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I Mass Flow Rate Problem

  1. Dec 29, 2016 #1
    Hi, I am not science student. Can any one help me to solve this?

    Density and velocity of a fluid is given as 920kg/m3 and 5m/s, this fluid is flowing through an area of 25cm2. Calculate the mass flow rate?

    I am in search of some easy unite conversion tricks also.
  2. jcsd
  3. Dec 29, 2016 #2


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    What have you done so far? Say converting cm2 to m2?
  4. Dec 29, 2016 #3
    So.... first we find the flow rate ....The area is 25cm2 ,(note this a square 5 by 5 cm , not 25 by 25 cm) = 25/100x100 M2 = 0.0025 square meters ....
    the fluid is moving at 5m/s through an area of 0.0025 square meters multiply together to get 0.0125 cubic meters flowing every second ...

    so if we multiply by density ... 0.0125 x 920 = 11.5 Kg/sec is the mass flow rate.

    Ebay is the best place to find 'easy unite conversion tricks' .... (ask a silly question)
  5. Dec 29, 2016 #4
    One of the "light bulb went on" moments for me was in chemistry class where I learned how cancellation of units was involved in unit conversions.

    For example, to convert from inches to feet, multiply the quantity of inches by the inch-to-feet conversion factor in the form of a ratio. The conversion factor is 12 inches per 1 foot. So that conversion factor can be written in a ratio in two ways - either 1 ft/12 in or 12 in/1 ft. The way to figure out which ratio to use is that we want to use the ratio that has the starting units in the denominator. So if we are converting, say, 144 inches into feet, we will use the ratio that has inches in the denominator so that the "in" units in the numerator and denominator will cancel each other out, leaving only the "ft" units in the numerator.
    So 144 in = (144 in)##(\frac {1 ft} {12 in})##
    Dividing 144 by 12 results in 12. And the "in" units in the numerator and denominator cancel each other out, leaving units of feet in the numerator.
    So the answer is 12 ft.
    If I want to convert back to inches, I use the other ratio: 12 in/1 ft so that the "ft" units in the numerator and denominator cancel out, leaving the "in" units.
    12 ft = (12 ft)##(\frac {12 in} {1 ft})## = 144 in

    Another example:
    Say we want to convert 15.3 cm3 to mm3.
    What is the conversion between centimeter and millimeter? It is 1 cm = 10 mm
    We are starting with cm3 in the numerator, so the ratio we use will need to have cm3 in the denominator and mm3 in the numerator so that the cm3 units cancel out, leaving only units of mm3 in the numerator. But how do we get cm3 and mm3? Well we have to start with the original conversion factor of 1 cm = 10 mm and cube both sides.
    So (1 cm)3 = (10 mm)3 --> 1 cm3 = 1000 mm3
    Then 15.3 cm3 = (15.3 cm3)##(\frac {1000 mm^3} {1 cm^3})## = 15300 mm3
    The cm3 units in the numerator and the denominator cancel out, leaving only mm3 units. And multiplying 15.3 by 1000 = 15300
    So 15.3 cm3 = 15300 mm3
  6. Dec 30, 2016 #5
  7. Dec 30, 2016 #6
    Thanks for the solution.
  8. Dec 30, 2016 #7
    Aaahh, I am grateful to you for providing such a useful examples. I can practice more now.
  9. Dec 30, 2016 #8
    Given parameters are,
    ρ = 920kg/m3, V = 5m/s and A = 25cm2 = 0.25m2
    The formula for mass flow rate is,
    m =ρVA
    m = 920×5×0.25 = 1150kg/s
  10. Dec 30, 2016 #9
    You can visit ashbox.com for any unit conversations.
  11. Dec 31, 2016 #10


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    If this ashbox site that you are advertising is as poor at unit conversion as you were in your previous post, then I will steer clear.
  12. Jan 2, 2017 #11
    Mr. boneh3ad,
    I think you have misconception. As AshBox is the most amazing site for many scientific conversion tools. Even more it saves time and gives accurate results. There are more than 200 tools are available for any math or science related problems.
  13. Jan 2, 2017 #12
    @Chadi B Ghaith, I think Mr. Boneh3ad was mainly trying to point out an error in your calculation in post #8.
    You wrote:
    It should be A = 25 cm2 = 0.0025 m2
  14. Jan 2, 2017 #13
    Oh yes, I accept that. I mistaken here. Thanks for informing me TomHart.
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