IWCF Well Control/IADC Well Cap Practice Questions, Quiz and Tests

Here you can find lots of practice questions, kill sheets, quizzes and tests for IWCF/IADC Well Control exam preparation. Take them check your preparation  and improve on it.

47 thoughts on “IWCF Well Control/IADC Well Cap Practice Questions, Quiz and Tests”

  1. Not much detail in this website. Questions are good. Answers are better. How to read and work the formula in DETAIL is best. Do you get the how to, or is this just another how to think course?

    1. Craig! This website is not for profit and its sole purpose is to help people in preparation of IWCF so if you will show interest we will help you as we helped zubin who got 100 in P&P and Naveen who also got 100 in P&P{ We can also provide guidance by Subsea offshore Engineer}
      and better will be you should join our forum : http://www.knowenergysolutions.com/forum/ and post your queries and topics on which you want more material. We are more than willing to help anyone in preparation of IWCF.

  2. Q1) IF we found that leak off test is not ok, then whats next step

    Ans 1) If Leak Off Test is not ok, this means cementing is not proper at casing shoe. You will have to go for cement repair. This can be done by cement squeez/plug job.

    Q2) IF we are running two mud pump at same spm, but spm of one mud pump is not remaining stable at same spm, what can be the reason?

    Ans 2) A very good question. Fluctuation or instability in pump SPM can tell you a lot about pump condition. Timely detection and fitting corrective action can save you rig time and money. Now coming to your question (actually I should dedicate a complete post for this, anyways in short)’unstable pump SPM’, fluctuations of pump SPM can be of two types. First, rapid fluctuations and second, slow and gradual fluctuations.

    Rapid Fluctuations: These are mainly due to mechanical problems which can be, worn out valve ring, valve, valve seat, piston ring and in worst case worn out wear plate and liner. You should stop the pump immediately and make a check for the things as mentioned in order. And if there are any indications of pump taking air then you should check the mud strainer in suction line.

    Gradual Fluctuations: These are mainly due to electrical and downhole problems. There may be problem in SCR, Motor Field or you are running near the load limit. Downhole problem like improper hole cleaning can also cause the SPM or pressure fluctuations. For electrical problems check the concerned electrical devices and for hole cleaning problem pump a lo-hi pill (low viscous mud followed by high viscous mud) and resume drilling after surface to bit time.

    Q3) In question 18 on your post http://www.knowenergysolutions.com/2012/03/iwcf-well-control-principles-and-practices-questions
    According to the formula given:
    13 x 0.052 x 0.0080 / (0.0744-0.0080) = 0.08144
    Now, 0.08144 X 15stnds X 88ft. = 107.508 p
    Yes, you got it right.

    Q4) What is the main factor that determines the rate of build-up to stabilized SIDPP & SICPP once the well is shut-in?
    a) Friction losses
    b)Gas migration
    c) Permeability

    Can someone plz answer and explain…

    Ans) For IWCF answer should be permeability because he is asking about initial pressure stablization (when well is just shut in) afterwards it would be due to gas migration. So if it shows up in exam mark permeability as correct answer as in IWCF answer key they use permeability as correct answer.

    Q5) One more practice question : Which of the following parameters might be affected by permeability of formation (select three answers)?
    a)SICP
    b)Bottom hole pressure
    c) SIDPP
    d)The time taken to stabilise the shut in pressures.
    e)Volume of influx

    Answer : a,d,e

    Q6) Select the good operating practices from the list (THREE) while drilling top hole section.
    a) Maintain as high overbalance as possible.
    b) Maintain high drilling rate
    c) Maintain mud density as low as possible.
    d) Continue to circulate when picking up for connection.
    e) Circulate at maximum rate to create highest possible ECD.
    f) Control drilling rate to avoid overloading the annulus with cuttings.

    Answer 6)
    I think answers are c,d,f because while drilling in top hole section : formation is weak so we have to keep mud density as low as possible, chances of shallow gas are there so continue to circulate when picking up for connection, controlling drilling rate to avoid overloading of annulus with cuttings.

    Q 7) Vertical well is drilled to 9000 ft. with 11.7 ppg mud.
    Pump speed: 70 spm.
    Pressure loss in surface equip.= 140 psi; press. loss in drill string = 800psi; press. loss in annulus = 100 psi; pressure loss at bit nozzle = 1400psi.

    a) Calculate BHP when mud is circulated at 70 SPM?
    b) Calculate eqvlt circulating density.

    BHP= hydrostatic pressure + annular losses
    = 0.052x 9000×11.7 + 100
    = 5575.6 psi
    ECD = APL/(.052*TVD) + mud wt
    = 100/(.052*9000) + 11.7
    = 11.913

    Q 8) When circulating out a kick in a deep well with a deep set casing shoe, the choke pressure approached MAASP while the influx is still in open hole. What is the most important action to take?

    Ans 8) You have to bleed through choke. SO that surface pressure goes below MAASP and ZUBIN can you like this website on facebook and ask your friends interested in IWCF to share their doubts so that we can create a group of interested people in IWCF field.

    In instrumentation section there will be questions on BOP’s, MGS, Inside BOp, FOSV, Rated working pressure,koomey control unit and may be 1-2 question on accumalator bottles. It will take us time to upload questions on instrumentation section but if you have any query you can post it here we will try our best to solve your queries.

    Q 10) All influx will be displaced from the hole at a pump rate considerably slower than that used while drilling. Select the correct reasons for this from the list below:

    One of the options is “To have minimum pressure being exerted on the open hole.”
    I think it will be one of the correct reasons, since slower pump rate means less annular pressure losses and hence less BHP. Is that correct?

    Ans10) Yes that is correct logic.!

    Q 11) While circulating out a kick, the pump speed is increased keeping SIDPP constant. What will happen to the BHP.
    a) Increase
    b) Decrease
    c) Remains Same
    (Plz explain the reason).

    Ans 10) According to me it should remain same as BHP = SIDPP + hydrostatic head of mud in string.
    As SIDPP is constant and IN iwcf we assume mud remains uniform in string so BHP should remain constant.

    people who asked question : Exactly what I thought. However the Transocean Q&A from where I got this question says it will DECREASE. I cannot understand how…

    Q11) A kick is being circulated out at 35 SPM, SIDP is 650 psi, SICP is 1050 psi. What would happen to BHP if pump is slowed down keeping SICP constant.
    a) Increase (b) Decrease (c) Stay the same
    Plz analyze…
    Reply
    Ans 11)

    Q12) One more gud question : Where is the mud loss first noticed during the time of drilling?
    a) on the mud pressure gauge
    b) on the return flow meter.
    c)In the mud pits P.V.T(Pit volume totalizer)
    d) From the loggers chart.

    Ans) on the return flow meter because it will tell instant change in the mud volume.

    Q13) Which of the following equipment may warn of increase in formation pressure while drilling overbalance? (3 ANSWERS)
    a) ROP meter
    b) Pump pressure guage
    c) Flow line temp guage
    d)SPM counter
    e) Gas Detector
    f) Return flow meter

    I think c,e,f. Plz confirm….

    Ans13) It is a) c) e) Return flow meter is a positive kick sign.See the difference between early warning sign and positive kick signs write if you dont get any material on them I will write more about it.

    Your Diagnosis of problems are correct
    Problem 1 : Nozzle plugging Symptoms : D/pipe Pr.: Increasing, Casing pressure :Constant
    Problem 2 : Choke plugged : D/ pipe pr.: Increses, Casing pressure: Increases
    Problem 3 : Choke Washout : D/ Pipe Pr : Decreases, Casing Pr : Decreases
    Problem 4 : Pump Failure : D/ Pipe Pr : Decreases, Casing Pr : Decreases ( No sound of pump in simulator)

    As while killing nozzle is plugged in question 1 so bottom hole pressure will remain same because we will calculate from annulus side because after nozzle plugging We hold Casing Pressure Constant and then bring pump down and Close Choke.Identify the problem and bring SIDP at original value. Keeping casing Pr. Constant bring pump to kill speed. Maintain new drill pipe pressure constant. Also change in reading in drill pipe pressure gauge is due to nozzle plugging ( As more pressure is lost in nozzle due to plugging) and as annular friction losses are same BHP will remain same,

    As choke is plugged in case 2 : Action to be taken are :
    Stop pump immediately and close HCR.
    Identify the problem bring the SIDP at original value.
    Line up alternate choke and HCR.
    Keeping casing Pr. Constant bring pump to kill speed.
    Maintain Drill pipe pressure constant.
    Because of choke plugging annular friction losses have increased by 200 psi in case 2 so BHP should increase by 200 psi.
    We are working on a kill sheet application which will help you in practicing kill sheet and also will post some questions on instrumentation section. In the meantime you can focus on BOP figures, koomey control unit figures,pipe ram figure according to course.

    Q14) Stack config: (from top to bottom) Annular=>> 5″ Pipe ram=>> Blind/Shear=>> 3 1/2″ Pipe ram =>> Drilling spool.
    With the drill string “hung-off” on the 31/2” rams, can the annular be repaired?
    Plz Explain “hung off”.
    Reply

    q15) With the drill string “hung-off” on the 31/2” rams, can the outer choke line valve be repaired?
    Yes/No

    q16) With the drill string “hung-off” on the 5” rams, can the annular be repaired?
    yes/no

    Ans 14) no as drilling spool is below 3 1/2″ ram outer choke line valve cant be repaired.

    Ans 15) yes as it is above on 5″ RAn from where drill string is hung off annular can be repaired.

    Ans 16) hung off means 3 1/2″ inch ram is closed and stringis hung on this ram BOP. so joints below it cant be repaired.
    Annular can be repaired.
    Modify message

  3. Q11 WHEN increase speed pump for circulating OUT kick and SIDPP constant that means is necessiry to open choke line for that annulus pressure loss increase so BHP increase .

    1. @BARCHI
      Yes, you have to open choke to maintain constant BHP while circulating out kick keeping SIDPP constant. In any case we have to keep BHP constant. If annular pressure loss increases BHP will increase and we have to open choke to reduce back pressure and maintain constant BHP.

      1. @BARCHI
        There is a similar IWCF question that tests if you understand the pump rate also affects the pressure losses in the choke, which have a direct effect upon the BHP.

        For example as you bring the pump speed up to kill rate, the DPP must be allowed to increase by the amount of the SCR. At same time the choke must be opened slowly sufficient to deduct the increase in choke friction. This is accomplished by holding the CP constant as we make the adjustments to the pump speed. This results in the BHP essentially remaining constant.

        Hence the question which goes something like this: What would be the effect upon BHP of holding the CP constant while increasing the pump speed? The answer is it will not affect BHP.

        Nevertheless, you obviously cannot hold the CP constant forever as it must be allowed to increase as the gas bubble is brought to surface. Holding the CP constant for more than a few minutes would result in letting in a second kick.

    2. @Barchi

      You are correct by saying the annular pressure losses will increase if you increase the pump speed; however, so too will all the other system circulating pressure losses. Obviously, to avoid allowing the pump pressure to increase, you must open the choke somewhat.

      Since the APL increase is only a very small component of all the circulating pressure losses that do not affect BHP, the net effect would be to lower BHP.

      For example, if you increase the pump rate and gain say an extra 200 psi on your drillpipe gauge, we can assume the SCR at the new pump speed would have been 200 psi greater than the original SCR. This is because ICP = SIDPP + SCR. Since the SIDPP does not change with pump speed, we will have reduced the BHP by 200 psi – APL increase, which may only be 10 – 20 psi, if we do not allow the DPP to increase by the additional 200 psi. In theory, the maximum you could reduce the DPP so as not to affect the BHP would only be by the 10 – 20 psi APL.

      To answer these type of questions, simply figure out what you would be doing with the choke: if you open it, you decrease BHP and if you close it you increase BHP.

    3. When we keep the observed casing pressure constant while changing the pumping rate, the bottom hole pressure remains the same. The observed pressure in the drill pipe should be allowed to drop to the true value as the pumping rate is reduced and the other way round as the pumping rate is increased. Note that the pressure you read at the casing side of the U tube is the difference between the formation pressure and the BHP plus any additional pressure applied so, if this pressure remains constant, it means the bottom hole pressure is constant.

  4. Pls can you post more equipment questions just like you did for pnp . Am preparing for my iwcf level 2 and i need some serious equipment question. Thanks

  5. This is a well control question……your drilling with a TDS 12,000 feet you have a washout in the drill string at 11,450 then you take a kick how do you control this matter? can some one give me the correct answer to this question. and pls. send to my email address.

    1. @ Rodger Cook

      The answer to this type of IWCF question will depend upon where the kick is in relation to the washout, mud type and influx type.

      The question I have seen tells us we have a gas kick below the washout.

      Assuming you have gas kick in a WBM and the kick is below the washout, our first action should be to evacuate the kick out of the wellbore. To do this, you need to use the Volumetric Method to allow the kick to migrate above the washout, while maintaining constant BHP. If you have no back-pressure valve above the washout, the process is easily accomplished by maintaining your DPP constant by bleeding off the extra DPP through the choke at intervals. Once the influx is above the washout, you can then circulate the kick to surface, using the Driller’s Method.

    1. Keep on visiting knowenergysolutions for some hard questions on P&P and equipment.Also, visit and contribute to unsual questions section

  6. Admin,
    I have a big problem with kill sheet, I try and try,I can’t get it,please advice me what to do?

  7. Hello Sir
    I have a question. Hope to see a reply soon
    Ques 1 If during shut in the SIDPP is increasing and inspite of kill mud insertion it keeps on increasing what could be the possible reason?

    1. If well is shut in and inspite of kill mud insertion as influx migrates upwards in annulus. influx pressure is pressure of resorvoir. Bottom hole pressure = Influx pressure + hydrostatic head of mud below influx in annulus which shows that bottom hole pressure has increased. And from u-tube effect SIDPP= BHP- hydrostatic head of mud in string. SO increase in Hydrostatic head due to Closed well upward migration of influx is negated by increase in hydrostatic head of string due to kill mud insertion but negation is less in this case.

      Like our website on facebook if you like answer. ALso if your query is not solved let us know we would try to make our answer more illustrative.

  8. Could you please send me the last two exam of IWCF for drilling supervisor level for surface BOP for P&P and equipment with answers , I will be highly appreciated.
    Best regards
    Nawaf
    Senior drilling supervisor

  9. Thanks a lot For this effort, I have a question, What’s the first thing to do if I get a kick in case I’m Driller and other case I’m Derrick man, Thanks in advance

  10. Good morning.
    many thanks for all the effort that you made to help drilling people to understand better.
    My quest is – using volumetric method to kill a kick (no DP in the hole) after bleeding the gas keeping formation pressure close to BHP ,what will be the next steps to kill the well? we know that using gas bubble we can make an overbalance on the bottom ,but if we will bleed all the gas we will be under balance. please explain.

  11. Dear rodger
    I was worked with you in Kuwait that time you r driller I am now working in n d c in abu dhabi

  12. please explain why the pump pressure(SPP)decreases when pumping a high-viscosity sweep and it gets to the bit then increase after it passes the bit.

  13. I have a question.
    Scenario
    Offshore operation – Deep water and therefore Choke Line Friction (CLF) needs to be accounted for in calculations for Initial Circulating Pressure (ICP) and Final Circulating Pressure (FCP).
    Lets say we record the Slow Circulating Rate (SCR) as normal by circulating 30 Strokes Per Minute (SPM) down the drill pipe, up the annulus and up through the riser and we get 250 psi of pump pressure. Then we take the CLF by closing the Annular, and circulating 30 SPM down the drill pipe, up the annulus and through the choke line with the choke WIDE OPEN (which is the actual flow path if there is a well control kill conducted) and we get 375 psi on the pump. Therefore 375 – 250 = CLF of 125 psi. This isn’t the normal procedure to take CLF because the CLF pressure will be applied to the circulating system including the open hole, but lets say we have a huge safety margin to our MAASP and it is company policy to simulate the actual well kill flow path.

    Now lets say we drill 1 foot and get a kick. The Shut in Drill Pipe Pressure (SIDPP) is 200 psi and the Shut in Casing Pressure (SICP) is 280 psi. Obviously we can use 30 SPM because the SICP is greater than the CLF.

    This is where I get confused. When we calculate ICP and FCP we use the SCR pressure through the riser (250 psi) in the calculations and we do this because we don’t want to impose the CLF pressure (125 psi) on the open hole and we assume we can eliminate the CLF from being applied to bottom by adjusting the choke. But how can this happen because the choke was WIDE OPEN when we took the CLF and we had 375 psi on the pump and now we are circulating through the exact same flow path at the exact same rate (30 SPM) with effectively no change in the well depth or geometry.

    If someone can answer this I will finally be able to sleep at night.

    1. Your concern is totally logical. Very few people think this much in exploration industry. I am pleased to answer your query.

      What we follow in general practice is because of some assumptions and approximations. To clear the confusion first look at what we generally do without considering choke. Please read step by step even if it looks trivial.

      (BHP’ = Theoretical bottom hole pressure
      BHP = practical bottom hole pressure
      ICP’ = Theoretical initial circulating pressure
      ICP = practical initial circulating pressure)

      1. We record SCR to know pressure loss while circulating at kill rate (like, 30 SPM, 40 SPM)

      2. One part of pressure loss is in drill string and another part is in annulus side.
      SCR = Drill String Pressure Loss (DPL) + Annular Pressure Loss (APL)

      3. At shut in condition well is under control.
      BHP = SIDPP + Drill String Hyd Press = SICP + Ann Hyd Press = Formation Pressure.

      4. During circulation for well killing, to maintain BHP, we should compensate for pressure loss in drill string by adding drill string part of SCR
      to SIDPP.
      Theoretically we should keep:
      ICP’ = SIDPP + DPL

      5. Instead in general practice is we add whole SCR to SIDPP. Here we assume that maximum part of the pressure loss is in drill string and the extra annulus side pressure loss is kept over BHP.

      6. Practically we maintain higher ICP than required:
      ICP = SIDPP + SCR = SIDPP + DPL + APL
      ICP = ICP’ + APL

      7. Theoretically BHP during circulation should be
      BHP’ = ICP’ – DPL+ Drill String Hyd Press = SIDPP + Drill String Hyd Press.

      8. But practically we maintain higher BHP than required
      BHP = ICP – DPL + Drill String Hyd Press = SIDPP + Drill String Hyd Press + APL
      BHP = BHP’ + APL

      Thus, in practice we maintain ICP higher by APL than required and in turn BHP also becomes higher by APL.

      On the Annular Side of the well:

      Theoretically, during circulation for well killing BHP should be same as at well shut in.
      BHP’ = SICP + Ann Hyd Press = Casing Press + Ann Hyd Press + APL
      Casing press = SICP – APL
      But, in practice, we maintain Casing Press = SICP
      It implies that in practice we are keeping BHP higher by the amount of APL than required. BHP = BHP’ + APL

      Now with Choke Line Friction (CLF) in offshore operations

      1. We do not take CLF while calculating ICP because for ICP calculation we are only concerned about the pressure losses in drill string.

      2. ONE IMPORTANT THING: Pressure gauge from which we take casing pressure is after CHOKE LINE but BEFORE CHOKE itself. We keep choke WIDE OPEN while taking SCR to just only get the CLF effect no CHOKE BACK PRESSURE.

      3. ICP is already more than required by the amount of APL, therefore, CLF is not added to the ICP.

      4. In offshore operations SCR is also recorded while circulating through choke because CLF value can be used where MAASP is critical.

      5. During circulation for well killing in offshore the excess pressure on annular side is from APL and CLF. If MAASP is critical we can reduce casing pressure by CLF safely without loosing control over the well.

      I hope this answered your question and now you can comfortably sleep at night.

  14. The questionnaires are really helpful and also the comments/questions that can be added. Recently I wanted to calculate the upforce from the well in accordance to the stringweight in relation to stripping. I came out with this formula : Fwp = 0.7854 * (Dp*Dp)*Ps
    example Fwp = 0.7854 * (5″x5″drillpipe) * (1200 psi)

    My question is; is 0.7854 a constant or do you know where it is coming from?

    1. Force = Area * Pressure
      Drill pipe has circular cross section. So, its area is same as that of a circle.
      Area of circle = (Pi/4)*(d*d)
      Pi = 3.1416
      Pi/4 = 0.7854
      Thus, the formula
      Fwp = 0.7854 * (Dp*Dp) * (1200 psi)

  15. someone told me that there are questions regarding adjustable/fixed chokes. what could be asked? how they are different or maybe how they are alike?

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