1. Home
  2. College physics
  3. Fluid statics
  4. Variation of pressure with depth

11.4 Variation of pressure with depth in a fluid

<< Chapter < Page
  College physics     Page 1 / 4
Chapter >> Page >
  • Define pressure in terms of weight.
  • Explain the variation of pressure with depth in a fluid.
  • Calculate density given pressure and altitude.

If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense.

Consider the container in [link] . Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure    is the weight of the fluid mg size 12{ ital "mg"} {} divided by the area A size 12{A} {} supporting it (the area of the bottom of the container):

P = mg A . size 12{P= { { ital "mg"} over {A} } } {}

We can find the mass of the fluid from its volume and density:

m = ρV . size 12{m=ρV} {}

The volume of the fluid V size 12{V} {} is related to the dimensions of the container. It is

V = Ah , size 12{V= ital "Ah"} {}

where A size 12{A} {} is the cross-sectional area and h size 12{h} {} is the depth. Combining the last two equations gives

m = ρ Ah . size 12{m=ρ ital "Ah"} {}

If we enter this into the expression for pressure, we obtain

P = ρ Ah g A . size 12{P= { { left (ρ ital "Ah" right )g} over {A} } } {}

The area cancels, and rearranging the variables yields

P = hρg . size 12{P=hρg} {}

This value is the pressure due to the weight of a fluid . The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation P = hρg size 12{P=hρg} {} represents the pressure due to the weight of any fluid of average density ρ size 12{ρ} {} at any depth h size 12{h} {} below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. [link] illustrates this situation.

A container with fluid filled to a depth h. The fluid’s weight w equal to m times g is shown by an arrow pointing downward. A denotes the area of the fluid at the bottom of the container and as well as on the surface.
The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.

Calculating the average pressure and force exerted: what force must a dam withstand?

In [link] , we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See [link] .) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be 1.96 × 10 13 N ).

Strategy for (a)

The average pressure P ¯ due to the weight of the water is the pressure at the average depth h ¯ of 40.0 m, since pressure increases linearly with depth.

Solution for (a)

The average pressure due to the weight of a fluid is

P ¯ = h ¯ ρg . size 12{P=hρg} {}

Entering the density of water from [link] and taking h ¯ size 12{h} {} to be the average depth of 40.0 m, we obtain

P ¯ = ( 40.0 m ) 10 3 kg m 3 9.80 m s 2 = 3.92 × 10 5 N m 2 = 392 kPa.

Strategy for (b)

The force exerted on the dam by the water is the average pressure times the area of contact:

F = P ¯ A . size 12{F= {overline {P}} A} {}

Solution for (b)

We have already found the value for P ¯ size 12{ { bar {P}}} {} . The area of the dam is A = 80.0 m × 500 m = 4.00 × 10 4 m 2 size 12{A="80" "." 0`m times "500"`m=4 "." "00" times "10" rSup { size 8{4} } `m rSup { size 8{2} } } {} , so that

F = ( 3.92 × 10 5 N/m 2 ) ( 4.00 × 10 4 m 2 ) = 1.57 × 10 10 N. alignl { stack { size 12{F= \( 3 "." "92" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } \) \( 4 "." "00" times "10" rSup { size 8{4} } `m rSup { size 8{2} } \) } {} #" "=1 "." "57" times "10" rSup { size 8{"10"} } `N "." {} } } {}

Discussion

Although this force seems large, it is small compared with the 1.96 × 10 13 N size 12{1 "." "96" times "10" rSup { size 8{"13"} } `N} {} weight of the water in the reservoir—in fact, it is only 0.0800% of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake—it depends only on its average depth at the dam. Thus the force depends only on the water’s average depth and the dimensions of the dam, not on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure.epth to balance the increasing force due to the increasing pressure.

Got questions? Get instant answers now!

Questions & Answers

what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
please i need answer
Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
good👍
Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
generally amnesi how long yrs memory loss
Kelu Reply
interpersonal relationships
Abdulfatai Reply
What would be the best educational aid(s) for gifted kids/savants?
Heidi Reply
treat them normal, if they want help then give them. that will make everyone happy
Saurabh
What are the treatment for autism?
Magret Reply
hello. autism is a umbrella term. autistic kids have different disorder overlapping. for example. a kid may show symptoms of ADHD and also learning disabilities. before treatment please make sure the kid doesn't have physical disabilities like hearing..vision..speech problem. sometimes these
Jharna
continue.. sometimes due to these physical problems..the diagnosis may be misdiagnosed. treatment for autism. well it depends on the severity. since autistic kids have problems in communicating and adopting to the environment.. it's best to expose the child in situations where the child
Jharna
child interact with other kids under doc supervision. play therapy. speech therapy. Engaging in different activities that activate most parts of the brain.. like drawing..painting. matching color board game. string and beads game. the more you interact with the child the more effective
Jharna
results you'll get.. please consult a therapist to know what suits best on your child. and last as a parent. I know sometimes it's overwhelming to guide a special kid. but trust the process and be strong and patient as a parent.
Jharna
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>
Practice Key Terms 1

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask