Question

A solar pond consists of a thin layer of fresh water floating on top of a denser layer of salt water. When the salty layer absorbs sunlight it warms up and much of that heat is held there by the insulating effect of the fresh water above it (without the fresh water, the warm salt water would rise to the surface and dissipate its heat to the atmosphere). A solar pond can easily be 100°C above the ambient temperature.

a. What is the maximum efficiency of a heat engine operating off of a 120°C solar pond on a 20°C day?

b. If a real engine is able to achieve half the efficiency of a Carnot engine, how many kilowatt hours of electricity could be generated per day from a 100 m × 100 m pond that captures and stores 50% of the 7 kWh/m² solar radiation striking the surface?

c. For a house that requires 500 kWh per (30-day) month, what area of pond would be needed ?

Answer 1

a. The maximum efficiency is 83.33%. b. 8,333 kWh of electricity could be generated per day. c. 39.26 m² of pond area would be needed for a house requiring 500 kWh per month.

Step-by-step explanation:

a. The maximum efficiency of a heat engine operating off of a 120°C solar pond on a 20°C day can be calculated using the equation:

Efficiency = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. In this case, the temperature difference would be (120 - 20) = 100°C. Plug in the values into the equation:

Efficiency = 1 - (20/120) = 0.83333333333333, or 83.33%.

b. If a real engine is able to achieve half the efficiency of a Carnot engine, the efficiency would be 0.5 * 0.83333333333333 = 41.67%. To calculate the amount of electricity generated per day from a 100 m × 100 m pond that captures and stores 50% of the 7 kWh/m² solar radiation, use the equation:

Electricity generated = Area of pond * Solar radiation * Efficiency

Electricity generated = (100 m * 100 m) * (7 kWh/m²) * (0.5) * (0.4167) = 8,333 kWh

c. To calculate the area of pond needed for a house that requires 500 kWh per (30-day) month, use the equation:

Area of pond = (Required electricity / (Solar radiation * Efficiency)) / Number of days

Area of pond = (500 kWh / (7 kWh/m² * 0.5 * 0.4167)) / 30 = 39.26 m²