1. FLUID FOW FOR CHEMICAL ENGINEERING
EKC 212
FLUID FOW FOR CHEMICAL ENGINEERING
(ALIRAN BENDALIR KEJURUTERAAN KIMIA)
CHAPTER 7 (PART 1)
Dr Mohd Azmier Ahmad
Tel:

2. Chapter 7 FLOW PAST IMMERSED BODIES
The situation where the solid is immersed in & surrounded by fluid.
Fluid may be at rest & solid moving through it; or solid may be at rest & fluid flowing past it or both may be moving.
Additional friction losses occur for flowing fluid which is not flowing parallel to the solid surface (e.g. sphere).

3. Goals Describe forces that act on a particle in a fluid.
Define and quantify the drag coefficient for spherical and non-spherical objects in a flow field.
Define Stokes’ and Newton’s Laws for flow around spheres.

4. Flow Around Objects
There are many processes that involve flow through a porous medium such as a suspension of particles:
Packed Bed Chemical Reactor
Food Industry
Oil Reservoirs

5. DRAG FORCE Force in the direction of flow.
When wall of the body is parallel with the direction of flow (e.g. thin flat plate) the only drag force is the wall shear, τw.
Shear drag is due to viscous friction as liquid flows past surface of object.
Form drag is due to liquid changing direction as it flows past the object.
Wall shear

6. WALL DRAG & FORM DRAG ON IMMERSED OBJECT
Fluid pressure acts in a direction normal to the wall.
Element of area, dA inclined at an angle of α to the direction of fluid flow.
Drag from the wall shear, Fw = τw sin α dA
Drag due to pressure, Ff = p cos α dA
Total drag on the entire body is ;
Wall shear & pressure

7. DRAG COEFFICIENT, Cd
Cd is similar to that of the friction factor in pipes.
Cd defined as ratio of drag force to the velocity head and density :
Fd = drag force, N
ρf = fluid density, kg/m3
Dp = particle diameter, m
Vo = fluid approach velocity, m/s
Ap = projected area
= L x Dp (for perpendicular cylinder)
= (π/4)Dp2 (for parallel cylinder & sphere)

8. Projected area
Drag coefficient & drag force depend on shape & orientation of body

9. FOR SPHERE IMMERSED IN A FLOWING LIQUID
At least 3 regions :
NRe < 1 (laminar); Cd = 24 / NRe
1 < NRe < 1000 (transition); Cd = 18 NRe-0.6
1000 < NRe < 2x105 (turbulent); Cd = 0.44
Applicable for cylinders and disks
For laminar flow:
………..(1)
Projected area for sphere :
………..(2)
Drag coefficient :
………..(3)

10. For small NRe < 1.0, Stoke’s law can be applied :
………..(4)
Combine (1-4):
………..(5)
The overall relationship between Cd & NRe for different type of object is shown in the following diagram.

11. Fig. 7.3 : Drag coefficients of typical shapes
The axis of the cylinder and the face of the disk are perpendicular to the direction of flow.

12. B : stagnation point, velocity = 0 C : separation point
NRe < 1.0 (Stoke’s law) : at low velocities, the sphere moves through the fluid by deforming it. The flow pattern behind & front is same.
NRe > 1.0, separation occurs at a point just forward of the equatorial plane. A wake is formed (covering the entire hemisphere) contribute to large form drag.
NRe > (turbulent), separation point moves toward the rear of the body and the wake shrinks. The Cd decrease to ~ 0.1.
Laminar flow
Turbulent flow
B : stagnation point, velocity = 0
C : separation point

13. Example 7.1
Air at 37.8oC and kPa flows past a sphere having diameter of 42 mm at a velocity of 23 m/s. What is the drag coefficient and the drag force on the sphere.
Given : ρair = kg/m3; μair = 1.90 x 10-5 Pa.s
Solution
From Fig. 7.3, when NRe = 5.78x104, Cd = 0.47

14. Example 7.2
Air at 37.8oC and kPa flows past a sphere having diameter of 30 mm at a velocity of 45 m/s. What is the drag coefficient and the drag force on the sphere (ρair = kg/m3; μair = 1.90 x 10-5 Pa.s).
Solution
From Fig. 7.3, when NRe = 8.08x104, Cd = 0.5

15. Example 7.3
Water at 24oC is flowing past a long cylinder at a velocity of 1.0 m/s in a large tunnel. The axis of the cylinder is perpendicular to the direction of flow. The diameter of the cylinder is 0.09 m. What is the force per meter length of the cylinder (ρwater = kg/m3; μwater = x 10-3 Pa.s).
Solution
From Fig. 7.3, when NRe = 9.817x104, Cd = 1.4

16. Example 7.4
Water at 24oC is flowing past a long cylinder at a velocity of 0.75 m/s in a large tunnel. The axis of the cylinder is perpendicular to the direction of flow. The diameter of the cylinder is 0.14 m. What is the force per meter length of the cylinder.
Given : ρwater = 1000 kg/m3; μwater = x 10-3 Pa.s
Answer:
From Fig. 7.3, when NRe = 1.15x105, Cd = 1.4
For 1 unit length,

17. Example 7.5
A cylindrical bridge pier 1 meter in diameter is submerged to a depth of 10m in a river at 20°C. Water is flowing past at a velocity of 1.2 m/s. Calculate the force in Newton on the pier.
Answer:
From Fig. 7.3, when NRe = 1.19x106, Cd = 0.35

18. WORK IN PAIR
1. Air at 37.8oC and kPa flows past a sphere having diameter of 3 mm at a velocity of 0.02 m/s. What is the drag coefficient and the drag force on the sphere (ρair = kg/m3; μair = 1.90 x 10-5 Pa.s)
2. Water at 24oC is flowing past 1 m long cylinder at a velocity of 0.75 m/s in a large tunnel. The axis of the cylinder is perpendicular to the direction of flow. The diameter of the cylinder is 0.12 m. What is the drag force of the cylinder (ρwater = 1000 kg/m3; μwater = x 10-3 Pa.s)

19. Minimize pressure drag by streamlining
Delay separation by streamlining for a pointed rear (e.g. air foil).
The streamline divides directly into 2 parts at point B (stagnation point). The velocity at this point is 0.
A perfect streamlined object would have no wake & no form drag.

20. Streamlining

21. Automobile Drag Scion XB Porsche
Cd = 1.0, A = 25 ft2, CdA = 25ft2
Cd = 0.28, A = 10 ft2, CdA = 2.8ft2
Drag force, Fd for Scion XB will be ~10 times larger than Porsche

22. Example 7.6
Calculate power required to overcome drag at 60 mph (26.8 m/s) and 120 mph (53.7 m/s).
Solution: Projected area = (H-G)W = ( )(1.775) = 2.5 m2
60 mph