Question 2 of 7: Material and Energy Balances — Steady and Unsteady State

Question 2: Material and Energy Balances — Steady and Unsteady State (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Cooling-Water Temperature Rise

Given. Electrical output and cooling-water data for the plant:

Find. The elevated river temperature just downstream of the cooling-water discharge.

Approach. An overall energy balance on the fuel input splits the total fuel energy into useful electrical output and waste heat, and a further split routes the waste heat between the stack and the cooling water; a sensible-heat balance on the cooling water then converts the heat it carries into a temperature rise.

1. Total fuel energy from plant efficiency. With $\eta = P_{elec}/Q_{fuel}$, $$Q_{fuel} = \frac{P_{elec}}{\eta} = \frac{10{,}000}{0.35} = 28{,}571.4\ \text{MW}.$$
2. Waste heat rejected. The balance of the fuel energy not converted to electricity is waste heat: $$Q_{waste} = Q_{fuel} - P_{elec} = 28{,}571.4 - 10{,}000 = \boxed{18{,}571.4\ \text{MW}}.$$
3. Split between stack and cooling water. 25% of the waste heat is stack loss and the remaining 75% is carried away by the cooling water: $$Q_{water} = 0.75\,Q_{waste} = 0.75(18{,}571.4) = \boxed{13{,}928.6\ \text{MW}}.$$
4. Cooling-water mass flow rate. With river density $\rho = 1000\ \text{kg/m}^3$, $$\dot{m} = \rho Q = (1000)(100) = 100{,}000\ \text{kg/s}.$$
5. Temperature rise from the sensible-heat balance. $Q_{water} = \dot{m} c_p \Delta T$, so $$\Delta T = \frac{Q_{water}}{\dot{m}\,c_p} = \frac{13{,}928.6\times10^{6}}{(100{,}000)(5000)} = \boxed{27.9\,{}^{\circ}\text{C}}.$$
6. Downstream temperature. Adding the rise to the inlet temperature, $$T_{downstream} = T_{in} + \Delta T = 10 + 27.9 = \boxed{37.9\,{}^{\circ}\text{C}}.$$

(ii) Carbonate Buffering of an Acid Spill

The lake's carbonate system rests on the coupled air–water–solid equilibria $$\text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq) + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3^{*} \rightleftharpoons \text{H}^+ + \text{HCO}_3^{-} \rightleftharpoons 2\text{H}^+ + \text{CO}_3^{2-}.$$ An acid spill that drives the lake to pH < 2 delivers a large excess of H⁺. That excess is first neutralized by the lake's existing alkalinity — the H⁺ reacts with bicarbonate and carbonate to form carbonic acid, which decomposes to dissolved CO₂ and ultimately outgasses to the atmosphere across the air–water interface, so the immediate response is a drop in alkalinity and a rise in dissolved/evolved CO₂ rather than an immediate collapse of pH. Once the dissolved bicarbonate/carbonate reservoir is exhausted, however, the buffering capacity is gone and further acid input drives pH down sharply — a classic titration-curve behaviour with a steep endpoint once alkalinity reaches zero.

If the lakebed or watershed contains limestone (CaCO₃), a second, slower buffering pathway operates: the excess carbonic acid dissolves the solid limestone, $$\text{CaCO}_3(s) + \text{H}_2\text{CO}_3^{*} \rightleftharpoons \text{Ca}^{2+} + 2\,\text{HCO}_3^{-},$$ which regenerates bicarbonate alkalinity over the timescale of weeks to months (limited by the limestone surface area and contact time), partially restoring the lake's acid-neutralizing capacity. A lake without a limestone-bearing watershed (e.g., on Canadian Shield granite) has little of this secondary buffering and is far more vulnerable to a sustained pH depression than one underlain by carbonate rock.

(iii) Unsteady, Non-Uniform Flow

Where depth and velocity vary with both time and position, the flow is governed by the unsteady, gradually-varied (Saint-Venant) equations — a continuity equation and a momentum (dynamic) equation written for the conduit or channel cross-section, expressed in terms of discharge or velocity and depth as functions of position $x$ and time $t$. Given the complete current state of the system (geometry, roughness, and the instantaneous $x$, $t$, velocity and depth/flow at every node), these coupled partial differential equations are integrated forward in time, most commonly by the method of characteristics (which converts the PDE pair into ordinary differential equations along characteristic curves whose slope is the local wave celerity, $c = \sqrt{gA/T}$ for open channels, or the acoustic wave speed for pressurized pipe transients/water hammer) or by an explicit/implicit finite-difference scheme (e.g., the Preissmann implicit scheme used in unsteady open-channel models such as HEC-RAS unsteady). Both approaches use the current state as the initial condition and propagate the solution forward using the boundary conditions (upstream inflow hydrograph, downstream stage or rating curve) to predict future depths, velocities and flow rates throughout the system.