The thermal design of heat exchanger is explained based on the heat quantity formula Q=mcΔt.

This formula is also learned in compulsory education as basic knowledge of heat.

You may think it’s too late, but thermal design in chemical plants is very important because it directly affects the cost.

This is an important factor that determines the competitiveness of a plant.

Let’s take a look at an actual design example to see how it is used in on-site design.

## Heat quantity Q=mcΔt is the basis of heat exchanger design

Let’s explain the formula Q=mcΔt for the amount of heat.

**Calorie = mass x specific heat x temperature difference**

There is a relationship.

This comes first because the definition of specific heat is **kJ/(kg・k)**.

In school, the definition of specific heat would have been given that way.

That’s fine in practice.

For the time being, let’s organize it in terms of dimensions.

amount of heat | Q. | kJ |

mass | m | kg |

specific heat | c | kJ/(kg・k) |

temperature difference | Δt | k |

If you’ve studied thermodynamics, you’ll notice that we routinely use division by time.

heat flow | Q. | kJ/s |

mass flow rate | m | kg/s |

specific heat | c | kJ/(kg・k) |

temperature difference | Δt | k |

In chemical plants, this heat flow rate and mass flow rate are used, but it is troublesome to call them with the flow rate.

I have used it in almost the same way as the heat formula .

## Specific heat c

The specific heat c is a fixed value (physical property value) and is not intentionally changed by designers.

The value of specific heat c varies depending on the material, but let’s organize only the major ones at the factory.

water | 1.0 |

iron | 0.1 |

organic solvent | 0.4 to 0.6 |

The unit is kJ/(kg・k)

As long as you stick with it, you’ll be fine for now.

Organic solvents would be more accurate to look at their individual specific heats.

## Calculation example for a plate heat exchanger

Here is an example using the heat formula.

Start with a simple plate heat exchanger.

A plate heat exchanger exchanges heat by flowing two fluids with different temperatures.

Let’s organize the information about the heat capacity formula by dividing it into 1 for the hottest and 2 for the coldest (with a subscript).

heat flow | Q_{1} | Q_{2} |

mass flow rate | m_{1} | m_{2} |

specific heat | c_{1} | c_{2} |

temperature difference | Δt_{1} | Δt_{2} |

For example, let’s say you want to flow water at 30°C at 100L/min and heat it to 60°C.

Here, hot water at 90°C is used as the heat medium.

What is the flow rate of hot water? That’s the design point.

You have to decide the flow rate and **the pipe diameter**.

### Calculation of heat quantity Q1

Let’s calculate the heat quantity Q_{1} first.

**Q _{1}** =100L/min*1kcal/kg/K*(60-30)K*1kg/L=

**3,000kcal/min**

This is easy, isn’t it?

### Calculation of flow rate m2

Next , I would like to determine the flow rate m_{2}, but the temperature difference Δt_{2} has not been determined.

We use Q_{1} =Q_{2} as a matter of course.

Since heat is exchanged, I think it is intuitively easy to understand.

A temperature difference of Δt_{2} is assumed here .

Let Δt_{2} =10.

This makes the calculation easy.

**m _{2}**=3,000/1/10=

**300L/min**

How much to set the temperature difference is actually a difficult problem.

If the temperature is not discharged outside the system like hot water or circulating water, it is safe to keep the temperature at around 5 to 10°C.

This is because, in the case of hot water, the load of the steam to warm the cold water, and in the case of circulating water, the load of the cooling tower must be considered in balance.

The more users there are, the more complicated it will be if the temperature difference settings are varied.

### Caliber calculation

Once the flow rate m_{2} is determined, let’s determine the pipe diameter.

It is good to use the standard flow velocity to decide.

Since the flow rate on the heating side and the cooling side is different, the diameter will also be changed.

I think the calculations up to this point are easy.

## Overall heat transfer coefficient (U value) of a plate heat exchanger

It is necessary to confirm the overall heat transfer coefficient for the design of the plate heat exchanger.

It is a major factor that determines the performance as a heat exchange device.

The following relational expression is used to design the overall heat transfer coefficient (U value).

**Heat quantity = overall heat transfer coefficient x heat transfer area x temperature difference**

Let’s say Q_{3}=UAΔT as a formula .

### Calorie Q3

Calculating the heat Q_{3} is easy_{.}

It is common to set Q_{1} =Q_{2} =Q_{3 .}

This becomes a problem of designing a plate heat exchanger with the minimum heat transfer area required to raise 100L/min of water from 30°C to 60°C.

**Q _{3} = 3,000L /min**

### Temperature difference ΔT

Strictly speaking, ΔT uses the logarithmic average temperature difference.

However, this calculation is somewhat cumbersome.

If so, you can take the average temperature difference.

Δt_{1} =45 (average of 60,30), Δt_{2} =85 (average of 90,80), so

**ΔT** = Δt_{2} – Δt_{1} =85-45= **40℃**.

For serious calculations, the logarithmic average temperature difference is used, but in practice, the arithmetic average temperature difference can often be used.

Another option would be to ask the manufacturer to design it for you. As an owner engineer, simple practical calculations are much more important.

### Overall heat transfer coefficient U

The overall heat transfer coefficient U is inherently complicated to calculate.

However, in practice, approximate values and actual values are used.

Since it is a plate heat exchanger, let’s set **U=10kcal/(m ^{2}・min・k).**

Detailed calculations to the manufacturer … (Omitted)

### Heat transfer area A

Once you have come this far, the calculation of the heat transfer area A is easy.

**A** =Q_{3} / UΔT ** ^{=}** 3,000/(10・40)=

**7.5m**

^{2}In practice, this value will be decided with some leeway.

### If the heat transfer area is large…

Consider the case where the heat transfer area is greater than 7.5m^{2}.

In this case, the desired result is fine.

The larger the heat transfer area, the greater the amount of heat that can be exchanged.

Originally, it would have been enough to raise the temperature to 60°C, but there is a possibility that it will rise to 65°C and 70°C.

If the outlet temperature is fixed at 60°C, a control will be applied to adjust the flow rate on the hot water side depending on the temperature.

In this way, the specifications of the equipment are set higher than the requirements, and the operating conditions are adjusted with some leeway.

### If the heat transfer area is small…

What should we think about when a device with a small heat transfer area A is installed?

For example , if the heat transfer area is 1m^{2}, the amount of heat exchanged will be reduced by the amount of the heat transfer area.

How much will it change?

**Q ^{3′}** =3,000/7.5=

**400kcal/min**

It is assumed that the temperature rise will be reduced by this amount.

30+400/100=30+4= **34℃** is considered as outlet water temperature.

The outlet temperature of the hot water is also reduced (assuming you do not change the outlet flow rate).

90-400/300=90-13=77℃

Let’s correct the temperature difference ΔT.

**ΔT’** =(90+77)/2-(34+30)/2=83-32= **51℃**

Textbooks are repetitive calculations.

As you do this calculation, you will start to feel that it is troublesome.

If ΔT changes, the amount of heat exchanged Q will change, and the fixed U value should also change.

Ordinarily, we would not consider the case where the equipment’s capacity is insufficient.

I hope you can understand that it is unexpectedly difficult to specify the operating conditions for the analysis of a problem device.

## reference

Knowledge of heat transfer is very important as a mechanical and electrical engineer.

It is important to study using the following books.

## Related article

Related information

## lastly

I introduced the design of a plate heat exchanger as an example of using the calorific formula Q=mcΔt in a chemical plant.

It also touches on ideas for realizing simple calculations unique to the site, such as temperature difference assumptions and comparisons with U values.

Even if you calculate seriously, it is surprisingly difficult to get consistency with the driving results.

I want to be able to quickly find an answer that does not fail in simple calculations.

Please feel free to post your worries, questions, and questions about the design, maintenance, and operation of chemical plants in the comments section. (Comments are at the bottom of this article.)

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