Question

An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate between the temperature limits of \(86^{\circ} \mathrm{F}\) at the ocean surface and \(41^{\circ} \mathrm{F}\) at a depth of \(2100 \mathrm{ft}\). About 13,300 gpm of cold seawater was to be pumped from deep ocean through a 40 -indiameter pipe to serve as the cooling medium or heat sink. If the cooling water experiences a temperature rise of \(6^{\circ} \mathrm{F}\) and the thermal efficiency is 2.5 percent, determine the amount of power generated. Take the density of seawater to be \(64 \mathrm{lbm} / \mathrm{ft}^{3}\)

Short answer

Using the given information, we have calculated the power generated by the OTEC power plant to be 17,895 kW. The key steps in solving this problem were converting temperatures to Kelvin, calculating the heat transfer rate, and then calculating the power generated using the given thermal efficiency.

Step-by-step solution

Convert temperature from Fahrenheit to Kelvin

To work with the given temperature constraints, we first need to convert the temperatures from Fahrenheit to Kelvin. To convert from Fahrenheit to Kelvin, use the following formula: $$ K = \frac{5}{9}(^{\circ}\mathrm{F} - 32) + 273.15 $$ Converting the temperatures: $$ T_{1} = 86^{\circ}\mathrm{F} = \frac{5}{9}(86 - 32) + 273.15 = 303.15\thinspace\mathrm{K} $$ $$ T_{2} = 41^{\circ}\mathrm{F} = \frac{5}{9}(41 - 32) + 273.15 = 278.15\thinspace\mathrm{K} $$

Calculate the heat transfer rate from the temperature rise

The heat transfer rate, \(Q\), from the temperature rise of the cooling water can be calculated as: $$ Q = m\thinspace c_p \thinspace\Delta T $$ Where \(m\) is the mass flow rate of seawater, \(c_p\) is the specific heat capacity of seawater (it can be assumed to be equal to the specific heat capacity of water which is approximately \(4.18\thinspace\mathrm{kJ/kgK}\)), and \(\Delta T\) is the temperature difference of the cooling water. First, we need to convert the given volume flow rate (13,300 gpm) to mass flow rate: $$ m = \rho\thinspace Q_v $$ Where \(\rho\) is the density of seawater converted to \(\mathrm{kg/m^3}\) and \(Q_v\) is the volume flow rate converted to \(\mathrm{m^3/s}\): $$ \rho = \frac{64\thinspace\mathrm{lbm/ft^3}}{0.06243} = 1025\thinspace\mathrm{kg/m^3} $$ $$ Q_v = \frac{13,300\thinspace\mathrm{gpm}}{264.172} = 50.34\thinspace\mathrm{m^3/s} $$ Now, we can calculate the mass flow rate: $$ m = (1025\thinspace\mathrm{kg/m^3})(50.34\thinspace\mathrm{m^3/s}) = 51548\thinspace\mathrm{kg/s} $$ The temperature rise of cooling water is given: $$ \Delta T = 6^{\circ}\mathrm{F} = \frac{5}{9}(6) = 3.33\thinspace\mathrm{K} $$ Now, we can calculate the heat transfer rate: $$ Q = (51548\thinspace\mathrm{kg/s})(4.18\thinspace\mathrm{kJ/kgK})(3.33\thinspace\mathrm{K}) = 715799\thinspace\mathrm{kW} $$

Calculate the power generated

The power generated can be found by multiplying the heat transfer rate by the thermal efficiency: $$ P = \eta \thinspace Q $$ Where \(\eta\) is the thermal efficiency, which is given as 2.5 percent (\(0.025\) in decimal form). Thus, the power generated is: $$ P = (0.025)(715799\thinspace\mathrm{kW}) = 17895\thinspace\mathrm{kW} $$ The power generated by the OTEC power plant is \(17,895\thinspace\mathrm{kW}\).