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Calculating pressure drop

  1. May 12, 2014 #1
    Hello. I have had an attempt at solving this pressure drop problem, but I have been told it is incorrect.

    I am very much a beginner in the area of fluid mechanics, so if you could please define any terms that you are using, that would be so helpful!

    I thought using the Hagen-Poiseuille would fit, as the question has given data that would fit into the equation and it also suggests that the pipe is laminar (smooth pipe). Although, it does not state whether the pipe has bends or is just straight. Does 'smooth pipe' also mean 'straight pipe'? Do I assume its straight?

    Why is my attempt wrong? What are the other possible ways to solve this/pressure drop?
    The lecturer I asked said that this method wasn't right, and that I should just calculate pressure multiplied by all the heads (pressure x heads). I still am confused by this though.

    Also, am I allowed to ask more questions relating to my given question in this thread?

    The question is below:

    1. The problem statement, all variables and given/known data

    Water at 20°C flows in a 35 cm diameter smooth pipe at a rate of 0.0476 m^3/s from the Goulburn River to the Holding tank. What is the frictional head lost* (in m) and corresponding pressure drop (in kPa) per 100m length of pipe?

    EDIT: *I didn't see this part of the question...

    Known data:
    • Water has a temperature of 20°C
    • Diameter of pipe is 35cm
    • Flow rate is 0.0476 m^3/s
    • Length of pipe is 100m

    2. Relevant equations

    Hagen-Poiseuille equation:

    558f317b3798844ad11293586bd69ce1.png

    Where:
    ΔP is the pressure loss
    L is the length of pipe
    μ is the dynamic viscosity
    Q is the volumetric flow rate
    d is the diameter
    Pi is the mathematical constant Pi

    3. The attempt at a solution

    Using:
    558f317b3798844ad11293586bd69ce1.png

    Where:
    ΔP is the pressure loss (Pa) = ?
    L is the length of pipe (m) = 100m
    μ is the dynamic viscosity (Pa*s) = 1.002 for Water @ 20°C
    Q is the volumetric flow rate (m^3/s) = 0.0476 m^3/s
    d is the diameter (m) = 0.35m
    Pi is the mathematical constant Pi


    Attempt:

    ΔP = (128(1.002 Pa*s)(100m)(0.0476 m^3/s))/pi(0.35)^4
    = 12949.786 Pa

    ΔP = 12.95 kPa/100m
    Pressure drop is 12.95 kPa per 100m length of pipe.

    Thank you very much in advance. I appreciate it.
     
    Last edited: May 12, 2014
  2. jcsd
  3. May 12, 2014 #2

    SteamKing

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    'Smooth pipe' does not mean 'straight pipe'. The absolute roughness of a pipe is a measure of the irregularities in the inner surface of the pipe, and different materials have different roughness values. The ratio of the roughness of the pipe material to the pipe inside diameter is an important factor in establishing how the fluid flows thru the pipe.

    http://www.enggcyclopedia.com/2011/09/absolute-roughness/

    I don't understand your last sentence. 'Head' is equivalent to 'pressure', just expressed in a different form. Multiplying head x pressure is meaningless.

    The Hagen-Poiseuille equation is valid only if the flow in the pipe is laminar. Do you understand how to determine if the flow is laminar? Have you calculated the Reynolds number for your flow problem?

    http://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

    To answer your original question, yes, there are other methods of calculating pressure drop, depending on whether you have laminar flow, fully turbulent flow, or something in between.

    You can ask as many questions as you want pertaining to your problem, if you don't understand everything posted in the responses.
     
  4. May 12, 2014 #3
    Thank you for your response. Oh ok so head is just another name for pressure. I see.


    No, how do I determine if a flow is laminar?

    I was reading a webpage saying that laminar flow is 'smooth' and 'turbulent' flow is 'rough'. Is this description true?

    http://en.wikipedia.org/wiki/Laminar_flow

    In order to solve this question, is the first step to find what type of flow it has?

    Do we have to calculate and obtain a value to see whether a flow is laminar or turbulent? Is this what the Reynolds equation is for?

    Sorry for all the questions
     
  5. May 12, 2014 #4

    SteamKing

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    The Reynolds number calculation is key to determining if your flow is laminar or not. Laminar flows have Reynolds numbers below about 2300. For circular pipe, the Reynolds No. is:

    Re = v*D*ρ/μ where,

    v = velocity of the flow, m/s
    D = inside diameter of the pipe, m
    ρ = density of the fluid, kg/m[itex]^{3}[/itex]
    μ = dynamic viscosity of the fluid, Pa-s

    'Laminar' flow is where the flow of the fluid occurs with no mixing between layers. As the flow increases velocity, mixing between layers will begin to occur, and the flow is said to become 'turbulent'.
     
  6. May 12, 2014 #5
    Thank you very much again.

    I tried the problem again earlier, it's a bit long though:

    ----

    $$ N_R = V D \rho/μ$$
    Where,
    N_R = Reynolds number
    V = Velocity of flow, 0.4947m/s
    D = Diameter of pipe, 0.35m
    ρ = Density of water at 20°C, 998 kg/m^3
    μ = Dynamic Viscosity, 1x10^-3 Pa*s

    Reynolds Number = ((0.4947 m/s)(0.35m)(998 kg/m^3))/1x10^-3 Pa*s
    = 172798.7

    As Reynolds number > 4000 the flow is turbulent

    ---

    As the flow is turbulent, I would need to find the relative roughness of the pipe.

    Relative Roughness = ε/D

    Where,
    ε = roughness of the 'smooth pipe', 0.0m (assuming 'smooth' is either plastic or glass)
    D = diameter of pipe, 35m

    Relative Roughness = 0.0m/35m
    = 0.0


    ---

    I have the Reynolds number (172798) and the Relative roughness (0.0).
    Using the Moody diagram, I get a frictional factor of around 0.027.

    Frictional Factor = 0.027

    ---

    I figured out that I cant use that Hagen-Poiseuille equation as it is only for laminar flow? I learned that for the flow to be laminar the Reynolds number < 2000.

    I'm using Darcy–Weisbach's equation for head loss.

    $$H_L = f (L/D) (V^2/2g)$$

    Where,
    H_L = Head Loss
    f = frictional factor, 0.027
    L = Length of pipe, 100m
    D = Diameter of pipe, 0.35m
    V = Velocity, 0.4947m/s '
    g = gravity due to acceleration, 9.81m/s^2

    H_L = 0.027 * (100m/0.35m) * (0.4947m/s)^2/(2*9.81 m/s^2)
    = 0.0962m

    Head Loss is 0.0962 metres.

    ---

    To find the pressure drop can I just convert the head loss (m) to pressure drop (kPa)?
    Since you said that head loss = pressure?

    Head Loss --> kPa

    P = SG * g * h

    Where,
    P = Pressure (kPa)
    SG = Specific Gravity of Water at 20°C, 0.998
    g = gravity due to acceleration, 9.81m/s^2
    h = Head Loss

    P = 0.998 * 9.81 m/s^2 * 0.0962m
    = 0.941 kPa/100m of pipe

    So the pressure drop is 0.941 kPa per 100m of pipe?
    Is this right? I still can't picture these measurements in real-life to know if it's close the right answer...is the pressure and head loss too small?


     
    Last edited: May 12, 2014
  7. May 12, 2014 #6

    SteamKing

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    Your numbers look OK to me.
     
  8. May 12, 2014 #7
    Ok thank you for your help! I appreciate it.
     
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