thermodynamics applications.

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Overview

Engineering reference data for Heat Loss Open Water Tanks in thermodynamics.

Key Formulas

First Law

ΔU=QW\Delta U = Q - W

Energy is conserved — heat added minus work done.

Ideal Gas Law

PV=nRTPV = nRT

Relates pressure, volume, and temperature of an ideal gas.

Heat Transfer

Q=mcΔTQ = mc\Delta T

Sensible heat transfer.

Carnot Efficiency

η=1TC/TH\eta = 1 - T_C/T_H

Maximum efficiency between two temperatures.

Variables

SymbolDescriptionUnit
UUInternal energyJ
QQHeatJ
WWWorkJ
PPPressurePa
VVVolume
TTTemperatureK

Heat Loss Data Table

13 rows
Estimated heat loss rates from open water tanks for still ambient air at 60°F (15.6°C).
Water Temperature(°F)
Evaporation Loss(Btu/ft²·hr)
Radiation Loss(Btu/ft²·hr)
Total Surface Loss(Btu/ft²·hr)
Bare Steel Wall Loss(Btu/ft²·hr)
Insulated Wall Loss (1 in.)(Btu/ft²·hr)
Insulated Wall Loss (2 in.)(Btu/ft²·hr)
Insulated Wall Loss (3 in.)(Btu/ft²·hr)
908050130501254
10016070230701586
110240903309019107
12036011047011023129
130480135615135271410
140660160820160311612
1508601801040180341813
16011002101310210382115
17013802351615235422316
18017402602000260462517
19021602902450290502719
20026803203000320532920
21032403603600360573122

Source: engineeringtoolbox.com

Worked Example Calculation

For an open water tank with:

  • Water temperature: 150°F
  • Liquid surface area: 10 ft²
  • Un-insulated bare steel wall area: 50 ft²

The total heat loss (Q) is calculated as:

1. Surface Heat Loss (Evaporation + Radiation): Using the total surface loss value for 150°F from the table (1040 Btu/ft²·hr): Qsurface=qsurface×Asurface=1040Btuft2hr×10 ft2=10400 Btu/hrQ_{\text{surface}} = q_{\text{surface}} \times A_{\text{surface}} = 1040 \frac{\text{Btu}}{\text{ft}^2 \text{hr}} \times 10 \text{ ft}^2 = 10400 \text{ Btu/hr}

2. Wall Heat Loss (Transmission): Using the bare steel wall loss value for 150°F from the table (180 Btu/ft²·hr): Qwalls=qwalls×Awalls=180Btuft2hr×50 ft2=9000 Btu/hrQ_{\text{walls}} = q_{\text{walls}} \times A_{\text{walls}} = 180 \frac{\text{Btu}}{\text{ft}^2 \text{hr}} \times 50 \text{ ft}^2 = 9000 \text{ Btu/hr}

3. Total Heat Loss: Q=Qsurface+Qwalls=10400+9000=19400 Btu/hrQ = Q_{\text{surface}} + Q_{\text{walls}} = 10400 + 9000 = 19400 \text{ Btu/hr}

Assumptions and Unit Conversions

Important Assumptions: The tabulated values are rough estimates for open tanks in still ambient air at 60°F (15.6°C). Actual losses are highly sensitive to environmental factors including air velocity, turbulence, moisture content, and water agitation. Surface losses, dominated by evaporation, increase dramatically with temperature.

Unit Conversions:

  • Temperature: T(°C)=59(T(°F)32)T(°C) = \frac{5}{9} (T(°F) - 32)
  • Heat Flux: 1 Btu/(ft2⋅hr)=3.1525 W/m21 \text{ Btu/(ft}^2\text{·hr)} = 3.1525 \text{ W/m}^2

Interactive Charts

Heat loss from open tanks or swimmingpools chart

References