Non circular duct flow

Non-Circular Ducts:

For flow in non-circular ducts or ducts for which the flow does not fill the entire cross-section, we can define the hydraulic diameter   Dh   as

where A = cross-sectional area of actual flow, P = wetted perimeter, i.e. the perimeter on which viscous shear acts

With this definition, all previous equations for the Reynolds number, Re, friction factor,  f,  and head loss,  hf ,  are valid as previously defined and can be used on both circular and non-circular flow cross-sections.

Minor Losses

In addition to frictional losses for a length  L  of pipe, we must also consider losses due to various fittings (valves, unions, elbows, tees, etc.).  These losses are typically expressed as

                                           
where

          hm = the equivalent head loss across the fitting or flow component

          V = average flow velocity for the pipe size of the fitting

Ki = the minor loss coefficient for given flow component; valve, union, etc.
See Sec. 6.7, Table 6.5, 6.6, Fig. 6.19, 6.20, 6.21, 6.22, etc.

i.e.     Loss Coefficient  
Table 6.5 shows minor loss K values for several common types of valves, fully open, and for elbows and tees. 

Figure 6.18 shows minor loss K values for several types of common valves.

Note that the K valves shown here are for the indicated fractional opening. Also, the fully open values may not be consistent with values indicated in Table 6.5 for fully open valves or for the valve of a particular manufacturer.  In general, use specific manufacturer’s data when available.

These and other sources of data now provide the ability to determine frictional losses for both the pipe and other piping/duct flow components.

The total frictional loss now becomes

or

These equations would be appropriate for a single pipe size (with average velocity V).   For multiple pipe/duct sizes, this term must be repeated for each pipe size. 

Key Point: The energy equation must still be used to determine the total
         head loss and pressure drop from all possible contributions.

Example  6.16

Water, r = 1.94 slugs/ft3 and  n = 1.1 E-5 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2–in diameter pipe with e/D = 0.001 having the indicated minor losses.  Compute the pump horsepower (motor size) required.

Writing the energy equation between points 1 and 2 (the free surfaces of the two reservoirs), we obtain

For this problem, the pressure (P1 = P2) and velocity (V1 = V2 = 0) head terms are zero and the equation reduces to

For a flow rate Q = 0.2 ft3/s we obtain

With  e/D = 0.001  and 

The flow is turbulent and Haaland’s equation can be used to determine the friction factor:

The minor losses for the problem are summarized in the following table:
Substituting in the energy equation we obtain

Note the distribution of the total loss between static, pipe friction, and minor losses.

The power required to be delivered to the fluid is given by

If the pump has an efficiency of 70%, the power requirements would be specified by

Solution Summary:
To solve a basic pipe flow pressure drop problem, use the following procedure:

Multiple-Pipe Systems

Basic concepts of pipe system analysis apply also to multiple pipe systems.  However, the solution procedure is more involved and can be iterative.

Consider the following:
a.  Multiple pipes in series
b.  Multiple pipes in parallel

Series Pipe System:

1.  The flow rate is the same through each pipe section. For incompressible flow, this is expressed as

     Q1  =  Q2  =  Q3  =  Q    or    D12V1  =  D22V2  =  D32V3

2.  The total frictional head loss is the sum of the head losses through the various sections.

Note:  Be careful how you evaluate the transitions from one section to the next.  In general, loss coefficients for transition sections are based on the velocity of the smaller section.
Example:  Given a pipe system as shown in the previous figure.  The total pressure drop is Pa – Pb = 150 kPa and the elevation change is Zb – Za = – 5 m.  Given the following data, determine the flow rate of water through the section.

Since the flow rate  Q  and velocity are the only remaining variables, the solution is easily obtained from a spreadsheet by assuming Q until DP = 150 kPa.

Thus it is seen that a flow rate of 10.17 m3/hr produces the indicated head loss through each section and a net total DP = 150 kPa.

A solution can also be obtained by writing all terms explicitly in terms of a single velocity, however, the algebra is quite complex (unless the flow is laminar), and an iterative solution is still required.  All equations used to obtain the solution are the same as those presented in previous sections.

Parallel Pipe Systems

1.  The total flow entering the parallel section is equal to the sum of the flow rates through the individual sections
2.  The total pressure drop across the parallel section is equal to the pressure drop across each individual parallel segment.

Note that if a common junction is used for the start and end of the parallel section, the velocity and elevation change is also the same for each section.  Thus, the flow rate through each section must be such that the frictional loss is the same for each and the sum of the flow rates equals the total flow.

For the special case of no kinetic or potential energy change across the sections, we obtain:

ht  =  ( hf  +  hm)1  =  ( hf  +  hm)2  =( hf  +  hm)3

and

QT  =  Q1  +  Q2  +  Q3

Again, the equation used for both the pipe friction and minor losses is the same as previously presented.  The flow and pipe dimensions used for the previous example are now applied to the parallel circuit shown above.
Example:  A parallel pipe section consists of three parallel pipe segments with the lengths and diameters shown below.  The total pressure drop is 150 kPa and the parallel section has an elevation drop of 5 m.  Neglecting minor losses and kinetic energy changes, determine the flow rate of water through each pipe section.

The solution is iterative and is again presented in a spreadsheet.  The net friction head loss of 20.3 m now occurs across each of the three parallel sections.

The strong effect of diameter can be seen with the smallest diameter having the lowest flow rate, even though it also has the shortest length of pipe.

Again the total flow is distributed between the three parallel sections such that the  head loss across each section is the same, in this case 20.3 m.

Bernoulli Obstruction Theory

Assume:
1.  Steady - flow      2.  incompressible flow          3.  no friction losses
4.  uniform velocity profiles at 1 & 2                      5.  Z1 = Z2

For these conditions, Bernoulli’s Equation and conservation of mass become:
               

Substituting and solving for V2, we obtain:

The volumetric flow rate is now given by

The above equation predicts the flow for ideal (frictionless) flow conditions.

Define:  Cd = discharge or flow coefficient

Can now write the equation for the actual flow rate as:

Define:  E  = Velocity of approach factor

      where   
  =  ratio of the throat-minimum flow diameter to the main flow diameter

Also:  define   Y =  expansion factor =  factor which accounts for compressibility effects of the flow ( usually small).  Y = 1   no compressibility effects,  Y < 1, some compressibility effects.

Therefore, the final equation now becomes:

or:

For gases, the Expansion Factor, Y, is given by the expression:

Orifice Flow Rate Analysis

We will now outline the procedure required to use a given thin plate orifice to determine the flow rate based on orifice geometry and measured pressure drop. Data:  D, b, d, P1g, DP, T1, Pbar 1.  Calculate orifice constants:  g, At, E g (fluid),    At = p d2/4,

2.  If fluid is a gas, calculate the Expansion factor,  Y.  Otherwise, set Y = 1.
                   
3.  Calculate the Discharge Coefficient

where

The factors F1 and F2 vary with pressure tap position.
corner taps:             F1 = 0                   F2 = 0
D & 1/2 D taps:       F1 = 0.4333          F2 = 0.47
Flange taps:                      D > 2.3 in                 
2.0 ≤ D ≤2.3 in   

4.  Calculate Mass Flow Rate

Where for ideal gases:   r  =  P1/RT1
5.  Calculate the actual volumetric flow rate