Lesson 4 - Convection

Forced convection occurs when heat is transferred by a fluid which is being moved by an external force. External forces include pumps, fans, and mixers.

In contrast to natural convection, forced convection does not rely on the density variation due to temperature. Thus, forced convection is often much stronger than natural convection.

A common example of forced convection is wind chill. Without any wind present, heat leaving your body due to convection and conduction creates a layer of warmer air just above your skin. This natural boundary layer provides a small amount of insulation for your body. When wind velocity is high enough, it starts to disrupt the thin boundary layer your body is creating, causing the warm air to be replaced by cooler air. The stronger the winds, the more disruptive it is to the warm boundary layer, resulting in a greater wind chill effect.

A common brewery example of forced convection is a heat exchanger. A plate heat exchanger is one system that can be used to exchange heat between two fluids. The fluids are pumped through channels formed between multiple heat exchanger plates. On one side of the plate is the cooler fluid and on the other side is a hotter fluid. Both the pumping speed and the design of the plates cause flows to be turbulent, which maximises the film heat transfer coefficient and the heat transfer rate between the fluids. Heat exchanger technology will be covered in more detail later in this unit.

Some typical values for the film heat transfer coefficients for forced convection are:

• Air, gases, and dry vapours: 10-1,000Wm^-2K^-1 

• Water and liquids: 50-10,000Wm^-2K^-1 

Let’s do an example of heat transfer due to forced convection. Let’s say we want to calculate the convective heat transfer between 85°C wort and a heat exchanger plate at 4°C. The plate is 0.3 m by 0.3 m. The film heat transfer coefficient is 2,000 Wm^-2K^-1.

First let’s recall the modified Fourier equation applied to heat transfer across the boundary layer between the solid and fluid interface and the bulk fluid:

We are being asked for the rate, not the rate per area, so let’s move the area to the other side of the equation. 

The area of heat transfer is the area of the plate. To calculate the area of the plate, we multiply the height by the width. 

Next, we need to assign values for T_w and T_∞. Recall that T_w is the tempearure of the wall. In this case that would be the temperature of the plate, which is 4°C. T_∞ is the temperature in the bulk fluid far away from the wall. In this example that would be the temperature of the wort, which is 85°C. 

Lastly, the film heat transfer (h) is given in the problem statement as 2,000 Wm^-2. We are now ready to enter the values and calculate. 

So, the wort is transferring 14,580 W of heat to the plate.

The result is positive when the fluid is gaining heat and negative when the fluid is losing heat.