Consider heating 100 kg of water from $20^\circ\text{C}$ to $80^\circ\text{C}$ in 5 minutes (300 seconds). The specific heat of water is $c_p = 4.2 \, \text{kJ/kg}^\circ\text{C}$ .

The power required for heating is calculated as:

$P = \frac{m \cdot c_p \cdot \Delta T}{\Delta t} = \frac{100 \, \text{kg} \times 4.2 \, \text{kJ/kg}^\circ\text{C} \times (80^\circ\text{C} - 20^\circ\text{C})}{300 \, \text{s}} = 84 \, \text{kW}$

Given absolute steam pressure of $220 \, \text{kPa}$ , the total enthalpy of steam is $h_g = 2711 \, \text{kJ/kg}$ .

The required steam flow rate is given by the steam injection formula:

$m_s = \frac{P}{h_g - (t \cdot c_p)} = \frac{84 \, \text{kW}}{2711 \, \text{kJ/kg} - (80^\circ\text{C} \times 4.2 \, \text{kJ/kg}^\circ\text{C})} = 0.035 \, \text{kg/s} = 127 \, \text{kg/h}$

The total steam consumption for the heating process is:

$q = m_s \cdot \Delta t = 0.035 \, \text{kg/s} \times 300 \, \text{s} = 10.5 \, \text{kg}$

This example illustrates the application of steam injection for direct heating, avoiding heat exchangers and fouling issues.

References

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Source: engineeringtoolbox.com