Given: `L=1.5" ft"`; `H=0.22" ft"`

Find: `Q="? " {:ft:}^3/s`

Q is a common term used to represent stream discharge and in the US, it is often reported as cubic feet per second. The units can be abbreviated as `{:ft:}^3/s " or " cfs`.

Solve: What do we "Know"?

A weir is a barrier installed in a stream that acts as a sort of mini-dam. It can be a variety of shapes but the most common is rectangular or v-notch. Each weir shape has an empirically-determined equation relating the dimensions of the weir, the depth of the water flowing over its base, and the stream's discharge. In this problem, you're given L, but the illustration below uses b. It's the same thing and a good example of how different people use different variables for the same term.

You may NOT know this at this time, but in Hydrology it's a common topic: the equation for a rectangular weir is `Q=3.33(L-0.2H)H^(3/2)` and will report a value in the given units cubed (to the third power).

`Q=3.33(1.5-0.2(0.22))0.22^(3/2)` 

`Q=0.50  cfs " or"  {:ft:}^3/s` 

Note that the answer is in cubic feet per second, which is a VOLUME past the weir per TIME. These units indicate a FLOW, an important distinction in Hydrology. Flow is volume per time! Think about other applications that include flow. For example, you can go to the hardware store and buy a pump that lists maximum gallons per minute or gallons per hour.

 The weir is a foot and a half wide (not very big) and only a few inches deep. Picture water flowing over such an opening. Do you think that is a lot or a little water? How much is a cubic foot? You might be surprised to know that there are almost seven and a half gallons in a cubic foot! Your answer suggests about 3-4 gallons of water are flowing over the weir every second. As you get more familiar with fluid dynamics and streams, you'll recognize that this answer is fine.  Good work!

Given: `M=3.3 xx 10^8" kg"`; `l=90" m"`; `w=90" m"` 

Find: `F="? N"`; `P="? KPa"`  

F is an obvious choice to represent Force. Newtons is the metric system unit of force. Remember that the metric system is the default system of units of science and accepted worldwide as such. A Newton is the equivalent of 1 kilogram times 1 meter per second squared `(1 (kg)/(m  s^2))`. In the US, many disciplines still cling to the English system, but as a student, you should continue to strive to move toward metric. The second part of this problem asks for the pressure applied, abbreviated P, in the units of Kilopascals, or a thousand Pascals. A Pascal is 1 Newton per square meter `(N/m^2)` or 1 kilogram per meter second squared `(1 (kg)/(m  s^2))`.

Solve: What do we "Know"?

In this problem, the foundation is described as 90 meters by 90 meters. The equation of Force is a simple one, F = Ma, or force equals mass times acceleration due to gravity. The equation for pressure is equally simple, P = F/A, or pressure equals force per unit area. We have everything we need for the first part, given the mass and we know that acceleration due to gravity is 9.81 meters per second squared `(9.81 m/s^2)`. However, that only gets us to force. To calculate pressure, we need the AREA, which here is simply the length times width of the concrete pad.

`A=90m * 90m`

`A=8100m^2`

`F=3.3 xx 10^8" kg" * 9.81 m/s^2` 

`F=3.2 xx 10^9 N` 

`P=(3.2 xx 10^9 N)/(8100 m^2)` 

`P=395","000  Pa` 

`P=400  KPa` 

Notice the change from Pascals to Kilopascals. While `4xx10^5` is not a terribly large number, it is easier to use kilopascals because it is less ungainly, especially if we are comparing like items. For example, if we were comparing the pressure exerted by the Pentagon in Washington, DC. It is a large building, but it is lower to the ground and spread over a larger area. But if we were comparing the pressure applied by a fly landing on your arm, we would need to re-think our choices!

Let's take the first part of the problem, force. A Newton is a kilogram times gravity, similar to "weight," or pounds, in the English system, though larger. In fact, a quick (and technically incorrect) comparison is that a Newton is about five times the weight of a pound. Our answer suggests that the Empire State Building is about three billion Newtons or about six hundred million pounds. That is a lot, but it seems like it's at least possible. Did your units work out? Certainly! For pressure, a kilopascal is about 7.5 mmHg, a common American unit for barometric pressure (the pressure of the air column above you). Under normal circumstances at sea level, New York City, barometric pressure is 760 mmHg. The Empire State Building is applying a pressure of about 3000 mmHg, which is also the equivalent of about 60 pounds per square inch on its foundation. That seems within the realm of possibility. Remember, we need to assume that the architects and builders were planning on the structure staying upright and not sinking for a long time. They had to engineer a foundation to withstand six hundred million pounds of pressure! Spreading that out to a reasonable 60 pounds per square inch seems like a good idea. Good work!

Given: `T=5" s"` 

Find: `L="?"  m`;  `U="?" m/s`  

We will use L as the term for wavelength, though you might also see the greek symbol lambda used frequently. Whereas T is an easy choice to represent time and L is obvious for length, U is less familiar to most as a representation of velocity. However, in Geology, U (or u) is commonly used for velocity because volume, another frequent term in Geology, also begins with V. 

Solve: What do we "Know"?

A question like this would be common in a Geomorphology course on a unit related to coastal processes. But it might also appear in an oceanography or hydrology course. The equation for waves, specifically "oscillatory" waves in open water is:

`L=(g*T^2)/(2*pi)`; `U=(g*T)/(2*pi)`

The g in the equation is gravity, and because the problem is in meters and seconds, it is appropriate to use 9.81 m/s².

`L=((9.81 m/s^2)*(5 s)^2)/(2*pi)`

`L=39 " m"`

`U=((9.81 m/s^2)*(5 s))/(2*pi)`

`U=7.8  m/s`

The wavelength calculated here is about 40 meters, or 130 feet. That might seem like a long wavelength, but remember that this is an open water wave, where wavelengths can be kilometers long, depending on wind speed and the great depths of the ocean and some lakes. But beach waves are commonly lapping up 3-5 meters apart. A calculation of 39 meters certainly falls within that range. For velocity, 7.8 m/s is quite fast, considering a small stream might flow at a rate of 1-2 m/s. However, consider open ocean water has no measurable friction with a bed, like a stream, and it is largely controlled by wind speeds. An average tropical storm, can have wind speeds in excess of 18 m/s. These solutions fall right in line with expected values. Good work!

Given: `u=1.5 m/s`;  `d=0.38  m`

Find: `Fr="?"`

The Froude number, applied in a shallow channel, is a ratio of flow velocity to flow "length". Froude number, named after William Froude, is dimensionless like most ratios, as we'll see below. Based on calculation of Froude, critical flow occurs at Fr = 1. Froude numbers less than 1 indicate "subcritical" and over 1 indicate "supercritical".

Solve: What do we "Know"?

The equation for a stream's criticality is simply:

`Fr=u/sqrt(g*d)`

The g in the equation is gravity, and because the problem is in meters and seconds, it is appropriate to use 9.81 m/s².

`Fr=(1.5 m/s)/sqrt((9.81 m/s^2)*(0.38  m))`

`Fr=0.78`

The Froude number here of 0.78 is definitely in the realm of reasonable. It is actually low, given the perspective of criticality. This flow is Subcritical. The flow is pretty shallow, a little over a foot and it isn't moving particularly fast at only about 2 m/s. However, if the flow velocity were 4 m/s, the Froude number would go over 1 and it would become supercritical and we would expect features like hydraulic jumps to form in the flow. Sediment entrainment and the development of bedforms along a stream are dependent upon the Froude ratio, so it is an important number to know when trying to understand sediment transport and analyzing a sedimentary depositional setting. Good work!

Given: `b=8" ft"`; `Deltah=1.33" ft"`;  `l=685" ft"`;  `K=251" ft/day"`;  `n=0.27` 

Many of these terms might seem odd or at least unfamiliar. Consider this problem as if you were taking a traditional Groundwater Hydrology course. Your instructor would likely help you recognize and define these different parameters. Conventionally, b is used to indicate aquifer thickness. `Deltah=h_2-h_1` (change in head) and l (length) are used as `(Deltah)/l` to indicate the hydraulic gradient. K is the hydraulic conductivity (from Darcy's Law) and n is the common variable used for porosity, sometimes with the subscript e for "effective porosity".

Find: `v="? " {:ft:}/(day)` 

Average linear velocity is simply a velocity term, often indicated by v or u, and sometimes with a subscript. But it will always have units of length per time such as m/s, ft/s, or ft/day.

Solve: What do we "Know"?

This problem is based on the fundamental equation, Darcy's Law, which states that flow (of flux) through a porous media (soil/rock/sediment) is inversely proportional to the hydraulic properties of the media, expressed as the hydraulic gradient. In other words, the more or less conductive a material is to water moving through it, the smaller or larger, respectively, will be the difference in head across distances. In terms of flux, we can use Darcy's Law like this: `Q=-K((Deltah)/(l))`

This is very important!  Notice that K is negative, which is the mathematical translation of being "inversely proportional". This negative sign accounts for the fact that the head (height of the groundwater) DROPS, or decreases, in the direction of flow, making `Deltah` also negative.

In terms of flow, Darcy's Law looks like this: `Q=-KA((Deltah)/(l))`

In this case, the A is the cross-sectional area that the water is flowing through. Because we are only given an aquifer thickness (one dimension) we can't calculate an area. However, we CAN substitute a "unit" width of 1, so that our area comes out in square length `(A = 8 ft^2)`. For this problem, you'll see that isn't necessary because we are solving for velocity.

The actual equation for average linear velocity, as asked for in this problem, is `v=q/n` and will report a value in the units of length per time. For this equation, `q` is Darcy's flux and it is related to `Q` (volume flux) as `q=Q/A`.

`q=Q/A` 

`q=(-KA(Deltah)/l)/A` 

Notice the areas (A) cancel each other out:

`q=-K(Deltah)/l` 

`q=-251 (ft)/(day)*((-1.33 "ft")/(685 "ft"))` 

`q=0.487 (ft)/(day)` 

`v=q/n`

`v=(0.487 (ft)/(day))/0.27` 

`q=1.8 (ft)/(day)` 

Groundwater moves more slowly than surface water in streams or even flowing over a surface like a road or the roof of your house. In fact, while a single rain drop might not travel very fast down a hillslope, it is probably moving down that slope in minutes to hours, depending on the length of the slope, its steepness, etc. But groundwater moves through porous media much more slowly. The forces acting upon it are much lower over longer distances and the water takes a "tortuous" path moving through pore spaces in soil, sediment, or rock. The term average LINEAR velocity takes into account the water's velocity over a straight LINE, too. So the answer here of 1.8 ft/day indicates that the water is moving less than 0.1 foot per hour. That might seem quite slow, but it IS a reasonable velocity for groundwater!