Stratified hot-water storage for dairy process heat
A dairy pasteurisation plant installs a 120 m³ stratified hot-water tank charged by a solar collector field. The tank operates between 95°C and 55°C. Water density is 975 kg/m³ and specific heat is 4.19 kJ/kg·K. The plant needs 2.5 MW of process heat for 2 hours each morning. The tank has 140 m² of external surface, an insulation overall heat-transfer coefficient of 4 W/m²·K, and sits in an ambient at 20°C. It stands idle for 16 hours between charging and use. The existing natural-gas boiler is 85% efficient, gas has an LHV of 36 MJ/m³ and costs INR 45/m³. The plant operates 300 cycles per year. The engineer must check the storage adequacy, quantify the standing loss and evaluate an insulation upgrade to U = 0.8 W/m²·K costing INR 1.2 million.
1. Calculate the usable thermal storage capacity between 95°C and 55°C and check the 2-hour process demand.
Read the worked answer
Tank mass = 120 m³ × 975 kg/m³ = 117,000 kg.
Temperature difference = 95 − 55 = 40 K.
Stored energy = 117,000 × 4.19 × 40 kJ = 19,609,200 kJ = 19.61 GJ.
Convert to MWh: 19.61 GJ/3.6 = 5.45 MWh.
Process demand = 2.5 MW × 2 h = 5.0 MWh.
The nominal capacity exceeds the demand by 0.45 MWh, or 9%. However, this assumes perfect stratification and full use of the 40 K span. Dead volume, mixing and the inability to draw down to 55°C at the outlet can easily consume the 9% margin. The tank is therefore only marginally adequate; a 10% usable-volume derating would make it slightly undersized. The design should either increase volume or verify the achievable drawdown with a stratification model.
2. Estimate the standing heat loss over 16 hours and express it as a fraction of storage capacity.
Read the worked answer
Use an average tank temperature of (95 + 55)/2 = 75°C, so ΔT to ambient = 75 − 20 = 55 K.
UA = 4 W/m²·K × 140 m² = 560 W/K.
Heat-loss rate = 560 × 55 = 30,800 W = 30.8 kW.
Energy lost over 16 h = 30.8 kW × 16 h = 492.8 kWh = 0.493 MWh.
Fraction of nominal capacity = 0.493/5.45 = 9.0%.
This is a significant loss: nearly one-tenth of the stored energy is lost while idle. Because the tank cools as it loses heat, the true average temperature and loss will be slightly lower, but the first-order estimate is adequate for screening. The loss also directly increases morning boiler firing, so it has a fuel cost.
3. Evaluate whether the insulation upgrade to U = 0.8 W/m²·K is justified by the gas saving.
Read the worked answer
New heat-loss rate = 0.8 × 140 × 55 = 6,160 W = 6.16 kW.
New loss over 16 h = 6.16 × 16 = 98.6 kWh = 0.355 GJ.
Old loss = 492.8 kWh = 1.774 GJ.
Saved heat per cycle = 1.774 − 0.355 = 1.419 GJ.
Boiler fuel required for that heat = 1.419/0.85 = 1.669 GJ.
Gas volume = 1.669 GJ/0.036 GJ/m³ = 46.4 m³.
Daily saving = 46.4 × INR 45 = INR 2,088.
Annual saving at 300 cycles = INR 626,400.
Simple payback = 1,200,000/626,400 = 1.9 years.
The upgrade is attractive on fuel savings alone, before any carbon or boiler-capacity benefit. The calculation assumes the tank completes 300 equivalent full cycles per year and that the insulation cost is fully incremental. Even with a 20% allowance for installation or reduced cycles, payback remains under about 2.5 years.