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Fruit processing
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Energy balance

Energy balance are used to account the needs of energy to run a process.

There are several different types of energy forms:

Indipendently to the form, is defined as:

energy is the capacity to perform work or heat.

The unit of measure of energy is joule (symbol: $\si{\joule}$, pronounce: "jul").

From the dymensional analysis, energy is defined as:

$$\text{Energy} = \text{Work} = \text{Force x Displacment} = M \cdot \frac{L}{t^2} \cdot L = M \cdot \frac{L^2}{t^2}$$

Thus, standard units for Energy are:

$$\si{\joule} = \si{\kilo\gram} \cdot \frac{\si{\meter^2}}{\si{\second^2}}$$

Thermal energy

To describe energy balance you should understand the meaning of:

  • Temperature
  • Sensible heat
  • Latent heat
  • Enthalpy
  • Specific heat

Temperature

The temperature of a material is the risult of all the forces that composed that material.

Temperature.
The temperature is the state of vibration of all the atoms and molecules that are composing that material.

Heat

Heat is a form of energy. Like the work, a body cannot contain heat. Heat is always a transient energy, which is passing from a hot body to a cold body.

There are two type of heat:

  • Sensible heat
  • Latent heat

The heat is often referred to a unit of mass. In such case, the enthalpy devided by mass unit is called enthalpy, which is the quantity of energy that a system of mass unit is able to exchange with the environment. It is measured as joule per kilogram (J/kg).

Sensible heat

Anytime two materials have a difference in temperture, the hot material loose heat towards the cold material. This transient energy is defined as sensible heat.

Sensible heat can be measured as:

$$q = m \cdot C_p \cdot \Delta{T} $$

where:

  • $q$ means heat (in joule)
  • $C_p$ is the specific heat (in J/kg℃)
  • $\Delta{T}$ is the temperature difference between the two bodies.

The specific heat is the amount of energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or Celsius degree. Thus, units of measure are J/kg℃

Example.
Determine the sensible heat required to bring 5 kg of water from 20 to 80℃

Solution.

Given:

  • $m = 5 \cdot kg$ (we may assume that no mass loss occurs because of evaporation during heating)
  • $C_p$ of water can be assumed constant within the temperature range from 0 to 100℃.
  • $\Delta T = 60℃$

Thus:

$$q = 5 \cdot kg \cdot 4.19 \cdot \frac{kJ}{kg \cdot ℃} \cdot (80 - 20)℃ = 1257 kJ$$

Latent heat

Latent heat is a form of energy that is gain or loss during a phase transition.

For instance, to bring water to boil, you need first to provide sensible heat that increase the water temperature up to 100℃. Afterwards, you need heat to transform all the water in steam. Such heat is called latent heat of vaporization. Pay attention that such form of heat is exchanged without changes to the temperature.

Enthalpy
Temperature Pression Volume Liquid Steam Latent heat
℃ kPa $\si{\meter^3\per\kilo\gram}$ $\si{\kilo\joule\per\kilo\gram}$ $\si{\kilo\joule\per\kilo\gram}$ $\si{\kilo\joule\per\kilo\gram}$
15 1.7051 77.026 62.99 2528.90 2465.91
30 4.246 32.894 125.79 2556.30 2430.51
45 9.593 15.258 188.45 2583.20 2394.75
60 19.940 6.671 251.13 2609.60 2358.47
75 38.58 4.131 313.93 2635.30 2321.37
90 70.14 2.361 376.92 2660.10 2283.18
100 101.325 1.6729 419.04 2676.10 2257.06
110 143.27 1.2102 461.30 2691.50 2230.20
120 198.53 0.8919 503.71 2706.30 2202.59
130 270.1 0.6685 546.31 2720.50 2174.19
140 316.3 0.5089 589.13 2733.90 2144.77

When a fluid is heated, we need to provide sensible heat. When there is a phase change, then we need to provde also the latent heat:

$$q = m \cdot C_p \cdot \Delta{T} + m \cdot \lambda$$

Example.
Determine the heat required to transform 5 kg of water at 20℃ in steam at 1 bar of pressure.

Solution.

Given:

  • $m = $ 5 kg (we may assume that no mass loss occurs because of evaporation during heating)
  • $C_p$ of water is constant and equal to 4.19 kJ/kg℃.
  • $\Delta T =$ 80℃
  • Latent heat of vaporization at 1 bar is $\lambda =$ 2257 kJ/kg

Thus:

$$q = 5 \cdot kg \cdot 4.19 \cdot \frac{kJ}{kg \cdot ℃} \cdot (100 - 20)℃ + 5 \cdot kg \cdot 2257 \cdot \frac{kJ}{kg} = (1676 + 11285) \cdot kJ = 13 MJ$$
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