[Dolphinsiu]

I get stucked when I do part (a), the following is the question

The Fanning friction factor for a pipe flow system is determined experimentally. The pipe used in this experiment is horizontal and 12 m long. IT has an inside diameter of 5.25 cm and no bends, fittings or diameter change over the 12 m length. Water at 16C flows through the pipe at a rate of 3.2 kg/s. The water density is 999 kg/m3. The viscosity of water is 1.11 x 10-3 Pas. The head loss due to friction over the 12 m pipe is 1.38 mH2O.

(a) Calculate Reynold number.

Here is what I feel strange in my calculation:

m = ρuA = constant

u = m/ρA = 3.2/999 (pi(0.0525)2)/4
             = 1.48 ms-1

Re = Duρ/μ = 0.0525 (1.48) (999)/ 1.11 x 10-3 = 69916

I feel very strange as Renold number is  < 2100: laminar flow > 4200 : turbulent flow. but it is not possible for Re > 10000. I do want to know which step that I do is wrong! Thank you!

[Montemayor]

You are trying to calculate the Reynolds Number, not the Reynold number.

Please explain what you mean by writing: “Renold number is < 2,100: laminar flow > 4,200 : turbulent flow”.  Your symbology is confusing and it's not clear as to what you are saying.

You state “but it is not possible for Re > 10000.  You are wrong; it is very possible for the Reynolds Number to be as great as 10,000 – and even greater!  Where do you obtain such an assertion?

Your answer for the Reynolds Number (Re) is correct.  To prove this to you, I calculated the Re as follows, in U.S. customary units:

D = 2.06693 inches = 0.172244 ft
v = 4.8546 ft/s
rho = 2.68519 lb/ft-h
mu = 62.36557 lb/ft3

Re = D*v*rho/mu = 69,619

I personally have resolved this very value every time I’ve employed 2” schedule 40 pipe in the past 47 years – and that’ s a lot of times! – and I know that a Re of 70,000 with a velocity of around 5 ft/s is what I would expect for a sound installation for water flowing at 50-60 gpm (which this is).

I don't know what else you may be doing that is wrong, but your Reynolds Number is correct.  I suspect your understanding of where the Reynolds Number originates and its use may be strange and not well understood.  You need more practice and research into the Reynolds to understand its significance and importance in fluid mechanics - that's what I think will cure your bad feelings about how you are doing your problem calculations.  With practice and experience you will learn to have the confidence you need.

Good Luck.

[Dolphinsiu]

That's what I attempt

(a) Write the total energy balance equation over 12 m pipe.
     Simplify the equation by deletion of energy terms in which the   
     change is negligible or equal to 0

     (P2 - P1) / ρ + 1/2 (u2^2 - u1^2) + g (z2 - z1) + hf = nWp
      g (z2 - z1) + hf = 0

(b) Calculate Reynolds number
     Q = 3.203 x 10-3 m^3/s,
     A = 2.165 x 10^-3 m^2,
     u = 1.48 m/s,
     Re = 69916

(c) Calculate the pressure drop over the 12 m pipe in N/m^2

     Delta p = ρgh = 999 . 9.81 . 12 = 117.6 kN/m^2

(d) Calculate the mechanical energy loss due to skin friction in J/kg

     Head loss due to friction = 1.38 m H2O
     Head loss = hf = pf / ρg
                      pf = ρghf = 999 . 9.81 . 1.38 = 13.524 kN /m^2

     Skin friction = hfs = pf/ρ = 13.524 / 999 = 13.54 J/kg

(e) 4 f (L/D) (u^2/2) = 13.54
                          f  = 13.54 / 2 (0.0525/12) 1/(1.48)^2
                             = 0.0135

Why the equation in (a) haven't been used? What step I get wrong? Thank you!

[Dolphinsiu]

Do anyone help?

[Dolphinsiu]

(d) Calculate the mechanical energy loss due to skin friction in J/kg

     Head loss due to friction = 1.38 m H2O
     Head loss = hf = pf / ρg
                      pf = ρghf = 999 . 9.81 . 1.38 = 13.524 kN /m^2

     Skin friction = hfs = pf/ρ = 13.524 / 999 = 13.54 J/kg

g (z2 - z1) + hf = 0
-117.6 + hs + h = 0
-117.6 + hs + 13.54 = 0
hs = 104.06 J/kg

(e) Calculate the Fanning friction factor.
 
     4 f (L/D) (u^2/2) = 104.06
                          f  = 104.06 / 2 (0.0525/12) 1/(1.48)^2
                             = 0.104

I have redone the question. Is my above caculation correct?