Complex Engineering Problem_Fluid Mechanics

Problem Statement

The shell and tube type heat exchangers are commonly used in process industries. One of the major problems occurs in heat exchangers is pressure drop. As a design engineers you have to consider design parameters such that minimum pressure occurs due to flow rate.

A heat exchanger used in process industry consists of a square duct with inside dimensions of 600  

mm by 600 mm and contains four tubes with outside diameters of 210 mm each (Figure 1). A liquid used to transfer heat through the tubes has a density of 970 kgm^-3 and a viscosity of 8 × 10^-4 Nsm^-2, which flows around the outside of the tubes at a rate of 100 liter per second.

(a)  Determine the pressure drop along the length of the duct by the heat transfer liquid if the absolute roughness of the steel pipes and duct is 0.0046 mm.

(b) What would be the change in pressure drop, if the duct’s cross section is considered as circular?

Solution(a)

Given:

Flow rate, 𝑄_𝑖 = 100 l/s = 0.1 m³/s

Density, ρ = 970 kg/m³

Viscosity, µ = 8 x 10^-4 Ns/m²

Absolute surface roughness, ε = 0.0046 x 10^-3 m = 4.6 x 10^-6

Assumed Length of Duct, L = 20 m

Diameter of Pipes = 0.21 m

Height and Breath of Square duct = 0.6 m

Figure 1: Duct cross section with tubes

Figure 2: 3D mode of Square Duct Cross Section with Tubes. Shaded area depicts flow field region

Assumptions:

1. The flow is steady, incompressible and fully developed.
2. There are no bends, valves, and connectors throughout the pipe.
3. The Piping section involves no working devices such as pump or turbines

Firstly, we will find average velocity of fluid flow:

Cross sectional Area, 
A = Area of Square duct – 4(Area of circular pipes)

= (Breadth x width) – 4(πD2/4)

= (0.6 x 0.6) – (π(0.21)2)

Ac = 0.2214 m2

V = (0.1 m³/s) ÷ (0.2214 m²) ↔ V = 0.45167 m/s

Now, we will find the Reynolds Number to determine type of flow:

wetted perimeter, p = 4(0.6 m) + 4(π*0.21) = 5.0389 m

Note: The wetted perimeter in Figure 1, is the total sum of perimeter outside the circular tubes and the perimeter on the inside of the square duct.

Where D is the Hydraulic Diameter, 
D₁ = 4A_c/p

= 4(0.2214 m²)/(5.0389)

D₁ = 0.17575 m

Re = (970 kg/m³ x 0.45167 m/s x 0.17575 m) ÷ (8 x 10^-4 Ns/m²) =

Re = 9.624 x 10⁴ > 4000, Thus the flow is Turbulent and we will have to find the relative roughness of the pipe,

ε/D₁ = 4.6 x 10^-6/0.17575 = 26.1735 x 10^-6

Note: As the Colebrook equation is implicit in f, and thus the determination of the friction factor requires some iteration unless an equation solver such as EES used.

The Moody chart and its equivalent Colebrook equation involve several uncertainties (the roughness size, experimental error, curve fitting of data, etc.), and thus the results obtained should not be treated as “exact.” It is usually considered to be accurate to +/-15 percent over the entire range

An approximate explicit relation for f was given by S. E. Haaland in 1983 as,

Note: The results obtained from this relation are within 2 percent of those obtained from the Colebrook equation

Thus we get, f = 0.018068

To find the Pressure drop,

= [0.018068*20*970 kg/m³*(0.45167 m/s)²]/2*0.17575 m

= 0.018068*6837.789

              ΔP = 203.436 Pa

Note: If the Length of the Duct were not assumed then we would have to find Pressure Drop Per unit Length,

ΔP/L = 10.17 kg/m²s²

Solution (b)

If the Duct were of circular shape as shown in Figure 2,

Given:

Flow rate, 𝑄_𝑖 = 100 l/s = 0.1 m³/s

Density, ρ = 970 kg/m³

Viscosity, µ = 8 x 10^-4 Ns/m²

Absolute surface roughness, ε = 0.0046 x 10^-3 m = 4.6 x 10^-6

Length of Duct, L = 20 m

Diameter of Pipes = 0.21 m

Diameter of Circular Duct = 0.84853 m

Figure 3 Circular Duct with tubes

Figure 4 3D mode of Circular Duct Cross Section with Tubes. Shaded area depicts flow field region

Assumptions:

1. The flow is steady, incompressible and fully developed.
2. There are no bends, valves, and connectors throughout the pipe that could disrupt the flow.
3. The Piping section involves no working devices such as pump or turbines.

Firstly, we will find average velocity of fluid flow:

Cross sectional Area, 
A = Area of Circular duct – 4(Area of circular pipes)

= (πD²/4) – 4(πD²/4)

= (0. 25π(0.848)²) – 4(0.25π(0.21)²)

A = 0.42694 m²

V = (0.1 m³/s) ÷ (0.42694 m²) ↔ V = 0.23422 m/s

wetted perimeter, p = (π*0.84853) + 4(π*0.21) = 5.3046 m

Note: The wetted perimeter in Figure 1, is the total sum of perimeter outside the circular tubes and the perimeter on the inside of the square duct.

Where D is the Hydraulic Diameter,
 D₁ = 4A_c/p

= 4(0.42694 m²)/(5.3046)

D₂ = 0.3219 m

Now, we will find the Reynolds Number to determine type of flow:

Re = (970 kg/m³ x 0.23422 m/s x 0.3219 m) ÷ (8 x 10^-4 Ns/m²)

Re = 9.1416 x 10⁴ > 4000, Thus the flow is Turbulent and we will have to find the relative roughness of the pipe,

ε/D₂ = 4.6 x 10^-6/0.3219 = 14.29 x 10^-6

Note: As the Colebrook equation is implicit in f, and thus the determination of the friction factor requires some iteration unless an equation solver such as EES used.

The Moody chart and its equivalent Colebrook equation involve several uncertainties (the roughness size, experimental error, curve fitting of data, etc.), and thus the results obtained should not be treated as “exact.” It is usually considered to be accurate to +/-15 percent over the entire range.

An approximate explicit relation for f given by S. E. Haaland in 1983 is as,

Note: The results obtained from this relation are within 2 percent of those obtained from the Colebrook equation

Thus we get, f = 0.01835

To find the Pressure drop,

= [0.01835*20*970 kg/m³*(0.23422 m/s)²]/2*0.3219m

ΔP = 129.51 Pa

Note: If the Length of the Duct were not assumed then we would have to find Pressure Drop Per unit Length,

ΔP/L = 6.4756 kg/m²s²

Conclusion

1. The Cross-Sectional Area, Hydraulic Diameter would almost be increased twice
2. The Average velocity will be halved
3. The Reynolds Number will be reduced but the flow will still remain Turbulent.
4. There is a significant decrease in Pressure drop across the circular duct cross section with pipes.
5. The Fluid flow in Duct Cross Section on Figure 2 is much better than that in Figure 1 because the no Corners in the flow region that could develop vortices and cause rotational flow fields.
6. The corners in the Square Duct Cross Section may get clogged more that may cause Disruptions in Flow.
7. As depositions accumulate on the surface of the Duct/Pipe (esp. sharp edges) the smoothness of the Duct/Pipe in altered either due to erosion, cavitation or rusting of walls.

Contributions:

SaeeD Ur Rehman_2017-ME-354 (Group Leader)

Generated the 3D model on Solid Edge with Dimensions of the required Duct cross-section and Circular Duct with pipes for FEM that may be required in Future and Imported (JPEG) file for representations in word.

Provided Assumptions, Calculations, Conclusions and Effective Reasoning throughout the solution of the CEP.

Analyzed and Solved the CEP, Documented the entire CEP Solution on Word in a well-organized way.

Appointed Different Tasks to all team members for Distribution of Work Load, Smooth Team work and Timely Submission of CEP Solution to the Concerned Teacher.

Ehtasham Maqsood_2017-ME-352

Proposed the use of an approximate explicit relation for Frictional Factor, f given by S. E. Haaland in 1983 instead of Colebrook Equation or Moody Chart and identified reason for using S. E. Haaland equation.

Helped in providing Relevant Assumptions and Application of Use in society.

Irfan_2017-ME-355

Helped in Calculations of Numerical Results and conclusion.

Helped in Applicable Equations with their background information, Type text in Bold, Italics, and Highlights for easy identification of Important Notes, Emphasis, Titles, and Results.

Rabi_2017-ME-351

Revised the Entire Document for suggesting relevant modifications and correction