9.3 Energy Resources

Step 1: Setup — Potential Energy of a Water Column

Consider a tidal basin of surface area $A_b$ impounded behind a barrage. At high tide the water surface inside is elevated by a tidal range $R$ above the low-tide level outside. As the basin empties through turbines, a thin horizontal slab of water at height $z$ above the exterior level has mass:

$$dm = \rho \, A_b \, dz$$

Step 2: Potential Energy of Each Slab

Each slab falls through height $z$ (its center of mass drops from $z$ to the exterior level). Its potential energy is:

$$dE = dm \cdot g \cdot z = \rho g A_b z \, dz$$

Step 3: Integrate Over the Tidal Range

Integrate from $z = 0$ (low-tide level) to $z = R$ (high-tide level):

$$E = \int_0^R \rho g A_b z \, dz = \rho g A_b \left[\frac{z^2}{2}\right]_0^R = \frac{1}{2}\rho g A_b R^2$$

Step 4: Average Power Over a Tidal Cycle

A semidiurnal tidal cycle has period $T \approx 12.42$ hours ($\approx 44712$ s). The energy is released once per ebb (or twice for ebb-flood generation). Average power for single-effect operation:

$$\bar{P} = \frac{E}{T/2} = \frac{\rho g A_b R^2}{T}$$

Step 5: Practical Power with Turbine Efficiency

Including turbine-generator efficiency $\eta_t$ and accounting for two ebb tides per day:

$$P_{\text{annual}} = \eta_t \cdot \rho g A_b \overline{R^2} \cdot \frac{n_{\text{cycles}}}{t_{\text{year}}}$$

Since the tidal range varies with spring-neap cycles, $\overline{R^2}$ must be averaged over the lunar month. Typically $\overline{R^2} \approx 0.5 R_{\text{spring}}^2$.

Step 6: Numerical Example (La Rance)

For La Rance: $A_b = 22 \times 10^6$ m$^2$, $R = 8$ m, $\rho = 1025$ kg/m$^3$:

$$E = \frac{1}{2}(1025)(9.81)(22 \times 10^6)(8)^2 \approx 7.1 \times 10^{12} \text{ J} = 7.1 \text{ TJ per cycle}$$

Step 1: Energy Density of a Monochromatic Wave

For a sinusoidal deep-water wave of amplitude $a$ and wave number $k$, the total energy per unit horizontal area (kinetic + potential) is:

$$E = \frac{1}{2}\rho g a^2$$

Step 2: Group Velocity in Deep Water

In deep water the dispersion relation is $\omega^2 = gk$. The phase speed is $c = \omega/k = g/\omega$ and the group velocity (energy propagation speed) is half the phase speed:

$$c_g = \frac{d\omega}{dk} = \frac{1}{2}\sqrt{\frac{g}{k}} = \frac{c}{2} = \frac{gT}{4\pi}$$

Step 3: Power Flux for a Regular Wave

The energy flux (power per unit crest width) is energy density times group velocity. Substituting $a = H/2$ (wave height $H = 2a$):

$$P = E \cdot c_g = \frac{1}{2}\rho g \left(\frac{H}{2}\right)^2 \cdot \frac{gT}{4\pi} = \frac{\rho g^2 H^2 T}{32\pi}$$

Step 4: Extend to Irregular Seas

For a wave spectrum $S(f)$, the significant wave height is $H_s = 4\sqrt{m_0}$ where $m_0 = \int S(f)\,df$. The energy period is $T_e = m_{-1}/m_0$. The power flux generalizes to:

$$P_w = \rho g \int_0^\infty c_g(f) S(f) \, df = \frac{\rho g^2}{64\pi} H_s^2 T_e$$

Step 5: Numerical Approximation

Substituting $\rho = 1025$ kg/m$^3$ and $g = 9.81$ m/s$^2$:

$$P_w = \frac{1025 \times 9.81^2}{64\pi} H_s^2 T_e \approx 490 \, H_s^2 T_e \;\; \text{W/m} \approx 0.49 \, H_s^2 T_e \;\; \text{kW/m}$$

For the North Atlantic with $H_s = 3$ m and $T_e = 9$ s: $P_w \approx 0.49 \times 9 \times 9 \approx 40$ kW/m.

Step 6: Capture Width and Device Efficiency

A wave energy converter of characteristic width $D$ captures power $P_{\text{cap}}$. The capture width ratio (hydrodynamic efficiency) is:

$$\eta = \frac{P_{\text{cap}}}{P_w \cdot D}, \quad \text{CW} = \eta D = \frac{P_{\text{cap}}}{P_w}$$

Point absorbers can have capture widths exceeding their physical size ($\eta > 1$) near resonance, analogous to the effective cross-section of an antenna.

Step 1: The Second Law and Carnot's Theorem

A heat engine operating between a hot reservoir at temperature $T_h$ and a cold reservoir at $T_c$ (both in Kelvin) receives heat $Q_h$ and rejects $Q_c$. By the second law, for a reversible engine the entropy change of the universe is zero:

$$\Delta S_{\text{universe}} = -\frac{Q_h}{T_h} + \frac{Q_c}{T_c} = 0 \quad \Longrightarrow \quad \frac{Q_c}{Q_h} = \frac{T_c}{T_h}$$

Step 2: Derive Maximum Efficiency

The work output is $W = Q_h - Q_c$. The thermal efficiency is:

$$\eta_{\text{Carnot}} = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_c}{T_h} = \frac{T_h - T_c}{T_h}$$

Step 3: Apply to OTEC Conditions

Tropical surface water: $T_h \approx 28°\text{C} = 301.15$ K. Deep water at 1000 m: $T_c \approx 4°\text{C} = 277.15$ K. The Carnot limit is:

$$\eta_{\text{Carnot}} = 1 - \frac{277.15}{301.15} = \frac{24}{301.15} \approx 7.97\%$$

Step 4: Practical OTEC Efficiency

Real OTEC systems achieve only 40–60% of the Carnot limit due to finite heat-exchanger temperature differences ($\Delta T_{\text{pinch}}$), pump parasitic loads (~30% of gross power), and working-fluid irreversibilities:

$$\eta_{\text{practical}} \approx (0.4\text{--}0.6) \times \eta_{\text{Carnot}} \approx 3\text{--}5\%$$

Step 5: Gross Power Output

The thermal power input is set by the warm-water mass flow rate $\dot{m}$, specific heat $c_p$, and the temperature drop across the evaporator $\Delta T_e$:

$$P_{\text{gross}} = \eta_{\text{practical}} \cdot \dot{m} c_p \Delta T_e$$

Step 6: Net Power and Parasitic Losses

Pumping cold water from 1000 m depth requires significant power. The net output is:

$$P_{\text{net}} = P_{\text{gross}} - P_{\text{pump,warm}} - P_{\text{pump,cold}} - P_{\text{aux}}$$

The cold-water pipe pumping power scales as $P_{\text{pump}} \propto \dot{m}_c g H / \eta_p$, where $H \approx 1000$ m is the pipe depth. Typically $P_{\text{net}} \approx 0.5 P_{\text{gross}}$, giving a net efficiency of $\sim 1.5\text{--}2.5\%$.