**Fundamentals of thermal calculations**

When determining the flow rate of the coolant (steam) and the size of the heating surface of the heat exchanger, the calculated equations of heat balance and heat transfer are usually compiled.

Total heat spent on heating, dissolving and evaporation of the product moisture, considering heat loss generally expressed by the formula (in Joules)

(1-9)

where Q_{1}, Q_{2}, Q_{3} - The relevant articles of the consumption of useful heat spent on heating, dissolving and evaporation of the constituent parts of the product, j;

Q_{п} - heat loss by the outer surface of the apparatus into the environment by radiation and convection, j.

When calculating the continuous flow of heat units in all respects shall be calculated in W (J / s) or J / h.

**Heat consumption for heating each component of the processed product is determined by the formula (in J)**

where G is the number of the corresponding component of the heating, kg;

c — specific heat capacity of the component, J / (kg * K);

t_{k} and t_{н}- final and initial component temperatures, ° С.

The heat capacity of the majority of products depends on the temperature. For example:

**specific heat of sugar = 1000 + 7,25t J / (kg * K) (1.11)**

**syrup specific heat = 1714 + 5,76t J / (kg * K). (1.12)**

The heat capacity of sugar solutions, including sugar syrup and caramel mass, depends on temperature and concentration. It can be calculated by the formula of V. V. Yanovsky [in J / (kg • K)]

**c = 4190 - (2514 — 7,540t) * a, (1.13)**

where a is the concentration of sugar in the solution, kg / kg.

The specific heat capacity of water in practical calculations can be taken equal to 4190 J / (kg • K) [1 kcal / (kg • deg)].

Heat consumption for the dissolution of crystals (eg, sugar) is determined by the formula (in J)

**Q _{2}= Gqк, (1-14)**

where G is the amount of product, kg;

qk is the latent heat of dissolution or crystallization of 1 kg of product equal to 4190 sugar j.

Heat consumption for the evaporation of moisture (in joules) is determined by the formula

**Q _{з} =D_{2}r, (1-15)**

where D_{2} - the amount of evaporated moisture, kg;

r — concealed heat of vaporization, J / kg; determined by the table of thermodynamic properties of steam depending on temperature or pressure (see Appendix).

The amount of moisture evaporated (in kg) when changing product concentration can be determined by solving equations solids balance

**G _{c.}_{в}=G_{1}a_{1}=G_{2}a_{2 } (1-16)**

and the material balance equation

Then (1-17) (1-18)

where G_{c.}_{в} - amount of dry substances in the product, kg;

G_{1} - the amount of product to be evaporated, kg;

G_{2} - quantity of the finished product, kg;

a_{1}- the initial content (concentration) of dry substances in the product, kg / kg;

а_{2} - final solids content in the finished product, kg / kg.

If the moisture evaporates from the surface of the solution without any appreciable change in its concentration, the

**D _{2} = 3600KF (р - φр^{1}) τ, (I-19)**

where *К* - coefficient of proportionality, depending on the air velocity and the physical properties of the evaporated product, kg / (m X NUMX-s * MPa);

*F* - evaporation surface area, м2;

τ is the duration of the evaporation process, s;

*р* - the vapor pressure of the vaporized product, MPa, at ambient temperature (determined according to the table of the application);

*р*'- elasticity of saturated vapors of the evaporated product, MPa, at ambient temperature (determined according to the table of the annex);

φ — relative air humidity (cf = 0,65-7-0,75).

The coefficient of proportionality *К* Water can be determined by the formula

*K*= 0,0745 (ʋр)^{0,8}(1-20)

where ʋ is the air velocity, m / s;

ρ is the air density, kg / m3.

Upon evaporation of water, depending on the air velocity proportionality factor K has the following meanings:

V | 0,5 | 1,0 | 1,5 | 2,0 |

К | 0,036 | 0,083 | 0,114 | 0,145 |

Heat losses to the environment through the outer wall by radiation and convection, the device can determine the formula (in watts)

**Qп=F _{a}α_{k}(t_{Art} t_{в}) (1-21)**

where F_{a} - area of the outer surface of the apparatus, m^{2};

α_{к}- heat transfer coefficient, W / (m X NUMX * K);

t_{Art} and t_{в}- temperature of the wall and ambient air, ° С.

The heat transfer coefficient (total), provided that the unit is indoors and t_{Ct} does not exceed 150 ° C, approximately

calculated according to the formula [in W / (m_{2} • K)]

**α _{к} - 9,76 + 0,07 (t_{Art} -t_{в}). (I-22)**

The amount of heating steam per cycle for a batch of devices in which the steam is condensed completely determined by the formula (in kg)

where Q_{common} - total heat consumption for one cycle, including losses to the environment from dy, j;

i_{1}^{"}и i_{1}'—Consistently the enthalpy of heating steam and condensate, J / kg (see annex).

All steam consumption for the same devices will be (in kg / h)

where τ is the cycle time, h

In tempering machines operating with an established thermal conditions, heating steam is consumed only for compensation of losses of heat into the environment. Its flow rate (kg / hr) is determined by the formula

where Q_{п} - heat loss to the environment, W;

i ”- heating steam enthalpy, J / kg;

i'— enthalpy of condensate, J / kg.

Steam consumption for continuous devices (in kg / s) is defined by the formula (1-23). But in this case, the total heat flow Q_{common} calculated in watts.

Consumption of heat transfer fluids (eg water) is determined by the formula (in kg / s)

where c is the specific heat capacity of the coolant, J / (kg-K);

t_{н} and t_{к}- the initial and final temperature of the coolant, ° C.

Area aids heat exchange surface is determined from the equation of heat transfer through the wall

** Q _{floor}=Fk_{cr}Δtτ (1-27)**

where the heat transfer surface area of the device (in m2)

The duration of the thermal process in the apparatus of periodic action (in s) will be

where Q_{floor} - consumption of useful heat in the apparatus, j;

F —the area of the heat exchange surface of the apparatus, m^{2};

k_{Wed} - average heat transfer coefficient, W / (m^{2}*K);

∆t is the average temperature pressure between the coolant and the heat receiving medium, ° С.

When calculating devices of continuous action, heat consumption is calculated in watts, in the formula (1-28) the duration of the process is taken as τ = 1с.

The average temperature difference ∆t depends on the nature of the thermal process. If during heat exchange between two streams, the initial and final temperatures of one stream are denoted by t_{1}"And t_{1}', and the second — through t_{2}'and t_{2}“, The process can be represented graphically for the cases of direct flow and backflow (Fig. 23).

Fig. 23. The graph of the temperature change of coolants: a - with forward flow; b - with countercurrent; in - - at condensation of the heating steam.

With forward flow and countercurrent, as well as at a constant temperature of one of the media, for example, during the condensation of heating steam (Fig. 23, c), the average temperature difference is defined as the average logarithm by the formula

here Δt_{б} and Δt_{м} - respectively, more or less temperature pressure between the coolants at the beginning and end of the heat exchange surface.

If <1,8, the average temperature difference may be determined as the arithmetic average

If instead of the formula (1-30), you can use the formula

The heat transfer coefficient of heating medium to be heated through the monolayer wall [W / (m^{2} • K)] is determined by the formula

(1-33)

where α_{1} - heat transfer coefficient from the heat carrier to the wall, W / (m X NUMX-K);

α_{2} - heat transfer coefficient from the wall to the heated medium, W / (m X NUMX-K);

s — wall thickness, m;

ƛ - coefficient of thermal conductivity of the wall material, W / (m * K).

When boiling the product in a batch apparatus due to changes in product concentration changes and the heat transfer coefficient, so the approximate calculations batch of devices should be the average heat transfer coefficient.

**Bases for siropovarochnoy station**

The required capacity of dispensers for supplying components of the syrup: sugar, molasses, water can be determined by jointly solving the material balance equations specified by the recipe, n servings of sugar and molasses in the syrup and the moisture balance equation that takes into account the moisture of the syrup, sugar and syrup.

The material balance equation for 1 hours for the case to be

** П=G _{sah}+G_{stalemate}+G_{water} (1-34)**

where P is the syrup capacity, kg / s;

G_{sah}, G_{stalemate}, G_{water} - respectively, the flow rate of sugar, pat and water supplied to the solvent, kg / s.

The proportion of sugar solids in the syrup and molasses according to the recipe

moisture balance equation for syrup having a certain humidity, would

** Ω _{с}=G_{sah}ω_{sah }+G_{stalemate}ω_{stalemate }+G_{water}ω_{water} _{ } (1-36)**

where ω_{с}, ω_{sah}ω_{time} ω_{water}- the moisture content of syrup, sugar and molasses, respectively; in the calculations, they can be taken within the following limits: ω_{с} = 16 ÷ 18%, or 0,16 — 0,18 kg / kg; ω_{sah} = 0,14 ÷ 0,15%, or 0,0014 — 0,0015 kg / kg ω_{stalemate}= 18 ÷ 22%, or 0,18 - 0,22 kg / kg.

Solving the last three equations and substituting in equation (1-36) instead of G_{stalemate} and G_{water} their expressions from equations (1-34) and (1-35), we obtain the required flow rate of sugar and, consequently, the performance of the dispenser (in kg / s)

According to the consumption of sugar, the consumption of molasses is determined from the equation of the proportions of sugar and molasses (1-35), and the consumption of water is determined from the material balance equation (1-34).

The total amount of heat required to heat the components of the syrup, dissolving sugar crystals and compensation solvent heat losses to the environment is determined by the formula (Watts)

where G_{j} - the number of components of the syrup supplied to the solvent, kg / s;

Δg_{j}- change in the enthalpy of the syrup components, J / kg;

G_{sah} - the amount of sugar supplied to the solvent, kg / s;

g_{k} - latent heat of dissolution of 1 crystals kg of sugar, j / kg (g_{к} = 4190)

Q_{П} - heat loss to the environment from radiation and convection (in W)

defined by the formulas (1-21) and (1-22).

It should be borne in mind that in the formula (1-38)

where G_{sah}, G_{stalemate}, G_{water} - consumption of sugar, molasses, water (determined by the above formulas), kg / s;

Δg_{sah}, Δg_{stalemate}, Δg_{water} - respectively, changes in the enthalpy of sugar, molasses and water at the initial and final temperatures, j / kg.

Enthalpy of said products (J / kg) in the initial and final temperature is defined as g_{Beg} = c_{н}t_{н} and g_{horse} - with_{к}t_{к}. To do this, the heat capacity of sugar and molasses is first calculated using the formulas (1-11) and (1-12) at the final (/ k) and initial (/ n) temperatures. In this case, the initial temperature of sugar will be the air temperature of the room from which it is supplied; the initial temperature of the supplied molasses within the limits of 55 — 60 ° С, water 70 — 80 ° С.

The final temperature of the syrup components will be the boiling point of the syrup, which is determined by the developed boiling temperature diagram of caramel syrups depending on the desired humidity of the caramel syrup ω_{с} and pressure p (Fig. 24) (in this case, for an open apparatus-solvent, the pressure is atmospheric - 100 kPa). For example, at a humidity of 16% syrup and atmospheric pressure, its boiling point according to the indicated schedule will be approximately 120 ° С.

When determining the parameters of heating steam, it should be borne in mind that the steam temperature should be approximately 15 — 20 ° С above the boiling point of the syrup; Thus, in this case, the temperature of the heating steam will be: t_{п} = 120 + 20 = 140 ° С.

The steam consumption for the solvent is determined by the formula (1-23), as for the apparatus of continuous action. When calculating the steam flow rate based on the adopted heating steam temperature, using the application table, first determine the required heating pressure p, and then use the same table to find the heating vapor enthalpy i ”_{1} and condensate I '_{1}_{.}

The surface area of the heating solvent is defined as the heating surface of the continuous apparatus, and only the useful heat is taken into account (without loss to the environment).

For this case the useful heat for the solvent of the formula (1-38) is (in watts)

Then the formula for determining the heating surface of the solvent will be (in m^{2}).

where k_{н}- heat transfer coefficient when heated, W / (m X NUMX-K) (you can take an average of k_{н} = 1500 ÷ 1740);

∆t is the average logarithmic temperature difference between the heat carrier (heating steam and the mixture - syrup, ° C; is determined by the formulas (1-30) and (1-31).

In our case

where Δt_{1} = t_{п} - t_{н}_{.cm }(Where t_{н}_{.cm }- initial average temperature of the mixture of syrup components);

Dt_{2} = t_{п}- t_{к}_{.cm }(Where t_{k.sm }- boiling point of syrup);

t_{п} - temperature of heating steam, ° С.

It should be borne in mind that the average temperature of the mixture (in this case, the mixture of syrup components - sugar, water and syrup) loaded into the solvent, is determined from the equation of the heat balance of the mixture or is given in simplified calculations.

heat balance equation for the mixture in this case would br follows:

or

where the average temperature of the mixture (in ° C)

where P is the amount of the mixture, kg / s;

Q_{sah}, Q_{stalemate}, Q_{water} - accordingly, the amount of heat introduced into the mixture by sugar syrup and water, W;

с_{cm}- specific heat capacity of the mixture, J / (kg * K).

The remaining symbols are met before.

The required motor power for water solvent mixer blades determined by the formula (1-6).

The geometric volume V (in m^{3}) Operating sugar atmospheric pressure of the solvent is determined by the formula

where G_{sah} and G_{water} - consumption of sugar and water, kg / h;

τ_{р} - the duration of dissolution, h (mp = 0,5-g-1,0); p is the density of the mixture of sugar and water, kg / m X NUMX;

ρ is the fill factor (<p = 0,7 - X 0,8).

The length of the coil in the hinge connected buses-1 station is determined based on the duration of dissolution of sugar

L = ʋ_{c}τ_{ρ }(1-46)

where ʋ_{c} - the average velocity of the mixture in the coil pipe, m / s (ʋ_{c} = 0,55 ÷ 0,65).

The diameter of the coil pipe d (in m) is found from the equation P hour stroke mixture through its cross-sectional area

hence

Bases for karamelevarochnoy station

To calculate the caramel station, it is first necessary to determine its performance taking into account possible caramel mass losses on all sections of the line. The approximate calculation sequence is as follows:

1.Opredelenie ready for hourly productivity caramel line with the time line for the cleaning equipment (kg / h):

where P_{cm} - given shift line capacity, kg per shift;

τ_{cm} - shift work time (h) minus approximately 15 min (0,25 h) for cleaning the line equipment.

2. Determining the amount of caramel mass processed on the line per hour for a given percentage of filling in the finished caramel (in kg / h),

where in_{н} - the specified content of the filling in the finished caramel,%.

Accordingly, the capacity of the equipment for preparing the filling for this line, i.e. the amount of fruit and berry filling supplied to the line, will be (in kg / h)

3. Determination of the hourly amount of caramel mass processed on the line in a dry matter, taking into account the desired moisture content of the caramel mass and the loss of dry substances (in kg / h)

where ω_{к}- given humidity of the finished caramel mass,%;

α is the caramel mass loss rate for dry matter on the line,% (taken approximately within 1,67 — 1,7%).

According to the formula (1-52) may be defined as the performance of individual sections or lines of machines and apparatus in view of the loss of product in the dry matter from the end portion of the line before or machines.

4. Determination of the hourly productivity of the caramel station by the caramel mass (in kg / h), taking into account the given humidity of the finished mass

5. The determination from the dry matter balance equation (1-16) of the syrup consumption, i.e. the amount of syrup that must be supplied from the syrup station to the vacuum coil. Since the concentration of any solution (in kg / kg) is

a = (100-ω) / 100

where ω is the humidity of the solution,%,

the solids balance equation for this case is

G_{c }(100 - ω_{с}) = G_{к} (100 - ω_{к}), Where the required amount of caramel syrup will be

Gc = Gk (100- ω_{к}) / (100 - ω_{с}) (1-54)

here ω_{с} - moisture content of caramel syrup,%.

Calculation of coil vacuum apparatus continuous conducted in the following order.

The equation of thermal balance for the coil vacuum unit with boiling caramel mass will

where G_{с}, G_{к} - the amount of syrup supplied for boiling and the finished caramel mass produced, kg / s;

с_{с} и с_{к} - specific heat capacity of syrup and caramel mass, J / (kg-K)

t_{c}, T_{k} - temperature of syrup and caramel mass, ° С;

i ”_{1}i '_{1} —Entalpia of heating steam and condensate, J / kg;

D_{2} - the amount of evaporated moisture (secondary steam), kg / s;

i_{2} - enthalpy of secondary steam, J / kg;

D is the heating steam consumption, kg / s;

Q_{п} - heat loss by the apparatus into the environment, watts.

The left side of the heat balance equation (1-55) expresses the heat parish:

G_{с},_{c}, T_{c} - heat introduced into the device by syrup, W;

Di_{1} - heat introduced into the apparatus by heating steam, W.

Members of the right-hand side of the equation indicate the article of this heat flow:

G_{k},_{k}, T_{k} - heat carried away with the finished caramel mass, W;

D_{2}i_{2} - heat carried away with secondary steam, W;

Di_{1}- heat carried away with condensate resulting from condensation of heating steam, W;

Q_{п} - heat leaving in environment (losses), W.

Consumption of heating steam for the machine (in kg / s) is determined from the heat balance equation (1-55)

caramel syrup temperature t_{с}Supplied to the coil unit is determined on the schedule (see. Fig. 24) depending on the desired moisture syrup at atmospheric pressure (cm. Solvent races).

The boiling point of the cooked candy mass t_{к} determined on the same schedule, depending on the desired final moisture caramel mass and vacuum in the vacuum chamber in the apparatus. In this residual pressure (kPa)

**ρ _{о} = 100 - B,** (I-57)

where B is the specified vacuum in the vacuum chamber of the apparatus, kPa.

The heat capacity of syrup_{с} and caramel mass with_{к} determined by the formula (1-13) heat capacity of sugar solutions.

The number of secondary steam (water evaporated) is determined from the material balance formula of the equation (1-18).

vapor Eytalpiya i_{2}”Is determined depending on the residual (absolute) pressure in the vacuum chamber according to the application table.

Heating steam enthalpy i_{1}”And condensate i_{1}'is determined by the same table, depending on the adopted pressure of the temperature of the heating steam.

The temperature of the heating steam supplied to the vapor space of the heating part of the coil vacuum apparatus should be 15 — 20 ° С higher than the boiling temperature of the caramel mass found earlier (practically, the temperature of the heating steam should be

limits 158 — 159 ° С, which corresponds to an overheating heating pressure up to 0,6 MPa). This must be borne in mind when determining the parameters of the heating steam.

The loss of heat to the environment unit Q_{п} determined by the formula (1-21) or take the experimental data.

Thus determined the value of all variables in the formula (1-56), calculate steam consumption.

The area of the heat transfer surface coil vacuum apparatus (in m^{2}) when boiling syrup is determined from the equation of heat transfer through the wall by the formula (1-28)

Q_{floor} - consumption of useful heat (excluding losses), W;

k is the heat transfer coefficient of the coil; established by trial. For approximate calculations, it can be taken equal depending on the diameter of the coil 350 - 1000W / (m_{2} • K);

∆t — average temperature difference between the heating steam, syrup and caramel mass, ° С; is determined by the formulas (1-30) and (1-31).

Having determined the diameter of the coil pipe by the formula (1-48) at the screen speed in the pipe ʋ = 1,0 m / s, the geometrical dimensions of the coil are determined from the found value of the heat exchange surface.

The length of the coil, specifying pipe diameter according to GOST, is calculated as follows (in m)

Where d_{н} - outer diameter of the coil pipe. The length of the coil is usually taken within 800 — 1000 of the coil pipe diameters.

Nominal diameter of the coil D_{Wed} = 680 mm and a pitch of the coil can be found round the corner lifting coil

In this case, 5 is taken to be 1,5 — 2,0 with? N- The length of a coil loop / (in m) will be

The number of turns of the coil

coil height (in m) will be

where h_{constr} - a constructive additive taking into account the height of stamped bottoms.

Case diameter of the heating portion (VM)

Finally, the diameter of the body of the heating part of the apparatus is taken to the nearest diameter of standard stamped bottoms. The geometric volume of the vacuum chamber of the apparatus is determined by the straight lines of its vapor space Rv [in m^{3}/ (h • m_{3})]

D_{2}- amount of secondary steam, kg / h;

ʋ_{2} - specific volume of secondary steam, m^{3}/ Kg;

V — volume of vacuum chamber, m^{3}.

At atmospheric pressure, Rv = 8000 m^{3}/ (M^{3} • h). Under vacuum, the vacuum chamber is Rv = 8000φ, where φ is a coefficient depending on the residual pressure values in the vacuum chamber (when boiling caramel masses is approximately 0,85).

Then from (1-64) the volume of the vacuum chamber (in m^{3}) will be

The inner diameter of the vacuum chamber housing d_{в} adopted for structural reasons, or depending on the diameter of standard stamped bottoms.

The height of the vacuum chamber body (in m) will

The wall thickness (in m) of the housing of the heating portion of the apparatus as a thin-walled cylindrical vessel operating under internal overpressure, calculated by the formula

where p is the pressure in the apparatus, MPa;

D_{в} - internal diameter of the case, m;

δ_{z}- permissible tensile stress, MPa;

φ is the strength coefficient of the weld (cf = 0,7 - g 0,8);

с - increase in corrosion, m.

Performance vacuum apparatus of the finished caramel mass (in kg / h) can be determined using the following form

where g_{с}= C_{с}t_{c} - enthalpy of syrup entering boiling, J / kg;

g_{k.}_{м }= c_{к}t_{к} - Enthalpy of the finished caramel mass, j / kg;

t_{п} - temperature of heating steam, ° С.

The condenser thermal mixing process takes place, which can be expressed by the following equation thermal balance (see. Diagram of ris.21)

from the cooling water flow into the mixing condenser is (in kg / s)

where D_{2} - the amount of condensed secondary steam, kg / s;

і_{2} - enthalpy of secondary steam, J / kg;

c is the specific heat capacity of water, J / (kg-K) (c = 4190);

t_{2Н} and t_{2K} - the initial and final temperature of the cooling water, ° С (final water temperature t_{2K} condensation is temperature).

The feed to the condenser cooling water W in an amount from an initial temperature t_{2Н} as drip down and condensing the steam is heated to a final temperature t_{2K}, which in direct-flow condensers on 5 — 6 ° С is lower than the temperature of the condensed vapor.

The internal diameter of the capacitor (vm) is determined by the formula

where ρ_{п} - steam density, kg / m 3;

ʋ - steam velocity in the condenser, m / s (ʋ = 20 ÷ 25).

The amount of air (in kg / s) pumped out of the condenser by a vacuum pump is determined by the formula

Air flow rate (in m^{3}/ c) from the condenser to the pump is determined by the formula

where G_{в} - the amount of incoming air, kg / s;

288 is the gas constant for air, J / (kg-K);

t_{в} - air temperature, ° С; for direct flow mixing capacitors TV = t_{2k} . Ie leaving water temperature from the condenser.;

р_{в} - partial air pressure, Pa.

Partial pressure of air (in Pa) can be defined by the formula

Р_{в} = P_{а}- R_{п} (I-74)

where p_{а} - absolute (residual) pressure in the vacuum chamber and condenser, Pa;

р_{п} - partial pressure of steam, Pa, which is taken to be equal to the pressure of saturated steam at air temperature.

In the vapor-air mixture in the condenser, the partial air pressure can also be determined from the equation

here

Vacuum Pump capacity to pump air-water mixture (in m_{3}/ No)

where the diameter of the pump piston (in m)

where p is the density of the air-water mixture, kg / m X NUMX;

s - piston stroke, m;

W - cooling water consumption, kg / s;

D_{2}- amount of condensate, kg / s;

V_{в} - the amount of exhaust air m3 / s;

n is the number of double strokes per minute;

ƛ_{0} - filling ratio (ƛ_{0} = 0,7 ÷ 0,8).

When determining the piston diameter by the stroke of the piston and the number of double strokes of the piston are set (according to the pump characteristic from the literature or reference data).

## 1 answer to “Basics of calculating heat exchangers and stations for the preparation of sugar syrups and caramel mass”

Hello, where is the reference bibliograph for the equation VV Yanovsky?

Thanks!