Fluid Mechanics Derivations

Minor Loss Equation

This section contains the derivation of the minor loss equation using the following figure as a reference. The derivation begins with a slightly simplified energy equation across the control volume shown. Our energy equation begins with hP and hT having been eliminated.

Example of flow expansion resulting in a minor loss

Fig. 25 This is the system we will use to derive the minor loss equation.

pinρg+zin+v¯in22g=poutρg+zout+v¯out22g+hL

Since the elevations at the center of the in and out control surfaces are the same, we can eliminate zin and zout. As we are considering such a small length of pipe, we will neglect the major loss component of head loss. Thus, hL=he+hf. The following three equations are all the same, simply rearranged to solve for he.

pinρg+v¯in22g=poutρg+v¯out22g+he
pinpoutρg=v¯out2v¯in22g+he
(16)he=pinpoutρg+v¯in2v¯out22g

This last equation has he as a function of four variables (pin,pout,vin, and vout); we would like it to be a function of only one. Thus, we will invoke conservation of momentum in the horizontal direction across our control volume to remove variables. The difference in momentum from the in point to the out point is driven by the pressure difference between each end of the control volume. We will be considering the pressure at the centroid of our control surfaces, and we will neglect shear along the pipe walls. After these assumptions, our momentum equation becomes the following:

Min,x+Mout,x=Fpin,x+Fpout,x
Such that:
Mx = momentum flowing through the control volume in the x-direction
Fpx = force due to pressure acting on the boundaries of the control volume in the x-direction

Recall that momentum is mass times velocity for solids, mv, with units of [M][L][T]. Since we consider water flowing through a pipe, there is not one singular mass or one singular velocity. Instead, there is a mass flow rate, or a mass per time indicated by m˙=ρQ, which has units of [M][T]. Therefore, the momentum for a fluid is ρQv¯. Applying the continuity Equation Q=v¯A, we get to the following equation for the momentum of a fluid flowing through a pipe which we will use in this derivation, M=ρv¯2A. The pressure force is simply the pressure at the centroid of the flow multiplied by the area the pressure is acting upon, pA.

To ensure correct sign convention, we will make each side of the equation negative for reasons discussed shortly. Since v¯in>v¯out, the left hand side will be MoutMin in order to be negative. The reduction in velocity from in to out causes an increase in pressure, therefore pinpout is negative. With these substitutions, the conservation of momentum equation becomes as follows:

MoutMin=pinpout
ρv¯out2Aoutρv¯in2Ain=pinAoutpoutAout

Note that the area term attached to pin is actually Aout instead of Ain, as one might think. This is because Aout=Ain. We chose our control volume to start a few millimeters into the larger pipe, which means that the cross-sectional area does not change over the course of the control volume.

Dividing both sides of the equation by Aoutρg, we obtain the following equation, which contains the very same pressure term as our adjusted energy equation above, Equation (16). This is why we chose a negative sign convention.

pinpoutρg=v¯out2v¯in2AinAoutg

Now, we combine the momentum, continuity, and adjusted energy equations:

Energyequation:he=pinpoutρg+v¯in2v¯out22g
Momentumequation:pinpoutρg=v¯out2v¯in2AinAoutg
Continuityequation:AinAout=v¯outv¯in

To obtain an equation for minor losses with just two variables, v¯in and v¯out.

he=v¯out2v¯in2v¯outv¯ing+v¯in2v¯out22g

Now we will combine the two terms. The numerator and denominator of the first term, v¯out2v¯in2v¯outv¯ing will be multiplied by 2 to become 2v¯out22v¯in2v¯outv¯in2g. The equation then looks like:

he=v¯out22v¯inv¯out+v¯in22g

Final Forms of the Minor Loss Equation

Factoring the numerator yields to the first ‘final’ form of the minor loss equation:

Firstform:he=(v¯inv¯out)22g

From here, the two other forms of the minor loss equation can be derived by solving for either v¯in or v¯out using the ubiquitous continuity Equation v¯inAin=v¯outAout:

Secondform:he=(1AinAout)2v¯in22g=Kev¯in22g,whereKe=(1AinAout)2
(17)Thirdform:he=(AoutAin1)2v¯out22g=Kev¯out22g,whereKe=(AoutAin1)2

You will often see Ke and Ke used without the e subscript, they will appear as K and K.

Being familiar with these three forms and how they are used will be of great help throughout the class. The third form is the one that is most commonly used.

We can convert the minor loss coefficient, Ke, into a flow contraction using the minor loss equation (see Equation (17))

(18)Ke=(1Πvc1)2

where Πvc is the ratio of the contracted area to the full expanded flow area.

(19)Πvc=AinAout

Given that Πvc is the ratio of the contracted area to the full expanded flow area we obtain the relationship between the contracted diameter, Dvc, and the expanded diameter, Dpipe.

(20)Dpipe=DvcΠvc

Solving Equation (18) for Πvc we obtain

(21)Πvc=1Ke+1

The extent of a flow contraction, Πvc can be estimated from the measured minor loss coefficient.