Fruit processing
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Lecture n. 04 - Heat exchangers


Mode of heat exchange

Heat is energy transferred due to a difference in temperature:

$$ q \propto \Delta{T} $$

There are three modes of heat transfer:

  1. conduction
  2. convection
  3. radiation

In a general heating process, all modes of heat transfer may occur at the same time. A good example is the heating of a tin can of water by using a Bunsen burner:

In particular:

Heat transfer by conduction

The heat transferred by conduction is given by Fourier's law. Considering only the simplest case of heat transfer by a single direction (i.e. along a homogeneous longitudinal plane), the Fourier's law of heat transfer can be written as:

$$ \frac{dq}{dt} = - k \cdot A \cdot \frac{dT}{dx} $$

where:

The negative sign is required to make the heat-flow positive when $T_1$ is higher than $T_2$.

The heat transfer by conduction can be easily examplified by looking at the cross section of a wall, where at the two sides there is a difference of temperature:

T1 T2 dx

The thermal resistance that create the gradient of temperature shown in the diagram above is given by:

$$ R = \frac{dx}{k \cdot A} $$

The Fourier's law becomes:

$$ \frac{dq}{dt} = - \frac{dT}{R} $$

Materials can be classified as insulators (high thermal resistance) or conductive (low thermal resistance). Examples of highly conductive materials are:

Material Thermal conductivity (W/ m $\cdot$ K) Application
Copper 400 Distilleries
Alluminium 200 Packaging
Silver 430 Electronics
Stainless steel 15 Heat exchangers

Instead, examples of insulating materials are:

Material Thermal resistance (W/K) Application
Cork 0.04 Building
Wood oak 0.20 Building
Wool 0.04 Insulating walls
Urethane foams 0.02 Insulating walls
Air 0.03 Building

Heat transfer by convection

Convection is the transfer of energy by conduction and radiation in moving, fluid media. The motion of the fluid is an essential part of convective heat transfer. In many cases, heat is transferred from one fluid, through a solid wall, to another fluid. Such transfer occurs typically in a heat exchanger. The amount of heat transferred by convection between a surface and a fluid is given by Newton's law of cooling:

$$ \frac{dq}{dt} = - h \cdot A \cdot \Delta{T} $$

where:

The heat transfer coefficient depends on the movement of the fluid. Examples of $h$ values are shown below:

Movement Laminar heat transfer (W/ m$^2 \cdot$ K) Application
Natural convection of air 5-10 Drier
Forced convection of air 25-300 Convective oven
Boiling water 1.500 - 57.000 Reactors
Condensing steam, film 5.700 - 17.000 Evaporator
Condensing steam, spray 30.000 - 100.000 Spray drier

Heat transfer between two fluids separated by a flat plate

This case is typical of the plate heat exchangers used for heating or cooling fluids. The problem is exemplified with the following diagram:

T1 T2 T3 T4 dx Wall thickness Convection Convection

The case represented by the diagram above is a typical example of heat transfer during cooking. We can observe four temperatures (from $T_1$ to $T_4$):

The heat transfer comprises two convective phenomena that are governed by the type of motion of the fluids, and by one conductive phenomena, which is governed by the material chosen for the heat exchange. The overall heat transfer equation that accounts for these three processes is the following: $$ \frac{dq}{dt} = \dfrac{\Delta{T_1} + \Delta{T_2} + \Delta{T_3}} {\dfrac{1}{h_1 \cdot A} + \dfrac{dx}{k \cdot A} + \dfrac{1}{h_2 \cdot A} } $$ In the case of cooking application, the aim is to maximize the heat transfer. Accordingly, the machine is designed to:

Often, manufacturers of food machinery provides the overall heat transfer coefficient (U), which is the value at the denominator of the previous equation:

$$ U = \dfrac{1}{\dfrac{1}{h_1 \cdot A} + \dfrac{dx}{k \cdot A} + \dfrac{1}{h_2 \cdot A} } $$

With the knowledge of such parameter, it is possible to use again he Fourier's law of heat transfer:

$$ \dfrac{dq}{dt} = A \cdot U \cdot \Delta{T} $$

where:

It should be noted that the term $\dfrac{dq}{dt}$ is the amount of heat transfer required to rise the temperature of the product up to a desired value. Such amount has been already estimated by the following equation: $$ \dfrac{dq}{dt} = m \cdot C_p \cdot \Delta{T} + m \cdot \Lambda$$ With the knowledge of the term $\dfrac{dq}{dt}$, which is generally known from energy balance, given the $U$ values from the manufacturer of the heat exchanger, it is possible to estimate the dimension of the heat exchanger ($A$).

Heat exchangers

Heat exchangers are devices used to heat or cool fluids. It is possible to classify heat exchangers in two classes:

  1. Indirect heat exchangers
  2. Direct heat exchangers

Indirect heat exchangers are the most common. The product is heated thanks to heat transfer from a hot fluid (generally hot water or steam), which is not in direct contact, but separated by a wall material.

Examples of indirect heat exchangers are:

  1. Plate heat exchangers
  2. Tubular heat exchangers
  3. Shell and tube heat exchangers
  4. Scraped surface heat exchangers

Direct heat exchangers heat the product by direct mixing with the hot fluid (generally steam). This causes a flash heating of the product, but also a dilution with the condensed water.

Flow in heat exchangers

The flow of the product and the hot/cold fluid in a indirect heat exchanger can be: The diagram below shows an example of tubular heat exchanger that is running with a counter current flow: Length of the heat exchanger Temperature

Instead, the diagram below shows an example of tubular heat exchanger that is running with a co-current flow:

Length of the heat exchanger Temperature

In general, the use of one or the other flow mode is determined by the type of application. Counter current heat exchangers offer the highest efficiency. With such configuration the heat exchange is always maximized. Moreover, the variation of temperature between the hot fluid and the product is minimized. This means that the heating is uniform along the process and as gentle as possible. Gentle heating is important to minimize the fouling mechanism.

However, co-curent heat exchangers are very important for applications where you need to quickly heat the product. With this mode of flow, as soon as the product enters into the heat exchanger, it is heated up very quickly thanks to the large difference of temperature between the hot fluid and the product. This condition is desired for evaporation process, where large amount of water must be removed from the product. The fact that along the heat exchanger the difference of temperature is reduced (i.e. lower heat transfer capacity) is also desired since, during evaporation, the product decresaes its heat capacity coefficient $C_p$. This means, in practice, that to rise its temperature along the evaporation process is required a progressively lower amount of energy. Conversely, if the same difference of temperature between the hot fluid and the product is maintained along the evaporation process, the drier product will easily burn at the contact with the metal plate, causing fouling.

For design purposes, the mode of flow affects the $\Delta{T}$ parameters. In particular for co-current heat exchangers, the changes of $\Delta{T}$ along the heating process can be relevant. Accordingly, to continue using the equation of heat transfer described before, it is often useful to express an average $\Delta{T}$. In case of counter-current heat exchangers, it is possible to estimate the $\Delta{T}$ with the arithmetic mean. Alternatively, it is very common to use a logarithmic mean temperature difference, often called as log mean temperature difference (LMTD). This is expressed as follow:

$$\Delta{T}_m = \dfrac{\Delta{T_2} - \Delta{T_1}}{\ln \left( \dfrac{\Delta{T_2}}{\Delta{T_1}} \right) } $$

where:

Heat recovery

Very often, the amount of energy required for heating or cooling a product can be reused. For instance, the product that exits from the het exchanger has a high temperature that can be conveniently used to heat the product entering to the heat exchanger. A typical example of this situation is displayed in the following diagram, where three heat exchangers are used to heat and cool the product:

T = 0℃ T = 10℃ IN T = 145℃ T = 145℃ T = 140℃ T = 100℃ T = 60℃ T = 5℃ (I) (II) (III)

Exercise

Design a sterilization process based on scheme shown above. The juice has a $C_p$ of 3.8 kJ/kg$\cdot$°C. Flow rate is 10 ton per hour. Initial temperature of the juice is 5°C. Sterilization temperature is 140°C. At the outlet of step (I), the temperature of the juice is cooled down to 45°C and, then, down to 5°C thanks to the use of ice-water. Ice-water enters into the heat exchanger at 0°C and exits at 20°C. The heating medium during the sterilization process is steam at 145°C. The overall heat transfer (U) is assumed as 3.000 W/m$^2 \cdot$°C

Determine:

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