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Steam Heating Process

Reference data and engineering information about steam heating process for material properties applications.

steamheatingprocess

Overview

Engineering reference data for Steam Heating Process in material science and properties.

Key Formulas

Stress

σ=FA\sigma = \frac{F}{A}

Force per unit area.

Strain

ε=ΔLL0\varepsilon = \frac{\Delta L}{L_0}

Change in length per original length.

Hooke's Law

σ=Eε\sigma = E \varepsilon

Stress proportional to strain in elastic region.

Thermal Expansion

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

Length change due to temperature.

Variables

SymbolDescriptionUnit
σ\sigmaStressPa
ε\varepsilonStrain
EEYoung's modulusPa
α\alphaThermal expansion coefficient1/°C
ΔT\Delta TTemperature change°C

Batch Heating Example

In non-flow applications, a fixed mass of product is heated by a steam coil or jacket.

Problem: Heat 50 kg of water (cp = 4.19 kJ/kg·°C) from 35°C to 100°C using steam at 5 bar (6 bar abs) over 20 minutes.

Calculations:

  1. Heat transfer rate: P=mcpΔTt=(50 kg)(4.19 kJ/kg⋅°C)(100°C35°C)(1200 s)=11.35 kWP = \frac{m \cdot c_p \cdot \Delta T}{t} = \frac{(50\ \text{kg}) \cdot (4.19\ \text{kJ/kg·°C}) \cdot (100°C - 35°C)}{(1200\ \text{s})} = 11.35\ \text{kW}

  2. Steam consumption rate: Using evaporation energy for steam at 6 bar abs, he=2085 kJ/kgh_e = 2085\ \text{kJ/kg}. ms=Phe=11.35 kW2085 kJ/kg=0.0055 kg/s=19.6 kg/hm_s = \frac{P}{h_e} = \frac{11.35\ \text{kW}}{2085\ \text{kJ/kg}} = 0.0055\ \text{kg/s} = 19.6\ \text{kg/h}

Continuous Flow Heating Example

In heat exchangers, the fluid is heated continuously as it flows over the steam-heated surface.

Problem: Heat water flowing at 3 l/s from 10°C to 60°C using steam at 8 bar (9 bar abs). Water density ≈ 1 kg/l.

Calculations:

  1. Heat flow rate: P=cpΔTm˙=(4.19 kJ/kg⋅°C)(60°C10°C)(3 kg/s)=628.5 kWP = c_p \cdot \Delta T \cdot \dot{m} = (4.19\ \text{kJ/kg·°C}) \cdot (60°C - 10°C) \cdot (3\ \text{kg/s}) = 628.5\ \text{kW}

  2. Steam consumption rate: Using evaporation energy for steam at 9 bar abs, he=2030 kJ/kgh_e = 2030\ \text{kJ/kg}. ms=Phe=628.5 kW2030 kJ/kg=0.31 kg/s=1115 kg/hm_s = \frac{P}{h_e} = \frac{628.5\ \text{kW}}{2030\ \text{kJ/kg}} = 0.31\ \text{kg/s} = 1115\ \text{kg/h}

Key Process Considerations

  • Direct vs. Indirect Heating: Direct steam injection transfers energy instantly, as all steam condenses. Indirect heating via a heat exchanger depends on the heat transfer coefficient and the temperature difference (ΔT\Delta T) between steam and the product.
  • Effect of Pressure: Increasing steam pressure increases its temperature, which can increase the ΔT\Delta T and improve heat transfer rates, potentially reducing heat-up time.
  • Efficiency Trade-off: Higher pressure steam may reduce heating time but could lead to higher heat losses. System configuration determines whether overall steam consumption increases or decreases.

References