I'd be happy to help you. I got my level 4 in 2019 using CBT and also have a PE in fire protection. The exam can be pretty tough.

Thanks I could use it

Question 1# If you want to know the number of standpipe risers required in a building ( exclude sg. ft. or a exemption for this ) There must be a fast and simple formula to determine the radius of travel for each riser mapped out for a building but in this scenario in my mind the known length and width of the building must be known. There was a question in another thread about this same question So am I over thinking this, how would one know the number of standpipes risers required in a building? Is it only the travel distance to determine the number of risers. The NFPA 20 7.1 states that the design of the system is governed by building height and so on. In the annex A.7.1 it states the building height determines the number of vertical zones. But where is the rest of the info. The IBC which I bought for this in chapter 905 and 905.3.9 in states every required stairwell it states 30 ft higher than lowest fire access level etc. But the question I had was how many standpipe risers in a building with a known sq. ft. Is there something in an IBC somewhere that states this many sg. ft. means this many stairwells or am missing the whole picture?

MGXFP

I am using firetech productions and the codes books. Unless you are referring to cognitive behavioral therapy as CBT, I am not familiar with it ?

Recent flow test of 1680 gpm was perform with a viscosity of 10ft/sec. What is the minimum size riser?

I pulled this question from a previous post but I remember a question like this on the test. I solved this as v=Qx0.4085/D^2 but I do not know how to rewrite the formula if the GPM and Velocity are known and the pipe size is not any thoughts anyone?

One at a time:

1. For standpipes, travel distance is 130 ft. for unsprinklered buildings [nfpa 14 7.3.2.2.1.2] and 200 ft. for sprinklered buildings [7.3.2.2.1.1]. Typically standpipes are in every stair and stairs are typically on the outside wall of a building. A 750 ft long building would need 3 standpipes if the building is sprinklered and 4 if not.

2. Firetech are probably your best study guide. CBT stands for Computer Based Training at a Pearson center. In the before times, exams were scheduled months in advance and were given in a pencil and paper format.

3. Use algebra to solve for D. D= sqrt(Qx0.4085/v)

Thanks for the help, I took firetech's class and passed two of their practice tests 97 and 90. Then took Nicet practice test and I got a 86.

I remember a few of the questions from the test and the other thread on this page is pretty close to what I saw. The conveyor one through me for a loop because I think there missing some of the details. But I solved the conveyor one as follows

What is the minimum design area for four (4) underground (1,000ft each) conveyors with 4 by 100 ft long belt.
Answer 4 x 100 =400 , 400 / .25 =1,600 not including hose stream allowance, Correct ?
Per NFPA 2019 13 26.34.1.3.2 (4)

This one stumps me, I thought you would use the loop calculation for this but I don't think that's right and I don't think you would use the k-factor formula for this because the GPM sounds to high for that. Is there another formula for junction points? I have the Pat Brock book but I cant seem to find anything called junction point. Any thoughts ?

Pipe “A” with a flow of 210 gpm and 36 psi and Pipe “B” with a flow of 760 gpm and 59 psi. What is the new gpm and psi at the junction point?


A sawmills total fire protection water demand is 1,300 GPM. The building fire protection is supplied by a welded steel suction tank and underground main loop. The fire protection alarms are not constantly monitored.

Answer 40,000.100,000.150,000.200,000
Correct Answer 200,000
Per NFPA 2019 19.3.3.1.2 Extra Hazard 90-120 minutes
Per NFPA 2019 19.3.3.1.3 the lower duration is only for constant monitor
Per NFPA 22 5.1.3.1 the 200,000 is correct due to 150,000 being to small
Equation 1,300 * 120 minutes =156,000 round up correct ?

Question my training material states when calculating tank volume , the total volume is to be based on the water capacity plus the air capacity. Does that mean that when sizing a tank you must add air in cubic feet to the total volume the only formula I am aware of is calculating what the air pressure should be in a tank. Pi=[(Pf+15)/A}-15. Is there another formula for accounting for the air capacity in the tank ? Example If the water demand in gallons is 10,000 gallons what size tank is required? Could someone explain this a little. I ask this based on a note in the firetech production Water based Layout Level III Part Two Water Supply Chapter Page 12 training module.

Matthew-for the tank sizing, look at NFPA13 (2013 Ed.) A.24.2.4.3

I took Level III about 5 years ago. I do not think I would have passed it with out Firetech's course. Now, if only they had a course for Level IV...

for the underground conveyor belts i read the following:
based on that, it seems three surfaces need protected so 3x4x100=1200sf. and you say there are 4 belts so 4 x 1200sf = 4800 sf. the minimum discharge would be 4800sf x 0.25gpm/sf = 1200gpm.

The keyword for this problem is imbalance. My first edition of Brocks hydraulics is equation 12.6.

200,000 gal. is most correct as 150,000 gal. is too small.

your last question is for a pressure tank. i've never done one, but if you follow the formula you should be fine. atmospheric tanks don't have pressure requirements or air volume requirements, just venting requirements so air can replace water that is discharged. The pressure tank must have capacity for the fire system which is the sprinkler demand times the duration (and plus hose if connected to the tank). I believe you as the designer get to choose the A value, the less air volume in the tank, the higher the pressure needs to be.

DennisR4, I took the Level 4 in Nov. 2019 and found a specific study guide unnecessary. If you follow the selected general references published by NICET you should be fine. The questions are harder but if you have 10 years experience you'll be ok. Personally I studied 2 hours a day for about a month and read through my old NFSA Levels 1-IV manual and old Firetech manuals and did good enough.

Thanks for the help so far i have a few more questions

What is the minimum design density for a pump room of an advanced light water reactor electric generating plant ?

NFPA13 2019 26.27.1.1 States The fire water supply shell be calculated on the basis of the largest expected flow rate for a period of 2 hours but not less than 300,000 gallons. So I read the annex but it doesn't say specifically about the minimum density. It's not listed in my hazards occupancy list by bldg. type under NFPA 13 2019 A.4.3.2-4.3.6 But I'm going to guess EH2 on this correct ? The answers are one of the following 0.15,0.19,0.25,0.30. I choose 0.30 but I had to guess and I would like to know how to do this. The minimum design density for EH2 is 2500 per NFPA 13 2019 19.3.3.1.1. Any help would be appreciated.

This one stumps me, I thought you would use the loop calculation for this but I don't think that's right and I don't think you would use the k-factor formula for this because the GPM sounds to high for that. Is there another formula for junction points? I have the Pat Brock book but I cant seem to find anything called junction point. Any thoughts ?

Pipe “A” with a flow of 210 gpm and 36 psi and Pipe “B” with a flow of 760 gpm and 59 psi. What is the new gpm and psi at the junction point?

For the pump room, i see 0.25gpm/sf over the entire fire area from NFPA 13 26.27.1.8 2019 ed.

Quote:

This one stumps me, I thought you would use the loop calculation for this but I don't think that's right and I don't think you would use the k-factor formula for this because the GPM sounds to high for that. Is there another formula for junction points? I have the Pat Brock book but I cant seem to find anything called junction point. Any thoughts ?

Pipe “A” with a flow of 210 gpm and 36 psi and Pipe “B” with a flow of 760 gpm and 59 psi. What is the new gpm and psi at the junction point?

Use the below to solve:

Quote:

The keyword for this problem is imbalance. My first edition of Brocks hydraulics is equation 12.6.
Qadj=(Ql)(sqrt(PH/PL))
Where Ql=calculated flow in low pressure line, Pl=lower pressure, Ph=higher pressure, Qadj=adjusted flow from low pressure line

I get 268.8 gpm using the K factor method or the method quoted above.

For the pump room, i see 0.25gpm/sf over the entire fire area from NFPA 13 26.27.1.8 2019 ed.


Quote:
This one stumps me, I thought you would use the loop calculation for this but I don't think that's right and I don't think you would use the k-factor formula for this because the GPM sounds to high for that. Is there another formula for junction points? I have the Pat Brock book but I cant seem to find anything called junction point. Any thoughts ?

Pipe “A” with a flow of 210 gpm and 36 psi and Pipe “B” with a flow of 760 gpm and 59 psi. What is the new gpm and psi at the junction point?


Your stating that because it's a fire pump room the fire pump room has it's own design density. But when I read this question it doesn't state that the room is under 400 SQFT. So I automatically think the the higher hazard would be at play here. NFPA 13 2019 19.2.6.1 (1) - (3) and A19.2.6.1 (3)

Question: When taking tests and certifications, is it customary when there is missing values / variable data that would be needed. Just answer the question has simply as possible? When I read this I thought higher hazard but maybe may answer would be wrong.

Thanks MGXFP (Automotive)

The imbalance formula is great , I found it in the 3rd edition of Pat Brock's book Equation 13.6, I cant believe I didn't see this before. Thank you

Ok, I spoke to soon, I just got done reading the steps for the imbalance but I'm not getting the same answer you had. So I must be missing a step

Use the below to solve:
Pipe “A” with a flow of 210 gpm and 36 psi and Pipe “B” with a flow of 760 gpm and 59 psi. What is the new gpm and psi at the junction point?
Quote:
The keyword for this problem is imbalance. My first edition of Brocks hydraulics is equation 12.6.
Qadj=(Ql)(sqrt(PH/PL))
Where Ql=calculated flow in low pressure line, Pl=lower pressure, Ph=higher pressure, Qadj=adjusted flow from low pressure line
I get 268.8 gpm using the K factor method or the method quoted above.


(Ql)*(sqrt(PH/PL) = 210 * (sqrt(59/36) = 210 * 1.28 = 268.8

K=Q/sqrt (P)= 760/sqrt of 59 = 98.94 , 210/ sqrt of 36 = 35, 35 + 98.94 = 133.94, 133.94/ sqrt of 59 = 17.44 ? I don't think I understand how you got the same answer using the k factor formula?

Question

An Urban water utility is providing 30 psi static pressure. A fire pump rated at 140 psi has been installed to supplement the pressure for a sprinkler system. The fire pump conforms with the max shut off head. What should the the sprinkler system pipe be hydrostatically tested at ?

190 Psi
200 Psi
220 Psi
276 Psi

140 * 1.4= 196 +30 = 226 + 50 + 276, I choose 276 Psi am I correct ?
Per NFPA 13 2019 7.1.2 ,28.2.1.3

MGXFP (Automotive) Is there a way to get a hold of you through Phone or Teams meeting ? I see your in Texas and I'm in Dallas. Would you be willing to tutor me a little my test is on Thursday and I feel like I still have a few things to wrap my brain around?

Based on hydrodynamics, only one pressure can be at one point at one time. The branch line with 59 psi at the fitting where the other branchline joins can't have 59 and 36 psi at that point. The higher pressure governs the flow of the lower flow branchline. The K factor of the smaller flow line is 35. The lower flow line must discharge more to balance to the higher pressure of 59 psi. So we use the K factor of the lower flow line to balance to the higher pressure point junction of 59 psi. 35*sqrt(59)=268.8 gpm.

The way you solved the problem the K factors of the two lines were added together, but we're not solving for two lines flowing at their respective different pressures but one line balanced to a higher pressure junction of another line.

You are correct, a fire pump of 140 psi at the max allowed churn of 140% will make 196 psi. Add the 30 psi from the municipal supply and the working pressure is 196 + 30 = 226. NFPA 13 requires the hydro test pressure be 200 psi or 50 psi above the working pressure, so 226 + 50 = 276 psi (they are ignoring the jockey pump which generally boosts the pressure by another 5-10 psi).

I'm available via Teams or phone. Do you know a way to send a private message on this forum? I hate to say i've been here 20 years and not sure how to do that.

Thanks for the information MGXFP!!!

I'm looking in to a way to send a private message. I sent a message through the contact us box. I have teams as well do you think it would be ok to post the link through here? I appreciate the help so far. I'll be studying all weekend on this.

.

Gentleman,

I want to say thank you all for your help especially MGXFP, thank you for help. So I took my Nicet Level III hydraulic Lay-out and passed. I wanted to say I have tips for those of you who are about to take the test. First of all pay attention to the calculator, Nicet's website has an example of the layout and model of the one their suppose to provide. My advise don't use the one on the computer, either bring one with you that is of the same model ( it's like 14.00 bucks ) in the event they are out of them. Also I was lucky I guess, was put into a room that was private with a window for the attendant to see me and the door was closed, it worked well since your flipping through those pages pretty fast and don't want to bother other test takers. I would ask to use that room if it's available. Also I had my tabs all over that book including Annex's, Chapters and other important items but I will say if you have to many tabs on there it might make it cumbersome.Please feel free to reach out if you need help, I'm not a PE or anything but I took my test last Thursday and it is a recent exam. Good luck out there and if I passed it, you can too.


PS : Do a search from youtube on Nicet Level III exam and watch the video of a guy out in Alabama, he had some good tips as well.

I was curious to see if someone could help me with some ESFR type questions that are on the test. I need help calculating those and I need help with understanding standpipes in a sprinklered building and an unsprinklered penthouse.

What do you need help understanding for ESFR systems?

standpipes in sprinklered and unsprinklered buildings have different requirements regarding spacing and criteria, see NFPA 14 for more info.