1. Assignment for term paper

According to the initial data for the course work, you must:

Determine the hydraulic losses of the evaporator circuit;

Determine the useful pressure in the natural circulation circuit of the evaporator stage;

Determine the operating circulation rate;

Determine the heat transfer coefficient.

Initial data.

Evaporator type - I -350

Number of pipes Z = 1764

Heating steam parameters: P p \u003d 0.49 MPa, t p \u003d 168 0 C.

Steam consumption D p \u003d 13.5 t / h;

Dimensions:

L 1 \u003d 2.29 m

L 2 = 2.36 m

D 1 = 2.05 m

D 2 \u003d 2.85 m

Drop pipes

Quantity n op = 22

Diameter d op = 66 mm

Temperature difference in steps t \u003d 14 o C.

2. Purpose and arrangement of evaporators

The evaporators are designed to produce distillate, which makes up for the loss of steam and condensate in the main cycle of steam turbine plants of power plants, as well as to generate steam for general plant needs and external consumers.

Evaporators can be used as part of both single-stage and multi-stage evaporative units for operation in the technological complex of thermal power plants.

As a heating medium, medium and low pressure steam from turbine extractions or ROU can be used, and in some models even water with a temperature of 150-180 °C.

Depending on the purpose and requirements for the quality of the secondary steam, the evaporators are manufactured with one- and two-stage steam flushing devices.

The evaporator is a vessel of a cylindrical shape and, as a rule, a vertical type. A longitudinal section of the evaporator plant is shown in Figure 1. The evaporator body consists of a cylindrical shell and two elliptical bottoms welded to the shell. Supports are welded to the body for fastening to the foundation. Cargo fittings (pins) are provided for lifting and moving the evaporator.

On the evaporator body, pipes and fittings are provided for:

Heating steam supply (3);

Removal of secondary steam;

Heating steam condensate drain (8);

Evaporator feed water supply (5);

Water supply to the steam washing device (4);

Continuous purge;

Draining water from the body and periodic purge;

Bypass of non-condensable gases;

Safety valve installations;

Installations of control and automatic control devices;

Sampling.

The evaporator body has two hatches for inspection and repair of internal devices.

Feed water flows through the manifold (5) to the flushing sheet (4) and downpipes to the bottom of the heating section (2). The heating steam enters through the branch pipe (3) into the annulus of the heating section. Washing the pipes of the heating section, the steam condenses on the walls of the pipes. Heating steam condensate flows down to the lower part of the heating section, forming an unheated zone.

Inside the pipes, first water, then the steam-water mixture rises to the steam-generating section of the heating section. Steam rises to the top, and water overflows into the annular space and falls down.

The resulting secondary steam first passes through the wash sheet, where large drops of water remain, then through the louvered separator (6), where medium and some small drops are trapped. The movement of water in the downpipes, the annular channel and the steam-water mixture in the pipes of the heating section occurs due to natural circulation: the difference in the densities of water and the steam-water mixture.

Rice. 1. Evaporation plant

1 - body; 2 - heating section; 3 - supply of heating steam; 4 - flushing sheet; 5 - feed water supply; 6 - louvered separator; 7 - downpipes; 8 - removal of heating steam condensate.

3. Determining the parameters of the secondary steam of the evaporation plant

Fig.2. Scheme of the evaporation plant.

The secondary vapor pressure in the evaporator is determined by the temperature difference of the stage and the flow parameters in the heating circuit.

At P p \u003d 0.49 MPa, t p \u003d 168 ° C, h p \u003d 2785 KJ / kg

Paparameters at saturation pressure Р n = 0.49 MPa,

t n \u003d 151 o C, h "n \u003d 636.8 KJ / kg; h "n \u003d 2747.6 KJ / kg;

The vapor pressure is determined from the saturation temperature.

T n1 \u003d t n - ∆t \u003d 151 - 14 \u003d 137 o C

where ∆t = 14°C.

At saturation temperature t n1 \u003d 137 about C vapor pressure

P 1 \u003d 0.33 MPa;

Steam enthalpies at P 1 \u003d 0.33 MPa h "1 \u003d 576.2 KJ / kg; h "1 \u003d 2730 KJ / kg;

4. Determination of the performance of the evaporation plant.

The performance of the evaporator plant is determined by the flow of secondary steam from the evaporator

D u = D i

The amount of secondary steam from the evaporator is determined from the heat balance equation

D ni ∙(h ni -h΄ ni )∙η = D i ∙h i ˝+ α∙D i ∙h i ΄ - (1+α)∙D i ∙h pv ;

Hence the flow of secondary steam from the evaporator:

D = D n ∙(h n - h΄ n )η/((h˝ 1 + αh 1 ΄ - (1 + α)∙h pv )) =

13.5∙(2785 – 636.8)0.98/((2730+0.05∙576.2 -(1+0.05)∙293.3)) = 11.5 4 t/h

where are the enthalpies of the heating steam and its condensate

H n = 2785 kJ/kg, h΄ n = 636.8 kJ/kg;

Enthalpies of secondary steam, its condensate and feed water:

H˝ 1 =2730 kJ/kg; h΄ 1 = 576.2 kJ/kg;

Feed water enthalpies at t pv = 70 o C: h pv = 293.3 kJ / kg;

Purge α = 0.05; those. 5 %. Evaporator efficiency, η = 0.98.

Evaporator capacity:

D u \u003d D \u003d 11.5 4 t / h;

5. Thermal calculation of the evaporator

The calculation is made by the method of successive approximation.

heat flow

Q = (D /3,6)∙ =

= (11,5 4 /3,6)∙ = 78 56.4 kW;

Heat transfer coefficient

k \u003d Q / ΔtF \u003d 7856.4 / 14 ∙ 350 \u003d 1.61 kW / m 2 ˚С \u003d 1610 W / m 2 ˚С,

where Δt=14˚C ; F \u003d 350 m 2;

Specific heat flux

q \u003d Q / F \u003d 78 56, 4 / 350 \u003d 22. 4 kW / m 2;

Reynolds number

Re \u003d q∙H / r∙ρ "∙ν \u003d 22, 4 ∙0,5725/(21 10 , 8 ∙9 1 5∙2,03∙10 -6 ) = 32 , 7 8;

Where is the height of the heat exchange surface

H \u003d L 1 / 4 \u003d 2.29 / 4 \u003d 0.5725 m;

Heat of vaporization r = 2110.8 kJ/kg;

Liquid density ρ" = 915 kg/m 3 ;

Kinematic viscosity coefficient at P n = 0.49 MPa,

ν = 2.03∙10 -6 m/s;

Heat transfer coefficient from condensing steam to the wall

at Re = 3 2 , 7 8< 100

α 1n \u003d 1.01 ∙ λ ∙ (g / ν 2) 1/3 Re -1/3 =

1.01 ∙ 0.684 ∙ (9.81 / ((0.2 0 3 ∙ 10 -6) 2 )) 1/3 ∙ 3 2, 7 8 -1/3 \u003d 133 78.1 W / m 2 ˚С ;

where at R p = 0.49 MPa, λ = 0.684 W/m∙˚С;

Heat transfer coefficient taking into account the oxidation of the pipe walls

α 1 \u003d 0.75 α 1n \u003d 0.75 133 78, 1 \u003d 10 0 3 3, 6 W / m 2 ˚С;

6. Determination of the circulation rate.

The calculation is carried out by a graph-analytical method.

Given three values ​​of the circulation rate W 0 = 0.5; 0.7; 0.9 m/s we calculate the resistance in the supply lines ∆Р sub and useful pressure ∆Р floor . According to the calculation data, we build a graph ΔР sub .=f(W) and ΔР field .=f(W). At these speeds, the dependences of the resistance in the supply lines ∆Р sub and useful pressure ∆Р floor do not intersect. Therefore, we again set the three values ​​of the circulation rate W 0 = 0.8; 1.0; 1.2 m/s; we calculate the resistance in the supply lines and the useful pressure again. The point of intersection of these curves corresponds to the operating value of the circulation rate. Hydraulic losses in the inlet part are made up of losses in the annular space and losses at the inlet sections of the pipes.

Annular area

F k \u003d 0.785 ∙ [(D 2 2 -D 1 2) -d 2 op ∙ n op ] \u003d 0.785 [(2.85 2 - 2.05 2) - 0.066 2 ∙ 22] \u003d 3.002 m 2;

Equivalent Diameter

D equiv \u003d 4 ∙ F to / (D 1 + D 2 + n d op ) π \u003d 4 * 3.002 / (2.05 + 2.85 + 22 ∙ 0.066) 3.14 \u003d 0.602 m;

Water speed in the annular channel

W k \u003d W 0 ∙ (0.785 d 2 vn ∙ Z / F k ) \u003d 0.5 ∙ (0.785 0.027 2 ∙1764/3.002) = 0.2598 m/s;

where the inner diameter of the pipes of the heating section

D vn \u003d d n - 2∙δ = 32 - 2∙2.5 = 27 mm = 0.027 m;

Number of heating section pipes Z = 1764 pcs.

The calculation is carried out in tabular form, table 1

Calculation of the circulation rate. Table 1.

p/n	Name, definition formula, unit of measure.	Speed, W 0 , m/s
	Water speed in the annular channel: W to \u003d W 0 * ((0.785 * d int 2 z) / F to), m / s	0,2598	0,3638	0,4677
	Reynolds number: Re \u003d W to ∙D eq / ν	770578,44	1078809,8	1387041,2
	Friction coefficient in the annular channel λ tr \u003d 0.3164 / Re 0.25	0,0106790	0,0098174	0,0092196
	Pressure loss during movement in the annular channel, Pa: ΔР to \u003d λ tr * (L 2 / D eq ) * (ρ΄W to 2 / 2) ;	1,29	2,33	3,62
	Loss of pressure at the inlet from the annular channel, Pa; ΔР in \u003d (ξ in + ξ out) * ((ρ "∙ W to 2) / 2), Where ξ in = 0.5; ξ out = 1.0.	46,32	90,80	150,09
	Pressure loss at the inlet to the pipes of the heating section, Pa; ΔР in.tr .=ξ in.tr .*(ρ"∙W to 2 )/2, Where ξ input.tr .=0.5	15,44	30,27	50,03
	Pressure loss during the movement of water in a straight section, Pa; ΔР tr \u003d λ gr * (ℓ but / d int ) * (ρ΄W to 2 / 2), where ℓ but -height of the lower unheated area, m. ℓ but = ℓ + (L 2 -L 1 )/2=0.25 +(3.65-3.59)/2=0.28 m,ℓ \u003d 0.25 - condensate level	3,48	6,27	9,74
	Downpipe losses, Pa; ΔР op = ΔР in + ΔР to	47,62	93,13	153,71
	Losses in an unheated area, Pa; ΔР but =ΔР in.tr .+ΔР tr .	18,92	36,54	59,77
	Heat flow, kW/m 2 ; G ext \u003d kΔt \u003d 1.08 ∙ 10 \u003d 10.8	22,4	22,4	22,4
	The total amount of heat supplied in the annular space, kW; Q k \u003d πD 1 L 1 kΔt=3.14∙2.5∙3.59∙2.75∙10= 691.8	330,88	330,88	330,88
	Increasing the enthalpy of water in the annular channel, KJ/kg; ∆h to \u003d Q to / (0.785∙d int 2 Z∙W∙ρ")	0,8922	0,6373	0,4957
	Economizer section height, m;ℓ ek \u003d ((-Δh to - - (ΔР op + ΔР but) ∙ (dh / dр) + gρ "∙ (L 1 - ℓ but ) ∙ (dh / dр)) / ((4g ext /ρ "∙W∙d ext )+g∙ρ"∙(dh/dр)), where (dh/dр)= \u003d Δh / Δp \u003d 1500 / (0.412 * 10 5) \u003d 0.36	1,454	2,029	2,596
	Losses in the economizer section, Pa; ΔР ek \u003d λ ∙ ℓ ek ∙ (ρ "∙ W 2) / 2	1,7758	4,4640	8,8683
15 15	Total resistance in supply lines, Pa; ΔР subv \u003d ΔР op + ΔР but + ΔР ek	68,32	134,13	222,35
	Amount of steam in one pipe, kg/s D "1 \u003d Q / z r	0,00137	0,00137	0,00137
	Reduced speed at the outlet of the pipes, m/s, W" ok \u003d D "1 / (0.785∙ρ"∙d int 2) \u003d 0.0043 / (0.785∙1.0∙0.033 2 ) \u003d 1.677 m / s;	0,83	0,83	0,83
	Average reduced speed, W˝ pr \u003d W˝ ok / 2 \u003d \u003d 1.677 / 2 \u003d 0.838 m / s	0,42	0,42	0,42
	Consumable steam content, β ok \u003d W˝ pr / (W˝ pr + W)	0,454	0,373	0,316
	Ascent rate of a single bubble in a stationary liquid, m/s W belly \u003d 1.5 4 √gG (ρ΄-ρ˝/(ρ΄)) 2	0,2375	0,2375	0,2375
	interaction factor Ψ vz \u003d 1.4 (ρ΄ / ρ˝) 0.2 (1- (ρ˝ / ρ΄)) 5	4,366	4,366	4,366
	Group speed of ascent of bubbles, m/s W* =W belly Ψ air	1,037	1,037	1,037
	Mixing speed, m/s W see p \u003d W pr "+ W	0,92	1,12	1,32
	Volumetric steam content φ ok \u003d β ok / (1 + W * / W see p )	0,213	0,193	0,177
	Driving head, Pa ΔR dv =g(ρ-ρ˝)φ ok L pairs, where L pairs =L 1 -ℓ but -ℓ ek =3.59-0.28-ℓ ek ;	1049,8	40,7	934,5
	Friction loss in the steam line ΔР tr.steam = \u003d λ tr ((L pairs / d int) (ρ΄W 2 /2))	20,45	1,57	61,27
	Pipe outlet loss ΔР out =ξ out (ρ΄W 2 /2)[(1+(W pr ˝/W)(1-(ρ˝/ρ΄)]	342,38	543,37	780,96
	Flow Acceleration Loss ΔР usk \u003d (ρ΄W) 2 (y 2 -y 1), where y 1 =1/ρ΄=1/941.2=0.00106 at x=0; φ=0 2 =((x 2 k /(ρ˝φ k ))+((1-x k ) 2 /(ρ΄(1-φ k )	23 , 8 51 0,00106 0,001 51	38 , 36 0,00106 0,001 44	5 4,0 6 0,00106 0,001 39
	W cm \u003d W˝ ok + W β k \u003d W˝ ok / (1+(W˝ ok / W cm )) φ k \u003d β k / (1+ (W˝ ok / W cm )) x k \u003d (ρ˝W˝ ok ) / (ρ΄W)	1 , 33 0, 62 0, 28 0 0,000 6 8	1 , 53 0, 54 0, 242 0,0005 92	1 , 7 3 0,4 8 0,2 13 0,000 523
	Useful pressure, Pa; ΔР floor \u003d ΔP dv -ΔP tr -ΔP vy -ΔP usk	663 ,4	620 , 8	1708 , 2

The dependency is built:

ΔP sub .=f(W) and ΔP floor .=f(W) , fig. 3 and find W p = 0.58 m/s;

Reynolds number:

Re \u003d (W p d int) / ν \u003d (0, 5 8 ∙ 0.027) / (0, 20 3 ∙ 10 -6) \u003d 7 7 1 4 2, 9;

Nusselt number:

N and \u003d 0.023 ∙ Re 0.8 ∙ Pr 0.37 \u003d 0.023 ∙ 77142.9 0.8 ∙ 1.17 0.37 \u003d 2 3 02, 1;

where the number Pr = 1.17;

Heat transfer coefficient from wall to boiling water

α 2 \u003d Nuλ / d ext = (2302.1∙0.684)/0.027 = 239257.2 W/m 2∙˚С

Heat transfer coefficient from the wall to boiling water, taking into account the oxide film

α΄ 2 \u003d 1 / (1 / α 2) + 0.000065 \u003d 1 / (1 / 239257.2) + 0.000065 \u003d 1 983 W / m 2 ∙˚С;

Heat transfer coefficient

K=1/(1/α 1 )+(d ext /2λ st )*ℓn*(d n /d ext )+(1/α΄ 2 )*(d ext /d n ) =

1/(1/ 1983 )+(0.027/2∙60)∙ℓn(0.032/0.027)+(1/1320)∙(0.027/0.032)=

17 41 W/m 2 ∙˚С;

where for Art.20 we have λst= 60 W/m∙aboutWITH.

Deviation from the previously accepted value

δ = (k-k0 )/k0 ∙100%=[(1 741 – 1603 )/1 741 ]*100 % = 7 , 9 % < 10%;

Literature

1. Ryzhkin V.Ya. Thermal power stations. M. 1987.

2. Kutepov A.M. and other Hydrodynamics and heat transfer during vaporization. M. 1987.

3. Ogai V.D. implementation of the technological process at thermal power plants. Guidelines for the implementation of the course work. Almaty. 2008.

Izm	Sheet	Dokum№	Sign	the date	KR-5V071700 PZ	Sheet
Fulfilled		Poletaev P.
Supervisor

When calculating the designed evaporator, its heat transfer surface and the volume of circulating brine or water are determined.

The heat transfer surface of the evaporator is found by the formula:

where F is the heat transfer surface of the evaporator, m2;

Q 0 - cooling capacity of the machine, W;

Dt m - for shell-and-tube evaporators, this is the average logarithmic difference between the temperatures of the refrigerant and the boiling point of the refrigerant, and for panel evaporators, the arithmetic difference between the temperatures of the outgoing brine and the boiling point of the refrigerant, 0 С;

is the heat flux density, W/m2.

For approximate calculations of evaporators, the heat transfer coefficient values ​​obtained empirically in W / (m 2 × K) are used:

for ammonia evaporators:

shell and tube 450 – 550

panel 550 – 650

for freon shell-and-tube evaporators with rolling fins 250 - 350.

The average logarithmic difference between the temperatures of the refrigerant and the boiling point of the refrigerant in the evaporator is calculated by the formula:

(5.2)

where t P1 and t P2 are the coolant temperatures at the inlet and outlet of the evaporator, 0 С;

t 0 - boiling point of the refrigerant, 0 C.

For panel evaporators, due to the large volume of the tank and the intensive circulation of the refrigerant, its average temperature can be taken equal to the temperature at the outlet of the tank t P2. Therefore, for these evaporators

The volume of the circulating coolant is determined by the formula:

(5.3)

where V R is the volume of the circulating coolant, m 3 / s;

с Р is the specific heat capacity of the brine, J/(kg× 0 С);

r Р – brine density, kg/m 3 ;

t Р2 and t Р1 – coolant temperature, respectively, at the entrance to the refrigerated space and exit from it, 0 С;

Q 0 - cooling capacity of the machine.

The values ​​of c Р and r Р are found according to the reference data for the corresponding coolant depending on its temperature and concentration.

The temperature of the refrigerant during its passage through the evaporator decreases by 2 - 3 0 С.

Calculation of evaporators for cooling air in refrigerators

To distribute the evaporators included in the chiller package, determine the required heat transfer surface according to the formula:

where SQ is the total heat gain to the chamber;

K - heat transfer coefficient of chamber equipment, W / (m 2 × K);

Dt is the calculated temperature difference between the air in the chamber and the average temperature of the coolant during brine cooling, 0 С.

The heat transfer coefficient for the battery is 1.5–2.5 W / (m 2 K), for air coolers - 12–14 W / (m 2 K).

Estimated temperature difference for batteries - 14–16 0 С, for air coolers - 9–11 0 С.

The number of cooling devices for each chamber is determined by the formula:

where n is the required number of cooling devices, pcs.;

f is the heat transfer surface of one battery or air cooler (accepted based on the technical characteristics of the machine).

Capacitors

There are two main types of condensers: water-cooled and air-cooled. In high-capacity refrigeration units, water-air-cooled condensers, called evaporative condensers, are also used.

In refrigeration units for commercial refrigeration equipment, air-cooled condensers are most often used. Compared with a water-cooled condenser, they are economical in operation, easier to install and operate. Refrigeration units with water-cooled condensers are more compact than those with air-cooled condensers. In addition, they make less noise during operation.

Water-cooled condensers are distinguished by the nature of the movement of water: flow type and irrigation, and by design - shell-and-coil, two-pipe and shell-and-tube.

The main type are horizontal shell-and-tube condensers (Fig. 5.3). Depending on the type of refrigerant, there are some differences in the design of ammonia and freon condensers. In terms of the size of the heat transfer surface, ammonia condensers cover a range from about 30 to 1250 m 2, and freon ones - from 5 to 500 m 2. In addition, ammonia vertical shell-and-tube condensers are produced with a heat-transfer surface area from 50 to 250 m 2 .

Shell and tube condensers are used in machines of medium and large capacity. Hot refrigerant vapors enter through pipe 3 (Fig. 5.3) into the annular space and condense on the outer surface of the horizontal pipe bundle.

Cooling water circulates inside the pipes under the pressure of the pump. The pipes are expanded in tube sheets, closed from the outside with water covers with partitions that create several horizontal passages (2-4-6). Water enters through pipe 8 from below and exits through pipe 7. On the same water cover there is a valve 6 for releasing air from the water space and a valve 9 for draining water during the revision or repair of the condenser.

Fig.5.3 - Horizontal shell and tube condensers

On top of the apparatus there is a safety valve 1 connecting the annular space of the ammonia condenser with the pipeline brought outside, above the roof ridge of the tallest building within a radius of 50 m. parts of the apparatus. From below, an oil sump with a branch pipe 11 for draining the oil is welded to the body. The liquid refrigerant level at the bottom of the casing is controlled by a level indicator 12. During normal operation, all liquid refrigerant should drain into the receiver.

On top of the casing there is a valve 5 for air release, as well as a branch pipe for connecting a pressure gauge 4.

Vertical shell-and-tube condensers are used in high-capacity ammonia refrigeration machines; they are designed for a heat load from 225 to 1150 kW and are installed outside the machine room without occupying its usable area.

Recently, plate-type capacitors have appeared. The high intensity of heat transfer in plate condensers, in comparison with shell-and-tube condensers, makes it possible, at the same heat load, to reduce the metal consumption of the apparatus by about half and to increase its compactness by 3–4 times.

Air capacitors are mainly used in machines of small and medium productivity. According to the nature of the movement of air, they are divided into two types:

With free air movement; such capacitors are used in machines of very low productivity (up to about 500 W) used in domestic refrigerators;

With forced air movement, that is, with blowing the heat transfer surface using axial fans. This type of condenser is most applicable in machines of small and medium capacity, however, due to the shortage of water, they are increasingly being used in machines of large capacity.

Air-type condensers are used in refrigeration units with stuffing box, sealless and hermetic compressors. Capacitor designs are the same. The condenser consists of two or more sections connected in series with coils or in parallel with collectors. Sections are straight or U-shaped tubes assembled into a coil with the help of coils. Pipes - steel, copper; ribs - steel or aluminum.

Forced air condensers are used in commercial refrigeration units.

Calculation of capacitors

When designing a condenser, the calculation is reduced to determining its heat transfer surface and (if it is water-cooled) the amount of water consumed. First of all, the actual thermal load on the capacitor is calculated.

where Q k is the actual thermal load on the capacitor, W;

Q 0 - compressor cooling capacity, W;

N i - indicator power of the compressor, W;

N e is the effective power of the compressor, W;

h m - mechanical efficiency of the compressor.

In units with hermetic or glandless compressors, the thermal load on the condenser should be determined using the formula:

(5.7)

where N e is the electric power at the compressor motor terminals, W;

h e - efficiency of the electric motor.

The heat transfer surface of the condenser is determined by the formula:

(5.8)

where F is the area of ​​the heat transfer surface, m 2;

k - heat transfer coefficient of the condenser, W / (m 2 × K);

Dt m is the average logarithmic difference between the condensation temperatures of the refrigerant and cooling water or air, 0 С;

q F is the heat flux density, W/m 2 .

The average logarithmic difference is determined by the formula:

(5.9)

where t in1 is the temperature of water or air at the inlet to the condenser, 0 C;

t v2 - temperature of water or air at the outlet of the condenser, 0 С;

t k - condensation temperature of the refrigeration unit, 0 С.

The heat transfer coefficients of various types of capacitors are given in Table. 5.1.

Table 5.1 - Heat transfer coefficients of capacitors

Irrigation for ammonia

Evaporative for ammonia

Air-cooled (with forced air circulation) for refrigerants

800…1000 460…580 * 700…900 700…900 465…580 20…45 *

Values to defined for a ribbed surface.

Where the evaporator is designed to cool liquid, not air.

The evaporator in the chiller can be of several types:

• lamellar
• pipe - submersible
• shell-and-tube.

Most often, those who wish to collect chiller by yourself, use a submersible - twisted evaporator, as the cheapest and easiest option that you can make yourself. The question is mainly in the correct manufacture of the evaporator, regarding the compressor power, the choice of the diameter and length of the pipe from which the future heat exchanger will be made.

To select a pipe and its quantity, it is necessary to use a heat engineering calculation, which can be easily found on the Internet. For the production of chillers with a capacity of up to 15 kW, with a twisted evaporator, the following diameters of copper pipes 1/2 are most applicable; 5/8; 3/4. Pipes with a large diameter (from 7/8) are very difficult to bend without special machines, so they are not used for twisted evaporators. The most optimal in terms of ease of operation and power per 1 meter of length is a 5/8 pipe. In no case should an approximate calculation of the length of the pipe be allowed. If it is not correct to make the chiller evaporator, then it will not be possible to achieve either the desired overheating, or the desired subcooling, or the boiling pressure of freon, as a result, the chiller will not work efficiently or will not cool at all.

Also, one more nuance, since the cooled medium is water (most often), the boiling point, when (using water) should not be lower than -9C, with a delta of no more than 10K between the boiling point of freon and the temperature of the cooled water. In this regard, the emergency low pressure switch should also be set to an emergency level not lower than the pressure of the freon used, at its boiling point of -9C. Otherwise, if the controller sensor has an error and the water temperature drops below +1C, the water will begin to freeze on the evaporator, which will reduce, and over time, reduce its heat exchange function to almost zero - the water cooler will not work correctly.

Details

Chiller calculation. How to calculate the cooling capacity or power of the chiller and correctly select it.

How to do it right, what should you rely on first of all in order to produce a quality product among the many offers?

On this page we will give some recommendations, listening to which you will come closer to doing the right thing..

Chiller cooling capacity calculation. Calculation of chiller power - its cooling capacity.

First of all, according to the formula in which the volume of the cooled liquid participates; change in the temperature of the liquid, which must be provided by the cooler; heat capacity of the liquid; and of course the time for which this volume of liquid must be cooled - cooling power is determined:

Cooling formula, i.e. formula for calculating the required cooling capacity:

Q\u003d G * (T1- T2) * C rzh * pzh / 3600

Q– cooling capacity, kW/h

G- volumetric flow rate of the cooled liquid, m 3 / hour

T2- final temperature of the cooled liquid, o С

T1- initial temperature of the cooled liquid, o C

C hw- specific heat capacity of the cooled liquid, kJ / (kg * o C)

pzh- density of the cooled liquid, kg / m 3

* For water C rzh *pzh = 4.2

This formula is used to determine necessary cooling capacity and it is the main one when choosing a chiller.

• Dimensional conversion formulas to calculate chiller cooling capacity:

1 kW = 860 kcal/hour

1 kcal/hour = 4.19 kJ

1 kW = 3.4121 kBtu/hour

Chiller selection

In order to produce chiller selection- it is very important to perform the correct preparation of the technical specifications for the calculation of the chiller, which involves not only the parameters of the water cooler itself, but also data on its location and the condition of its joint work with the consumer. Based on the calculations performed, you can - select a chiller.

Don't forget what region you are in. For example, the calculation for the city of Moscow will be different from the calculation for the city of Murmansk, since the maximum temperatures of the two cities are different.

PAbout the tables of parameters of water-cooling machines, we make the first choice of a chiller and get acquainted with its characteristics. Further, having on hand the main characteristics of the selected machine, such as:- chiller cooling capacity, the electrical power consumed by it, whether it contains a hydromodule and its supply and pressure of liquid, the volume of air passing through the cooler (which heats up) in cubic meters per second - you can check the possibility of installing a water cooler on a dedicated site. After the proposed water cooler satisfies the requirements of the technical specifications and most likely will be able to work on the site prepared for it, we recommend that you contact the specialists who will check your choice.

Chiller selection - features that must be considered when selecting a chiller.

Basic site requirementsfuture installation of a water cooler and the scheme of its work with the consumer:

• If the planned place is indoors, then is it possible to provide a large exchange of air in it, is it possible to bring a water cooler into this room, will it be possible to serve it in it?
• If the future location of the water cooler is outdoors - will it be necessary to operate it in the winter, is it possible to use non-freezing liquids, is it possible to protect the water cooler from external influences (anti-vandal, from leaves and tree branches, etc.)?
• If the temperature of the liquid to which it must be cool below +6 o C or she is above + 15 about C - most often this temperature range is not included in the quick selection tables. In this case, we recommend contacting our specialists.
• It is necessary to determine the flow rate of the cooled water and the required pressure, which the water cooler hydronic module must provide - the required value may differ from the parameter of the selected machine.
• If the temperature of the liquid needs to be lowered by more than 5 degrees, then the scheme for direct cooling of the liquid with a water cooler is not applied and calculation and completion of additional equipment is required.
• If the cooler will be used around the clock and all year round, and the final temperature of the liquid is high enough - how appropriate would it be to use a unit with ?
• In the case of using high concentrations of non-freezing liquids, an additional calculation of the capacity of the water cooler evaporator is required.

Chiller selection program

For your information: it gives only an approximate understanding of the required cooler model and compliance with its technical specifications. Next, you need to check the calculations by a specialist. In this case, you can focus on the cost obtained as a result of the calculations. +/- 30% (in cases with low-temperature models of liquid coolers - the indicated figure is even higher). Optimal the model and cost will be determined only after checking the calculations and comparing the characteristics of different models and manufacturers by our specialist.

Chiller selection Online

You can do it by contacting our online consultant, who will quickly and technically justify the answer to your question. Also, the consultant can perform based on the briefly written parameters of the terms of reference chiller calculation online and give an approximately suitable model in terms of parameters.

Calculations made by a non-specialist often lead to the fact that the selected water cooler does not fully correspond to the expected results.

The Peter Kholod company specializes in integrated solutions for providing industrial enterprises with equipment that fully meets the requirements of the terms of reference for the supply of a water cooling system. We collect information to fill in the terms of reference, calculate the cooling capacity of the chiller, determine the optimally suitable water cooler, check with the issuance of recommendations for its installation on a dedicated site, calculate and complete all additional elements for the operation of the machine in a system with a consumer (calculation of an accumulator tank, a hydronic module, additional, if necessary, heat exchangers, pipelines and shut-off and control valves).

Having accumulated many years of experience in calculations and subsequent implementation of water cooling systems at various enterprises, we have the knowledge to solve any standard and far from standard tasks associated with numerous features of installing liquid coolers at an enterprise, combining them with production lines, setting up specific equipment operation parameters.

The most optimal and accurate and accordingly, the determination of the model of the water cooler can be done very quickly by calling or sending an application to the engineer of our company.

Additional formulas for calculating the chiller and determining the scheme for connecting it to a cold water consumer (chiller power calculation)

• The formula for calculating the temperature when mixing 2 liquids (the formula for mixing liquids):

T mix= (M1*S1*T1+M2*S2*T2) / (S1*M1+S2*M2)

T mix– temperature of the mixed liquid, o С

M1– mass of the 1st liquid, kg

C1- specific heat capacity of the 1st liquid, kJ / (kg * o C)

T1- temperature of the 1st liquid, o C

M2– mass of the 2nd liquid, kg

C2- specific heat capacity of the 2nd liquid, kJ / (kg * o C)

T2- temperature of the 2nd liquid, o C

This formula is used if a storage tank is used in the cooling system, the load is not constant in time and temperature (most often when calculating the required cooling capacity of the autoclave and reactors)

Chiller cooling capacity.

Moscow..... Voronezh..... Belgorod..... Nizhnevartovsk..... Novorossiysk..... Yekaterinburg..... in Rostov-on-Don..... Smolensk..... Kirov..... Khanty-Mansiysk..... Rostov-on-Don..... Penza..... Vladimir..... Astrakhan..... Bryansk..... Kazan..... Samara..... Naberezhnye Chelny..... Ryazan..... Nizhny Tagil..... Krasnodar..... Tolyatti..... Cheboksary..... Volzhsky..... Nizhny Novgorod Region..... Nizhny Novgorod..... Rostov-on-Don..... Saratov..... Surgut..... Krasnodar region..... in Rostov-on-Don..... Orenburg..... Kaluga..... Ulyanovsk..... Tomsk..... Volgograd..... Tver..... Mari El Republic..... Tyumen..... Omsk..... Ufa..... Sochi..... Yaroslavl..... Eagle..... Novgorod region.....

Task 1

The hot product flow leaving the reactor must be cooled from the initial temperature t 1n = 95°C to the final temperature t 1k = 50°C, for this it is sent to a refrigerator, where water is supplied with an initial temperature t 2n = 20°C. It is required to calculate ∆t cf in the conditions of co-current and counter-flow in the refrigerator.

Solution: 1) The final temperature of the cooling water t 2k in the condition of co-current movement of heat carriers cannot exceed the value of the final temperature of the hot coolant (t 1k = 50°C), therefore, we take the value t 2k = 40°C.

Calculate the average temperatures at the inlet and outlet of the refrigerator:

∆t n cf = 95 - 20 = 75;

∆t to cf = 50 - 40 = 10

∆tav = 75 - 10 / ln(75/10) = 32.3 °C

2) The final temperature of the water in the countercurrent flow will be the same as in the direct flow of heat carriers t 2k = 40°C.

∆t n cf = 95 - 40 = 55;

∆t to cf = 50 - 20 = 30

∆tav = 55 - 30 / ln(55/30) = 41.3°C

Task 2.

Using the conditions of problem 1, determine the required heat exchange surface (F) and the flow rate of cooling water (G). Hot product consumption G = 15000 kg/h, its heat capacity C = 3430 J/kg deg (0.8 kcal kg deg). Cooling water has the following values: heat capacity c = 4080 J / kg deg (1 kcal kg deg), heat transfer coefficient k = 290 W / m 2 deg (250 kcal / m 2 * deg).

Solution: Using the heat balance equation, we obtain an expression for determining the heat flux when the cold coolant is heated:

Q \u003d Q gt \u003d Q xt

whence: Q \u003d Q gt \u003d GC (t 1n - t 1k) \u003d (15000/3600) 3430 (95 - 50) \u003d 643125 W

Taking t 2k \u003d 40 ° C, we find the flow rate of the cold coolant:

G \u003d Q / c (t 2k - t 2n) \u003d 643125 / 4080 (40 - 20) \u003d 7.9 kg / s \u003d 28,500 kg / h

Required heat transfer surface

for forward flow:

F \u003d Q / k ∆t cf \u003d 643125 / 290 32.3 \u003d 69 m 2

with countercurrent:

F \u003d Q / k ∆t cf \u003d 643125 / 290 41.3 \u003d 54 m 2

Task 3

In production, gas is transported through a steel pipeline with an outer diameter d 2 \u003d 1500 mm, wall thickness δ 2 \u003d 15 mm, thermal conductivity λ 2 \u003d 55 W / m·deg. Inside the pipeline is lined with fireclay bricks, the thickness of which is δ 1 = 85 mm, thermal conductivity λ 1 = 0.91 W/m·deg. The heat transfer coefficient from the gas to the wall α 1 = 12.7 W / m 2 · deg, from the outer surface of the wall to air α 2 = 17.3 W / m 2 · deg. It is required to find the coefficient of heat transfer from gas to air.

Solution: 1) Determine the inner diameter of the pipeline:

d 1 \u003d d 2 - 2 (δ 2 + δ 1) \u003d 1500 - 2 (15 + 85) \u003d 1300 mm \u003d 1.3 m

average lining diameter:

d 1 cf \u003d 1300 + 85 \u003d 1385 mm \u003d 1.385 m

average pipe wall diameter:

d 2 cf \u003d 1500 - 15 \u003d 1485 mm \u003d 1.485 m

Calculate the heat transfer coefficient using the formula:

k = [(1/α 1) (1/d 1) + (δ 1 /λ 1) (1/d 1 sr)+(δ 2 /λ 2) (1/d 2 sr)+( 1/α 2)] -1 = [(1/12.7) (1/1.3) + (0.085/0.91) (1/1.385)+(0.015/55) (1/1.485 ) + (1 / 17.3)] -1 \u003d 5.4 W / m 2 deg

Task 4

In a single-pass shell-and-tube heat exchanger, methanol is heated with water from an initial temperature of 20 to 45 °C. The water flow is cooled from 100 to 45 °C. The tube bundle of the heat exchanger contains 111 tubes, the diameter of one tube is 25x2.5 mm. The flow rate of methyl alcohol through the tubes is 0.8 m/s (w). The heat transfer coefficient is equal to 400 W/m 2 deg. Determine the total length of the tube bundle.

Let us define the average temperature difference of heat carriers as the average logarithmic.

∆t n cf = 95 - 45 = 50;

∆t to cf = 45 - 20 = 25

∆tav = 45 + 20 / 2 = 32.5°C

Let us determine the mass flow rate of methyl alcohol.

G cn \u003d n 0.785 d int 2 w cn ρ cn \u003d 111 0.785 0.02 2 0.8 \u003d 21.8

ρ cn \u003d 785 kg / m 3 - the density of methyl alcohol at 32.5 ° C was found from the reference literature.

Then we determine the heat flux.

Q \u003d G cn ​​c cn (t c cn - t n cn) \u003d 21.8 2520 (45 - 20) \u003d 1.373 10 6 W

c cn \u003d 2520 kg / m 3 - the heat capacity of methyl alcohol at 32.5 ° C was found from the reference literature.

Let us determine the required heat exchange surface.

F \u003d Q / K∆t cf \u003d 1.373 10 6 / (400 37.5) \u003d 91.7 m 3

Let us calculate the total length of the tube bundle from the average diameter of the tubes.

L \u003d F / nπd cf \u003d 91.7 / 111 3.14 0.0225 \u003d 11.7 m.

Task 5

A plate heat exchanger is used to heat the flow of 10% NaOH solution from 40°C to 75°C. The consumption of sodium hydroxide is 19000 kg/h. Water vapor condensate is used as the heating agent, its consumption is 16000 kg/h, the initial temperature is 95°C. Take the heat transfer coefficient equal to 1400 W / m 2 deg. It is necessary to calculate the main parameters of the plate heat exchanger.

Solution: Find the amount of heat transferred.

Q \u003d G p with p (t k p - t n p) \u003d 19000/3600 3860 (75 - 40) \u003d 713 028 W

From the heat balance equation, we determine the final temperature of the condensate.

t to x \u003d (Q 3600 / G to c to) - 95 \u003d (713028 3600) / (16000 4190) - 95 \u003d 56.7 ° C

с р,к - heat capacity of solution and condensate found from reference materials.

Determination of average temperatures of heat carriers.

∆t n cf = 95 - 75 = 20;

∆t to cf = 56.7 - 40 = 16.7

∆tav = 20 + 16.7 / 2 = 18.4°C

We determine the cross section of the channels, for the calculation we take the mass velocity of the condensate W c = 1500 kg/m 2 ·sec.

S \u003d G / W \u003d 16000/3600 1500 \u003d 0.003 m 2

Assuming the channel width b = 6 mm, we find the width of the spiral.

B = S/b = 0.003/ 0.006 = 0.5 m

Let us refine the channel section

S \u003d B b \u003d 0.58 0.006 \u003d 0.0035 m 2

and mass flow rate

W p \u003d G p / S \u003d 19000 / 3600 0.0035 \u003d 1508 kg / m 3 s

W to \u003d G to / S \u003d 16000 / 3600 0.0035 \u003d 1270 kg / m 3 s

The determination of the heat exchange surface of a spiral heat exchanger is carried out as follows.

F \u003d Q / K∆t cf \u003d 713028 / (1400 18.4) \u003d 27.7 m 2

Determine the working length of the spiral

L \u003d F / 2B \u003d 27.7 / (2 0.58) \u003d 23.8 m

t = b + δ = 6 + 5 = 11 mm

To calculate the number of turns of each spiral, it is necessary to take the initial diameter of the spiral based on the recommendations d = 200 mm.

N \u003d (√ (2L / πt) + x 2) - x \u003d (√ (2 23.8 / 3.14 0.011) + 8.6 2) - 8.6 \u003d 29.5

where x \u003d 0.5 (d / t - 1) \u003d 0.5 (200/11 - 1) \u003d 8.6

The outer diameter of the spiral is determined as follows.

D = d + 2Nt + δ = 200 + 2 29.5 11 + 5 = 860 mm.

Task 6

Determine the hydraulic resistance of heat carriers created in a four-pass plate heat exchanger with a channel length of 0.9 m and an equivalent diameter of 7.5 10 -3 when butyl alcohol is cooled with water. Butyl alcohol has the following characteristics: consumption G = 2.5 kg/s, speed W = 0.240 m/s and density ρ = 776 kg/m 3 (Reynolds criterion Re = 1573 > 50). Cooling water has the following characteristics: flow rate G = 5 kg/s, speed W = 0.175 m/s and density ρ = 995 kg/m 3 (Reynolds criterion Re = 3101 > 50).

Solution: Let's determine the coefficient of local hydraulic resistance.

ζ bs = 15/Re 0.25 = 15/1573 0.25 = 2.38

ζ in \u003d 15 / Re 0.25 \u003d 15/3101 0.25 \u003d 2.01

Let's specify the speed of movement of alcohol and water in fittings (we take d pcs = 0.3m)

W pcs \u003d G bs / ρ bs 0.785d pcs 2 \u003d 2.5 / 776 0.785 0.3 2 \u003d 0.05 m / s less than 2 m / s, therefore, can be ignored.

W pcs \u003d G in / ρ in 0.785d pcs 2 \u003d 5/995 0.785 0.3 2 \u003d 0.07 m / s less than 2 m / s, therefore, can be ignored.

Let us determine the value of hydraulic resistance for butyl alcohol and cooling water.

∆Р bs = xζ ( l/d) (ρ bs w 2 /2) \u003d (4 2.38 0.9 / 0.0075) (776 0.240 2 / 2) \u003d 25532 Pa

∆Р in = xζ ( l/d) (ρ in w 2 /2) \u003d (4 2.01 0.9 / 0.0075) (995 0.175 2 / 2) \u003d 14699 Pa.