How Do You Calculate the Necessary Collector Area for a Solar Water Heater?

[Ishida52134]

Homework Statement

In a solar water heater, energy from the Sun is gathered by rooftop collectors, which circulate water through tubes in the collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes. Assuming that 80% of the incident solar energy gets lost (i.e; it does not go into heating the water), what collector area is necessary to take water from a 200-L tank and raise its temperature from 20 C to 40 C in 1.0 h? The intensity of incident sunlight is 700 W/m^2.

Homework Equations

Q = mcdelta t

delta t = iR

The Attempt at a Solution

 

[SammyS]

Welcome to PF.

What have you tried?

Where are you stuck?

 

[Ishida52134]

I've tried applying those formulas and combinations of them. I don't really understand the problem.

 

[SammyS]

What is the definition of power?

How much energy does it take to raise the temperature of 1 kg of H₂O by 1°C

 

[asteriatic]

To solve this problem, we can use the equation Q = mcdelta t, where Q is the energy absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and delta t is the change in temperature. We can also use the equation delta t = iR, where delta t is the change in temperature, i is the intensity of incident sunlight, and R is the collector area.

First, we need to calculate the energy absorbed by the water. We know that the specific heat capacity of water is 4.186 J/gC, so we can convert the volume of water (200 L) to mass using the density of water (1 g/mL). This gives us a mass of 200,000 g. Plugging in the values, we get Q = (200,000 g)(4.186 J/gC)(40 C - 20 C) = 16,744,000 J.

Next, we can calculate the change in temperature using the equation delta t = iR. We know that the intensity of incident sunlight is 700 W/m^2, and we want to raise the temperature by 20 C in 1.0 h, which is equivalent to 3600 seconds. Plugging in the values, we get 20 C = (700 W/m^2)(R)(3600 s), which gives us a collector area of 0.007 m^2.

In conclusion, to raise the temperature of water in a 200-L tank from 20 C to 40 C in 1.0 h with an 80% energy loss, we would need a collector area of 0.007 m^2.