This heat transfer coefficient calculator will help you determine the **overall heat transfer coefficient** or **film coefficient**. This parameter is vital to most heat transfer calculations and **insulation**, especially for building walls. For instance, if a designer wishes to reduce the heat transfer via a building wall or heat exchanger, they **add several layers of insulation**.

This tool uses the heat transfer coefficient equation while giving you the option of adding **up to 10 layers to your wall** and returning the **thermal resistance and overall heat transfer coefficient** for the stacked structure.

The heat transfer coefficient is a function of **wall thickness, thermal conductivity, and the contact area of the wall**. The tool considers free convection on either side of the wall while performing the calculations. Different types of convection and flow geometries also exist using the Nusselt number. Read on to understand what the heat transfer coefficient is and how to use the heat transfer coefficient formula.

π If you are unsure about *what thermal conductivity is*or how to calculate it, you can look at Omni's thermal conductivity calculator to learn more about this topic!

## Overall heat transfer coefficient β Concept of thermal resistance

**What is heat transfer coefficient?** β It is a measure of **how well a wall or structure conducts heat**. In other words, the **ratio of heat transfer through unit area and temperature difference**. This ratio is the **combined value for all the layers in the structure**. It is measured in $\text{W/m}^2\text{K}$W/m2K or $\text{BTU/(h}\cdot^\circ\text{F}\cdot\text{ft}^2)$BTU/(hβ
βFβ
ft2). The heat transfer coefficient formula considering `n`

layers in a structure is:

$\quad \frac{1}{U_t} = \frac{1}{A} \sum_{i=1}^{n} \frac{L_i}{k_i}$Utβ1β=A1βi=1βnβkiβLiββ

where:

- $U_t$Utβ β
**Heat transfer coefficient**; `A`

β Contact**area**;`L`

β**Thickness**of $\text{n}^\text{th}$nth layer; and`k`

β**Thermal conductivity**of layer material.

**Did you know?**

The **lower** the value of heat transfer coefficient, the **better the insulation** provided by the structure and vice versa.

* Concept of thermal resistance* β It is the

**resistance of a material**against

**heat flow or conductance**. In other words, thermal resistance is the ratio of the temperature difference and heat conducted through a medium. It is analogous to Ohm's law, which is:

$\quad I = \frac{V_1 - V_2}{R_e}$I=ReβV1ββV2ββ

where:

- $I$I β
**Current**; - $V_1 - V_2$V1ββV2β β
**Voltage difference**; and - $R_e$Reβ β
**Resistance**.

Here, the **current is the rate of heat transfer, $Q$Q, voltage difference being the temperature difference, $T_1 - T_2$T1ββT2β**. The **electrical resistance corresponds to thermal resistance $R_t$Rtβ** (check Ohm's law calculator). Therefore the thermal resistance equation becomes,

$\quad Q = \frac{T_1 - T_2}{R_t}$Q=RtβT1ββT2ββ

The units of thermal resistance are `K/W`

or `Β°C/W`

. You can also relate thermal resistance with the overall heat transfer coefficient as:

$\quad R_t = \frac{1}{U_t} = \frac{L}{k}$Rtβ=Utβ1β=kLβ

For several layers stacked one after another, the thermal resistance equation is:

$\quad R_t = \frac{1}{A} \sum_{i=1}^{n} \frac{L_i}{k_i}$Rtβ=A1βi=1βnβkiβLiββ

The above case applies to `conduction`

-only. However, when the inner and outer surfaces of walls are exposed to air or any other fluids, another factor needs to be considered. In that case, the **convective heat transfer coefficient** $h$h is used to determine the convective resistance of the medium. Such that:

$\quad R_\text{conv} = \frac{1}{hA}$Rconvβ=hA1β

To find out the overall heat transfer coefficient, we add the convective resistance to the conductive resistance, $R_t$Rtβ:

$\quad\footnotesize R = \frac{1}{U} = \frac{1}{A} \left [ \frac{1}{h_i} + \sum_{i=1}^{n} \frac{L_i}{k_i} + \frac{1}{h_o}\right ]$R=U1β=A1β[hiβ1β+i=1βnβkiβLiββ+hoβ1β]

where:

- $h_i$hiβ β
**Convective heat transfer coefficient**of fluid, inner surface; and - $h_o$hoβ β
**Convective heat transfer coefficient**of fluid, outer surface.

The units of convective heat transfer coefficient are equal to those of the heat transfer coefficient, i.e., $\text{W/m}^2\cdot\text{K}$W/m2β K or $\text{BTU/(h}\cdot^\circ\text{F}\cdot\text{ft}^2)$BTU/(hβ βFβ ft2).

π‘ If you'd like to read more about heat transfer calculations applied to heat exchangers, visit the LMTD calculator or the effectiveness NTU calculator.

## How to calculate heat transfer coefficient or film coefficient

The calculator has `two modes`

:

- Conduction only; and
- Conduction with convection on both sides.

In addition to this, it begins with a single layer of material, on which you can stack or remove layers by using the `add`

or `remove`

button.

To find the overall heat transfer coefficient and thermal resistance:

- Select the
**mode of heat transfer**, say,`conduction only`

. - Enter the
**area**of contact, $A$A. - Insert the
**initial thickness**of the wall, $L_0$L0β. - Fill in the
**thermal conductivity**of the wall material, $k_0$k0β. - Add more layers using the
`Add`

button at your convenience. - Repeat the steps
`3`

and`4`

for all layers. - The calculator gives
**thermal resistance**and**overall heat transfer coefficient**as per the configuration.

**Stacked walls**

You can use this tool to find the heat transfer coefficient and thermal resistance for a structure by stacking up to 11 layers.

## Example: Using the heat transfer coefficient calculator

Find the overall heat transfer coefficient of a window having 2 layers of `2 mm`

thick glass with a gap of `5 mm`

filled with air in between. Take contact area as $1.2 \ \text{m}^2$1.2m2. Use the following properties:

- Convective heat transfer coefficient of air, inner $h_i = 10 \text{ W/m}^2\cdot\text{K}$hiβ=10W/m2β K;
- Convective heat transfer coefficient of air, outer $h_o = 40 \text{ W/m}^2\cdot\text{K}$hoβ=40W/m2β K;
- Thermal conductivity of air, $k_{air} = 0.026 \text{ W/m}\cdot\text{K}$kairβ=0.026W/mβ K; and
- Thermal conductivity of glass, $k_{glass} = 0.78 \text{ W/m}\cdot\text{K}$kglassβ=0.78W/mβ K.

To find the thermal resistance and overall heat transfer coefficient:

- Select the
**mode of heat transfer**,`conduction and convection (on both sides)`

. - Enter the
**area**of contact, $A = 1.2 \text{ m}^2$A=1.2m2. - Insert the
**convective heat transfer coefficient**for inner surface, $h_i = 10 \text{ W/m}^2\cdot\text{K}$hiβ=10W/m2β K. - Fill in the details of initial layer as $L_0 = 2 \text{ mm}$L0β=2mm and $k_0 = 0.78\text{ W/m}\cdot\text{K}$k0β=0.78W/mβ K.
- Use the
`Add`

button to insert a 2nd layer. - Insert the properties of the 2nd layer as $L_1 = 5 \text{ mm}$L1β=5mm and $k_1 = 0.026\text{ W/m}\cdot\text{K}$k1β=0.026W/mβ K.
- Use the
`Add`

button to insert a 3rd layer. - Enter the properties of the 3rd layer as $L_2 = 2 \text{ mm}$L2β=2mm and $k_2 = 0.78\text{ W/m}\cdot\text{K}$k2β=0.78W/mβ K.
- Insert the
**convective heat transfer coefficient**for outer surface, $h_o = 40 \text{ W/m}^2\cdot\text{K}$hoβ=40W/m2β K. - Using the thermal resistance calculator:

$\quad\scriptsize\begin{align*}R &= \frac{1}{A} \left [ \frac{1}{h_i} + \frac{L_0}{k_0} + \frac{L_1}{k_1}+ \frac{L_2}{k_2} + \frac{1}{h_o}\right ] \\\\R &= \frac{1}{1.2} \Big [ \frac{1}{10} + \frac{0.002}{0.78} + \frac{0.005}{0.026}+ \frac{0.002}{0.78} \\ &\qquad \quad+ \frac{1}{40}\Big ] \\\\R &= 0.2687 \ \text{Β°C/W}\end{align*}$RRRβ=A1β[hiβ1β+k0βL0ββ+k1βL1ββ+k2βL2ββ+hoβ1β]=1.21β[101β+0.780.002β+0.0260.005β+0.780.002β+401β]=0.2687Β°C/Wβ

- The overall heat transfer coefficient is:

$\quad \ \ \scriptsize U = \frac{1}{R} = \frac{1}{0.2687} = 3.722 \text{ W/m}^2\cdot\text{K}$U=R1β=0.26871β=3.722W/m2β K

## FAQ

### What is heat transfer coefficient?

It is the **ratio of heat flow through a unit area and temperature difference**. The heat transfer coefficient measures how well a structure conducts heat. If the value of this proportionality constant is low, it means the material is a better insulator.

### How do I calculate heat transfer coefficient?

To calculate heat transfer coefficient:

**Divide**the**thickness**of the first layer with the**thermal conductivity**of the medium.**Repeat**the previous step for all layers and**add**them together.- Find the
**reciprocal of convective heat transfer for the inner surface**and**add**it to the sum. - Find the
**reciprocal of convective heat transfer for the outer surface**and**add**it to the sum. - Find the
**reciprocal of the resultant**to obtain the heat transfer coefficient.

### What is thermal resistance?

It is the **resistance offered by the medium against the heat flow through it**. The thermal resistance is also the reciprocal of the overall heat transfer coefficient. It is desirable for insulating-type materials like cotton and wool, whereas it is undesirable for conductors.

### How do I calculate thermal resistance?

To calculate thermal resistance:

**Divide**the**thickness**of the first layer with the**thermal conductivity**of the medium.**Repeat**the previous step for all layers and**add**them together.- Find the
**reciprocal of convective heat transfer for the inner surface**and**add**it to the sum. - Find the
**reciprocal of convective heat transfer for the outer surface**and**add**it to the sum to obtain the thermal resistance.

Alternatively, you can find the reciprocal of **overall heat transfer coefficient** to find the thermal resistance.