Please help me to calculate drag force and lift force

[mizisyafie]

Homework Statement

Hello guys, I am here and really need your help.
In the figure below, the water flow from inlet and then out at outlet with velocity, v = 0.71 m/s. The objective is to calculate find the drag force and lift force act on fishing lure. I have done Ansys Fluent analysis but I need to validate the result from Ansys with calculation analysis. I really stuck about how the way to calculate this. I hope you guys can help me to show the way to calculate.
Here is the result executed from Ansys CFD. In the Ansys, I use pressure based mode.

Result from CFD Ansys:

Coefficient of Drag, Cd = 0.02
Coefficient of Lift, Cl = -0.0068

For x-component (Drag Force)

Pressure Drag = 0.03239 N
Coefficient of Pressure, Cp = 0.013
Viscous Drag = 0.01334 N
Coeeficient of Friction, Cf = 0.0059
Total Drag = 0.04574

For y-component (lift Force)

Pressure Drag = -0.016
Coefficient of Pressure, Cp = -0.0073
Viscous Drag = 0.0012
Coeeficient of Friction, Cf = 0.0005
Total Lift = -0.015

The picture below is clearly show the situation:



Given data :

Data of Fluid (Water):

Density of water = 998.2 kg/m^3
Viscosity = 0.001003 kg/m-s
Velocity, v = 0.71

Boundary Condition Data:

Length of X = 0.97676 m
Length of Y =0. 22656 m
Length of Z = 0.225 m
Volume of Water = 0.049774 m^3
Surface Area = 0.98861 m^2

Fishing Lure Data

Mass of Fishing Lure, m : 0.01874 kg
Volume of Fishing Lure, V : 0.000018196 m^3
Surface Area (Wetted Area), A = 0.0050176 m^2
Fishing Lure Body Length, L=0.095 m

Homework Equations

The equation that i used to calculated this is F=ma=Drag but the percentage error is very large so that the validation is not acceptable. Now i am very stuck because i am confusing whether to put mass in equation F=ma which is mass of fishing lure or mass of water?

pressure drag + friction drag = drag force

I think the equation above can help me but since i am very slow in my class :)..
My problem is also

1)i do not know how to find pressure drag
2)if i want to find lift force, what equation i want to use?

The Attempt at a Solution

I really need you guys to help me in anyway to explain me in how the exactly i must do to calculate this problem.

 

[mizisyafie]

I will post my calculation just take a few minutes..i have very bad internet connection at this time..sorry guys

 

[rude man]

You haen't given enough geometrical data on the lure. Lures have odd shapes, including usually a foil shape (unequal top and bottom surface lengths) to give lift like an aeroplane wing. And then there's the shape of the cross-section the water sees.

Except for standard shapes, drag data is usually empirically derived.

 

[mizisyafie]

I tried to use this formula

acceleration taken from experimental, a = 0.03 ms^2

So, F=ma=Drag

m(a + g)= Drag

(0.01874 kg)(0.03 ms^2 + 9.81 ms^2) = 0.8035 N

Lift Force

Lift=W - Fb

Where Fb is (spesific weight)(volume of fishing lure)

= (0.01874kg)(9.81) - (9.789)(0.000018196m^3)

= 0.00568 N


I just used the simple equation.But the problem is the drag force and lift force that i got from calculation is not same or precise with the simulation result. am i missing something??or this equation is not ok at all??

1)is it possible i put the F=ma is from mass of water??

 

[mizisyafie]

You haen't given enough geometrical data on the lure. Lures have odd shapes, including usually a foil shape (unequal top and bottom surface lengths) to give lift like an aeroplane wing. And then there's the shape of the cross-section the water sees.

Except for standard shapes, drag data is usually empirically derived.

I mean in Ansys, i just used all the surface area as wetted area and not cross-section. The lure is also have lip on front of mouth portion..but honestly i also confusing about the Area that i must use. can you correct me if i am wrong??

 

[rude man]

I tried to use this formula

acceleration taken from experimental, a = 0.03 ms^2

So, F=ma=Drag

m(a + g)= Drag

(0.01874 kg)(0.03 ms^2 + 9.81 ms^2) = 0.8035 N

That should be drag = m(g-a) if a is acceleration measured vertically in the water.

Lift Force

Lift=W - Fb

Where Fb is (spesific weight)(volume of fishing lure)

= (0.01874kg)(9.81) - (9.789)(0.000018196m^3)

= 0.00568 N

I make it
L = W - B where
W = .01874*9.81 = 0.1838 N = weight
B = 998.2*1.8196e-5 = 0.01816 N = buoyancy force
So L = .1838 N - .0182 N = 0.1656 N = lift force

Your mistake here appears to be in both terms of your Fb equation, and it's very large.

Hopefully that rather larger error, when corrected, will approximate your Ansys simulation. I'm not familiar with Ansys.

I just used the simple equation.But the problem is the drag force and lift force that i got from calculation is not same or precise with the simulation result. am i missing something??or this equation is not ok at all??

1)is it possible i put the F=ma is from mass of water??

 

[mizisyafie]

hello buddy, sorry for late reply, still doing revision on this..i forgot to mention that the lure has angle = 14 degree...so it should be

x component

F cos 14 = Drag

then m(g-a)cos 14 = drag am i right??

but for lift, since calculation on x component has cos 14 so for y component

F sin 14 + Fb + Lift - W = 0

so, Lift = W-fb-F sin 14

but i just want to confirm for y component, F = mg sin 14 right??

and another one thing is for your calculation on Fb, did u missing on gravity acceleration = 9.81??

Fb = pgV.

Thanks buddy

 

[rude man]

hello buddy, sorry for late reply, still doing revision on this..i forgot to mention that the lure has angle = 14 degree...so it should be

x component

F cos 14 = Drag

then m(g-a)cos 14 = drag am i right??

but for lift, since calculation on x component has cos 14 so for y component

F sin 14 + Fb + Lift - W = 0

so, Lift = W-fb-F sin 14

but i just want to confirm for y component, F = mg sin 14 right??

and another one thing is for your calculation on Fb, did u missing on gravity acceleration = 9.81??

Fb = pgV.

Thanks buddy

You are absolutely right. I forgot the g term = 9.81 in my expression for B. So actually

B = 998.2*1.8196e-5*9.81 = 0.1781 N
and L = W - B = 0.1838 - 0.1781 = 0.00565 N
So there is very little L, in other words, in the absence of water velocity the lure would barely sink.

But L = W - B is correct. The thing is that L is a function of your 14 degree angle coupled with the velocity of the water stream.

Fcos14 can't be right. If the angle = 0 deg. there would be no lift, and drag would be due solely to the lure's cross-section facing the stream. As the angle increases you get more lift and more drag. Above 45 deg. you would get still more drag but less lift. At 90 deg. you would get max. drag but zero lift.

 

[mizisyafie]

thanks for your explanation, okay, in this fishing lure, the lure will retrieve so that the lure will going down to dive..so the lift force will indivate negative sign as show the lure is going down...this is my opinion.

For F=m(g-a)---->a is measured vertically or horizontally??

However, i also want to calculate based on Reynold Number...i stuck on how to find Form drag. As we know Form Drag + Friction Drag = Total Drag..but to find lift force, is it also posibble to find using the equation right?? form drag + friction drag = Total Lift..?

I am very sorry because my english is so bad but i hope you can understand of what i want to explain to you.

About friction drag, i will attached something about to find friction drag using one of formula that i am not familiar.