In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is 28.0% (that is, 72% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 450 L of water in the tank from 18°C to 50°C in 2.8 h when the intensity of incident sunlight is 520 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.
To find the collector area necessary to raise the temperature of 450 L of water in the tank from 18°C to 50°C in 2.8 hours, we need to calculate the total energy required and then divide it by the energy received per square meter.
Step 1: Convert the volume of water from liters to kilograms:
Given:
Density of water = 1.00 g/cm^3
1 L = 1000 cm^3
Volume of water = 450 L
450 L * 1000 cm^3/L = 450,000 cm^3
450,000 cm^3 * 1.00 g/cm^3 = 450,000 g
450,000 g * 1 kg/1000 g = 450 kg
Therefore, the mass of water is 450 kilograms.
Step 2: Calculate the energy required to raise the temperature of the water:
Given:
Specific heat of water (c) = 4186 J/kg·K
Change in temperature (ΔT) = 50°C - 18°C = 32°C
Time taken (t) = 2.8 hours = 2.8 * 3600 seconds = 10,080 seconds
Energy required (Q) = mass * specific heat * change in temperature
Q = 450 kg * 4186 J/kg·K * 32°C = 5,651,200 J
Step 3: Calculate the incident solar energy received:
Given:
Incident sunlight intensity = 520 W/m^2
Time taken (t) = 2.8 hours = 10,080 seconds
Incident energy received (E) = intensity * collector area * time
E = 520 W/m^2 * A * 10,080 s
Step 4: Calculate the collector area:
Given:
Efficiency of the system = 28% = 0.28
Total energy required = energy received * efficiency
5,651,200 J = 520 W/m^2 * A * 10,080 s * 0.28
Simplifying the equation:
A = 5,651,200 J / (520 W/m^2 * 10,080 s * 0.28)
A ≈ 7.08 m^2
Therefore, the collector area necessary to raise the temperature of 450 L of water in the tank from 18°C to 50°C in 2.8 hours is approximately 7.08 square meters.
To find the collector area necessary to raise the temperature of water in the tank, we can use the following steps:
Step 1: Calculate the energy required to heat the water in the tank.
The energy required can be calculated using the following equation:
Energy = mass × specific heat × temperature change
Given:
Mass of water (m) = 450 L
Density of water = 1.00 g/cm^3 = 1000 kg/m^3 (since 1 cm^3 = 1 mL = 1 g)
Specific heat of water = 4186 J/kg·K
Initial temperature (T1) = 18°C = 18 + 273 = 291 K
Final temperature (T2) = 50°C = 50 + 273 = 323 K
Energy = 1000 kg × 4186 J/kg·K × (323 K - 291 K)
Step 2: Calculate the incident solar energy.
Incident solar energy can be calculated using the following equation:
Solar energy = incident sunlight intensity × time
Given:
Incident sunlight intensity = 520 W/m^2
Time (t) = 2.8 h = 2.8 × 3600 s
Solar energy = 520 W/m^2 × 2.8 × 3600 s
Step 3: Calculate the energy received by the collector (taking efficiency into account).
Energy received by the collector = incident solar energy × efficiency
Energy received by the collector = (520 W/m^2 × 2.8 × 3600 s) × 0.28
Step 4: Calculate the collector area.
Collector area can be calculated using the following equation:
Collector area = energy required / energy received by the collector
Collector area = Energy / (incident solar energy × efficiency)
Now, plug in the previously calculated values to find the collector area.