06 / CHAPTER 12
Thermal studio: warming a basin
Estimate heating, evaporation, and environmental losses in a water feature.
ANCHOR EQUATION
Basin energy balance
In symbols: mc dT/dt = Qin − Qloss
A practical thermal model balances supplied and lost heat.
A warming basin is an energy balance
Water temperature changes according to net heat rate divided by thermal storage capacity. The useful model separates supplied power from evaporation, convection, radiation, and heat carried away by replacement water.
Integrate the lumped balance
For basin mass m and heat capacity c, m c dT/dt = P_in − P_loss(T). If P_loss is approximately constant, ΔT = (P_net t)/(mc). If loss is linear, P_loss = hA(T − T_a), then dT/dt = [P_in − hA(T − T_a)]/(mc), approaching a steady temperature where input equals loss.
A 500 kg basin receives 2 kW net for 30 minutes. ΔT = 2000(1800)/[500(4180)] = 1.72 K. If evaporation removes 1 kg/h, latent loss is about 628 W, reducing the net heating rate substantially.
Lumped temperature needs fast internal mixing
The model assumes one basin temperature. Stratification or a localized heater requires a field model. Evaporation also changes mass, so a long-duration estimate should update m and include incoming water temperature.
Work it: For m = 300 kg, c = 4180 J/(kg·K), P_in = 1500 W, and constant losses 900 W, calculate the time for a 5 K rise. Then estimate the effect of an added 0.5 kg/h evaporation loss.