Problem 2: Filling a Tank Problem You\'re designing the measuring system for a c
ID: 1713208 • Letter: P
Question
Problem 2: Filling a Tank Problem You're designing the measuring system for a cylindrical water tank in which you will be determining the available water vohume by the depth through the center. The formula is created by modifying the classic volume equation with Calc2-Fu. at's a concept called a cylindrical segment.") 2R (the h is our d (aside: "Ooo! How would you now estimate the error in measuring the vohane of the tank?) For a 3-ft diameter tank that's 6 ft long, at what depth will the tank be 10% full (at which point you will need to send a warning system to the resource manager This one can give you more than 1 answer but the ones that won't work will be obvious throug a little deductive reasoning. You'll want to get the max volume of the cylinder from your geometry notes from high school.. or Wolfram Mathworld. Root (today, Root is a verb) the equation where your target will be 10% of the full vohume of the sphere. Also graph it between credible your credible ranges (from an 1. 2. empty to full tank). For this one use the Secant method. Document why you chose your starting first guesses 3.Explanation / Answer
Diameter of cylinder = 3 ft
Radius (R) = 1.5 ft
Length (L) of cylinder = 6 ft
Volume of the cylinder = R2 L
10% volume of the cylinder = 0.1 R2 L
Expression of available water volume by depth through the center of cylinder:
V(L, R, d) = L[R2 cos-1((R-d)/R)) – (R-d)(2Rd-d2)0.5]
We have to find the depth at which the cylinder will be 10% full.
So,
0.1 R2 L = L[R2 cos-1((R-d)/R)) - (R-d)(2Rd-d2)0.5]
=> R2 cos-1((R-d)/R)) - (R-d)(2Rd-d2)0.5 - 0.1 R2 = 0
=> 1.52 cos-1((1.5-d)/1.5)) - (1.5-d)(2x1.5xd-d2)0.5 - 0.1 1.52 = 0
=> 2.25 cos-1((1.5-d)/1.5)) - (1.5-d)(3d-d2)0.5 - 0.225 = 0
We will use Secant Method for finding the depth ‘d’ that satisfies the above equation.
f(d) = 2.25 cos-1((1.5-d)/1.5)) - (1.5-d)(3d-d2)0.5 - 0.225
For carrying out the iterative process to get a converged result, the following expression is used in secant method.
Xi+1 = Xi – f(Xi) (Xi - Xi-1) / (f(Xi) – f(Xi-1))
We will take X0 = 0.2R = 0.3 and X1 = 0.25R = 0.375 as the initial guesses because as per intuition, these can be considered as good guesses for the value of depth for 10% volume of cylinder.
Iteration No. (i)
Xi
Expressions used for finding Xi
0
0.3
X0 = 0.3
1
0.375
X1 = 0.375
2
0.478915
X2 = X1 – f(X1) (X1 – X0) / (f(X1) – f(X0))
Where X1 = 0.375, X0 = 0.3,
f(X0)=-0.33898, f(X1)=-0.19688
3
0.469
X3 = X2 – f(X2) (X2 – X1) / (f(X2) – f(X1))
Where X2 = 0.478915, X1 = 0.375,
f(X1)= -0.19688, f(X2)=0.020768
4
0.469425
X4 = X3 – f(X3) (X3 – X2) / (f(X3) – f(X2))
Where X3 = 0.469, X2 = 0.478915,
f(X2)= 0.020768, f(X3)=-0.00093
5
0.469427
X5 = X4 – f(X4) (X4 – X3) / (f(X4) – f(X3))
Where X4 = 0.469425, X3 = 0.469,
f(X3)= -0.00093, f(X4)=-3.8E-06
6
0.469427
X6 = X5 – f(X5) (X5 – X4) / (f(X5) – f(X4))
Where X5 = 0.469427, X4 = 0.469425,
f(X4)= -3.8E-06, f(X5)=7.03E-10
The iterative process converges to 0.469427 in six iterations.
So, the d=0.469427 satisfies the equation
f(d) = 2.25 cos-1((1.5-d)/1.5)) - (1.5-d)(3d-d2)0.5 - 0.225
Therefore, the depth at which tank will be 10% full is 0.469427 ft.
Iteration No. (i)
Xi
Expressions used for finding Xi
0
0.3
X0 = 0.3
1
0.375
X1 = 0.375
2
0.478915
X2 = X1 – f(X1) (X1 – X0) / (f(X1) – f(X0))
Where X1 = 0.375, X0 = 0.3,
f(X0)=-0.33898, f(X1)=-0.19688
3
0.469
X3 = X2 – f(X2) (X2 – X1) / (f(X2) – f(X1))
Where X2 = 0.478915, X1 = 0.375,
f(X1)= -0.19688, f(X2)=0.020768
4
0.469425
X4 = X3 – f(X3) (X3 – X2) / (f(X3) – f(X2))
Where X3 = 0.469, X2 = 0.478915,
f(X2)= 0.020768, f(X3)=-0.00093
5
0.469427
X5 = X4 – f(X4) (X4 – X3) / (f(X4) – f(X3))
Where X4 = 0.469425, X3 = 0.469,
f(X3)= -0.00093, f(X4)=-3.8E-06
6
0.469427
X6 = X5 – f(X5) (X5 – X4) / (f(X5) – f(X4))
Where X5 = 0.469427, X4 = 0.469425,
f(X4)= -3.8E-06, f(X5)=7.03E-10
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