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thesis 1 A horizontal vessel with 2:1 S.E. heads the volume is
 * $$ V = V_{2:1 S.E. head} f(h) = \frac{\pi}{6}h^2(3R-h) + L[R^2 \cos^-1 \frac({R-h}{R}) - a(R-h)]$$

Surface area
 * $$ S = S_{2:1 S.E. head} f(h) = L(2R \cos^-1 \frac({R-h}{R})) + 2(2.168 Rh)$$

Title: Correct Volume and Surface Area Mensuration of Partially Filled Spheres and Spheroids ABSTRACT: Liquid storage equipment in the shape of spheres and horizontal vessels (with hemi-spherical or semi-elliptical heads) require volume capacity and surface area measurement for process and safety requirements. Liquid volume is necessary to determine the process capacity of the vessel and maximum operating time period. Surface area of the partially filled portion of the storage vessel is required to determine heat transfer from fire to the wetted area of the vessel. This article provides the equations necessary to calculate the partial volume and surface area, as well as the proofs of the equations. This article also demonstrates that the equations found in many textbooks for partial volume are incorrect.

SPHERICAL CAPS:

Sperical caps and segments: volume and area
 * $$V = \frac{4}{3} \pi R^3$$
 * $$A = 4 \pi R^2$$

Two common Engineering applications If the radius of the base of the cap is $$a$$, and the height of the cap is $$h$$, then the volume of the spherical cap is
 * $$ V = \int \pi a^2 dh $$.

Per the Pythagorean theorem, in the upper hemisphere of the diagram,
 * $$R^2 = (R - h)^2 + a^2$$

and in the lower hemisphere
 * $$R^2 = (h - R)^2 + a^2$$

therefore in both hemispheres
 * $$a^2 = (2Rh - h^2)$$ and $$a = \sqrt{h (2R - h)}$$. This is substituted in the volume integral as:
 * $$ V = \pi \int (2Rh - h^2) dh $$,
 * $$ V = \pi Rh^2 - \frac{\pi}{3}h^3 $$, and
 * $$ V = V f(h) = \frac{\pi}{3}h^2(3R-h) $$ which is valid for h from 0 to 2R.

Another form of the volume expression is incorrectly based on rearranging $$a^2$$ algebraically. That is,
 * $$ 2Rh = a^2 + h^2 $$ and $$ R = \frac{a^2}{2h} + \frac{h}{2} $$.

Replacing R in $$ V = \frac{\pi}{3}h^2(3R-h) $$ gives
 * $$ V = V f(h, a) = \frac{\pi}{6}h(3a^2+h^2) $$ (incorrectly applied).

The latter equation, $$ V f(h,a) $$, while prevalent in most textbooks[references], is incorrect. We know that volume is a 3rd order function. The equation $$ V f(h) $$ is a third order polynomial with respect to $$h$$. The equation $$ V f(h,a) $$, is a 6th order equation with respect to $$h$$. This is because $$a$$ is a dependent variable of $$h$$. That is, both equations $$ a = R \sin (h \pi) $$ and $$a = \sqrt{h (2R - h)}$$ are both valid expressions for $$a$$ as a function of $$h$$.

Consider a spherical segment from plane $$a$$ to plane $$a$$. The segment volume calculated by integration textbooks[references] report the volume of the segment as
 * $$ V = V f(h, a, b) = \frac{\pi}{6}h(3a^2+3b^2+h^2) $$ (incorrectly applied).

Similar to the volume equation $$ V = V f(h, a) $$, $$ V = V f(h, a, b) $$ is also a 6th order equation. Instead of integration from plane $$a$$ to plane $$b$$, the segment volume is correctly calculated by subtraction of $$ V f(h) $$ at plane $$b$$ from $$ V f(h) $$ at plane $$a$$.
 * $$ V_{a-b} = V f(h, a) - V f(h, b) $$.

Right angle cones and conical frustums

Surface area calculation: The surface area of a spherical cap or segment is calculated from surface of revolution equation:
 * $$ S = 2 \pi \int_1^2 y d s $$, where $$ y = f(h) = a $$ and $$ ds = \sqrt{1 + ( y')^2 } dh $$

where the curve of the surface of revolution is rotated about the axis perpendicular to the plane of the spherical segment or cap. As previously demonstrated, $$a = \sqrt{h (2R - h)}$$. Therefore, $$(a')^2 = \frac{(R-h)^2}{a^2}$$. Substituting
 * $$ S = 2 \pi \int_{h_a}^{h_b} a \sqrt{1 + \frac{(R-h)^2}{a^2}} dh $$,
 * $$ S = 2 \pi \int_{h_a}^{h_b} \sqrt{h(2R-h) + R^2 - 2Rh + h^2} dh $$, and
 * $$ S = 2 \pi R \int_{h_a}^{h_b} dh $$, and
 * $$ S = 2 \pi R (h_a-h_b) = 2 \pi R h $$.

This formula appears counter-intuitive. That is, the formula demonstrates that the surface area of a spherical segment or cap varies linearly with the height of the segment. This formula is also the same formula as a cylinder. This appears incorrect because a sphere has a curved surface such that one would expect a sinusoidal relationship between height of the cap or segment and surface area. Consider a cylinder and sphere both of radius $$R$$ and height $$h$$. We can see that the surface area is the same, yet the sphere's rotated plane is curved toward the center and the cylinder's rotated plane is vertical. Imagine that the curved surface of revolution of the sphere is peeled back toward the flat surface of revolution of the cylinder. We would observe empty cuts along the length of the sphere. At the same time, however, we would also observe the sphere's surface would extend beyond $$h$$ of the cylinder. The surface extension equals the empty cuts, such that the curved surface area of a sphere is linear against the height of the sphere.

appears to be a first order function such that area is a second order function cap

2:1 SE head

Consider a spheroid vs. a sphere. The equation of a sphere is $$x^2 + y^2 + z^2 = R^2$$ and a spheroid $$\frac{x^2 + y^2}{r_1^2} + \frac{z^2}{r_2^2} = R^2$$, such that dimension $$r_1$$ equals radius $$R$$. The volume of a segment and the surface area of a segment of a spheroid which is oriented so that the segmental planes are perpendicular to $$r_1$$
 * $$ V = V_spheroid f(h) = \frac{\pi}{3}h^2(3R-h) \frac{r_2}{R} $$
 * $$ S = S_spheroid f(h) = \frac{h}{R}[2 \pi R^2 + \pi (r_2)^2 e_1] $$, where
 * $$ e_1 = \frac{1}{e}\ln(\frac{1+e}{1-e}) $$ and : $$ e \equiv \sqrt{1 - (\frac{r_2}{r_1})^2} $$.

These formulas for volume and surface area have applications to calculate the liquid filled volume and wetted surface area of a horizontal cylindrical vessel with hemispherical or semi-elliptical (S.E.) heads. For the special case of a 2:1 S.E head (an oblate spheroid which $$ r_2 = \frac{R}{2} $$), the liquid filled volume calculation becomes
 * $$ V = V_{2:1 S.E. head} f(h) = \frac{\pi}{6}h^2(3R-h) \frac{1}{2} $$

for EACH head and surface area
 * $$ S = S_{2:1 S.E. head} f(h) = 2.168 Rh$$

for EACH head.

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 * $$ V = V f(h) = \frac{\pi}{3}h^2(3R-h) \frac{r_2}{R} $$ which is valid for h from 0 to 2R for BOTH heads and $$ V = V f(h) = \frac{\pi}{6}h^2(3R-h) \frac{r_2}{R} $$ valid for EACH head.

the surface area calculation reduces to The ellipticity of an oblate spheroid, defined as $$r_1 > r_2$$, is
 * $$ e \equiv \sqrt{1 - (\frac{r_2}{r_1})^2} $$.

The surface area is calculated around the surface of revolution about the axis perpendicular to r_2:
 * $$ S = 2 \pi \int_{-r_2}^{+r_2} y d s $$, where $$ y = f(h) = r_1 e $$ and $$ ds = \sqrt{1 + ( y')^2 } dh $$.  At $$r_1$$ = $$R$$.  Integration produces the linear formula:
 * $$ S = 2 \pi R^2 + \pi (r_2)^2 e_1 $$, where $$ e_1 = \frac{1}{e}\ln(\frac{1+e}{1-e}) $$.


 * $$ V_1 = \frac{\pi}{3}h^2(3R-h) $$
 * $$ V_2 = \frac{\pi}{6}h(3a^2+h^2) $$
 * $$ V f(h) = \frac{\pi}{3}h^2(3R-h) $$
 * $$ V f(h,a) = \frac{\pi}{6}h(3a^2+h^2) $$

ha to hb. By

This equation, similar to $$ V f(h,a) $$, is also incorrect. The incorrect textbook method is to integrate from $$ R-h $$ to $$ h$$ with respect to the radius of each plane. That is, along the vertical axis of rotation $$ r^2 = R^2 - z^2 $$ per Pythagorean Theorem, where $$ r $$ is the radius of either the $$ a $$ or $$ b $$ plane and $$ R $$ is the radius across the center of the sphere. This method defines $$ d $$ as $$ R-h $$, $$ h $$ as $$ d+h $$ and integrates along the defined bounds:
 * $$ V = \int_d^{d+h} \pi r^2 dz $$.
 * $$ r^2 = R^2 - z^2 $$
 * $$ V = \pi \int_d^{d+h} R^2 dz - \pi \int_d^{d+h} z^2 dz $$
 * $$ V = \pi [R^2 z - \frac{1}{3}z^3]_d^{d+h} $$
 * $$ z_d^{d+h} = (d+h)-d = h $$
 * $$ (z^3)_d^{d+h} = (d+h)^3 - d^3 $$
 * $$ (z^3)_d^{d+h} = d^3+3d^2 h + 3dh^2 +h^3 - d^3 $$
 * $$ V = \pi [R^2 h - \frac{1}{3}(3d^2 h + 3dh^2 +h^3) ] $$

Consider the relationships
 * $$ a^2 = R^2 - d^2 $$,
 * $$ b^2 = R^2 - (d+h)^2 = R^2 - d^2 - 2dh - h^2 $$, and
 * $$ d = \frac{a^2 - b^2 - h^2}{2h} $$.

These give
 * $$ R^2 = a^2 + \frac{(a^2 - b^2 - h^2)^2}{4h^2} $$ and substitute
 * $$ V = \pi h [R^2 - d^2 h - dh^2 -\frac{1}{3}h^3) ] $$ to
 * $$ V = \frac{\pi}{6} h (3a^2 + 3b^2 + h^2) $$.

This then indicates  From the shape of the sphere we can observe that the change in volume of the cap from 0 to D Both equations for V are valid for h from 0 to D.
 * $$ a = R \sin (h \pi) $$

and so an alternative expression for the volume is

Per the Pythagorean theorem,
 * $$ a^2 = R^2 - (R-h)^2 $$ and $$ a^2 = 2Rh - h^2 $$. Then,
 * $$V = \frac {\pi h^2}{3} (3r-h)$$

$$ R^2 = a^2 + (R-h)^2 $$ and $$ R^2 = a^2 + R^2 -2Rh + h^2 $$. Then,


 * $$V = int_ \frac{\pi h}{6} (3a^2 + h^2)$$


 * $$V = \frac{\pi h}{6} (3a^2 + h^2)$$

The relationship between $$h$$ and $$r$$ is irrelevant as long as 0 ≤ $$h$$ ≤ $$2R$$. The blue section of the illustration is also a spherical cap.

The parameters $$a$$, $$h$$ and $$r$$ are not independent:


 * $$R^2 = (R-h)^2 + a^2 = R^2 +h^2 -2Rh +a^2$$
 * $$R = \frac {a^2 + h^2}{2h}$$.

Substituting this into the area formula gives:


 * $$A = 2 \pi \frac{(a^2 + h^2)}{2h} h = \pi (a^2 + h^2)$$

Note also that in the upper hemisphere of the diagram,
 * $$R^2 = (R - h)^2 + a^2$$

and in the lower hemisphere
 * $$R^2 = (h - R)^2 + a^2$$

hence in either hemisphere
 * $$a = \sqrt{h(2R-h)}$$

and so an alternative expression for the volume is
 * $$V = \frac {\pi h^2}{3} (3r-h)$$


 * $$h = R - \sqrt{R^2 - a^2}$$

and in the lower hemisphere
 * $$h = R + \sqrt{R^2 - a^2}$$; hence in either hemisphere
 * $$a = \sqrt{h(2R-h)}$$ and so an alternative expression for the volume is


 * $$V = \frac {\pi h^2}{3} (3r-h)$$.

and the curved surface area of the spherical cap is


 * $$A = 2 \pi r h.$$


 * $$A_f = A_1 - A_2 = \pi b_1 s_1 - \pi b_2 s_2]$$
 * $$A_f = A_1 - A_2 = \pi[ b_1 s_1 - b_2 s_2]$$
 * $$A_f = \pi[ b_1 s_1 - b_2 s_2]$$
 * $$b_1 = h_1\tan\theta$$, is the area of the base
 * $$s_1 = \sqrt{b_1^2 + h_1^2}$$


 * $$s_1 = \sqrt{(b_1^2 - b_1^2)+ (h_1-h_2)^2}$$
 * $$A_f = \pi (b_1 + b_2) s$$


 * $$V = \frac{1}{3}A_B H$$

For a right circular cone, the surface area $$A$$ is
 * $$A =\pi b^2 + \pi b s\,$$

where
 * $$s = \sqrt{b^2 + h^2}$$

is the slant height. The first term in the area formula,
 * $$A = \pi b s\,$$
 * $$\pi b^2$$, is the area of the base, while the second term,
 * $$\pi b s$$, is the area of the lateral surface.
 * $$b = h\tan\theta$$, is the area of the base
 * $$b = h \tan \theta$$, is the area of the base

$$A = \pi b^2 s\,$$
 * $$ V = \int_0^H r^2 \pi \, dh $$

where r is the radius of the cone at height h measured from the apex:


 * $$ r= R \frac{h}{H} $$

$$A = \pi R S\,$$ $$A = \theta2 R S\,$$


 * $$ V = \int_0^H r^2 \pi \, dh $$

where r is the radius of the cone at height h measured from the apex:


 * $$ r= R \frac{h}{H} $$

Thus:


 * $$ V = \int_0^H \left[R \frac{h}{H}\right]^2 \pi \, dh $$

Thus:


 * $$V = \frac{1}{3} \pi R^2 H. $$

For a right circular cone, the surface area $$A$$ is
 * $$A =\pi b^2 + \pi b s\,$$  where   $$S = \sqrt{R^2 + H^2}$$   is the slant height.

The first term in the area formula, $$\pi b^2$$, is the area of the base, while the second term, $$\pi b s$$, is the area of the lateral surface.

A right circular cone with height $$h$$ and aperture $$2\theta$$, whose axis is the $$z$$ coordinate axis and whose apex is the origin, is described parametrically as
 * $$F(s,t,u) = \left(u \tan s \cos t, u \tan s \sin t, u \right)$$

where $$s,t,u$$ range over $$[0,\theta)$$, $$[0,2\pi)$$, and $$[0,h]$$, respectively.

H2SO4 &rarr; 2 H+ + SO42- gives: H2SO4 → 2 H+ + SO42-