It has been proposed to use the thermal gradient of the ocean to drive a heat engine. Suppose that at a certain location the water temperature is $22°\mathrm{C}$at the ocean surface and $4°\mathrm{C}$at the ocean floor.

(a) What is the maximum possible efficiency of an engine operating between these two temperatures?

(b) If the engine is to produce $1\mathrm{GW}$of electrical power, what minimum volume of water must be processed (to suck out the heat) in every second?

The maximum possible efficiency of heat engine.

Given:

The temperature of hot reservoir $=22°\mathrm{C}$

The temperature of cold reservoir $=4°\mathrm{C}$

Formula:

The expression for the efficiency of heat engine is as follows:

${\mathrm{e}}_{\max }=1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}$

Here, ${\mathrm{T}}_{\mathrm{c}}$is temperature of cold reservoir

${\mathrm{T}}_{\mathrm{h}}$is temperature of hot reservoir.

Calculation:

The temperature of hot reservoir Kelvin is:

role="math" localid="1648159148570" $$\begin{gathered} {\mathrm{T}}_{\mathrm{h}}=273+22=295\mathrm{K} \\ {\mathrm{T}}_{\mathrm{c}}=273+4=277\mathrm{K} \end{gathered}$$

Now Substituting the values of $T_c$and $T_h$in the above expression

$$\begin{gathered} e_{\max }=1-\frac{277\mathrm{K}}{295\mathrm{K}} \\ =0.061 \\ =6.1\backslash \% \end{gathered}$$

Hence, the maximum possible efficiency of heat engine is$6.1\%$

Minimum volume of water that must be processed in every second to produce$1\mathrm{GW}$of electrical power.

Since the temperature of ocean water drops as the engine extracts heat from it, the engine's efficiency varies.

The temperature differential between the cold and hot reservoirs would be as follows:

$22°\mathrm{C}-4°\mathrm{C}=18°\mathrm{C}$

The temperature difference between the cold and warm water is equivalent to half of the temperature difference at equilibrium.

That is

$\Delta \mathrm{T}=\frac{18°\mathrm{C}}{2}=9°\mathrm{C}$

The average temperature of reservoirs is $18°\mathrm{C}$and $9°\mathrm{C}$Thus, the efficiency of heat engine is

$$\begin{gathered} \mathrm{e}=1-\frac{(9+273)\mathrm{K}}{(18+273)\mathrm{K}} \\ =0.0309 \\ =3.09\% \end{gathered}$$

The heat energy removed from each kilogram of the warm water is

${\mathrm{Q}}_{\mathrm{c}}={\mathrm{mC}}_{\mathrm{w}}\Delta \mathrm{T}$

Where,

$\mathrm{m}$is mass of the water,

${\mathrm{C}}_{\mathrm{w}}$is the specific heat of water, and

$\Delta \mathrm{T}$is change in temperature.

Now Substitute the values of $\Delta T$and $C_w$respectively

$$\begin{gathered} \frac{\mathrm{Q}}{\mathrm{m}}=\left(4186\mathrm{J}/\mathrm{kg}°\mathrm{C}\right)\left(9°\mathrm{C}\right) \\ =37.674\times {10}^3\mathrm{J}/\mathrm{kg} \end{gathered}$$

But, the efficiency of the engine is $3.09\%$.

Thus, the heat energy produced per kilograms is as follows:

$=1.164\times {10}^4\mathrm{J}/\mathrm{kg}$

The work done per each second is called its power.

$\mathrm{P}=\frac{{\displaystyle \mathrm{W}}}{{\displaystyle \mathrm{t}}}$

Here, t is the time interval.

Substitute $1GW$for P

$\begin{array}{r}\frac{W}{t}=1GW\left(\frac{{10}^{9}W}{1GW}\right)\\ \frac{W}{t}={10}^{9}W\\ \frac{W}{t}={10}^9\mathrm{J}\cdot {\mathrm{s}}^{-1}\end{array}$

Thus, the total mass of the water per second is as follows:

$\frac{\mathrm{m}}{\mathrm{t}}=\frac{{10}^9\mathrm{J}·{\mathrm{s}}^{-1}}{1.164\times {10}^4\mathrm{J}·{\mathrm{kg}}^{-1}}=8.6\times {10}^4\mathrm{kg}·{\mathrm{s}}^{-1}$

The total amount of water in cubic meters per second is

$8.6\times {10}^4\mathrm{kg}·{\mathrm{s}}^{-1}\left(\frac{{10}^{-3}{\mathrm{m}}^3}{1\mathrm{kg}}\right)=86{\mathrm{m}}^3·{\mathrm{s}}^{-1}$

Hence the total amount of water in cubic meters per second would be

$86{\mathrm{m}}^3·{\mathrm{s}}^{-1}$.