0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass. Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan
0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass. Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan
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
Related questions
Question
i dont really understand the explanation of part b or how they used trigonometry to get that. can you please help.
Transcribed Image Text:0
h
So the volume of the trough is
V = Lh²/tan 0
h =
V tan
L
So
p = pg
V tan
L
Our results show that the pressure only depends on depth (h) and is not affected by the shape of the
vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your
pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same
amount of water into a wide glass.
Transcribed Image Text:Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The
troughs are filled with water to absorb the emitted radiation.
Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure
below.
a) Express the pressure at the bottom of the triangular and the rectangular troughs as a
function of the depth of the fluid (h).
b) Express the pressure at the bottom of the triangular and the rectangular troughs as a
function of the volume of fluid in the trough (V).
Part (a)
The stagnation pressure is related to the depth according to
p = pgh
This is the same for both the square and the triangular troughs.
Part (b)
We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the
equation for p, so get p in terms of V.
For the rectangular trough
For the triangular one:
The area of a triangle is
V = wLh
V
h =
WL
p = pg
\wL
h²
A =
tan
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