A Theme Park site is being designed. It will occupy a rectangular piece of land 800m by 600m. There are to be 4 particularly noisy rides. The ‘Drumbeat’ uses music which reaches a level of 95dB(A). The ‘Bigdrop’ generates screams which can reach 98dB(A). The ‘Fastride’ uses a engine which has a noise level of 101dB(A). The bend on the driving track generates noise levels up to 99dB(A).
Name the four corners of the rectangle A, B, C and D. The long sides are AB and CD, and the short sides are BC and DA.
Assume that there are noise sensitive receivers at the three following points:-
A house at point B
A first aid point at the midpoint along the side CD.
A school 20m from the midpoint of the line AB.
The noisy rides must be separated by distances of at least 60m.
You are looking for a site layout which will have the lowest noise levels at the three noise sensitive receivers. Note that it is the highest of the three noise levels which matters; you are looking for a layout which achieves the lowest value for the highest of the three noise levels. You should include with your answer any graphs you produce in your search for the lowest noise levels.
You will need to use the following equations:-
Lw= Lp+20log10(r)+8
Where Lw is the sound power level (dB(A)) at the source And Lp is the sound pressure level (dB(A))at a distance r(in metres) from the source.
The sound pressure levels must be added using the following formula:-
Ltotal=10log10(Summation10^(L/10)). [Under the summasation there is a j written].
I have no idea what to do! Please help!!!
LJMU i bet u
sorry didn't mean to lol but it is lol
Peroleum engineering
To find the site layout that will have the lowest noise levels at the three sensitive receivers, we need to calculate the sound pressure levels at each of these points for different possible arrangements of the rides. We will compare these levels using the given equation Ltotal = 10log10(Summation10^(L/10)).
Let's start by calculating the sound power level (Lw) for each ride using the equation Lw = Lp + 20log10(r) + 8, where Lp is the sound pressure level at a distance r from the source.
1. Drumbeat (Lw = 95dB(A)):
Since there is no specific distance mentioned, we can assume the distance is zero (at the source itself). So Lp = Lw - 20log10(0) - 8 = 95dB(A) - infinity - 8 = 95dB(A).
2. Bigdrop (Lw = 98dB(A)):
Let's assume Bigdrop is located at point A. We need to calculate the sound pressure level at the three sensitive receivers.
- Point B (house) is at the corner of the site, so the distance from point A to B is 800m.
Lp(B) = Lw - 20log10(r) - 8 = 98dB(A) - 20log10(800) - 8.
- Midpoint of CD (first aid point):
The midpoint of CD is (400, 600). The distance from point A to the midpoint is the hypotenuse of a right-angled triangle with sides 400m and 600m. Using Pythagorean theorem, we can calculate the distance.
distance = √(400^2 + 600^2)
Lp(first aid point) = Lw - 20log10(r) - 8 = 98dB(A) - 20log10(distance) - 8.
- School 20m from the midpoint of AB:
The school is located on AB, 20m away from the midpoint. We can calculate the distance from the midpoint of AB to the school using the Pythagorean theorem.
distance = √(20^2 + (600/2)^2)
Lp(school) = Lw - 20log10(r) - 8 = 98dB(A) - 20log10(distance) - 8.
3. Fastride (Lw = 101dB(A)):
Let's assume Fastride is located at point C. We need to calculate the sound pressure level at the three sensitive receivers.
- Point B (house):
The distance from point C to B is the length of AB, which is 800m.
Lp(B) = Lw - 20log10(r) - 8 = 101dB(A) - 20log10(800) - 8.
- Midpoint of CD (first aid point):
Since Fastride is located on CD, the distance from point C to the midpoint is half the length of CD, which is 300m.
Lp(first aid point) = Lw - 20log10(r) - 8 = 101dB(A) - 20log10(300) - 8.
- School 20m from the midpoint of AB:
We can use the same calculation as before since the distance to the school doesn't depend on the position of Fastride.
Lp(school) = Lw - 20log10(r) - 8 = 101dB(A) - 20log10(distance) - 8.
4. Bend on the driving track (Lw = 99dB(A)):
Let's assume the bend is located at the midpoint of BC and D (since it's not specifically mentioned). We need to calculate the sound pressure level at the three sensitive receivers.
- Point B (house):
The distance from the bend to B is half the length of BC, which is 300m.
Lp(B) = Lw - 20log10(r) - 8 = 99dB(A) - 20log10(300) - 8.
- Midpoint of CD (first aid point):
The distance from the bend to the midpoint is the same as from the bend to D, which is 600m.
Lp(first aid point) = Lw - 20log10(r) - 8 = 99dB(A) - 20log10(600) - 8.
- School 20m from the midpoint of AB:
We can use the same calculation as before since the distance to the school doesn't depend on the position of the bend.
Lp(school) = Lw - 20log10(r) - 8 = 99dB(A) - 20log10(distance) - 8.
Now, we can calculate the total sound pressure level (Ltotal) at each sensitive receiver by using the formula Ltotal = 10log10(Summation10^(L/10)). Add up the calculated sound pressure levels (Lp) for each ride at each point, and then take the logarithm.
By considering different arrangements and calculating the Ltotal values, you can compare the noise levels at the three sensitive receivers and determine the layout with the lowest highest noise level.