Transient Orifice Flow: Instructions Properly label the iterations. Take note of the tolerance value and criteria. Determine the height of the fluid in the reservoir at time, t = 2.5 seconds, given that the velocity at the outfall, v(t) = 3 m/s, the acceleration due to gravity, g = 9.81 m/s² and the length of the pipe to outfall, L = 1.5 meters. A friendly reminder: The equation involved Reservoir v(t) V2gh tanh V2gh h water: 2L Pipe is hyperbolic tangent. This is not tan (h)(t/2L * sqrt(2gh)) - v (1) Hint: Transform the equation to a function of form: f(h) = 0 Solve MANUALLY using BISECTION AND REGULA-FALSI METHODS, starting at xjeft = 0.1, x-iaght = 1, ɛ = 0.001 and |f(xnew)l < ɛ Note: We are expected to get a value ofh around 0.47. The number of significant digits that we will be reporting depends on the tolerance value and iteration criterion set.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Transient Orifice Flow:
Instructions
Properly label the iterations. Take note of the tolerance value and criteria.
Determine the height of the fluid in the reservoir at time, t = 2.5 seconds,
given that the velocity at the outfall, v(t) = 3 m/s, the acceleration due to
gravity, g = 9.81 m/s² and the length of the pipe to outfall, L = 1.5 meters.
A friendly reminder: The equation involved
Reservoir
v(t)
V2gh
tanh (V2gh
h
water
2L
Pipe
is hyperbolic tangent. This is not tan (h)(t/2L * sqrt(2gh))
Hint: Transform the equation to a
(1)a
function of form: f(h) = 0
Solve MANUALLY using BISECTION AND REGULA-FALSI METHODS,
starting at xeft = 0.1, xight = 1, ɛ = 0.001 and |f (xnew)] < ɛ
Note:
We are expected to get a value of h around 0.47. The number of significant digits that we will be reporting depends on the tolerance value and iteration criterion set.
Transcribed Image Text:Transient Orifice Flow: Instructions Properly label the iterations. Take note of the tolerance value and criteria. Determine the height of the fluid in the reservoir at time, t = 2.5 seconds, given that the velocity at the outfall, v(t) = 3 m/s, the acceleration due to gravity, g = 9.81 m/s² and the length of the pipe to outfall, L = 1.5 meters. A friendly reminder: The equation involved Reservoir v(t) V2gh tanh (V2gh h water 2L Pipe is hyperbolic tangent. This is not tan (h)(t/2L * sqrt(2gh)) Hint: Transform the equation to a (1)a function of form: f(h) = 0 Solve MANUALLY using BISECTION AND REGULA-FALSI METHODS, starting at xeft = 0.1, xight = 1, ɛ = 0.001 and |f (xnew)] < ɛ Note: We are expected to get a value of h around 0.47. The number of significant digits that we will be reporting depends on the tolerance value and iteration criterion set.
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