How Mathcad Rocks: Mining (Part II)

mathcad mining

Alright, so last time, I started to blog about using Mathcad in the mining industry.  So let’s talk about some more about calculations that a mining engineer might make in a typical day.

Ventilation is a huge part of mining.  As mentioned in the previous blog, it’s sometimes referred to as “the lifeblood of a mine.”  Ventilation is not a problem specific to mining.  In fact, I recognized a lot of the upcoming equations from my fluids class.  Figuring out the pressure needed to circulate air into/through a mine shaft is no different from figuring out the pressure of a fluid as it flows through a pipe.  Different dimensions/shapes and different fluids, but same underlying principles.

Here’s an equation from Mining Engineering Analysis (by Christopher Bise) relating the pressure loss due to friction as air travels through an airway. 

mining04_headLoss

Note that the equation specifies the units of each variable.  The result, a measurement of pressure, is given in inches of water.  There is also a mysterious 5.2 factor in the denominator.  Now at first glance, one might think this is similar to the empirical relationship used to calculate the pillar strength mentioned in Part I of this blog.  The 5.2 is really a conversion factor that keeps the specified units balanced.  It also includes the density of water and gravitational acceleration to give the result in inches of water.  Like we showed in Part I of the blog, we can input the parameters with any unit we want.  Then using Mathcad’s built-in unit conversion, we can divide out the specified units with each term.

mining07_headLossUnits

Those of you keen-eyed readers might point out that the coefficient of friction K should technically have units as well (lb*min2/ft4).  I defined my K unitless here, but you get the idea.  Another way to approach this problem is using the Darcy-Weisbach Equation.

mining05_pressureLossDarcy

This equation is a bit different.  Previously, the volumetric flow rate Q was used.  Here we use V.  Of course, the two are related (Q=A*V).  The hydraulic diameter is used (D=4*O/A). And a friction factor f is used instead of K.  This friction factor is unitless.  You will note that the density of air shows up in this equation.  Interestingly, K and ρair have the same units, so you can think of K as a combination of the friction factor (f) and air density (ρair).  To calculate the head loss (in inches of water) from the pressure, we just divide out the weight density of water.

mining06_headLossDarcy

In my opinion, the Darcy-Weisbach form is cleaner to use and eliminates the need to strip out units.  There is a slight discrepancy between the two results as I used an approximate relationship to define the friction factor f from the K value.

Taking this one step further, we can relate the pressure P and flow rate Q by defining flow resistances R.

mining08_flowResistance

Let’s say we want to determine the pressure drop as air flows down a duct and around a bend.  From the example above, the resistance of flowing down a straight duct is:

mining09_flowResistanceTube

We can calculate the equivalent flow resistance of the bend by looking up the shock factor (X) associated with the bend angle and duct (airway) dimensions.

mining10_flowResistanceBend

The total resist is just the sum of the two and we can quickly calculate the total pressure loss.

mining11_flowResistanceTotal

You may have noticed that calculating pressures and flow rates in pipes is very similar to calculating voltage and current in electric circuits.  Imagine that you have a network of ducts and you want to determine the pressure drops between the intake and return shafts.  To make things more interesting, add a fan in one of the branches.  In the following example (take from the SME Mining Engineering Handbook), we have a schematic of a simple ventilation network.

mining12_ventNetwork

For the sake of this exercise, let’s say we have already figured out the flow resistances coming from each branch.  We can apply Kirchoff’s First and Second Laws to this network just as we would with an electric circuit.  The equations can be plugged into a solve block:

mining13_ventNetworkKirchoff

And the solution falls right out!  Solve blocks have been covered in other blogs, YouTube videos, PTC Communities submissions, etc., so I won’t talk about them at length here.  But I do want to point out that we can specify units in a solve block and get results with corresponding units.

mining14_ventNetworkSolution

Alright, now that we’ve talked about ventilation calculations, let’s move on to hoisting systems.  I’m going to pick a simple example problem again from Bise’s Mining Engineering Analysis.  In this problem, we’re asked to look at the power output from a motor as it hoists a car and load through a specified slope.  We need to consider the weight of the car, load, and even rope.  Various values – such as type of rope, type of drum, wheel/rope friction, motor efficiency – need to be looked up to complete the calculations.  But at the end of the day, this problem essentially reduces to a basic physics problem.  A force is needed to pull some mass at some specified rate.

I’m going to skip through the problem setup.  You can go through it in detail by downloading the worksheet (link at bottom).  The interesting plots are the following:

mining15_hoistAccel

For this particular problem, we specify acceleration, cruising (constant speed), and deceleration values and durations.  We can integrate the acceleration to get velocity:

mining16_hoistVelocity

And once more to get position:

mining17_hoistPosition

Next, we determine the amount of mass that is being hauled at any given time, as we’ll need that to determine the force required.

mining18_hoistForce

We see two “drops” in the force.  This corresponds with the two acceleration changes.  The force drops steadily throughout the entire duration as the mass of the rope is decreasing.  Once we have the force, we can multiply it by the velocity to get the power.

mining19_hoistPower

In Bise’s solution, the power profile is generated by breaking up each phase in a piece-wise fashion.  Being able to define acceleration and mass as functions of time allow us to determine the power at any point in time and therefore plot the power profile much more easily.

So there you have have it.  I think I’ve written as much as I know about what I learned on mining.  As I said before, most of these examples are not specific to mining and can be found in many disciplines.  I guess that’s the beauty of science and engineering.  Once you learn the fundamentals, you can easily apply it to any problem in any industry.

Links to the Mathcad worksheets I created on basic ventilation, ventilation network, and hoist design.

About Roger Yeh @ PTC

Roger is a mechanical engineer by training. He studied non-Newtonian fluid mechanics (specifically the extensional flow of polymer solutions) in graduate school. Since graduating, he has dabbled in systems engineering (integrating hardware and software components on a radar program) and test and measurement applications (across all industries). He joined PTC in March 2011 as an application engineer in the Mathcad Business Unit.
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One Response to How Mathcad Rocks: Mining (Part II)

  1. Pingback: How Mathcad Rocks: Mining (Part I) | PTC

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