In today’s lesson, we will discuss the following topics
1. Changes in Calculations: Noncircular Ducts
2. Flow Between Parallel Plates
3. Flow Between Parallel Plates – Laminar Flow Solution
4. Flow Between Parallel Plates – Turbulent Flow Solution
5. Example – Find Head Loss and Pressure Drop
6. Concentric Annulus Flow
7. Example – Find Reservoir Level through an Annulus
8. Example – Find Pressure Drop in other Cross-Sections
Note: All variables and their units are given in the first lesson plan for this course, linked here.
Changes in Calculations: Noncircular Ducts
There are some calculation changes that are made when a duct is non-circular. A non-circular duct is just an area (eg. a pipe) that does not have a circular cross-section, like parallel plates. Let’s go through the differences when we have flow in non-circular ducts. There are specific equations later in this lesson that discuss flow between parallel plates specifically though, but let’s go through general equations can be used with all types of non-circular ducts:
(a) If it is turbulent flow, the Logarithm Law velocity profile OR the hydraulic diameter (discussed in a previous lesson) can be used rather than solving the exact equations of continuity and momentum as you would for laminar flow.
(b) Cross sectional area is not pi*R^2 and the cross sectional perimeter is not 2*pi*R. Rather, a value called the length scale, as shown below, replaces R in the equations, which is now called Rh (hydraulic radius). The wetted perimeter includes all surfaces that the shear stress acts on (eg. In an annulus, which looks like a donut as shown below, we would include both the inside and outside perimeter of the circle):
(c) Shear stress (tau) is now taken as an average value around the perimeter of the duct. The equation for the darcy friction factor was discussed in this previous lesson, but now the equation takes the average shear stress tau (that is what the bar on top means). Note that NCD = Noncircular duct.
The head loss equation (discussed in the same lesson plan as the darcy friction factor linked above) is now equal to:
(d) The hydraulic diameter is defined as Dh, just like hydraulic radius was used instead of regular radius. Dh is equal to:
(e) The effective diameter, Deff, is equal to the following. It is a correction to the diameter and is similar to the hydraulic diameter:
Flow Between Parallel Plates
One of the most common flows in a non-circular duct is that between parallel plates, such as this image below. Since the fluid is not flowing through a circular pipe, it is in this section for non-circular ducts:
The length (that is used in the Reynolds Number, L), otherwise known as the hydraulic diameter in this case can be taken as this equation:
The pressure gradient is also constant, and is equal to:
Flow Between Parallel Plates – Laminar Flow Solution
When there is laminar flow between the parallel plates, there are specific equations that apply more so than the ones above, and the same applies to turbulent flow. These equations include the average velocity u, volume flow rate Q, volume flow rate per unit area V, wall shear stress Tw, head loss hf, laminar friction factor f lam and effective diameter Deff.
At a Reynolds Number of about >= 2000, flow transitions to turbulent.
Flow Between Parallel Plates – Turbulent Flow Solution
There are certain equations that can be used for turbulent flow between parallel plates.
(a) The Logarithm Law, discussed in a previous lesson plan, can be used to find the average velocity u:
(b) The average velocity V can be found:
(c) The friction factor f can be found, which is actually derived from the equation in part (b):
(d) The effective diameter Deff, which is almost the same as the hydraulic diameter Dh, can be calculated through the following equations:
Example – Find Head Loss and Pressure Drop
To solve this problem, I have written out Steps 1-4 below. Note that when calculating the Reynolds number, the L (length) used in the equation is the hydraulic diameter, Dh, in this case. Also note that we didn’t have any specific equations for pressure drop with turbulent flow between parallel plates, but we could use the pressure drop equation of p*g*h, which is a common equation in fluid flow (and which I used for part 4). The final answers are boxed, as well as the Reynolds Number I had calculated. I also made sure everything was in the correct units! The Moody Chart I used for this question is in the textbook or in this previous lesson plan linked here.
Concentric Annulus Flow
Consider flow through a concentric annulus, with the fluid flowing in the outer tube. Although it looks like a circle, it is considered a non-circular duct because the fluid is only in the outer part of the circle.
The velocity, volume flow, friction factor and hydraulic diameter are given by the following equations. The friction factor could also be rewritten as so, with the effective
Example – Find Reservoir Level through an Annulus
To solve this question, let’s write down the given variables, and go backwards. What equation will we use to find height? The energy equation works well here because it contains the variable z which is the height. This question indicates points 1 and 2 in the picture above, and the energy equation is taken at two different points. This is a hint that you should use the energy equation! I have set up the equation below and cancelled out terms that are not needed.
Now let’s find the variables that we need to fill into this equation:
Example – Find Pressure Drop in other Cross-Sections
To solve this question, we will need to use the same pressure drop equation we used in one of the examples above. But in this pressure drop equation, we don’t know the head loss due to friction, hf. To find hf, we need the friction factor f. But to find f, we need to know the Reynolds Number! So let’s calculate all of this below.
Before we start, one thing to note that is that we will use Table 6.4, given below, to find the ratio of the side lengths depending on our geometry. We have a square, so the side lengths will be in a 1:1 ratio. This means b/a = 1.0, and we will look at column 2 to obtain a corresponding value of 56.91. This is substituted into the equation given earlier in this lesson for the effective diameter, which would be equal to D,eff = (64/56.91)*Dh.
Now let’s go onto the steps to find the pressure drop:
