How much steam is needed to exchange steam for hot water?

The amount of steam required to heat water with steam depends on many factors, including the steam pressure, temperature, target temperature of the hot water, heat transfer efficiency, and heat loss of the system. The following is a simplified calculation method, assuming that the steam releases mainly latent heat and there is no significant heat loss:

1. Basic formula:

   Q=mhot water⋅c⋅ΔT+mhot water⋅L(if hot water needs to be vaporized)Q = m_{\text{hot water}} \cdot c \cdot \Delta T + m_{\text{hot water}} \cdot L \quad (\text{if hot water needs to be vaporized})Q=mHot water​⋅c⋅ΔT+mHot water​⋅L(If hot water needs to be vaporized)

   in:

   • $Q$: Total calorie requirement (kilojoules, kJ)
   • mHot waterm_{\text{Hot water}} mHot water​: Hot water mass (kg), 1 cubic meter of hot water ≈ 1000 kg (assuming density is approximately 1 g/cm³)
   • $c$: The specific heat capacity of water is about 4.18 kJ/(kg·℃)
   • $\Delta T$: Hot water temperature rise value (℃)
   • $L$: Latent heat of vaporization of water (kJ/kg, negligible if vaporization is not involved)

2. Heat provided by steam： The heat released by steam mainly comes from itsLatent heat of condensation(latent heat of vaporization) and possibleSensible heat.For example:

   • The latent heat of vaporization of saturated steam (0.1 MPa, 100°C) is approximately 2257 kJ/kg.
   • After the steam condenses, if the temperature continues to drop, sensible heat will be released (calculated according to the specific heat capacity of liquid water).

   Steam volume calculation:

   msteam = Q steam latent heat + sensible heat (if any)m_{\text{steam}} = \frac{Q}{\text{steam latent heat} + \text{sensible heat (if any)}}msteam​=Latent heat of steam+Sensible heat (if any)Q​

3. Simplified example: Assumptions:

   • It is necessary to heat 1 cubic meter (1000 kg) of water from 20°C to 80°C (without vaporizing).
   • The steam is 0.1 MPa saturated steam, and only the latent heat (2257 kJ/kg) is considered.
   • Heat exchange efficiency is 100% (efficiency loss needs to be considered in practice).

   Calorie requirements:

   $$Q=1000\cdot 4.18\cdot (80-20)=250,800\text{kJ}$$

   Required steam volume:

   m steam = 250,8002257≈111 kg m_{\text{steam}} = \frac{250,800}{2257} \approx 111 \, \text{kg}msteam​=2257250,800​≈111kg

   Therefore, approximately 111 kg Steam.

4. Practical considerations:

   • Heat transfer efficiency: In the actual system, the heat exchanger efficiency may be 80%-95%, and additional steam volume is required.
   • Steam state: The latent heat and sensible heat of high-pressure or superheated steam are different, and it is necessary to look up a table (such as a steam table) based on specific parameters.
   • Heat loss： Heat dissipation from pipes and equipment will increase steam demand.
   • Condensate recovery：If the heat of condensed water is recovered, the steam consumption can be reduced.