In the winter, a dependable heating system is essential to keeping our homes comfortable. However, have you ever given any thought to the internal mechanisms of your heating system? Hydraulic resistance is an important factor that is sometimes disregarded. Similar to water flowing through a pipe, your system may find it difficult to distribute heat efficiently if there is excessive resistance. Comprehending hydraulic resistance is essential for enhancing the efficiency of your heating system and guaranteeing a cozy atmosphere throughout the year.

Imagine your heating system as an intricate web of radiators, valves, and pipes that work together to distribute heat throughout your house. The flow of water through these pipes is hampered by impediments, much like traffic on a busy road. We refer to this type of resistance as hydraulic resistance. It is the force that prevents water from flowing freely and can cause uneven heating, decreased efficiency, and increased energy costs.

In your heating system, there are several factors that affect hydraulic resistance. A number of factors come into play, including the diameter and length of the pipes, the quantity of bends and fittings, the kind of valves used, and even the viscosity of the heating fluid. Imagine it as a maze with twists and turns; the more complicated the path, the more difficult it is for water to flow through.

What makes hydraulic resistance important, then? That being said, consider your heating system as a garden hose. Water flow decreases or stops completely if you kink or crimp the hose. This also holds true for your heating system. A home with excessive hydraulic resistance may have less comfortable conditions overall, less water flow, and higher pump pressure.

Thankfully, there are techniques to reduce hydraulic resistance and maintain the efficiency of your heating system. In order to lower resistance and raise system efficiency overall, homeowners and HVAC specialists can take preventative measures such as appropriately sizing pipes, minimizing bends, and optimizing pump settings. You can improve the efficiency and lifespan of your home heating system by making wise decisions by being aware of the significance of hydraulic resistance.

Component | Hydraulic Resistance |

Radiators | Varies depending on size and design; typically moderate resistance |

Pipes | Depends on material, length, and diameter; can have significant resistance |

Valves | Minimal resistance, but poorly maintained valves can increase resistance |

- Hydraulic calculation of the heating system, taking into account pipelines
- An example of a hydraulic calculation of a two -pipe gravitational heating system
- Video on the topic
- Dressing water in the pipeline
- Local hydraulic resistances
- What does hydraulic resistance look like, a short test
- "Boiler warriors" hydraulic resistance of the heating system .
- Video 10 TM-4.2 hydraulic resistance
- The first hydraulic calculation.
- Practical lesson of the hydraulic calculation of the heating system

## Hydraulic calculation of the heating system, taking into account pipelines

Heating system layout featuring an open expansion tank and pumping circulation.

The primary hydraulic parameters—the hydraulic resistance of the reinforcement and pipelines, the coolant’s flow rate and speed, the table, and the program—will be utilized in all computations. There is a full relationship between such parameters. You must rely on this when doing computations.

Example: if you increase the speed of the heat carrier, the hydraulic resistance of the pipeline will also increase. If the heat carrier consumption is increased, the coolant rate and hydraulic resistance can increase at the same time. The larger the diameter of the pipeline will be, the smaller the speed of the coolant and hydraulic resistance. Based on the analysis of such relationships, it is possible to turn a hydraulic calculation into an analysis of reliability and efficiency parameters of the entire system, which can help reduce the costs of materials that are used. It is worth remembering that hydraulic characteristics are not constant, which can help nomograms with.

Hydraulic calculation of the water heating system. The flow rate of the coolant

Potential layout for the two-pipe heating system in the future.

The thermal load that the coolant must withstand while heat is transferred from the heat generator to the heating device will determine the coolant flow rate. This criterion includes a program and table.

The hydraulic calculation entails figuring out the coolant flow rate relative to the specified area. A plot with a constant diameter and a steady heat carrier consumption will be used in the calculation section.

An example of a brief calculation will contain a branch that includes 10 kilowatt radiators, while the flow rate of the coolant is calculated on the transfer of thermal energy at 10 kW. In this case, the calculated section is a section from the radiator, which is the first in the branch to the heat generator. However, this is only on condition that such a section will be characterized by a constant diameter. The second section will be located between the first and second radiators. If in the first case the transfer consumption of 10-kilovatt energy of heat is calculated, then in the second section the amount of energy that is calculated will be 9 kW with a possible gradual decrease as such calculations are carried out.

System of natural circulation for heating.

The hydraulic resistance of the supply and reverse pipelines will be computed concurrently.

In order to determine the coolant flow rate for such heating, the hydraulic calculation uses the following formula for the calculated site:

G Uch is equal to (3.6*q Uch)/(c*(t r-t o)), where q q is the site’s calculated heat load (in BT). The specific heat capacity for water (C), which is 4.2 kJ (kg*° C), is constant in this example. The temperature of the coolant in the heating system is denoted by t r, and the temperature of the cold coolant is denoted by t o. Calculating the heating gravitational system hydraulically: The coolant’s flow rate

Design of the distributors’ heat supply system.

It is recommended to use a threshold value of 0.2-0.26 m/s for the minimum coolant speed. Reduced speed may cause the coolant to release too much air, which could result in air plugs. The heating system will then completely or partially fail as a result of this. In relation to the upper threshold, the coolant velocity ought to be between 0.6 and 1.5 m/s. The pipeline will not be able to produce hydraulic noises if the speed does not increase above this indicator. Experience has shown that the ideal speed range for heating systems is between 0.4 and 0.7 m/s.

The parameters of the pipeline materials in the heating system must be considered if a more precise computation of the coolant speed range is required. More precisely, internal pipeline surfaces will require a roughness coefficient. For instance, the ideal coolant speed for steel pipelines is between 0.26 and 0.5 m/s. A polymer or copper pipeline may allow for a 0.26–0.7 m/s speed increase. To be safe, you should carefully review the recommended speed listed by heating system equipment manufacturers.

The material of the pipelines used in the heating system, specifically the coefficient of roughness of the inner surface of the pipeline, will determine a more precise range for the coolant velocity, which is advised. For instance, it is advised to follow the coolant speed range of 0.26 to 0.5 m/s for steel pipelines. From 0.26 to 0.7 m/s for polymer and copper (polyethylene, polypropylene, metal-plastic pipelines). If the manufacturer’s recommendations are available, it makes sense to follow them. Pressure loss in the hydraulic resistance calculation of the heating gravitational system

The distributor "3"’s schematic for the heating system.

The total of all losses for hydraulic friction and local resistances is known as pressure losses in specific areas, which are also referred to as "hydraulic resistance." The following formula can be used to calculate such an indicator, which is measured in PA:

Ruuch is equal to r * l + ((p * v2)/2) * e3, where V is the coolant speed (measured in m/s), p is the coolant density (measured in kg/m³), r is the pipeline pressure loss (measured in pa/m), l is the estimated pipeline length on the site (measured in m), and E3 is the total of all coefficients of local resistances in the equipped area and shut-off-regulating reinforcement.

The total resistance of the calculation sections is the general hydraulic resistance. The following table (image 6) is included in the data. Calculating the hydraulics of a gravitational heating system with two pipes: Selecting the primary branch

Pipeline hydraulic calculation.

For the two-pipe system, you must choose the ring of the most loaded riser via the heating device below if the hydraulic system is defined by the coolant passing through it.

In the event that the heat carrier in a two-pipe system moves in a dead end, you will need to choose a lower heating device ring for the highest load from the farthest risers.

In the case of a horizontal heating structure, the lowest floor’s most heavily loaded branch must be selected for ring placement.

Go back to the contents table.

In understanding the hydraulic resistance of your home"s heating system, it"s crucial to grasp its impact on efficiency and comfort. Simply put, hydraulic resistance refers to the force that slows down the flow of hot water through pipes, radiators, and other components of your heating setup. This resistance can lead to uneven heating, higher energy bills, and strain on your system over time. By recognizing and managing hydraulic resistance, you can optimize your heating system"s performance, ensuring even warmth throughout your home while maximizing energy efficiency and minimizing maintenance costs. Understanding how to mitigate hydraulic resistance through proper design, maintenance, and equipment selection is key to keeping your home cozy and your energy bills manageable.

## An example of a hydraulic calculation of a two -pipe gravitational heating system

Computation of the distributors’ heat supply system.

The horizontal two-pipe heating system comprises two distinct heating systems: the supply of heat to distributors (between distributors and thermal point) and the heating from distributors (between heating devices and the distributor). The heating devices of the system are linked to the heating system through the use of a distributor.

The heating system design is typically composed of distinct circuits:

- scheme of heating systems from distributors;
- Scheme of the heat supply system of distributors.

An illustration would be a hydraulic calculation for a two-pipe heating system in a two-story administrative building with lower wiring. The integrated top-end is where the heat supply is organized.

The available source data are as follows:

- The calculated load of the heat of the heating system: q Zd = 133 kW.
- Heating system parameters: t g = 75 ° C, t o = 60 ° C.
- The calculation consumption of the coolant in the heating system: V Co = 7.6 m³/h.
- The heating system is attached to the boilers through a hydraulic horizontal separator.
- The automation of each boiler supports the constant temperature of the heat carrier at the output of the boiler: t g = 80 ° C throughout the year.
- At the input of each distributor, an automatic pressure retail regulator is designed.
- The system of heat supply of the distributors is made of steel water and gas pipes, the heating system from distributors – from metal -polymer pipes.

It is necessary to install a pump with rotation speed control for this two-pipe heating system. To select a circulating pump, you must ascertain the pressure (p n, kPa) and supply values (v n, m³/h).

The pump supply and the heating system’s computed consumption are the same:

7.6 m3/h is V n = v Co.

The total of the following factors determines the necessary pressure, p n, which is equivalent to the estimated loss of heating pressure, a P CO:

- Losses of pressure from OA P CHD.With.T.
- Loss of pressure of the heating system from OA P distributors.from.
- Pressure losses in the distributor a P.

A p co = oa p = P n.+ OA P Uch.from + a p rebt + with.

In order to compute OA P Uch.With.T and OA P.You should run the heat supply system diagram and the heating circuit from the distributor "3" using the circulation calculation ring.

The thermal loads of the Q4 rooms (calculated heat-related losses) must be distributed on the heating system diagram from the "3" distributor in accordance with the heating devices, which are compiled by distributors. The calculation scheme also shows the distributors’ thermal loads.

Both boilers or just one of them can operate, depending on the necessary heat production from combustion (during the spring and summer months). Every boiler features an independent circulation circuit equipped with a P1 pump. Within this circuit, the coolant will maintain a consistent flow rate and temperature of T G = 80 ° C for a duration of one year.

A two-position temperature controller that regulates the power of the P2 pump allows the water supply in a boiler 2 to have a temperature of T G = 55 °C. The coolant circulation will supply a pump with electronic control P3 during heating. Using the observant electronic regulator 11, the heating system’s supply water temperature varies based on outside air temperature, influencing the three-way control valve.

The first direction can be used to perform the hydraulic calculation of the distributors’ heat supply system. You must select a ring through a loaded heating device of the most loaded distributor "3" in order to determine the main circulation ring.

Using a nomogram, the diameters of the main heat pipeline sections d Y, mm are chosen, with a water speed of 0.4–0.5 m/s.

The nomogram’s nature of use is illustrated by a table (section No. 1) with the formula g Uch = 7581 kg/h/h. It is advised to stick to a certain RUS REAST loss of friction and nothing more. Nomograms define Pa pressure on local resistances Z as a function z = f (oae). A table is included in the hydraulic calculation results.

For every segment of the main circulation ring, the coefficients of local OAE resistances should be added up as follows:

- Plot No. 1 (beginning from the pressure pipe of the P3 pump, without a check valve): sudden narrowing, sudden expansion, valve, OAE = 1.0 + 0.5 + 0.5 = 2.0;
- Plot No. 2: the tee of the branch, oae = 1.5;
- Plot No. 3: passing tee, divert, OAE = 1.0 + 0.5 = 1.5;
- Plot No. 4: passing tee, divert, OAE = 1.0 + 1.0 = 2.0;
- Plot No. 2: the tee on the anti -fluid, OAE = 3.0;
- Plot No. 1 to the jumper of the admixture: sudden narrowing, sudden expansion, valve, divert, oah = 1.0 + 0.5 + 0.5 + 0.5 = 2.5;
- Plot No. 1a from the jumper of the admixture to the suction pipe of the P3 pump, without a valve, without a filter: a hydraulic separator in the form of a sudden narrowing and sudden expansion, two withdrawals, two gate valves, OAE = 1.0 + 0.5 + 0.5 + 0, 5 = 2.5.

For the check valve D Y = 65 mm, GHC = 7581 kg/h on site No. 1, the manufacturer’s monogram should be used to calculate the valve resistance, which is as follows:

The bandwidth value at site No. 1a, K V = 55 m3/h, should be used to calculate the filter resistance, d = 65 mm.

(G | k v) pf = 0.1 for A. (7581 /55) 2 = 1900 PA. 2 = 0.1.

The three-way valve’s standard size is chosen, establishing the necessary value as follows: k v = (2 g… 3 g), or k v > 2. 7.58 = 15 m3/h.

We take the valve with d = 40 mm and k v = 25 m3/h.

It will encounter resistance from:

G | k v; A P CL = 0.1 2 is 0.1. (25) * (7581 /5) = 9200 PA.

As a result, the following represents the loss of heat distributor pressure supply:

OA P.With.T equals 21.5 kPa (21514 PA).

The remaining portion of the heat distributor supply is calculated using the chosen pipeline diameters in the same manner.

Using the most loaded heating device Q PR = 1500 W (branch “B”), you should choose the calculated main circulation ring for calculating the OA P Uch.With.T heating system from the “3” distributor.

The first direction is used to calculate the hydraulics.

A nomogram for metal-polymer pipes is used to choose the diameters of the thermal pipeline sections (d y, mm), and the water speed is limited to 0.5–0.7 m/s.

The figure (example of plots No. 1 and No. 4) illustrates the nomogram’s type of use. It is advised to stick to a certain RUS REAST loss of friction and nothing more.

The formula for loss of pressure on resistance z, P is z = f (oae).

It is essential to comprehend the hydraulic resistance of your home’s heating system in order to maintain effective and efficient heating during the winter months. The force that prevents water from passing through pipes, valves, and other heating system components is known as hydraulic resistance. Homeowners can make educated decisions regarding system design, upkeep, and upgrades by understanding this idea.

One important lesson to learn is that reducing hydraulic resistance can result in major energy savings as well as increased comfort. A high resistance means that the pump has to work harder to circulate the hot water, which could lead to uneven heating distribution and increased energy consumption. Homes can get better performance and save money on energy costs by optimizing the system to lower resistance.

Hydraulic resistance management is heavily dependent on routine maintenance. Over time, the system may experience increased resistance due to various factors like corrosion, scale buildup, and debris accumulation. Homeowners can reduce these problems and guarantee their heating system works effectively by arranging routine inspections and cleaning or replacing parts as needed.

Moreover, hydraulic resistance must be taken into account when designing a new heating system installation or upgrades. Selecting pipes, valves, and other parts that are the right size can reduce resistance and increase system efficiency. Speaking with a qualified HVAC specialist can guarantee that the system is built to operate at its best and offer insightful advice.

In conclusion, homeowners who want to maintain effective and efficient heating in their homes must have a thorough understanding of hydraulic resistance. Through appropriate system design, routine maintenance, and strategic upgrades, homeowners can minimize resistance and increase overall comfort levels, operating costs, and energy efficiency all during the heating season.

## Video on the topic

### Dressing water in the pipeline

### Local hydraulic resistances

### What does hydraulic resistance look like, a short test

### "Boiler warriors" hydraulic resistance of the heating system .

### Video 10 TM-4.2 hydraulic resistance

### The first hydraulic calculation.

### Practical lesson of the hydraulic calculation of the heating system

**What type of heating you would like to have in your home?**