Problem 4.1 An incompressible liquid with density p is contained in a tank with rigid, vertical walls. The tank is open to the atmosphere on the top and divided into two sections, Side A and Side B. The height h of the liquid and the cross-sectional or foot-print area A on each side of the tank (area perpendicular to the coordinate h) is shown in the figure. There is a hole in the partition and the liquid can flow freely between the two sides. Assume that the system is the mass of liquid in both tanks. hA AA AB a) Draw a system diagram. b) Write an equation for the mass of the system, msys, in terms of the given information. One approach is to find mâys,ª and Msys‚ß separately and then add them together to find msys. Your answer should only contain symbols. c) Using your result from Part (a), calculate the first derivative of the system mass with respect to time: dmsys/dt. Your answer should include two derivatives dh/dt and dhp/dt. What is the physical meaning of these two terms and why aren't there more derivatives? [FYI –The derivative dmsys/dt is called the time rate of change of the mass inside the system.] d) If the system mass is a constant, determine dh/dt, the time rate of change of the liquid level in Tank A, when hè, the liquid level in Tank B, is rising at the rate of 5 cm/s. Assume AA = 0.5 m² and AB = 0.2 m². Ans: d) -2.25 cm/s ≤ A ≤ -1.50 cm/s dhA dt hB
Problem 4.1 An incompressible liquid with density p is contained in a tank with rigid, vertical walls. The tank is open to the atmosphere on the top and divided into two sections, Side A and Side B. The height h of the liquid and the cross-sectional or foot-print area A on each side of the tank (area perpendicular to the coordinate h) is shown in the figure. There is a hole in the partition and the liquid can flow freely between the two sides. Assume that the system is the mass of liquid in both tanks. hA AA AB a) Draw a system diagram. b) Write an equation for the mass of the system, msys, in terms of the given information. One approach is to find mâys,ª and Msys‚ß separately and then add them together to find msys. Your answer should only contain symbols. c) Using your result from Part (a), calculate the first derivative of the system mass with respect to time: dmsys/dt. Your answer should include two derivatives dh/dt and dhp/dt. What is the physical meaning of these two terms and why aren't there more derivatives? [FYI –The derivative dmsys/dt is called the time rate of change of the mass inside the system.] d) If the system mass is a constant, determine dh/dt, the time rate of change of the liquid level in Tank A, when hè, the liquid level in Tank B, is rising at the rate of 5 cm/s. Assume AA = 0.5 m² and AB = 0.2 m². Ans: d) -2.25 cm/s ≤ A ≤ -1.50 cm/s dhA dt hB
Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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![Problem 4.1
An incompressible liquid with density p is contained
in a tank with rigid, vertical walls. The tank is open to
the atmosphere on the top and divided into two
sections, Side A and Side B.
The height h of the liquid and the cross-sectional or
foot-print area A on each side of the tank (area
perpendicular to the coordinate h) is shown in the
figure. There is a hole in the partition and the liquid
can flow freely between the two sides.
Assume that the system is the mass of liquid in both
tanks.
hA
AA
a) Draw a system diagram.
b) Write an equation for the mass of the system, msys, in terms of the given information.
One approach is to find mys, and msys, separately and then add them together to find
msys. Your answer should only contain symbols.
Ав
c) Using your result from Part (a), calculate the first derivative of the system mass with
respect to time: dmsys/dt. Your answer should include two derivatives dh/dt and
dhp/dt. What is the physical meaning of these two terms and why aren't there more
derivatives? [FYI -The derivative dmsys/dt is called the time rate of change of the mass
inside the system.]
d) If the system mass is a constant, determine dh/dt, the time rate of change of the liquid
level in Tank A, when hg, the liquid level in Tank B, is rising at the rate of 5 cm/s. Assume
A₁ = 0.5 m² and AB = 0.2 m².
Ans: d) -2.25 cm/s ≤
dhA
dt
< -1.50 cm/s
hB](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6abec00c-0df5-4c95-bf6e-fd605cb914b4%2Ff77772d6-57cc-4a7a-9d87-e9d41ffe8c93%2Fu7pyqp_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 4.1
An incompressible liquid with density p is contained
in a tank with rigid, vertical walls. The tank is open to
the atmosphere on the top and divided into two
sections, Side A and Side B.
The height h of the liquid and the cross-sectional or
foot-print area A on each side of the tank (area
perpendicular to the coordinate h) is shown in the
figure. There is a hole in the partition and the liquid
can flow freely between the two sides.
Assume that the system is the mass of liquid in both
tanks.
hA
AA
a) Draw a system diagram.
b) Write an equation for the mass of the system, msys, in terms of the given information.
One approach is to find mys, and msys, separately and then add them together to find
msys. Your answer should only contain symbols.
Ав
c) Using your result from Part (a), calculate the first derivative of the system mass with
respect to time: dmsys/dt. Your answer should include two derivatives dh/dt and
dhp/dt. What is the physical meaning of these two terms and why aren't there more
derivatives? [FYI -The derivative dmsys/dt is called the time rate of change of the mass
inside the system.]
d) If the system mass is a constant, determine dh/dt, the time rate of change of the liquid
level in Tank A, when hg, the liquid level in Tank B, is rising at the rate of 5 cm/s. Assume
A₁ = 0.5 m² and AB = 0.2 m².
Ans: d) -2.25 cm/s ≤
dhA
dt
< -1.50 cm/s
hB
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