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Thread: Hydrostatic Pressure

  1. #1
    Newbie Demen's Avatar
    Feb 2008
    Saint Paul, MN, USA

    Hydrostatic Pressure (Hydrostatic Force)

    Hello. I am actually an active member of a science forum, and have gotten help learning from its math members for a good three or four months, and am impossibly grateful to them for their time, however that help has recently seemed to drop off. I now find myself stuck in the application of integration to hydrostatic pressure. I apologize if it seems unethical of me to post a link to another website, but I have done a great deal of posting describing my problem in this science forum's math section. This is the link to my thread where 13 posts down I begin to describe where I have been stuck for two weeks.

    (website removed by Demen)

    I would be very grateful if I could get some help so I can understand what I am doing wrong. I will surely repay your effort by sharing my time here, on the MHF to help answer calculus questions of the fellow members.
    Last edited by Demen; Mar 3rd 2008 at 08:45 AM. Reason: website recieved too much traffic
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    You should consider posting your problem as opposed to a link one has to scroll down to find. Some may be put off by it and won't bother.

    Anyway, since we have a nice rectagular flat vertical surface, the force on the side of the aquarium would be:

    $\displaystyle \int_{a}^{b}{\rho}h(x)w(x)dx$

    $\displaystyle 124.8\int_{0}^{3}xdx=561.6 lbs$

    Is that what you got?.

    Also, one would think that the force on the bottom would just be the weight of the water assuming the tank is full.

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  3. #3
    Newbie Demen's Avatar
    Feb 2008
    Saint Paul, MN, USA
    My problem is actually not that which you describe. My problem is discovering why I always end up with an incorrect answer when calculuating the hydrostatic pressure on a two dimensional plate that has a width which varies with the depth. Keep in mind that the author of my book assumes that I am not using anything in multivariable calculus.

    My explaination is four thread pages long. That is the reason I posted a link. I seem to be having internet trouble at the moment, but when I get a chance I'll break off a few sections here to make it more readable, although I suspect only someone that is as crazy as me is going to actually read it since it is so long.
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  4. #4
    Newbie Demen's Avatar
    Feb 2008
    Saint Paul, MN, USA
    The book I am learning out of is Calculus Early Transcendentals 5/E Volume 1 written by James Stewart. The first section of my post is an explanation of the idea that I am trying to be able to perform myself. The second section is a description of the specific problems that I would like to solve. The third section is my work on the problems of the second section. For more details, please refer to the link which I have posted above. Happy reading.

    8.3 Applications to Physics and Engineering

    Among the many applications of integral calculus to physics and engineering, we consider two here: force due to water pressure and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and then evaluate the resulting integral.

    Hydrostatic Pressure and Force

    Deep-sea divers realize that water pressure increases as they dive deeper. This is because the weight of the water above them increases.

    In general suppose that a tin horizontal plate with the area A square meters is submerged in a fluid density p kilograms per cubic meter at a depth d meters below the surface of the fluid as in Figure 1.

    Figure 1

    The fluid directly above the plate has volume V = Ad, so its mass is m = pV = pAd. The force exerted by the fluid on the plate is therefore

    F = mg = pgAd

    where g is the acceleration due to gravity. The pressure P on the plate is defined to be the force per unit area

    P = F/A = pgd

    The SI unit for measuring pressure is newtons per square meter, which is called a pascal (abbreviation: $\displaystyle \frac{1N}{m^2} = 1 Pa $). Since this is a small unit, the kilopascal (kPa) is often used. For instance, because the density of water is $\displaystyle p = \frac{1000 kg}{m^3}$, the pressure at the bottom of a swimming pool 2 m deep is

    $\displaystyle P = pgd = (\frac{1000 kg}{m^3})( \frac{9.8 m}{s^2})(2 m)
    = 19,600 Pa = 19.6 kPa$

    An important principal of fluid pressure is the experimentally verified fact that at any point in a liquid the pressure is the same in all directions. (A diver feels the same pressure on nose and both ears.) Thus, the pressure in any direction at a depth d in a fluid with mass density p is given by

    $\displaystyle P = p g d = \delta d$

    Note: When using U.S. Customary units, we write $\displaystyle P = p g d = \delta d$, where $\displaystyle \delta = pg $ is the weight density (as opposed to p, which is the mass density.) For instance the weight density of water is $\displaystyle \delta = \frac{62.5 lb}{ft^3}$.

    This helps us determine the hydrostatic force against a vertical plate or wall or dam in a fluid. This is not a straightforward problem because the pressure is not constant but increases as the depth increases.

    EXAMPLE 1 A dam has the shape of the trapezoid shown in Figure 2.

    Figure 2

    The height is 20 m, and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.

    Figure 3a

    SOLUTION We choose a vertical x-axis with origin at the surface of the water as in Figure 3(a). The depth of the water is 16 m, so we divide the interval [0,16] into sub-intervals of equal length with endpoints $\displaystyle x_{i}$ and we choose $\displaystyle x_{i}^\ast \in [ x_{i-1}, x_{i} ]$. The ith horizontal strip of the dam is approximated by a rectangle with height Δx and width $\displaystyle w_i$, where, from similar triangles in Figure 3(b),

    Figure 3b

    $\displaystyle \frac{a}{16 - x_i^\ast} = \frac{10}{20}$

    $\displaystyle a = \frac{16 - x_i^\ast}{2} = 8 - \frac{x_i^\ast}{2}$

    and so
    $\displaystyle w_i = 2(15 + a) = 2(15 + 8 - \frac{1}{2}x_i^\ast) = 46 - x_i^\ast$

    If $\displaystyle A_i$ is the area of the ith strip, then

    $\displaystyle A_i \approx w_i \Delta x = (46 - x_i^\ast)\Delta x$

    If Δx is small, then the pressure $\displaystyle P_i$ on the ith strip is almost constant and we can use Equation 1 to write

    $\displaystyle P_i \approx 1000gx_i^\ast$

    The hydrostatic force $\displaystyle F_i$ acting on the ith strip is the product of the pressure and the area:

    $\displaystyle F_i = P_i A_i \approx 1000gx_i^\ast ( 46-x_i^\ast ) \Delta x$

    Adding these forces and taking the limit as $\displaystyle n\rightarrow\infty$, we obtain the total hydrostatic force on the dam:

    $\displaystyle F = \lim_{n\rightarrow\infty} \sum_{i=1}^n 1000 g x_i^\ast ( 46 -x_i^\ast ) \Delta x$
    $\displaystyle = \int_0^{16} 1000 g x ( 46 - x ) dx$
    $\displaystyle = 1000 ( 9.8 ) \int_0^{16} ( 46 x - x^2) dx$
    $\displaystyle = 9800 \left[ 23 x^2 - \frac{x^3}{3} \right]_0^{16}$
    $\displaystyle =4.43 \times 10^7 N$

    EXAMPLE 2 Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft if the drum is submerged in water 10 ft deep.

    Figure 4

    SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that the origin is placed at the center of the drum. Then the circle has a simple equation. $\displaystyle x^2 + y^2 = 9$. As in Example 1 we divide the circular region into horizontal strips of equal width. From the equation of the circle, we see that the length of the ith strip is $\displaystyle 2 \sqrt{ 9 - ( y_i^\ast)^2}$ and so its area is

    $\displaystyle A_i = 2 \sqrt{ 9 - ( y_i^\ast)^2} \Delta y $

    The pressure on this strips is approximately

    $\displaystyle \delta d_i = 62.5 ( 7 - y_i^\ast ) $

    and so the force on the strips is approximately

    $\displaystyle \delta d_i A_i = 62.5 ( 7 - y_i^\ast ) 2 \sqrt{ 9 - ( y_i^\ast)^2 } \Delta y $

    The total force is obtained by adding the forces on all the strips and taking the limit:

    $\displaystyle F = \lim_{n\rightarrow\infty} \sum_{i=1}^n 62.5 ( 7 - y_i^\ast ) 2 \sqrt{ 9 - ( y_i^\ast)^2 } \Delta y $
    $\displaystyle = 125 \int_{-3}^3 ( 7 - y ) \sqrt{ 9 - y^2} dy$
    $\displaystyle = 125 \displaystyle{\cdot} 7 \int_{-3}^3 \sqrt{ 9 - y^2} dy - 125\int_{-3}^3 y \sqrt{ 9 - y^2} dy $

    The second integral is 0 because the integrand is an odd function(see Theorem 5.5.7). The first integral can be evaluated using the trigonometric substitution y = 3 sin Θ, but it's simpler to observe that it is the area of a semicircular disk with radius3. Thus

    $\displaystyle F = 875 \int_{-3}^3 \sqrt{ 9 - y^2} dy = 875 \displaystyle{\cdot} \frac{1}{2} \pi (3)^2 $
    $\displaystyle = \frac{7875 \pi}{2} \approx$ 12,370 lb

    #3 to #9: A vertical plate is submerged in water and has the indicated shape. Expain how to approximate the hydrostatic force against the end of the tank by a Riemann sum. Then express the force as an integral and evaluate it.

    The book answers for the odd numbered questions are:
    3) 6000 lb
    5) $\displaystyle 6.5 \times 10^6 N $
    7) $\displaystyle 3.5 \times 10^4 N $
    9) $\displaystyle 1000 g \pi r^3 N $

    Being that I don't believe I am having difficulty at all with Riemann sums and that it might be too time consuming to create pictures, load them on the net, and reference them here, I'll just skip this part for you guys.

    3) This rectangular plate is measured in feet, so we measure our P as $\displaystyle \delta d$ or as 62.5d. When we enter this into our integral, since we will be integrating with respect to depth, we would write this element as 62.5x. The next step is to find a way to express the width in terms of depth. In this particular problem, the width is constant no matter the depth, so we can just enter a 6 for this element. The last element of our integral will of course be our dx that represents the height of an infinitely small strip of our rectangular plate. Also, since the top of our plate is submerged 2 feet under water and the bottom of our rectangle is 6 feet under water, we will be integrating from 2 to 6. Putting all this information together we have this integral:
    $\displaystyle \int_a^b ( p g x ) ( w ) dx $
    $\displaystyle \int_2^6 (62.5 x) ( 6 ) dx $
    $\displaystyle 375 \left[ \frac{x^2}{2} \right]_2^6$
    $\displaystyle 375 \left[ 18 - 2 \right] $
    $\displaystyle ( 375 ) ( 16 ) $
    $\displaystyle 6000 lb $

    This is the correct answer for problem 3.

    4) This triangle plate is measured in feet so we use 62.5x for our pressure. We can express the width as $\displaystyle 4 - \frac{4}{3} ( x - 1 ) $. This makes our integral:
    $\displaystyle \int_1^4 (62.5 x)(4 - \frac{4}{3} ( x - 1 ) ) dx$

    which I've evaluated to:
    750 lb

    5) This semicircle is measured in meters so we use 9800x for our pressure. To find an expression for the width, I've taken the formula for the circle, $\displaystyle x_1^2 + y_1^2 = 10^2$

    and first figured that if x (not $\displaystyle x_1 $) is going to be the depth within the integral then:
    $\displaystyle y_1 = -x$

    so then I could substitute x for $\displaystyle y_1$ and solve for $\displaystyle 2x_1$ which I figure to be the width:
    $\displaystyle 2x_1 = 2 \sqrt{100 - x^2}$

    Now that I can express both the pressure and area within the integral, I can construct it:
    $\displaystyle \int_0^{10} (9800x)( 2 \sqrt{100 - x^2} ) dx $

    which I've evaluated to be:
    $\displaystyle 1.039 \times 10^7 N $

    $\displaystyle P = 1000 g x $
    $\displaystyle w = \frac{4}{5} x$
    $\displaystyle F = \int_0^5 (1000 g x) ( \frac{4}{5} x ) dx $
    $\displaystyle F = \int_0^5 800 g x^2 dx $
    $\displaystyle F = \frac{800 g}{3} \left[ x^3 \right]_0^5 $
    $\displaystyle F = \frac{ 100000 g }{3} N $

    $\displaystyle P = 62.5 x$
    $\displaystyle w = 12 + 8x $
    $\displaystyle F = \int_0^8 (62.5 x) (12 + 8x ) dx $

    I've evaluated this to be:
    109333 lb

    $\displaystyle P = 1000 g x $
    $\displaystyle w = b + \frac{x ( a - b ) }{h} $
    $\displaystyle F = \int_0^h (1000 g x ) ( b + \frac{ x ( a - b ) }{h} ) dx $

    I've evaluated this to be:
    $\displaystyle F = \frac{500}{3} g h^2 (2a + b) N $

    $\displaystyle P = 1000 g x $
    $\displaystyle w = 2 \sqrt{ 2 r x - x^2 }$
    $\displaystyle F = \int_0^2r (1000 g x ) (2 \sqrt{ 2 r x - x^2 } ) dx $

    I've evaluated this to be:
    $\displaystyle 2000 g \pi r^3 N $


    In conclusion I suspect that I am getting incorrect answers whenever the width of a plate varies with the depth, however I can't really be sure where my mistakes are being made. I've been stuck on this idea for about two to three weeks. Any help would be greatly appreciated.
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