A storage tank contains a liquid at depth y where y = 6 when the tank is hall lüll. Liquid withdrawn at a constant flow rate Q to meet demands. The contents are resupplied at a sinuso- rate 3Q sin'(t), The outflow is not constant but rather depends on the depth. For this case, the differential equ for depth is shown below. Some variable values are A = 1200 m, Q = 500 m2 /d, and a (force function) = 300. Arrange the variables into the Mathematical Model format: stating which variable(s) is denendent independent, etc.

A storage tank contains a liquid at depth y where y = 6 when the tank is hall lüll. Liquid withdrawn at a constant flow rate Q to meet demands. The contents are resupplied at a sinuso- rate 3Q sin'(t), The outflow is not constant but rather depends on the depth. For this case, the differential equ for depth is shown below. Some variable values are A = 1200 m, Q = 500 m2 /d, and a (force function) = 300. Arrange the variables into the Mathematical Model format: stating which variable(s) is denendent independent, etc.

Name: A storage tank contains a liquid at depth y where y= 0 when the tank
Uploaded: 2024-03-20T06:56:55.000Z
Channel: Bartleby
Description: A storage tank contains a liquid at depth y where y= 0 when the tank is half full. Liquid iswithdrawn at a constant flow rate Q to meet demands. The contents are resupplied at a sinusoidalrate 3Q sin'(t),The outflow is not constant but rather depend

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

The laminar flow of a fluid with a constant viscosity, 4, inside a channel is governed by the following boundary value problem (strong form): d'u dp for y E (0, h) "dy dr u = 0 at y = 0 and y = h vhere u is the fluid velocity and is the pressure drop in the direction of the flow. a) Derive the weak form of the boundary value problem described above.
G1/ in the Pipe (A) is equal (PA= 25 Lit15), the diameter is equal ( DA - 75mm, Do = Fig.shown, the Flow rate in 6o mm, Dg = Dc%=30 mm), and the velocit Pipe (D) is equal (U0=5m/s), the relation between flow rate in Pipe (B) and pipe (c) %3D Find the Value af . 2- Un, UB, Uc. water Os = 30 mm input A QA - 25 Lit/sec = 75 mm fo mm %3D Inpud = A outPut = B,C,D accumule tion =o.
1. The deformation of the fluid will continue as long as the 2. When designing gates, vessels, dams or other bodies that are submerged in a liquid, it is important not only to estimate the - 3. The surface tension is the tensile force per -- in any direction along the surface the surface tension. 4. The value of - specifies whether the fluid can be assumed compressible or incompressible. 5. Archimedes' principle indicates- 6. A truck is filled with gasoline. If the container (tank) on the truck undergoes a - - is applied. - but also to specify its location on the body. then the liquid surface will begin to rotate clockwise about the center of the container and will eventually maintain a fixed tilted position.
2. A storage contains liquid at depth y where y-0 when the tank is half full, as shown below. Liquid is withdrawn at a constant flow rate Q to meet demands. The contents are resupplied at a sinusoidal rate 3Qsin2(t). The storage tank has a diameter D=5m. a. Beginning from conservation of mass, formulate an equation for the change in dV TD2 depth of water as a function of time. Hint: dt din - 9out and V = 4 b. Use Euler's method to solve for the depth y, show the iterative equation for the Euler's method, and use Q=5 m3/s to complete the table below: t (s) y (m) 0.5 1.0
For the following questions, Circle the correct answer Q1. We have a steady fully rough turbulent flow of water with constant density in a cylindrical pipe of constant diameter. The height of the uniform roughness elements in the pipe is k.. . . For a fully developed rough turbulent flow in the pipe, the condition ku. -> 70 is the same as k,> 66,. [True/False] V avg The wall shear stress T-pfU/8. It is applicable only for steady fully rough turbulent flow in the pipe. [True/False] The shear stress distribution is linear with the radial distance from the pipe centerline only for laminar flow. [True/False] The head loss per unit length is quadrupled when the discharge is doubled. [True/False] The piezometric pressure drops uniformly along the pipe because the flow is steady along the pipe [True/False]
The pressure drop (Ap) test is carried out using a pipe configuration as illustrated below: Manometer 1 Manometer 2 straight pipe D= 2R R= radius in pipe The pipe data and the flowing fluid are as follows: Pipe: D = 1 cm; L= 100 cm. Fluid: Water, with density (A) = 1000 kg/m"; absolute viscosity (u) = 0.001 kg/im.s); Experimental data is shown as shown in the following table: Task: Ja. Plot the graph of the pressure as a function of the average velocity (V.v). b. Based on the equation for laminar flow in the pipe as follows: Ap = 32VuL, Vavg (m/s) Ap (Pa) 0,001 0,002 0,005 0,01 0,02 0,04 0,06 0,08 0,1 0,12 0,15 0,30 0,62 1,61 3,10 6,10 12,10 20,10 26,00 32,50 38,90 47,20 D Compare the experimental results in the table with the results of calculations using the above equation. Leave a comment. Note: Ap = p1-p2. c. The coefficient of friction (f) in the pipe is formulated as follows: f- 2DAD PL(V.) plot (plot) this distribution of fas a function of the Reynolds number (Re). Re is…
At a point in turbulent flow the mean kinetic energy of the turbulent velocity fluctuations is k = 0.9 m² s-2, and the known values of the turbulent velocity fluctuations are U - 0.4 m² s-2, 'u What is the value of -0.5 m² s-2. = value of ...21 U 3? Give your answer to two decimal places.
The velocity distribution in a 0.02 m diameter horizontal pipe conveying carbon tetrachloride (specific gravity = 1.59, absolute viscosity = 9.6 x 10-6 Pa sec) is given by the parabolic equation: v(r)=0.01(0.12- r?), where v(r) is the velocity in (m/s) at a distance r in (m) from the pipe center. What is discharge? O a. 3.13 x-8 m3/s O b. None of the mentioned O c. 1.047 x 108 m3/sec O d. 4.97 x 109 m3/sec
At t = 0 the tank shown in fig. is full up to top level, write the differential equation for the function of depth with respect to time. Tank 2 m 2 m conical 1 m 1 m cylinder r = 0.05 m Choose T or F F Case 1: y> 1 case 2: y< 1 m boundary condition at t=tm ,y=1 TI(1)2 dy/dt =0 -cd* t *( 0.05)2 * (2gy)1/2 I(y+1)2 dy/dt =o - cd* n *( 0.05)2 * (2gy)1/2
G1/ In the fig.shown, the Flow rate Pipe (A) is equal ( QA = 25 Lit15), the diameter is equal ( DA = 75 mm, D, = De = Dc =30mm), and the velocity in Pipe (D) is equal (Uo=5 m/s), the relation between flow rate in in 6o mm, Pipe (B) and pipe (c) GB =3 Qc. Find the Value of '. Re, Qc, Qo. 2- UA, UB, Uc: Pe = 30 mm water Pc = 30mm imput QA = 25 2i+/sec OA= 75 mm = 6o mm input = A outPut = B, C, D accumulation = o.
Solving fluid dynamic problem, flow through a cylinder, a uniform gradient only applied in z direction, If r is the radial polar co-ordinate, derive the velocity v_z(r) for steady flow at low Reynolds number. The result is about pressure gradient divide by 4 times viscosity and multiply by the difference of square of two distances. Do I need to convert distance into to radial polar co-ordinate in the result? How to consider this result in cylindrical coordinate?
The following date are provided for a pneumatic displacement gage shown in the figure: Ambient pressure = Pa Orifice Flow di 2 d, = 0.75 mm d2 = 1.50 mm Orifice Diameter: %3D Outlet Diameter: The relationship for the pressure ratio, r, and the displacement, x, for a pneumatic displacement gage is given by: P2 - Pa r = P1- Pa d2 = 1.1– 2.00 d? 0.4