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Questions to do with crankshaft of an engine

  1. Dec 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A cylinder in a truck engine are shown in
    Figure Q8.1 below. The piston slides without friction in the cylinder of diameter
    150 mm. The combustion chamber is subjected to an internal pressure of
    50 bar gauge as a result of fuel combustion without any leakage past the piston.
    The connecting rod AB has a length of 200 mm. It is linked to the piston at A
    and to the crank-shaft at B. The crank-shaft OB has a revolution radius of
    75 mm

    1 - Calculate the resulting instantaneous force on the piston if downward force is 80kN about the crankshaft axis of rotation O..
    2 - Calculate the angle OAB.
    3 - Calculate the projection of the force transmitted by the piston to connecting rod AB.
    4 - Isolate crank pin OB and calculate the moment resulting from the force transmitted by connecting rod AB.
    Consider an anti-clockwise rotation as positive.
    5 - Calculate the rotational velocity of the engine crankshaft if engine is running at 1500rpm.
    6 - What is the crankshaft power output of a 6 cylinder engine at this running
    condition if the average torque produced on the crankshaft at a certain
    engine load in one cylinder at 1500 rpm is 170 Nm? piston.png Please find diagram below.

    2. Relevant equations


    3. The attempt at a solution
    Are these answers right?

    1 - 50BAR = 5,000,000Pa, P=F/A so F=PA.
    A = pi r^2, r= 0.075 so area = 0.01767m^2.
    F=5,000,000 x 0.01767 = 88350N

    2 - Length AB = 0.2m OB= 0.075m
    therefore OAB= sin (0.075/0.2) = 22.02 degrees.

    3 - Projection of force, 80,000 = adjacent, therefore Force on AB = 80,000/cos(22) = 86,282.8N

    4 - Moment on OB from AB = 80,000 x 0.075 = 6000Nm as only vertical force of 80,000N can act on OB?

    5 - Rotational velocity = 2pi x f.
    1500rpm = 25Hz, therefore Rotational velocity = 2pi x 25 = 157.08 rad/s

    6 - Torque = 170Nm, Power = Work/time
    Work = 170 x 1500 x 6 = 1,530,000N for all 6 cylinders per minute
    Power = 1,530,000/ 60 seconds = 25,500W

    Thank you, I hope these are mostly right. I'm not sure about last one.
     
    Last edited by a moderator: Dec 22, 2014
  2. jcsd
  3. Dec 22, 2014 #2

    CWatters

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    Your answer for 1) seems ok but appears to ignore the bit I've underlined below.

    What does that bit mean anyway? What exactly is a "downward force about a crankshaft axis of rotation" ? Is that the exact wording?
     
  4. Dec 22, 2014 #3
    Hi,
    This was the exact working from the test:
    "Calculate the instantaneous torque produced by a downwards piston force
    of 80 kN about the crankshaft axis of rotation O at the position shown in
    Figure Q8.1"
    I have no idea, it seemed confusing
     
  5. Dec 22, 2014 #4

    CWatters

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    :)

    So your answer to Q1 is incorrect. They aren't asking you to calculate the force on the piston. They are telling you it's about 80KN and are asking you to calculate the torque on the crankshaft.
     
  6. Dec 22, 2014 #5
    Actually, this is very misleading.
    We have a webpage where we do the test, and it says calculate the FORCE, but it also gives us a test paper and refers us to the question to answer which gives us the details of the situation. In the test paper it says torque not force, so I'm not sure which one we are supposed to do
     
  7. Dec 22, 2014 #6

    NascentOxygen

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    Staff: Mentor

    Perhaps clearer:

    Calculate the instantaneous torque about the crankshaft axis of rotation O (at the position shown in
    Figure Q8.1) which would be produced by a downwards piston force of 80kN.

    http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
  8. Dec 22, 2014 #7
    So was my method correct?
     
  9. Dec 22, 2014 #8

    CWatters

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    I think you will have to ask for clarification. If that's not possible I would answer as you have done with the force = 88,350N.

    Questions 2, 3 and 4 appear to rely on the answer to question 1 being a force. They lead on naturally from one to the other.

    If the answer to Question 1 was the torque then you would need to work out the answers to questions 2,3 and 4 first in order to solve Question 1.
     
  10. Dec 22, 2014 #9
    I just sent the lecturer an email asking to clarify. Questions 2, 3 and 4 say assume it is 80,000N though?
     
  11. Dec 23, 2014 #10
    The first question is:
    Calculate the resulting force on the piston from the pressure in the
    combustion chamber.
    So was what I did for this and the following question right?

    So on the first one you do not need to assume downwards force is 80kN, but on the second question onwards you do.
     
  12. Dec 23, 2014 #11
    So if it's from the pressure I believe I have done it correctly.
    However, question 2 says to find the angle OAB from the question in the test paper: "
    Calculate the instantaneous torque produced by a downwards piston force
    of 80 kN about the crankshaft axis of rotation O at the position shown in
    Figure Q8.1." So I am unsure if I did this correctly, as force doesn't seem to affect the angle it is the length of the crankshaft that dictates the angle?
    thanks.
     
  13. Dec 23, 2014 #12

    CWatters

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    OAB = 22 degrees is correct.

    I don't think that's correct. If the force from the piston is FP acting vertically then the force acting in the direction of the crank is

    FpCos(22)

    Not sure about that. I would use the force in the direction of the connecting rod (eg FPCos(22)) and this diagram to calculate the tangential distance...
    Tangential.jpg
     
  14. Dec 23, 2014 #13
    What about the first question?
    Thank you for the help! Why is it cos(22)? isn't the force downwards the adjacent, because it looked like the connecting rod AB would be the hypotenuse?
    Q4 - I don't quite understand what it is I am supposed to do for this one, which force component am I finding exactly?
    Thanks
     
    Last edited: Dec 23, 2014
  15. Dec 23, 2014 #14
    Actually, for Q4. I could find the angle ABO, then use the force of the connecting rod Fp cos (22) to find the adjacent of the triangle in the diagram?
     
  16. Dec 23, 2014 #15
    Just looked at the question 4 again it says: Calculate the projection of the force transmitted by the piston to connecting rod AB.
    So you are finding the force on AB not the crank. Is the 80kN force by the piston not going to be acting as the adjacent as the angle is 22 degrees between the 80kN and AB? so connecting rod AB is going to be the hypotenuse?
     
  17. Dec 23, 2014 #16

    CWatters

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    I answered that above when I wrote this earlier... "I think you will have to ask for clarification. If that's not possible I would answer as you have done with the force = 88,350N."

    When you resolve a force (eg FP) into two components the components need to be at right angles or they aren't independent of each other. (If they aren't orthogonal then one component would itself have a component in the direction of the other.)

    In this problem we want one of the components to be in the direction of the con rod so the other will be at right angles to the con rod. If you draw that you get this..

    ConRod.jpg

    Cos(22) = Fconrod/Fp
    or
    Fconrod = FpCos(22)
     
    Last edited: Dec 23, 2014
  18. Dec 23, 2014 #17

    CWatters

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    I'm getting really confused with the question numbers. In the original post Q4 says...

    That's asking you to calculate the moment/torque that the con rod gives the crankshaft. For that see the diagram in post #12.

    Torque = Fconrod * tangential distance

    From above

    Fconrod = FpCos(22)

    From the diagram in #12..

    Tangential distance = LcrankCos(22)

    So

    Torque = FpCos(22) * LcrankCos(22)
    =Fp*LcrankCos(22)2

    Fp is either 88,350N (if you calculate it from the pressure) or 80,000N (if you use the value the question tells you to assume). Hopefully your tutor will clarify that issue.
     
  19. Dec 23, 2014 #18
    I know it is very confusing for me too, I'm sorry at the mess of this. It's stupid how they've done this test with the different set of questions it has made it very complicated.

    The first question said to find the resultant force from the pressure, so I'm pretty sure I've done that right.
    Then for question involving the diagram you just sent, why is it a right angle between the connecting rod and crank shaft? because it makes sense how that is the hypotenuse in that case but in the diagram it is perpendicular to the cylinder walls.

    I'm really sorry about the confusion, I'm so confused you must be getting annoyed with me!
     
  20. Dec 23, 2014 #19
    I'm gonna try and start fresh:
    The questions we are having troubles with:
    1:
    Calculate the projection of the force transmitted by the piston to connecting rod AB.
    Give your answer in Newtons
    2:
    Isolate crank pin OB and calculate the moment resulting from the force transmitted by connecting rod AB.
    Consider an anti-clockwise rotation as positive.
    Give your answer in N.m

    Both these questions are linked to a single question from the test paper link we have been given which says "
    Calculate the instantaneous torque produced by a downwards piston force
    of 80 kN about the crankshaft axis of rotation O at the position shown in
    Figure Q8.1." So we have to answer the original question from this question.

    I hope this clears things up?
    Thank you for putting up with this!
     
  21. Dec 24, 2014 #20

    CWatters

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    If you mean this diagram..
    Tangential.jpg

    It's not a right angle between the connecting rod and crank. The crank is horizontal in my diagram. Take a look at how torque is calculated. Only the tangential component of the force contributes to torque.

    There are two ways to work that out but both give the same answer. In the first you work out the effective length of the crank (the tangential distance) in the second you work out the component of the force that is tangential to the crank. Both methods give the same answer... but perhaps the second is easier to understand?

    Tangential1.jpg Tangential2.png
     
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