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\(\bullet\) Solar collectors. A well-insulated house of moderate size in a temperate climate requires an average heat input rate of 20.0 \(\mathrm{kW}\) . If this heat is to be supplied by a solar collector with an average (night and day) energy input of 300 \(\mathrm{W} / \mathrm{m}^{2}\) and a collection efficiency of \(60.0 \%,\) what area of solar collector is required?

Short Answer

Expert verified
The required solar collector area is approximately 111.11 m².

Step by step solution

01

Understand the Problem

We need to find the area of a solar collector required to meet the average heat input rate of 20.0 kW, given that the solar collector has an average energy input of 300 W/m² and a collection efficiency of 60%.
02

Calculate the Effective Energy Input

Calculate the effective energy collected per square meter by considering the collection efficiency. Since the collector has an efficiency of 60%, the effective energy input per square meter is given by \(0.60 \times 300\, \text{W/m}^2\).
03

Effective energy input calculation

The effective energy input is \(0.60 \times 300\, \text{W/m}^2 = 180\, \text{W/m}^2\). Thus, each square meter of the solar collector effectively captures 180 W.
04

Determine Total Area Required

To find the total collector area required, divide the total energy needed (20.0 kW) by the effective energy input per square meter (180 W/m²). Convert the 20.0 kW to watts for consistency, resulting in 20,000 W.
05

Total Area Calculation

The required area \(A\) is \(\frac{20,000\, \text{W}}{180\, \text{W/m}^2}\).
06

Calculate the Area

Perform the division to find \(A\). So, \(A = \frac{20,000}{180} \approx 111.11\, \text{m}^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Input Rate
The heat input rate refers to the amount of energy required per unit of time to maintain a comfortable environment in the house. In this context, a heat input rate of 20.0 kW is necessary to keep the house warm. But what exactly does this mean? Well, when we say 20.0 kW, we're talking about an energy flow of 20,000 watts continuously supplied to the house.
It is important because it represents the demand that the solar collector needs to meet to ensure the house stays warm, regardless of weather conditions. Understanding how much energy the house demands helps in determining what size of a solar collector is necessary to meet this need effectively. This helps us ensure that our system design will be capable of keeping the house comfortable at all times.
Energy Input Calculation
Energy input calculation plays a crucial role in designing an effective solar collector system. So, why do we need this calculation? Let's take it step by step. First, we look at the average energy input the solar collector receives, which is 300 W/m² in this case. This means that each square meter of the solar collector can capture 300 watts of solar energy under optimal conditions.
However, solar collectors are not perfectly efficient; there are always some losses, which is why we factor in collection efficiency. The formula for effective energy input becomes:
  • Effective energy input = Energy input per square meter × Efficiency
For this problem, multiplying 300 W/m² by the collection efficiency of 60%, gives us an effective energy input of 180 W/m². This tells us how much useful energy each square meter of the solar collector can actually capture.
Solar Collector Area
Calculating the solar collector area is key to ensuring that the right amount of energy is harvested to meet the home's needs. Now, knowing that each square meter captures 180 W effectively, we can determine how much area is needed to collect enough energy to match the house’s heat demand of 20,000 watts (or 20 kW).
Here's how the calculation works: we divide the total energy requirement of the house by the effective energy per square meter of the collector. So, we take:
  • Total area required (A) = Total energy needed ÷ Effective energy per square meter
Substituting the given values: \[A = \frac{20,000\, \text{W}}{180\, \text{W/m}^2} \approx 111.11\, \text{m}^2\] This calculation ensures you have the right amount of surface area to capture the necessary energy to keep your house comfortable.
Collection Efficiency
Collection efficiency is a measure of how well a solar collector converts available solar energy into usable energy. Not all the energy hitting the collector can be converted into heat for the home. Therefore, efficiency accounts for real-world losses and inefficiencies. In this example, the collector's efficiency is 60%.
This means that 60% of the solar energy that hits the collector is transformed into heat. It's critical to evaluate efficiency to predict how much solar energy will be effectively utilized.
  • This efficiency rate factors into the calculation by determining the effective energy input. Thus, knowing your collector's efficiency helps plan the system to ensure you get the energy you need.
A higher efficiency means more energy is captured, which could reduce the collector area required, ultimately affecting the design and cost of your solar system.

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Most popular questions from this chapter

Entropy change due to driving. Premium gasoline pro- duces \(1.23 \times 10^{8} \mathrm{J}\) of heat per gallon when it is burned at a temperature of approximately \(400^{\circ} \mathrm{C}\) (although the amount can vary with the fuel mixture). If the car's engine is 25\(\%\) efficient, three- fourths of that heat is expelled into the air, typically at \(20^{\circ} \mathrm{C}\) . If your car gets 35 miles per gallon of gas, by how much does the car's engine change the entropy of the world when you drive 1.0 mile? Does it decrease or increase it?

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(\mathrm{K}\) and 300 \(\mathrm{K}\) . (a) If the engine receives 6.45 \(\mathrm{kJ}\) of heat energy from the reservoir at 520 \(\mathrm{K}\) in each cycle, how many joules per cycle does it reject to the reservoir at 300 \(\mathrm{K} ?\) (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

.. You are designing a Carnot engine that has 2 mol of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

\(\cdot\) In one cycle, a freezer uses 785 \(\mathrm{J}\) of electrical energy in order to remove 1750 \(\mathrm{J}\) of heat from its freezer compartment at \(10^{\circ} \mathrm{F}\) . (a) What is the coefficient of performance of this freezer? (b) How much heat does it expel into the room during this cycle?

An experimental power plant at the Natural Energy Labo- ratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water tempera- tures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\) , respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and \(\mathrm{L} / \mathrm{h}\) .

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