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.
The intensity of sun light falling on the earth = I'
Let area of the plate = A
Power received by the plate, P = I'A
Efficiency of the over all system = 28 %
The power given to the tank is given by
P = 26% * I' A = 0.28 I'A...........(1)
m= mass of 450 L of water
S =specific heat of water = 4186 J/kg·K
4.186 J/g.K
ΔT = temparature difference = 32 deg C
The energy needed to raise the temp of water from 18 to 50 C is
Q = mSΔT = J
4.186*(450*10^3)*32
=60278400 J
power in put needed
P ' = Q / 1.0 * 60*60 sec
P =60278400 / 2.8*60*60 sec
P' = 5980 W -----(2)
From the equations (1) and (2) we have
0.28 I'A = 5980
A =5980/(0.28*520)
A = 41 m^2
To find the collector area required to raise the temperature of the water in the tank, we need to calculate the total energy needed to heat the water and then determine the collector area based on the efficiency of the system.
1. Calculate the mass of water in the tank:
- Volume of water = 450 L
- Density of water = 1.00 g/cm³ = 1000 kg/m³
- Mass of water = Volume * Density = 450 L * 1000 kg/m³ = 450,000 kg
2. Calculate the energy needed to raise the temperature of the water:
- Initial temperature of water = 18°C
- Final temperature of water = 50°C
- Change in temperature = Final temperature - Initial temperature = 50°C - 18°C = 32°C
- Specific heat of water = 4186 J/kg·K
- Energy needed = Mass of water * Specific heat * Change in temperature
= 450,000 kg * 4186 J/kg·K * 32°C
3. Determine the incident energy from the sunlight:
- Intensity of incident sunlight = 520 W/m²
- Time = 2.8 hours
- Incident energy = Intensity of sunlight * Collector area * Time
4. Calculate the efficiency of the overall system:
- Efficiency = 0.28 = 28%
5. Set the incident energy equal to the energy needed and solve for the collector area:
- Incident energy = Energy needed / Efficiency
- Intensity of sunlight * Collector area * Time = Mass of water * Specific heat * Change in temperature / Efficiency
- Collector area = (Mass of water * Specific heat * Change in temperature * Time) / (Intensity of sunlight * Efficiency)
Now we can substitute the given values and calculate the collector area:
Collector area = (450,000 kg * 4186 J/kg·K * 32°C * 2.8 h) / (520 W/m² * 0.28)
After calculating the expression above, you will get the required collector area in square meters.
To solve this problem, we need to calculate the energy required to raise the temperature of the water and then determine the area of the collector needed to provide that energy.
Step 1: Calculate the mass of the water.
The density of water is given as 1.00 g/cm3. The volume of water is given as 450 L. We can convert volume to mass using the formula:
mass = density x volume
First, convert the volume from liters to cubic meters:
volume = 450 L = 0.45 m^3
Now, calculate the mass:
mass = 1.00 g/cm3 x 1000 kg/m^3 x 0.45 m^3
mass = 450 kg
Step 2: Calculate the energy required to raise the temperature.
The specific heat of water is given as 4186 J/kg·K, and the change in temperature is from 18°C to 50°C.
The equation to calculate the energy is:
energy = mass x specific heat x temperature change
temperature change = final temperature - initial temperature
= 50°C - 18°C = 32°C
Now, calculate the energy required:
energy = 450 kg x 4186 J/kg·K x 32 K
energy = 596,256,000 J
Step 3: Calculate the incident solar energy.
The intensity of incident sunlight is given as 520 W/m2, and the time is given as 2.8 h.
The equation to calculate the incident solar energy is:
incident solar energy = intensity x area x time
Now, calculate the incident solar energy:
incident solar energy = 520 W/m2 x area x 2.8 h
incident solar energy = 1456 W·h/m2 x area
Step 4: Calculate the usable energy from the incident solar energy.
The overall system efficiency is given as 28%, which means 72% (100% - 28%) of the incident solar energy is lost.
The equation to calculate the usable energy is:
usable energy = incident solar energy x efficiency
Now, calculate the usable energy:
usable energy = 1456 W·h/m2 x area x 0.28
usable energy = 407.68 W·h/m2 x area
Step 5: Equate the usable energy to the energy required.
Since the usable energy is the energy provided by the solar collector and the energy required is the energy needed to raise the water temperature, we can equate these two values to find the required collector area.
407.68 W·h/m2 x area = 596,256,000 J
To convert watt-hours (W·h) to joules (J), multiply by 3600:
407.68 W·h/m2 x area = 596,256,000 J x 3600
Simplifying:
area = (596,256,000 J x 3600) / (407.68 W·h/m2)
area ≈ 5,273,920,000,000 / 407.68
area ≈ 12,941,123.06 m2
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 h is approximately 12,941,123.06 square meters.