steam plant operating on the Rankine cycle is supplied by a solar plant with a solar collector area of 40 000 m2. The plant produces

5 MW of power. The efficiency of the collectors is 65%. The Rankine efficiency is 32%.
Calculate the:
10.1 solar power generated
10.2 time it will take to produce 93,75 MWh of solar energy
10.3 irradiance of the region.

1 answer

To calculate the requested components, we'll use the following definitions and relationships in thermodynamics and solar energy:

10.1 Solar Power Generated

The solar power generated can be calculated using the formula:

\[ P_{\text{solar}} = A \cdot I \cdot \eta_c \]

Where:

  • \(P_{\text{solar}}\) is the solar power generated (W)
  • \(A\) is the collector area (m²)
  • \(I\) is the solar irradiance (W/m²)
  • \(\eta_c\) is the efficiency of the collectors (decimal)

10.2 Time to Produce 93.75 MWh of Solar Energy

First, we need to convert the energy requirement from MWh to Wh:

\[ E = 93.75, \text{MWh} = 93.75 \times 10^6, \text{Wh} \]

The time required to generate this energy can be calculated with:

\[ t = \frac{E}{P_{\text{solar}}} \]

10.3 Irradiance of the Region

The solar irradiance can be derived from the power output of the collectors:

Rearranging the solar power equation gives:

\[ I = \frac{P_{\text{solar}}}{A \cdot \eta_c} \]

Now let’s perform the calculations step-by-step.

Step 1: Calculate Solar Power Output Given:

  • \(\eta_c = 0.65\)
  • \(A = 40,000, m^2\)

We can plug \(P_{\text{solar}}\) into the equation using the Rankine plant's output power and efficiency.

Using the overall efficiency of the solar to electrical is given by:

\[ \eta = \eta_c \cdot \eta_r \]

Where:

  • \(\eta_r\) is the Rankine cycle efficiency (given as 0.32).

The overall efficiency becomes:

\[ \eta = 0.65 \cdot 0.32 = 0.208 \]

Now using the output power from the Rankine cycle:

\[ P_{\text{solar}} = \frac{P_{\text{out}}}{\eta} = \frac{5\text{ MW}}{0.208} = \frac{5 \times 10^6, W}{0.208} \approx 24,038,462, W \text{ or } 24.04, MW \]

Step 2: Calculate Solar Irradiance Now we can find the solar irradiance:

\[ I = \frac{P_{\text{solar}}}{A \cdot \eta_c} = \frac{24,038,462, W}{40,000, m^2 \cdot 0.65} \approx \frac{24,038,462, W}{26,000, m^2} \approx 924.56, W/m^2 \]

Step 3: Calculate Time to Produce 93.75 MWh of Solar Energy Substituting into the equation for \(t\):

First, find \(P_{\text{solar}}\) again to find time, but taking into account the Rankine output as previously calculated:

\[ t = \frac{E}{P_{\text{out}}} = \frac{93,750,000, Wh}{5,000,000, W} \approx 18.75, hours \]

Summary of Results

  • 10.1 Solar power generated: \(24.04, MW\)
  • 10.2 Time it will take to produce \(93.75, MWh\) of solar energy: \(18.75, hours\)
  • 10.3 Irradiance of the region: \(924.56, W/m²\)