## Tuesday, October 25, 2011

### MrsDrPoe: The First Law of Thermodynamics, Part IV

Happy Thesis Tuesday!  Today we'll continue our discussion of the head form of the First Law of Thermodynamics with an example problem:

Water is moved from one large reservoir to another at a higher elevation.  The loss of available energy associated with 2.5 ft*ft*ft/s being pumped from sections (1) to (2) is 30.5 |V|*|V|/g, where |V| is the average velocity of water in the 8 in inside diameter piping involved.  Determine the amount of shaft power required to pump the water.

Given: Q = 2.5 ft*ft*ft/s, losses = 30.5 *|V|*|V|/g, zdelta = 50 ft, D = 8 in

Find: shaft power required

Assumptions: steady, laminar, 1D, incompressible, constant average velocity, and constant properties

Solution:

For this case, we do not need to consider the continuity equation if we set our problem up correctly.  So we will start with the head form of the energy equation:

(pout/gamma) + 0.5*|V|*|V|/g + zout = (pin/gamma) + 0.5*|V|*|V|/g + zin + hP - hL

We will chose "in" as a point at the surface of the lower tank and "out" as a point at the surface of the upper tank, which tells us:

pin = 0, pout = 0 since both surfaces are exposed to the atm
vin = 0, vout = 0 since we can assume that the water on the surfaces is still

zin = 0, zout = zdelta setting z = 0 at the surface of the bottom tank

In putting these values into the equation, we have:

zout = hP - hL where hL = losses

Since our equation for the losses is given as a function of the average velocity, we know that these losses are primarily due to the friction in the pipe.  We can determine this velocity since we know both the flow rate and the pipe diameter.  NOTE: Q = |V|*A is only valid for use with uniform velocity or average velocity.  We know that since there are friction losses, mu is not zero, our velocity is not uniform, so in order to use this equation, we must use average velocity, which we are given.

|V| = 4*Q/(pi*D*D) = 7.162 ft/s

hL = 30.5*|V|*|V|/g = 48.625 ft

Solving for hP:

hP = zout + hL = 98.625 ft
(since this answer is positive, work is introduced INTO the flow)

Now we can determine the necessary power from the pump head:

Wreq = hP*gamma*Q = 20,850 W = 27.958 hp

And that's the head form of the equation.  Next week, we'll look at the differential form of the First Law!