Hello, this is Muhaba Lee Ji-yeon. In this session, we will look at the trends in the questions for the 2024 2nd round practical exam for fire equipment technicians in the mechanical field. We will select and solve only two problems of the new type. In terms of scoring, we have 14 points in the fluid dynamics, 6 points in fire extinguishing equipment and automatic fire extinguishing devices, 49 points in water-based fire extinguishing facilities, 17 points in gas-based fire extinguishing facilities, no questions were asked in evacuation facilities, and fire activity facilities and other facilities related to combustion facilities had a problem worth 14 points. So, we need to focus a bit on water-based fire extinguishing facilities, as they account for nearly 50% of the total score. Let's take a look at what kind of problems were asked in fluid dynamics. First, there was a question about defining the phenomenon of cavitation and writing preventive measures.

Once a problem is presented, and then it was presented again in succession. Usually, in the practical exam, questions about the capillary phenomenon were not often asked, but they were consecutively presented. Next, a problem about calculating the pressure difference between the venturi and the orifice was given. Oh, many people might have found this a bit difficult. Problems related to the venturi and orifice meters have been frequently presented these days. I emphasized during the theory class to make sure to memorize the formulas. So, if you had memorized the formulas, you could practically solve the problem. However, when it comes to pressure difference, it may feel a bit difficult for no reason. I will solve this problem together later. Usually, questions about calculating flow rate or specific gravity are often asked, but this seemed to be a slightly new type of problem. Next, determining the occurrence of the Coanda effect.

A problem has arisen. It is a question that requires comparing the effective inhalation dose and the required inhalation dose to determine whether a joint phenomenon occurs. So, three questions like this came up, and next, we will look at a problem related to the digestive system and automatic fire extinguishing devices. In total, two questions were given, and in one of the questions related to the capacity unit of the digestive system, the specific target material was lodging facilities. Problems related to lodging facilities were not a common type that appeared in the practical skills test, but it came out this time. However, those of you who have been studying with me will remember. We memorized lodging facilities as B&B operators and innkeepers. So, if you calculated in units of 1 square meter or more, you would get the correct answer. Next, a short-answer question about residential kitchen automatic fire extinguishing devices came up. However, there are many new parts in the part where the gap is slightly noticeable.

It was open, related to the installation target, and so on, so it seems a bit new. Now, let's move on and take a look at the sewage treatment facilities. Many problems were raised in the sewage treatment facilities. First, a problem came out regarding the calculation of pressure flow in the water pump. It was a somewhat difficult problem. However, the problem is not a new one but was a problem that was given in 2015. So, is this not in the recent year's problem book? No, it is not. In the supplementary section at the back of our 7-year problem book, there are many good problems from the past 7 years, including new types of problems or good problems from previous years. If you consider it as a problem included in the appendix, you will see that it was included there.

In the appendix, there was a question, and in our first session, there was also a high-scoring question. So, the appendix is doing a significant job. Next, let's go down and see. A problem related to the fixed gun outlet of a foam extinguishing system came out, a classic problem, and a problem related to the foam generator also appeared. Next, there was a problem calculating the number of large drop sprinkler heads installed in a Lexic warehouse with head installation. You should consider this problem as a new one because from January 1, 2024, the fire protection performance criteria and technical standards for warehouse facilities were established and implemented. Therefore, if you solved the problem according to these revised laws, it would be correct. In the case of the distance problem set, all have been updated with the revised laws at the problem set house. So, please think that the problems are presented exactly as they are in the problem set book. The answer will be the same. Later, we will solve this together. Next, let's move on.

A problem related to digestion tanks came up. Next, there were questions about performance tests, piping symbols, and driving rules. Also, a question about calculating the number of heads in a sprinkler system came up. By doing this, a total of 49 points are allocated. Moving on to gasification facilities, there are two questions worth a total of 17 points. The first question was about surface fires in carbon dioxide fire suppression systems, and the next one was about powder fire suppression systems. Both are somewhat traditional types of questions that have been frequently asked. Next, in fire extinguishing facilities and other facilities, there were two questions about fire prevention facilities. I will show you. Regarding fire prevention facilities, calculating the discharge amount and a related question about the principle of conservation of mass were also included. When calculating the discharge amount, in cases where the area of the floor is less than 400 square meters, there was a difference.

The problems in the book are very similar. However, here, the unit of pressure is not in millimeters of mercury but in kilopascals. Next, another question came up about the ignition method of machine number 16. So, if you can explain types 1, 2, and 3, it will be the correct answer. By checking the overall trend of the questions, we can see that there were some difficult questions that seemed hard to solve. However, entering the passing zone didn't seem too difficult. I will calculate the scores. I have marked in yellow the questions that you may have found a bit challenging. But these questions are completely wrong, to the extent that you cannot even get partial credit.

Even if everything is wrong, when you add up all the yellow color questions, it's a total of 31 points. So, if we have a total of 60 points, even if 31 points are deducted, it's still a passing score. Even if everything here is wrong and I didn't write down the measures for the phenomenon, I only got a little deduction. That's -2 points. And then, coming down, there is a problem related to the fixed foam outlet in the foam extinguishing equipment. Even if I got one sub-question wrong partially, it's -3 points. Like this, even if there are partial deductions for getting one wrong, currently, even if it's 5 or 6 points deducted, it seems like we are still within the passing range. So, in fact, there were difficult and high-level questions, but it seems okay to pass. In reality, this round was like that. So, let's finish the trend of the questions up to this point.

I will start by solving two problems of the new type. Let's try the first problem. I have brought a problem related to the venturi tube and pressure difference. Now, in a venturi tube like the one in the picture, water is flowing with a flow rate of 5.6m³/min. When water is given, you should always consider the density and specific weight given. Also, as it comes down, the inner diameter is 36cm and at the throat, the inner diameter is 13cm. It is stated that a venturi meter of this type is installed. Calculate the pressure difference P1 - P2. However, the discharge coefficient between the venturis is given as 0.85, and it is stated that friction losses on the pipe walls are not considered. Take a look at the picture below. At this moment, there is a thick pipe with a diameter of 36cm and it narrows down like this.

It will be 13cm. At this point, the problem is to find the pressure difference between point 1 and point 2. So, how can we solve this? Please don't think about the Venturi effect. If we are just given a pipe and asked to find the pressure difference, we don't need to consider friction losses. Right? We can apply Bernoulli's equation. So, I have two solutions prepared. First, let's try solving it with Bernoulli's equation. When we solve it with Bernoulli's equation, it will be like this: Gamma * P1 + 2 * G * v1^2 + z1 = Gamma * P2 + 2 * v2^2 + z2. If we consider this as a horizontal pipe, z1 and z2 will be canceled out. So what remains here is Gamma * P1 + 2 * v1^2 and Gamma * P2 + 2 * v2^2. Since what I want to find here is the pressure difference...

I will pass this on, in other words, concentrate all the pressure on one side and all the velocity with a free stream on the other side. Now, let's rearrange this equation. If we rearrange this into terms of velocity, it will be gamma times P1 - P2. Let's rearrange this again to 2 times V2 squared - V1 squared. Then, to find P1 - P2, we just need to pass this term over. So, by passing P1 - P2, it will be gamma times the whole 2 times V2 squared - V1 squared. So, this will result in V1 squared, right? That's how it will be. Then, here, when we calculate the pressure difference, since gamma is the specific weight of water, if this value is not given separately, we can just put in the acceleration due to gravity, 9.8 m/s squared. So, what do we need to know? We need to find V2 and V1, and then plug in V2 squared and V1 squared accurately.

If you pay attention, it should be fine. Now, the key is how to calculate the flow rate. How can we calculate the flow rate? The flow rate is here. There is a horizontal pipe here. Water is flowing. So, what comes to mind? Continuity equation. Q1 and Q2 are equal. Then, if we solve it as qn AV, it should work. But what is given here? What conditions are stated? Let me go up there and take a look. The discharge coefficient here is 0.85. It is stated here. The discharge coefficient mentioned here can be seen as the discharge coefficient. In other words, you should look at the CD value, which is the coefficient value. If you have studied fluid mechanics practically, you may understand what I am talking about. However, the discharge coefficient, simply put, is related to Qn AV. Where does the CD value attach? It attaches in front.

That's right. So, when we calculate the V value, how should we do it? Let me give you V sub n. CDA divided by Q, right? By doing this, we can calculate V1 and V2. So, let's go. To calculate V1, we divide Q by A. Here, the CD value is attached like this. Then, how can we change Q and N? Right. If we put 4D squared, it will be 4Q divided by Pi D squared. So, the CD value is attached in front. Alright, let's try putting the formula here. What is 4 times the value of Q? It's 5.6m^3. But the unit is in cubic meters, right? It should be converted to cubic centimeters. It should be divided by 60. So, 5.6 divided by 60, and then the CD value is 0.85. Next is Pi times D squared. At this point, we will calculate V1. So, if we take 36cm and convert it to meters, it will be 0.36m. Let's put this in here.

If we square 0.36, we will get the value of v1. Then, if we change 0.36 below to 0.13 and calculate by replacing only 0.13 squared, we can also obtain the value of V2. By calculating v1 and V2 in this way, we can see that v1 is immediately 1.079 and V2 is 8.273 m/s. Then, we can calculate the pressure difference here, which is P1 - P2 N. We will multiply gamma by 2 times V2 squared minus v1 squared, and substitute v1 and V2 accordingly. In this case, since we are looking for the value in kilopascals, we will put gamma as K/m^3. So, if we put 9.8 m, which is the specific weight of water, we can continue with the rest of the calculations.

I will get 33.64 K. So, we can obtain the value like this. If you didn't solve it this way, some of you might say that. Let's take a look at the second solution method. We can apply the Venturi meter flow formula. Do you remember this? Regarding the Venturi meter flow formula, qn = AV. Looking at the given conditions, we had the value of Q, the value of CD, the discharge coefficient, and the narrowed section of the pipe. So, we could find the cross-sectional area of each pipe, and you know the acceleration due to gravity, 9.8. Then, what is h here? That's right, h directly relates to the pressure difference. Typically, when we learn about the Venturi meter, we often study this part extensively. At that time, there are two fluids...

They were attached, so there are two fluids, the light gamma divided by the heavy gamma minus 1, we added this together. At this time, the light one is floating on top, and the heavy one is sitting at the bottom, right? So, by doing this, we looked at the formula like this, but the problem is, how many fluids are there? Right now, there is one. It's just passing through the tube, just a route, right? So, you just need to put it in as it is. So here, you just need to put it up to the route, and for the classification below, it needs to go into the formula as it is. So, if you simply substitute in the formula, you can fill in all the values except for HM. Then you can find the water flow rate for the pressure difference. Once you know the water flow rate for this pressure difference, what should we convert the metal into? Right, by pressure.

Well, because it's a gamma h, in order for us to calculate in pascal, we just need to put the value of gamma h directly. If you put 9.8 kN/m^3 into gamma, it should be fine. So, if you multiply 9.8 by h, that will be the result. So, let's try substituting the formula as it is. Let's redraw the diagram we can see above. One part is 36 cm, and the other part is 13 cm, right? 13 cm. Then, since we have given the diameter to us, it would be much easier to calculate d1, d2, and r. In other words, A1 is 4/πd1^2, and A2 is 4/πd2^2, so if you simplify it, d1^2 divided by d2^2, you can square the whole thing. So, when you distribute this, it becomes 4^2. So, let's solve it with the formula. Let's substitute the values directly. At this point, the value of Q is in units of m^3.

Let's enter Secro. So, given the conditions, it is written as 5.6m^3, but please write it as 60m^3. The CD value output coefficient is 0.85, right? Let's put 0.85. Multiply by the square root of 1. So, d1 is 36cm, which means 0.36m. Next is D2, right? 13cm, which is 0.13cm, right? This will be to the power of 4. This divided by a A2 window, so if it's A2, you should put the narrow part's area here. Put in 4π * 0.13 squared. Then, move on to square root 2, which is 2 * 9.8 * h. This is the variable hga we are trying to find. If you use the calculator's solve function as it is, once you find hl, let's see how much it is. It turns out to be 3.432. So, since h is 3.43, we can find gamma h. So, if you put in 9.8 and calculate, it will be 33.63 K. This is how you get the correct answer. So,

Using the Bernoulli equation we found earlier, it was 33.64. And when we solve it using the Venturi meter formula, it becomes 33.63. Although there is a very slight error, this is due to the units. As we did with Bernoulli earlier, we rounded it to three digits precisely like this due to rounding. You can consider this as a difference that occurs due to intermediate calculation results. They are the same values because where did this formula for fluid flow rate come from? Right, it was derived from the Bernoulli equation. So, ultimately, the same numerical values come out. Therefore, the second method of solving is actually a much more concise form in the calculation process. By solving it this way using Venturi, and then moving on, let's solve the second warehouse facility problem. It consists of a single protection zone with a width of 15m and a height.

At a height of 26m, there is a 7m high Lexic warehouse storing special flammable materials. Now, they said they installed Large Drop sprinkler heads. So, what would be the horizontal distance? Originally, for warehouses with Large Drop sprinkler heads installed, the horizontal distance would just be the warehouse facility. Then, the horizontal distance would be below 2.1m. However, if the stored items are special flammable materials, then it should be enhanced to below 1.7m. In this case, if the warehouse is a fire facility, it would be below 2.3m. This is all specified in our warehouse fire safety technical standards. Now, going back, if you want to install Large Drop sprinkler heads in a square shape, in the corresponding warehouse...

The total number of sprinkler heads, please solve this problem. The architectural structure is a non-combustible structure, and we will not consider anything other than the given conditions. So, how should we solve this? The width is 15m, the length is 26m, and the height is 7m. Then, this is a rack warehouse, right? It was given. First, how many heads should be installed on the ceiling, both horizontally and vertically, to find the total number of heads to be installed on the ceiling. But that's not the end. In the case of a rack warehouse, besides installing on the top, we also need to install within 3m below. So, since it was mentioned that it is 7m here, we need to install more here. If we placed one layer on the top, then it means we need to install below 3m.

If we say we've gone one step further with the installation, there is 4m left here. So, since there is 4m left, we need to lay one more step, right? If we install less than 3m here, you should consider installing a total of 3 rows. In other words, if we say the height is 7m, then we install less than 3m. If we divide by 3, how much will it be? It will be 2 point something, right? So, here, it must be rounded up, we can't be short. Therefore, please consider installing 3 rows. Then, how many heads to install on the ceiling horizontally and vertically, you can calculate by multiplying the number of heads horizontally and vertically. After calculating, since we need to install a total of three rows, multiply by 3 to get the correct answer. Now, let's start. If it's 15 horizontally and 26 vertically, let's draw it like this.

Let's see. The width here is 15m, and the height is 26m. Like this. So, in that case, let's take a look at how many heads need to be installed on the width. At this time, what is the horizontal distance value? It should be below 1.7m, up to a maximum of 1.7m. Then, what about the forward design, the distance S between our heads? Right, it will be 2r times cosine 45°. If we substitute 1.7m into the R value and calculate, let me show you the result. It comes out as 2.44m. So, the distance between the heads is precisely 2.44. Now, if we need to know how many to install along the width of 15m, we just need to divide by S. Let me calculate this. When we divide, what comes out? Let me show you. It shows 6.24 units. So, we can't have a fraction, we need to round up, which gives us a total of seven units. Then, using the same method, for the 26m, we do the same thing with S.

Divide it by the value. So, when we divide it by the head distance, how many points did we get? It always rounds up. So, a total of 11 comes out. Therefore, the number of heads that need to be installed on the ceiling is 7 multiplied by 11, which totals 77. This is our Lexi warehouse. It should be installed under 3 meters. Therefore, we will install them in rows of three, totaling how many? 77 heads will be installed in total. So, if we multiply 77 by 3, the result will be 231. This is how you get the answer. However, I have one more thing to mention. During the practical field lecture, many people asked questions. They mentioned that it is also covered in another book. In the case of the Lexi warehouse, specifically in 2.3.1.1, when installing warehouse facilities, the sprinkler equipment is large-droplet.

Install the sprinkler head in a wet manner, as it is now presented. It says to install it like this, but in addition to this, it also says to install a large drop-type sprinkler head at a height of less than 3m. So, I brought this according to the fire safety technical standards. Besides this, since it says to install it at less than 3m, install one at the top and one more. This is in addition to the installation, so the total height of the warehouse is 7m, and it says to install it at less than 3m. Since it shows 2 points, add 3 + 1, so you need to install a total of four rows. This is how you solve it, right? Some people asked questions about this, but this is not it. If you draw it again, it can be easily explained. Install it at the top first, and then install it here. If you install it...

If it wasn't this way, we would have installed it on top like this. However, in addition to this, we were asked to install more. So, apart from this, install one more below 3M here, and install another one below 3M once again. So, if we install in addition to this, it will be a total of three. You need to explain it like this. Also, we discussed about the valve related matters. Then, we will wrap up the issue. What I want to mention again is that in this case, the problem of calculating the number of large drop sprinkler heads came up. But in the next session, there will definitely be a question about calculating the amount in Suwon. So, if we look at it a bit more, the number of installations of large drop sprinklers should be the highest in the defense zone. Here, the number of installations should be such that it exceeds the amount calculated by multiplying 3.2m by 3.

I will do that. And if it was a Lexic warehouse, you need to multiply the installation quantity by 9.6 cubic meters. Please make sure to remember that. So, how many minutes is this exactly? You might be curious. First of all, in the case of large dropping sprinkler heads, the water amount per head is 160 liters. So, if you reverse calculate with 160 liters, what happens? It's a concept of 20 minutes out of 60 minutes. Let me explain. In the case of a general warehouse, if you multiply the installation quantity by 160 liters per minute for a 20-minute basis, it will be 3.2 cubic meters. And if it was a Lexic warehouse, it will be 9.6 cubic meters, as provided. In other words, it will be the installation quantity multiplied by 160 liters per 60 minutes. 60 minutes are crucial. By doing this, the water amount in Suwon will definitely be a topic in the future, so I hope those preparing for the next session will memorize it. At this time,

If I may point out one more important thing, when it comes to the number of installations, it should be calculated as 30 or more installations. So, just like our regular spring cluster, you can consider the standard number. Got it? So, in this way, the problems we prepared end here. Alright, everyone, you have worked so hard taking the exam, and I sincerely hope you get great results. Fighting!

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24년 소방기사 기계 2회차 실기 출제경향 및 신유형 문제 미리보기!

24년 소방기사 기계 2회차 실기 출제경향 및 신유형 문제 미리보기!

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