# 3.5: Problems

## Section 3.2

1. In Figure $$\PageIndex{1}$$ we have illustrated a body in the shape of a sphere located in the center of tube. The flow in the tube is laminar and the velocity profile is parabolic as indicated in the figure. Indicate how the shape of the sphere will change with time as the body is transported from left to right. Base your sketch on a cut through the center of the sphere that originally has the form of a circle. Keep in mind that the body does not affect the velocity profile.

2a. If the straight wire illustrated in Figure $$\PageIndex{2a}$$ has a uniform mass per unit length equal to $$\xi_{ o}$$, the total mass of the wire is given by

$mass = \xi_{ o} L \nonumber$

If the mass per unit length is given by $$\xi (x)$$, the total mass is determined by the following line integral:

$mass = \int_{x = 0}^{x = L}\xi (x) dx \nonumber$

For the following conditions

$\xi (x) = \xi_{ o} + \alpha \left(x - \frac{1}{ 2} L\right)^{2} \nonumber$

$\xi_{ o} = 0.0065 \ kg/m, \quad \alpha = 0.0017 \ kg/m^{ 3} , \quad L = 1.4 \ m \nonumber$

determine the total mass of the wire.

2b. If the flat plate illustrated in Figure $$\PageIndex{2b}$$ has a uniform mass per unit area equal to $${ \psi }_{ o}$$, the total mass of the plate is given by

$mass = \psi_{o} A = \psi_{o} L_1 L_2 \nonumber$

If the mass per unit area is given by $$\psi (x, y)$$, the total mass is determined by the area integral given by

$mass = \int_{A}\psi dA = \int_{y = 0}^{y = L_{ 2} } \int_{x = 0}^{x = L_{ 1} }\psi (x, y) dx dy \nonumber$

For the following conditions

$\psi (x, y) = \psi_{ o} + \alpha xy \nonumber$

$\psi_{ o} = 0.0065 \ kg/m^{ 2} , \quad \alpha = 0.00017 \ kg/m^{ 4} , \quad L_{1} = 1.4 \ m, \quad L_{2} = 2.7 \ m \nonumber$

determine the total mass of the plate.

2c. If the density is a function of position represented by

$\rho = \rho_{ o} + \alpha \left(x - \frac{1}{ 2} L_{1} \right) + \beta \left(y - \frac {1}{ 3} L_{2} \right)^{2} \nonumber$

develop a general expression for the mass contained in the region indicated by

$0 \leq x \leq L_{ 1} \quad 0 \leq y \leq L_{ 2} \quad 0 \leq z \leq L_{ 3} \nonumber$

## Section 3.3

3. To describe flow of natural gas in a pipeline, a utility company uses mass flow rates. In a 10 inch internal diameter pipeline, the flow is 20,000 $${ lb}_{ m} /{ hr}$$. The average density of the gas is estimated to be 10 kg/m$$^{3}$$. What is the volumetric flow rate in ft$$^{3}$$/s ? What is the average velocity inside the pipe in m/s?

4. For the coating operation described in Example $$3.3.1$$, we have produced an optical fiber having a diameter of 125 micrometers. The speed of the coated fiber at the take-up wheel is 4.5 meters per second and the desired thickness of the polymer coating is 40 micrometers. Assume that there is no change in the polymer density and determine the volumetric flow rate of the coating polymer that is required to achieve a thickness of 40 micrometers.

5. Slide coating is one of several methods for continuously depositing a thin liquid coating on a moving web. A schematic of the process is shown in Figure $$\PageIndex{3}$$. In slide coating, a liquid film flows down an inclined plate (called the slide) owing to a gravitational force that is balanced by a viscous force. In a subsequent course in fluid mechanics, it will be shown that the velocity profile upstream on the slide is given by

${ v}_{x} = \frac{\rho g h^{2} \sin \theta }{ \mu } \left[{y /h} - \frac {1}{ 2} \left({y / h} \right)^{2} \right] , \quad \text{ on the slide} \label{1a}$

Here $$y$$ is the distance perpendicular to the slide surface and $$h$$ is the thickness of the liquid film. Variations of the velocity, $${ v}_{x}$$, across the width of the slide can be ignored. In a steady operation all the liquid flowing down the slide is picked up by a vertical web moving at a constant speed, $$U_{ o}$$. Far downstream on the moving web, the velocity profile is given by

${ v}_{x} = U_{ o} - \frac{\rho g b^{2} }{ \mu } \left[{y / b} - \frac {1}{ 2} \left({y / b} \right)^{2} \right] , \quad \text{ on the moving web} \label{2a}$

and this velocity profile is illustrated in Figure $$\PageIndex{3}$$. Here one must note that the coordinate system used in Equation \ref{2a} is different from that associated with Equation \ref{1a}. In slide coating operations, the system is operated in a manner such that

$\frac{\rho g b^{2} }{ \mu } << U_{ o} \label{3a}$

and Equation \ref{2a} and be replaced with the approximation given by

${ v}_{x} = U_{ o} , \quad \text{ far downstream on the moving web} \label{4a}$

In this problem you are asked to carry out the following steps in the analysis of the slide coating process.

(a) Demonstrate that the flow on the slide can be expressed as

${ v}_{x} = 3 \langle { v}_{x} \rangle \left[{y / h} - \frac {1}{ 2} \left({y / h} \right)^{2} \right] \label{5a}$

To accomplish this, make use of Equation $$(3.3.6)$$ in the form

$Q_{ entrance} = \int_{A_{entrance} }\left|\mathbf{v} \cdot \mathbf{n}\right| dA , \quad \left|\mathbf{v} \cdot \mathbf{n}\right| = { v}_{x} \label{6a}$

and apply Equation $$(6)$$ of Example $$3.2.1$$.

(b) Construct an appropriate control volume and develop a macroscopic balance that will allow you to determine the thickness of the liquid film, $$b$$, on the moving web during steady operation.

6. One method for continuously depositing a thin liquid coating on a moving web is known as slot die coating and the process is illustrated in Figure $$\PageIndex{4}$$. In slot die coating, the liquid is forced through a slot of thickness $$d$$ and flows onto a vertical web moving at a constant speed, $$U_{ o}$$. The velocity profile in the slot is given by

${ v}_{x} = 6 \langle { v}_{x} \rangle \left[{y / d} - \left({y / d} \right)^{2} \right] \label{1b}$

and this profile is illustrated in Figure $$\PageIndex{4}$$. Variations of the velocity, $${ v}_{x}$$, across the width, $$w$$, of the slot can be ignored. In a steady operation all the feed liquid to the slot die is picked up by the web. As in the case of slide coating (see Problem 5), the fluid velocity on the moving web can be approximated by

${ v}_{x} = U_{ o} , \quad \text{ far downstream on the moving web} \label{2b}$

1. Select an appropriate control volume and construct the macroscopic balance that will allow you to determine the thickness, $$b$$, of the liquid film on the moving web during steady operation.
2. If far downstream on the moving web all the liquid in the coated film of thickness, b, moves at the velocity of the web, $$U_{ o}$$, determine the thickness of the coated film.
3. If the gap, $$d$$, of the die slot changes by 10% and the average velocity remains constant, how much will the final thickness, $$b$$, of the coated film change?

7. If the delivery charge for the propane tank described in Example $$3.3.2$$ is $37.50, and the cost of the next largest available tank is$2500 (for a 2.2 cubic meter tank), how long will it take to recover the cost of a larger tank?

8. The steady-state average residence time of a liquid inside a holding tank is determined by the ratio of the volume of the tank to the volumetric flow rate of liquid into and out of the tank, $$\tau = \mathscr{V}/Q$$. Consider a cylindrical tank with volume $$\mathscr{V} = 3 \ m^{ 3}$$ and an input mass flow rate of water of 250 kg/minute. At steady-state, the output flow rate is equal to the input flow rate. What is the average residence time of water in the tank? What would be the average residence time if the mass flow rate of water is increased to 300 kg/minute? What would be the average residence time if a load of 1.2 $$m^{3}$$ of stones is dropped into the holding tank?

9. Mono Lake is located at about 6,000 ft above sea level on the eastern side of the Sierra Nevada mountains, and a simple model of the lake is given in Figure $$\PageIndex{5a}$$. The environment is that of a high, cold desert during the Winter, a thirsty well during the Spring runoff, and an cornucopia of organic and avian life during the Summer. Mono Lake is an important resting place for a variety of birds traveling the flyway between Canada and Mexico and was once the nesting place of one-fourth of the world’s population of the California gull.

The decline of Mono Lake began in 1941 when Los Angeles diverted the water from four of the five creeks flowing into the lake and sent 56,000 acre-feet per year into the Owens River and on to the Los Angeles aqueduct. By coincidence, the surface area of the lake in 1941 was 56,000 acres. The fall of Mono Lake was apparently10 secured in 1970 with the completion of a second barrel of the already-existing Los Angeles aqueduct from the southern Owens Valley. This allowed for a 50% increase in the flow, and most of this water was supplied by increased diversions from the Mono Basin. To be definitive, assume

that the export of water from the Mono Basin was increased to 110,000 acre-feet per year in 1970. Given the conversion factor

$1 \text{ acre-foot} = 43,560 \ ft^{ 3} \nonumber$

one finds that $$4.79\times 10^{9}$$ $${ ft}^{ 3}$$ of water are being removed from the Mono Basin each year. In 1970 the surface area of the lake was $$185\times 10^{6} { m}^{ 2}$$ and the maximum depth was measured as 50 m. If the lake is assumed to be circular with the configuration illustrated in Figure $$\PageIndex{5a}$$ we can deduce that the tangent of $$\theta$$ is given by $$\tan \theta = h(t) / r(t) = 6.52\times 10^{-3}$$.

In this problem you are asked to determine the final or steady-state condition of the lake, taking into account the flow of water to Los Angeles. The control volume to be used in this analysis is illustrated in Figure $$\PageIndex{5b}$$ and one needs to know the rate of evaporation in order to solve this problem. The rate of

evaporation from the lake depends on a number of factors such as water temperature, salt concentration, humidity and wind velocity, and it varies considerably throughout the year. It appears that the rate of evaporation from Mono Lake is about 36 inches per year. This represents a convenience unit and in order to determine the actual mass flux, we write

$\dot{m}_{ 2} = \left\{\begin{array}{c} \text{ mass flow rate owing} \\ \text{ to evaporation}\end{array}\right\} = \rho_{\ce{H2O}} \beta \left(\begin{array}{c} \text{ surface area} \\ \text{ of the lake} \end{array}\right) \nonumber$

in which $$\beta$$ is the convenience unit of inches per year. This parameter should be thought of as an average value for the entire lake, and for this problem $$\beta$$ should be treated as a constant.

## Section 3.4

10. A cylindrical tank having a diameter of 100 ft and a height of 20 ft is used to store water for distribution to a suburban neighborhood. The average water consumption (stream 2 in Figure $$\PageIndex{6}$$) during pre-dawn hours (midnight to 6 AM) is about 100 m$$^{3}$$/hr. From 6 AM to 10 AM the average water consumption increases to 500 m$$^{3}$$/hr, and then diminishes to 300 m$$^{3}$$/hr from 10 AM to 5 PM. During the night hours, from 5 PM to midnight, the average consumption falls even lower to 200 m$$^{3}$$/hr. The tank is replenished using a line (stream 1) that delivers water steadily into the tank at a rate of 1120 gal/min. Assuming that the level of the tank at midnight is 3 m, plot the level of the tank for a 24 hour period.

11. The 7$$^{th}$$ Edition of Perry’s Chemical Engineering Handbook11 gives the following formula to compute the volume of liquid inside a partially filled horizontal cylinder:

$V = \pi L R^{2} - \frac{L R^{2} }{ 2} \left(\alpha - \sin \alpha \right) \label{1c}$

Here $$L$$ is the length of the cylinder and $$R$$ is the radius of the cylinder. The angle, $$\alpha$$, illustrated in Figure $$\PageIndex{7a}$$, is measured in radians. In this problem we wish to determine the depth of liquid in the tank, $$h$$, as a function of time when the net flow into the tank is $$Q$$. This flow rate is positive when the tank is

being filled and negative when the tank is being emptied. The depth of the liquid in the tank is given in terms of the angle $$\alpha$$ by the trigonometric relation

$h = R + R\cos \left({\alpha / 2} \right) \label{2c}$

It follows that $$h = 0$$ when $$\alpha = 2\pi$$ and $$h = 2R = H$$ when $$\alpha = 0$$.

#### Part I.

Choose an appropriate control volume and show that the macroscopic mass balance for a constant density fluid leads to

$\frac{d\alpha }{ dt} = \frac{2Q}{ LR^{2} \left(\cos \alpha - 1\right)} , \quad Q = Q_{in} - Q_{out} \label{3c}$

#### Part II.

Given an initial condition of the form

IC $\alpha = \alpha_{ o} , \quad t = 0 \label{4c}$

show that the implicit solution for $$\alpha (t)$$ is given by

$\sin \alpha (t) - \alpha (t) = \sin \alpha_{ o} - \alpha_{ o} + \frac{2Q t}{ LR^{2} } \label{5c}$

This equation can be solved using the methods described in Appendix B in order to determine $$\alpha (t)$$ which can then be used in Equation \ref{2c} to determine the fluid depth, $$h(t)$$.

#### Part III.

In Figure $$\PageIndex{7b}$$ values of $$\alpha (t)$$ are shown as a function of the dimensionless time, $$\left|Q\right|t/\pi LR^{2}$$ for $$\alpha_{ o} = 0$$. The curve shown in Figure $$\PageIndex{7b}$$ represents values of $$\alpha (t)$$ when the tank is being drained, while the curve shown in Figure $$\PageIndex{7c}$$ represents the values of $$\alpha (t)$$ when the tank is being filled.

The length and radius of the tank under consideration are given by

$L = 8 \ ft , \quad R = 1.5 \ ft \label{6c}$

##### Part IIIa.

Given the conditions, $$Q = - 0.45$$ gal/min and $$\alpha = 0$$ when $$t = 0$$, use Figure $$\PageIndex{7b}$$ to determine the time required to completely drain the tank.

##### Part IIIb.

If the initial depth of the tank is $$h = 0.6{ ft}$$ and the net flow into the tank is $$Q = 0.55$$ gal/min, use Figure $$\PageIndex{7c}$$ to determine the time required to fill the tank. While Figure $$\PageIndex{7c}$$ has been constructed on the basis that $$\alpha = 2\pi$$ when $$t = 0$$, a little thought will indicate that it can be used for other initial conditions.

12. A cylindrical tank of diameter $$D$$ is filled to a depth $$h_{ o}$$ as illustrated in Figure $$\PageIndex{8}$$. At $$t = 0$$ a plug is pulled from the bottom of the tank and the volumetric flow rate through the orifice is given by what is sometimes known as Torricelli’s law12

$Q = C_{d} A_{ o} \sqrt{2\Delta p / \rho } \label{1d}$

Here $$C_{d}$$ is a discharge coefficient having a value of 0.6 and $$A_{ o}$$ is the area of the orifice. If the cross-sectional area of the tank is large compared to the area of the orifice, the pressure in the tank is essentially hydrostatic and $$\Delta p$$ is given by

$\Delta p = \rho g h \label{2d}$

where $$h$$ is the depth of the fluid in the tank. This leads to Torricelli’s law in the form

$Q = C_{d} A_{ o} \sqrt{2gh} , \quad \text{ hydrostatic conditions} \label{3d}$

Use this information to derive an equation for the depth of the fluid as a function of time. For a tank filled with water to a depth of 1.6 m having a diameter of 20 cm, how long will it take to lower the depth to 1 cm if the diameter of the orifice is 3 mm?

13. The system illustrated in Figure $$\PageIndex{9}$$ was analyzed in Example $$3.4.2$$ and the depth at a single specified time was determined using the bisection method. In this problem you are asked to repeat the type of

calculation presented in Example $$3.4.2$$ applying methods described in Appendix B. Determine a sufficient number of dimensionless times so that a curve of $$x = {\beta \sqrt{h} / \alpha }$$ versus $${\theta = \beta^{2} t / 2\alpha }$$ can be constructed.

Part (a). The bi-section method

Part (b). The false position method

Part (c). Newton’s method

Part (d). Picard’s method

Part (e). Wegstein’s method

14a. When full, a bathtub contains 25 gallons of water and the depth of the water is one foot. If the bathtub is filled with water from a faucet at a flow rate of 10 liters per minute, how long will it take to fill the bathtub?

14b. Suppose the bathtub has a leak and water drains out of the bathtub at a rate given by Torricelli’s law (see Problem 12)

$Q_{leak} = C_{d} A_{ o} \sqrt{2gh} \label{1e}$

Here $$h$$ is the depth of water in the bathtub, and $$C_{d}$$ represents the discharge coefficient associated with the area of the leak in the bathtub, $$A_{ o}$$. Since neither $$C_{d}$$ nor $$A_{ o}$$ are known, we express Equation \ref{1e} as

$Q_{leak} = k\sqrt{h} \label{2e}$

in which $$k = C_{d} A_{ o} \sqrt{2g}$$. To find the value of $$k$$ we have a single experimental condition given by

$Q_{leak} = 3.16\times 10^{-5} m^{3} /s , \quad h = 0.10m \label{3e}$

Given the experimental value of $$k$$, assume that the cross section of the bathtub is constant and determine how long it will take to fill the leaky bathtub.

15. The flow of blood in veins and arteries is a transient process in which the elastic conduits expand and contract. As a simplified example, consider the artery shown in Figure $$\PageIndex{10}$$. At some instant in time, the inner radius has a radial velocity of 0.012 cm/s. The length of the artery is 13 cm and the volumetric flow rate at the entrance of the artery is 0.3 cm$$^3$$/s. If the inner radius of the artery is 0.15 cm, at the particular instant of time, what is the volumetric flow rate at the exit of the artery?

16. A variety of devices, such as ram pumps, hydraulic jacks, and shock absorbers, make use of moving solid cylinders to generate a desired fluid motion. In Figure $$\PageIndex{11}$$ we have illustrated a cylindrical rod entering a cylindrical cavity in order to force the fluid out of that cavity. In order to determine the force acting on the cylindrical rod, we must know the velocity of the fluid in the annular region. If the density of the fluid can be treated as a constant, the velocity can be determine by application of the macroscopic mass balance and in this problem you are asked to develop a general representation for the fluid velocity.

17. In Figure $$\PageIndex{12}$$ we have illustrated a capillary tube that has just been immersed in a pool of water. The water is rising in the capillary so that the height of liquid in the tube is a function of time. Later, in a course on fluid mechanics, you will learn that the average velocity of the liquid, $$\langle { v}_{z} \rangle$$, can be represented by the equation

$\underbrace{ 2\sigma / { r}_{ o} }_{\text{capillary force}} - \underbrace{ \rho gh }_{\text{gravitational force}} = \underbrace{ \frac{8\mu \langle { v}_{z} \rangle h}{ r_o^{2} } }_{\text{viscous force}} \label{1f}$

in which $$\langle { v}_{z} \rangle$$ is the average velocity in the capillary tube. The surface tension $$\sigma$$, capillary radius $$r_{ o}$$, and fluid viscosity $$\mu$$ can all be treated as constants in addition to the fluid density $$\rho$$ and the gravitational constant $$g$$. From Equation \ref{1f} it is easy to deduce that the final height (when $$\langle { v}_{z} \rangle = 0$$) of the liquid is given by

$h_{\infty } = {2\sigma / \rho g r_{ o} } \label{2f}$

In this problem you are asked to determine the height $$h$$ as a function of time13 for the initial condition given by

I.C. $h = 0, \quad t = 0 \label{3f}$

Part (a). Derive a governing differential equation for the height, $$h(t)$$, that is to be solved subject to the initial condition given by Equation \ref{3f}. Solve the initial value problem to obtain an implicit equation for $$h(t)$$.

Part (b). By arranging the implicit equation for $$h (t)$$ dimensionless form, demonstrate that this mathematical problem is identical in form to the problem described in Example $$3.4.2$$ and Problem 13.

Part (c). Use the bisection method described in Example $$3.4.2$$ and Appendix B1 to solve the governing equation in order to determine $$h(t)$$ for the following conditions:

$\frac{\mu }{ \rho } = 0.010 \ cm^{ 2}/s, \quad g = { 980} \ cm / { s}^{ 2}, \quad r_{ o} = 0.010 \ cm , \quad \sigma = { 70 } \ dyne/cm , \quad \rho = { 1} \ g/cm^{ 3} \nonumber$

If a very fine capillary tube is available ($$r_{ o} = 0.010$$ $$cm$$), you can test your analysis by doing a simple experiment in which the capillary rise is measured as a function of time.

18. In Figure $$\PageIndex{13}$$ we have illustrated a cross-sectional view of a barge loaded with stones. The barge has sprung a lead as indicated, and the volumetric flow rate of the leak is given by

$Q_{leak} = C_{d} A_{ o} \sqrt{g (h-h_{i} )} \nonumber$

Here $$C_{d}$$ is a discharge coefficient equal to 0.6, $$A_{ o}$$ is the area of the hole through which the water is

leaking, $$h$$ is the height of the external water surface above the bottom of the barge, and $$h_{i}$$ is the internal height of the water above the bottom of the barge. The initial conditions for this problem are

I.C. $h = h_{ o} , \quad h_{i} = 0 , \quad t = 0 \nonumber$

and you are asked to determine when the barge will sink. The length of the barge is $$L$$ and the space available for water inside the barge is $$\varepsilon HwL$$. Here $$\varepsilon$$ is usually referred to as the void fraction and for this particular load of stones $$\varepsilon = 0.35$$.

In order to solve this problem you will need to make use of the fact that the buoyancy force acting on the barge is

$\text{ buoyancy force } = (\rho gh) wL \nonumber$

where $$\rho$$ is the density of water. This buoyancy force is equal and opposite to the gravitational force acting on the barge, and this is given by

$\text{ gravitational force } = mg \nonumber$

Here $$m$$ represents the mass of the barge, the stones, and the water that has leaked into the barge. The amount of water that has leaked into the barge is given by

$\text{ volume of water in the barge } = \varepsilon h_{i} (wL) \nonumber$

Given the following parameters:

$w = 30 \ ft , \quad L = 120 \ ft , \quad A_{o} = 0.03 \ ft^{2} , \quad h_{o} = 8 \ ft , \quad H = 12 { ft} \nonumber$

determine how long it will take before the barge sinks. You can compare your solution to this problem with an experiment done in your bathtub. Fill a coffee can with rocks and weigh it; then add water and weigh it again in order to determine the void fraction. Remove the water (but not the rocks) and drill a small hole in the bottom. Measure the diameter of the hole (it should be about 0.1 cm) so that you know the area of the leak, and place the can in a bathtub filled with water. Measure the time required for the can to sink.

19. The solution to Problem 9 indicates that the diversion of water from Mono Lake to Los Angeles would cause the level of the lake to drop 19 meters. A key parameter in this prediction is the evaporation rate of 36 in/year, and the steady-state analysis gave no indication of the time required for this reduction to occur. In this problem you are asked to develop the unsteady analysis of the Mono Lake water balance. Use available experimental data to predict the evaporation rate, and then use your solution and the new value of the evaporation rate to predict the final values of the radius and the depth of the lake. You are also asked to predict the number of years required for the maximum depth of the lake to come within 10 cm of its final value. The following information is available:

Table $$\PageIndex{1}$$

Year Surface Area Maximum Depth
prior to 1941 56 000 acres 181.5 ft.
1970 45 700 acres 164.0 ft.

During these years between 1941 and 1970 the diversion of water from Mono Lake Basin was 56,000 acre-ft. per year and in 1970 this was increased to 110,000 acre-ft. per year. The additional water was obtained from wells in the Mono Lake Basin, and as an approximation you can assume that this caused a decrease in $$\dot{m}_{4}$$ (see Figure $$\PageIndex{5b}$$) by an amount equal to 54 000 acre-ft. per year. Your analysis will lead to an implicit equation for $$\beta$$ and the methods described in Appendix B can be used to obtain a solution. Develop a solution based on the following methods:

Part (a). The bisection method

Part (b). The false position method

Part (c). Newton’s method

Part (d). Picard’s method

Part (e). Wegstein’s method

20. During the winter months on many campuses across the country, students can be observed huddled in doorways contemplating an unexpected downpour. In order not to be accused of idle ways, engineering students will often devote this time to the problem of estimating the speed at which they should run to their next class in order to minimize the unavoidable soaking. This problem has such importance in the general scheme of things that in March of 1973 it became the subject of one of Ann Landers’ syndicated columns entitled “What Way is Wetter?” Following typical Aristotelian logic, Ms. Landers sided with the common sense solution, “the faster you run, the quicker you get there, and the drier you will be.” Clearly a rational analysis is in order and this can be accomplished by means of the macroscopic mass balance for an arbitrary moving control volume.

In order to keep the analysis relatively simple, the running student should be modeled as a cylinder of height $$h$$ and diameter $$D$$ as illustrated in Figure $$\PageIndex{14}$$. The rain should be treated as a continuum with the mass flux of water represented by $$\rho\mathbf{v}$$. Here the density $$\rho$$ will be equal to the density of water multiplied by the volume fraction of the raindrops and the velocity $$\mathbf{v}$$ will be equal to the velocity of the raindrops. The velocity of the student is given by $$\mathbf{w}$$, and both $$\mathbf{v}$$ and $$\mathbf{w}$$ should be treated as constants. You should consider the special case in which $$\mathbf{v}$$ and $$\mathbf{w}$$ have no component in the $$y$$-direction, and you should separate your analysis into two parts. In the first part, consider $$\mathbf{i}\cdot \mathbf{v} > 0$$ as indicated in Figure $$\PageIndex{14}$$, and in the second part consider the case indicated by $$\mathbf{i} \cdot \mathbf{v} < 0$$. In the first part one needs to consider both $$\mathbf{i} \cdot \mathbf{v} > \mathbf{i} \cdot \mathbf{w}$$ and $$\mathbf{i} \cdot \mathbf{v} < \mathbf{i} \cdot \mathbf{w}$$. When $$\mathbf{i} \cdot \mathbf{v} > \mathbf{i} \cdot \mathbf{w}$$, the runner gets wet on the top and back side.

On the other hand, when the runner is moving faster than the horizontal component of the rain, the runner gets when on the top and the front side. When $$\mathbf{i}\cdot\mathbf{v} <0$$ the runner only gets wet on the top and the front side and the analysis is somewhat easier. In your search for an extremum, it may be convenient to represent the accumulated mass in a dimensionless form according to

$\frac{m(t) - m_{ o} }{ \rho LDh} = \mathscr{F}\left(parameters\right) \nonumber$

in which $$t$$ is equal to the distance run divided by $$\mathbf{i}\cdot \mathbf{w} = u_{ o}$$, and $$m_{ o}$$ is the initial mass of the runner.

1. This result is not applicable to bodies moving at velocities close to the speed of light. See Hurley, J.P. and Garrod, C. 1978, Principles of Physics, Houghton Mifflin Co., Boston.↩
2. The definition, physical significance and evaluation of line integrals, area integrals and volume integrals is an important part of every calculus text. Review of prerequisite course material is normal and to be expected.↩
3. Denn, M.M. 1980, Continuous Drawing of Liquids to Form Fibers, Ann. Rev. Fluid. Mech. 12, 365 - 387.↩
4. Stein, S.K. and Barcellos, A. 1992, Calculus and Analytic Geometry, McGraw-Hill, Inc., New York.↩
5. See Sec. 17.1 of Stein, S.K. and Barcellos, A. 1992, Calculus and Analytic Geometry, McGraw-Hill, Inc., New York↩
6. . See Sec. $$2.4$$ for a discussion of convenience units.↩
7. Sandler, S.I. 2006, Chemical, Biochemical, and Engineering Thermodynamics, 4$$^{th}$$ edition, John Wiley and Sons, New York.↩
8. See Sec. $$2.1.1$$ for a discussion of molecular mass.↩
9. Rouse, H. and Ince, S. 1957, History of Hydraulics, Dover Publications, Inc., New York.↩
10. Details concerning the fight to save Mono Lake are available at http://www.monolake.org, and a description of a similar problem at the Aral Sea is available at www.worldsat.ca/image_gallery/aral_sea.html.↩
11. Perry, R. H., Green, D. W., and Maloney, J. O. 1997, Perry's Chemical Engineer' Handbook, 7$$^{th}$$ Edition, McGrawHill Books, New York,↩
12. Rouse, H. and Ince, S. 1957, History of Hydraulics, Dover Publications, Inc., New York↩
13. Levich, V.G. 1962, Physicochemical Hydrodynamics, Prentice-Hall, Inc., Englewood Cliffs, N.J.↩