Students accomplished a series of hands-on experiments on the EPS, working collaboratively within their groups. Lab began with students familiarizing themselves with the physical structure of the EPS and its components. A brief comparison was then made between the physical structure of the EPS and its block diagram.
As a group, students learned how to check the EPS battery pack's voltage, as well as charge the battery pack. Students then experimentally observed voltage fluctuations due to variations of light exposure to the solar arrays. Measurements were then done by students to confirm voltage values at several voltage sources throughout the EPS module.
Lab session concluded with Software Functionality Test and Solar Panel Power Input Test. Software Functionality Test allowed students to interact with the EPS via control panel on a computer. Solar Panel Power Input Test simulated eclipse scenario (with battery pack supplying power) followed by sun exposure scenario (with battery pack charging).
Wei-hann's Group: Dana
By danarose(oops... I had this posted in the previous discussion)
Problem 3)
In order for a satellite to successfully collect solar energy, it must be in view of the sun. Since planetary motion is cyclical, this occurs periodically and poses a challenge to the EPS. Now we need to know how much time a satellite will be in the sun and how long it will need to be powered by energy stored in the battery. It is advantageous that the sun is a free and abundant energy source, and that solar panels are lightweight, flat and therefore easy to incorporate.
Problem 4)
A way to use less power on the dark side is to run systems in a conservation mode. Systems function normally while in view of the sun and the battery stores excess energy to be used on the dark side.
Philip Truong's Answers to Questions 3 and 4
By philipqJames' Group: Elizabeth, Pamela, Philip
By ElizabethEnergy [Wh] = 10.00 V x 0.7 A x (90 min. - 30 min. eclipse period)
Energy [Wh] = 7 Wh
2) 7 Watt Hours/2 Watts = 3.5 Hours
Elizabeth Viernes' Critical Thinking Response
By Elizabeth3) As an engineer, what kinds of advantages/challenges does solar power collection present?
Wei-Hann's Group: John
By jfurumoProblem 3)
Solar power is advantageous because the sun is an unlimited source of energy, practically speaking. For all its drawbacks on Earth, there are much fewer in space. For example, there are no clouds to block the sun in space. As long as a satellite is between the Earth and the Sun, baring a solar eclipse, there will be plenty of power to be had. Solar cells are relatively light components that, due to their shape, can be put on the outer surfaces of the spacecraft, requiring little extra room or consideration for the structural team. They are self-sustaining devices that do not need constant input of extra materials to produce power.
That being said, there are some drawbacks to solar cells. When a satellite is eclipsed by the earth or other heavenly body, the lack of sunlight will render solar cells useless. In order to remedy this, batteries will usually be incorporated into a satellite that will be charged by the solar cells when in direct sunlight. This way, when the satellite is out of contact with the sun, the batteries can be used to run the different subsystems of the spacecraft. Also, solar cells are very inefficient, with a maximum efficiency of around 30%.
Problem 4)
To maximize satellite potential, a satellite should be programmed to conserve as much energy as possible when on the dark side of the earth. Any subsystems or payloads that draw a lot of power should be utilized when direct sunlight is available. For example, imaging or photographing that draws a lot of power should be done when the satellite is in direct sunlight. Communication and signal transmission should also be done during the day, if possible. Another consideration would be to space out power consuming processes, trying not to overload the EPS at any one time.
Wei-Hann's Group:
By jfurumoProblem 1)
From data collected during our Solar Array Function Test of the EyasSat kit, we determined a peak power production. For this test, the calculation was as follows:
Energy (Wh) = Voltage (V) * Current (mA) * Time (hr)
6.1V * 39mA * 1 hr = .2379 Wh
Problem 2)
A 2 Watt radio could operate on this amount of energy for (.2379 Wh)/(2 W) = .11895 hr = 7.137 min
Pamela Toshi's answers to questions 3 & 4
By ptoshi3) An advantage to solar power is that the satellite doesn't have to rely solely on battery power. With solar power, batteries can be recharged and reused. The downside to using solar power is that the satellite cannot rely solely on solar power because of the eclipse phase. The highly efficient solar panels are expensive and all are relatively fragile.
4) The satellite can reduce energy consumption during the eclipse period if it has a way of detecting that no light is being collected by the solar panels. When there is no light, subsystems that use a lot of energy can be paused until there is light exposure again. Also, solar panels can be placed on the satellite so that maximum exposure to the sun can be achieved.
Alex Gao's Answers to 3 & 4
By Alex Gao3) As of now, the most efficient solar cells are only about 40% efficient. These “high efficiency” cells are also very costly, and would certainly be a big factor when considering budget constraints. Solar cells are also quite fragile, another thing that must be considered, especially due to the extreme conditions the satellite can be put under. Solar cells also provide a source of renewable energy, however, and can extend the life of a satellite in orbit, although solar power will not be available at all times (ex. sun off and eclipse periods).
4) Since solar cells are relatively inefficient, you should design the satellite in a way that allows for maximum exposure of its solar cells. If you know when sun off and eclipse periods are, you could design the satellite to put high-energy consuming tasks (ex. using cameras and sensors, transmitting information) on hold and instead have the satellite perform low energy tasks in order to avoid overstressing the battery.
Josh's Group: Matthew and Alex Problem 1 and 2
By brownmkProblem 1) We used a computer program to collect data as we charged the batteries with a DC voltage generator. Using the data we collected, determine the amount of electrical energy that can be collected during peak power point of a 90min orbit with a 30min eclipse per orbit.
E = V*A*t = 5.9v*29.7a*1h = 175.23 Wh
Probelm 2) Based on how much energy you calculated, how long can you run a 2W radio?
t = E/P = 175.23/2 = 87.615 hours.
Matthew Brown Answers to Question 3 and 4
By brownmk3. Solar power collection is advantageous because it allows you to generate power for your satellite while the sun is in view of your satellite. It gives you a way to sustain optimal working conditions barring any unforeseen circumstances (ie asteroid colliding with it and demolishing it). It is more compact than other forms of energy generation and therefore easier to utilize in smaller satellites. It also converts solar energy directly into electrical energy which makes it convenient to store and use.
Some of the disadvantages of solar power collection is there are definite sun-on and sun-off periods so power generation isn't happening all the time. As a result you have to plan for these times to make sure that you have enough power when you aren't generating power.
4. If you know how much power you can generate during a sun on period, how long your sun on and off periods are, and how much your average power consumption and peak power consumptions are you can design your satellite to work using a specified solar cell. If you generate power at a certain voltage you can transform the power into a lower amperage current in order to reduce your i^2*r consumption losses to a minimum.
James' Group: Elizabeth, Pamela, Philip
By Elizabeth1) Based on the data you collected in the "Solar Array Function Test", if your solar array operates at the peak power point, how much electrical energy could be collected in one 90-minute orbit?
Answer:
Energy [Wh] = Voltage [V] x Current [A] x Time [Hours]
= 10.00 V x 0.7 A x (90 min. -30 min. eclipse period)
= 7 Wh
2) How long could this power a 2-watt radio?
Answer:
7 Watt Hours/2 Watts = 3.5 Hours