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While setting the task of conducting a technical calculation of heat-exchange equipment, input data of the heat transfer medium (flow rate, initial and final temperature, physico-chemical properties) should be known. The missing values are determined during the thermal calculation.

The aim of the thermal calculation is to determine the main technical data of heat exchange equipment, i.e. thermal load, heat transfer medium flows, average temperature difference of the heat transfer media, heat transfer coefficient. These parameters are calculated based on the heat balance equation.

Below is an example of a general calculation of heat exchange equipment.

In heat exchangers, heat energy is transferred from one process stream (heating medium) to another, resulting in the heating or cooling.

Q = Q_{г} = Q_{х}

Q – quantity of heat transferred or received by the heat transfer medium [W],

Wherefrom:

Q_{г} = G_{г}c_{г}·(t_{гн} – t_{гк}) и Q_{х} = G_{х}c_{х}·(t_{хк} – t_{хн})

Where

G_{г}_{,}_{х} – hot and cold heat transfer medium flow rates [kg/h];

с_{г}_{,}_{х} – heat capacity of hot and cold heat transfer media [J/kg·deg];

t_{г}_{,}_{х н} – initial temperature of hot and cold heat transfer media [°C];

t_{г}_{,}_{х к} – final temperature of hot and cold heat transfer media [°C];

It should be taken into account that the amount of transmitted/ received heat depends on the aggregate state of the heat transfer media. If, in the course of heat exchange process, it does not change, then the above mentioned formula is used for calculation. If one (or both) heat transfer medium changes its aggregate state (steam heating), the following formula is used to calculate the amount of transmitted or received heat:

Q = Gc_{п}·(t_{п} – t_{нас})+ Gr + Gc_{к}·(t_{нас} – t_{к})

Where

r – heat of condensation [J/kg];

с_{п}_{,}_{к} – specific heat capacities of steam and condensate [J/kg·deg];

t_{к} – condensate temperature at the unit outlet [°C].

If condensate does not cool, the first and third terms are excluded from the right side of the equation, and then it takes the following form:

Q_{гор} = Q_{конд} = Gr

The flow of the heat transfer media can be determined as follows:

G_{гор} = Q/c_{гор}(t_{гн} – t_{гк}) или G_{хол} = Q/c_{хол}(t_{хк} – t_{хн})

In case of steam heating, its consumption is defined by the formula:

G_{пара} = Q/ Gr

Where

G – flow rate of corresponding heat transfer medium [kg/h];

Q – amount of heat [W];

с – specific heat of heat transfer media [J/kg·deg];

r – heat of condensation [J/kg];

t_{г}_{,}_{х н} – initial temperature of hot and cold heat transfer media [°C];

t_{г}_{,}_{х к} – final temperature of hot and cold heat transfer media [°C].

The difference between heat transfer media is a driving force of the heat exchange process. Since the temperature of the streams changes as they pass the path, the temperature difference also changes, so it is customary to use an averaged value for the calculation. The average temperature difference for the direct and counter flows is calculated as the average logarithmic value of:

∆t_{ср} = (∆t_{б} - ∆t_{м}) / ln (∆t_{б}/∆t_{м})

Where ∆t_{б}, ∆t_{м} – greater and smaller average temperature difference of the heat transfer media at the unit inlet and outlet.

The same formula, with the addition of the correction factor ∆t_{ср} = ∆t_{ср} ·f_{попр}, is used during the cross and mixed flows of heat transfer media.

Heat transfer coefficient can be determined as follows:

1/k = 1/α_{1} + δ_{ст}/λ_{ст} + 1/α_{2} + R_{заг}

In the equation:

δ_{ст} – wall thickness [mm];

λ_{ст} – coefficient of thermal conductivity of the wall material [W/m·deg];

α_{1,2} – heat transfer coefficients of the inner and outer sides of the wall [W/m^{2}·deg];

R_{заг} – coefficient of wall contamination.

Design calculation of heat exchange equipment is divided into the preliminary and detailed calculation.

The preliminary calculation aims to choose approximate values of the heat transfer coefficient from reference materials, and to determine heat exchange surface and the flow path cross-section area for heat transfer media.

The approximate heat exchange surface is calculated as follows:

F = Q/ k·∆t_{ср} [m^{2}]

The flow path cross-section area of the heat transfer media is determined from the formula:

S = G/(w·ρ) [m^{2}]

Where

G – heat transfer medium flow [kg/h];

(w·ρ) – mass flow rate of the heat transfer medium [kg/m^{2}·s].

For calculations, the flow velocity is assumed based on the type of heat transfer medium:

Type of heat transfer medium |
Flow velocity, m/s |

Viscous liquids | <1 |

Low-viscous liquids | 1-3 |

Dusty gases | 5-10 |

Pure gases | 10-15 |

Saturated steam | 30-50 |

Based on the preliminary design calculation, one or more heat exchangers, which satisfy the required heat exchange surface, are selected. Then, detailed structural and thermal calculations are carried out under specified conditions for the selected units.

During design calculations, additional indicators are determined for different heat exchangers.

E.g., for shell-and-tube heat exchangers, a length or a number of tubes is to be found.

l = F/ πdn

Where:

l – tube length [m];

n – number of tubes [pcs];

F – required heat exchange surface [m^{2}];

d – tube diameter [m];

Typically, during shell-and-tube heat exchanger calculations, the number and diameter of tubes is determined from reference materials.

The inner diameter is determined as follows:

D_{вн} = s (b-1) + 4d_{п}

Where:

D_{вн} – inner diameter of the heat exchanger [m];

s – tube pitch [m] (to be assumed from 1.2 to 1.5 d_{н});

d_{п} – tube outer diameter [m];

b – number of tubes [m] (b = 2a-1, where *a* is the number of tubes on the side of the largest hexagon);

Then, the tube and shell area is determined:

S_{тр} = (πd^{2}_{в} /4) n_{х}

Where:

S_{тр} – tube space area [m^{2}];

d^{2}_{в} – tube inner diameter [m];

n_{х} – number of tubes in one pass;

S_{мтр} = (π/4) (D^{2} - nd^{2}_{п})

Where:

S_{мтр} – shell space area [m^{2}];

D – shell inner diameter [m];

d_{в} – tube outer diameter [m];

n – number of tubes in one pass;

In a specific case, when longitudinal partitions are placed in the shell side to increase the heat exchange intensity, the area will be determined as follows:

S_{мтр} = (π/4) (D^{2} - nd^{2}_{п}/ N)

Where:

N – number of passes when divided by partitions;

During the design calculation of coil heat exchangers, the total length of the coil as well as the number of turns and sections are determined.

L = F/ πd_{р}

Where:

L – total length of the coil [m];

d_{р} – design diameter of the coil tube [m];

n = L/ πd_{р}

Where:

n – number of turns;

Knowing the heat transfer medium flow rate and velocity in the coil tube, the number of coil sections can be determined:

m = V_{сек}/(π/4)d^{2}w

Where:

V_{сек} – flow rate [kg/h];

d – coil tube diameter [m];

w – heat transfer medium velocity in the coil tube [m/s];

In the spiral heat exchanger calculations, such data as the cross-section of channels as well as width, length, pitch, number of turns and outer diameter of the spiral, are determined.

S = G/W

Where

S – channel cross-section [m^{2}];

G – heat transfer medium flow [kg/h];

W – heat transfer medium mass velocity [kg/m^{2}·sec].

When process streams pass through heat exchange equipment, a loss of head or pressure, caused by the hydraulic resistance of the units, occurs.

The general formula for calculating the hydraulic resistance, created by heat exchangers, is as follows:

∆Р_{п} = (λ·(l/d) + ∑ζ) · (ρw^{2}/2)

Where

∆p_{п} – pressure loss [Pa];

λ – coefficient of friction;

l – tube length [m];

d – tube diameter [m];

∑ζ – sum of local resistance coefficients;

ρ – density [kg/m^{3}];

w – flow velocity [m/s].

**Problem 1**

A hot product flow, leaving the reactor, should be cooled from the initial temperature t_{1}_{н} = 95°C to the final temperature t_{1}_{к} = 50°C; for this purpose, it is directed to a refrigerator, where water with the initial temperature of t_{2}_{н} = 20°C is supplied. Please calculate ∆t_{ср} for the direct flow and counter flow conditions in the refrigerator.

Solution: 1) Since the final temperature of the cooling water t_{2}_{к} for the direct flow of the heat transfer media can not exceed the value of the final temperature of the hot heat transfer medium (t_{1}_{к} = 50°C), so let’s assume that t_{2}_{к} = 40°C.

Let’s calculate average temperatures at the refrigerator inlet and outlet:

∆t_{н} _{ср} = 95 - 20 = 75;

∆t_{к} _{ср} = 50 - 40 = 10

∆t_{ср} = 75 - 10 / ln(75/10) = 32.3 °C

2) For the counter flow conditions, let’s assume the final water temperature the same as for the direct flow of the heat transfer media, i.e. t_{2}_{к} = 40°C.

∆t_{н} _{ср} = 95 - 40 = 55;

∆t_{к} _{ср} = 50 - 20 = 30

∆t_{ср} = 55 - 30 / ln(55/30) = 41.3°C

**Problem 2**

Using the conditions of Problem 1, please determine the required heat exchange surface (F) and the cooling water flow (G). The hot product flow G = 15000 kg/h and its heat capacity C = 3430 J/kg·grad (0.8 kcal·kg·deg). Cooling water parameters are as follows: heat capacity c = 4080 J/kg·grad (1 kcal·kg·grad), heat transfer coefficient k = 290 W/m^{2}·grad (250 kcal/m^{2}*deg).

Solution: Using the heat balance equation, we will obtain an expression for determining a heat flux when cold heat transfer medium is heated:

Q = Q_{гт} = Q_{хт}

Wherefrom: Q = Q_{гт} = GC (t_{1}_{н} - t_{1}_{к}) = (15000/3600)·3430·(95 - 50) = 643125 W

Assuming t_{2}_{к} = 40°C, we will find the flow rate of the cold heat transfer medium:

G = Q/ c(t_{2}_{к} - t_{2}_{н}) = 643125/ 4080(40 - 20) = 7.9 kg/s = 28 500 kg/h

Required heat exchange surface

In case of a direct flow:

F = Q/k·∆t_{ср} = 643125/ 290·32.3 = 69 m^{2}

In case of a counter flow:

F = Q/k·∆t_{ср} = 643125/ 290·41.3 = 54 m^{2}

**Problem 3**

At a plant, gas is transported through a steel pipeline with an outer diameter d_{2} = 1500 mm, wall thickness δ_{2} = 15 mm, thermal conductivity λ_{2} = 55 W/m·deg. From the inside, the pipeline is lined with fireclay bricks, whose thickness is δ_{1} = 85 mm, thermal conductivity λ_{1} = 0.91 W/m·deg. The coefficient of heat transfer from gas to the wall is α_{1} = 12.7 W/m^{2}·deg; from the outer surface of the wall to air is α_{2} = 17.3 W/m^{2}·deg. Please find the coefficient of heat transfer from gas to air.

Solution: 1) Let’s determine the inner diameter of the pipeline:

d_{1} = d_{2} - 2·(δ_{2 }+ δ_{1}) = 1500 - 2(15 + 85) = 1300 mm = 1.3 m

Average lining diameter:

d_{1 }_{ср} = 1300 + 85 = 1385 mm = 1.385 m

Average pipeline wall diameter:

d_{2 }_{ср} = 1500 - 15 = 1485 mm = 1.485 m

Let’s calculate the coefficient of heat transfer by the formula:

k = [(1/α_{1})·(1/d_{1}) + (δ_{1}/λ_{1})·(1/d_{1 }_{ср})+(δ_{2}/λ_{2})·(1/d_{2 }_{ср})+(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} = 5.4 W/m^{2}·grad

**Problem 4**

A single-pass shell-and-tube heat exchanger heats methanol with water from the initial temperature of 20 to 45°C. The water flow is cooled from 100 to 45°C. A tube bundle of the heat exchanger contains 111 tubes, the diameter of one tube is 25x2.5 mm. The velocity of the methanol flow through the tubes is 0.8 m/s (w). The heat transfer coefficient is 400 W/m^{2}·deg. Please, determine the total length of the tube bundle.

Solution:

Let’s determine the average temperature difference of the heat transfer media as the average logarithmic value.

∆t_{н ср }= 95 - 45 = 50;

∆t_{к ср }= 45 - 20 = 25

∆t_{ср} = 50 + 25 / 2 = 37.5°C

Then, let’s determine the average temperature of the heat transfer medium flowing through the tube side space.

∆t_{ср} = 45 + 20 / 2 = 32.5°C

Let’s determine a mass flow of methanol.

G_{сп} = n·0.785·d_{вн}^{2}·w_{сп}·ρ_{сп} = 111·0.785·0.02^{2}·0.8· = 21.8

ρ_{сп} = 785 kg/m^{3} – methanol density at 32.5°C, the value is taken the reference literature.

Then, let’s determine the heat flux.

Q = G_{сп}с_{сп} (t_{к} _{сп} – t_{н} _{сп}) = 21.8·2520 (45 – 20) = 1.373·10^{6} W

c_{сп} = 2520 kg/m^{3}– heat capacity of methanol at 32.5°C, the value is taken the reference literature.

Let’s determine the required heat exchange surface.

F = Q/ K∆t_{ср} = 1.373·10^{6}/ (400·37.5) = 91.7 m^{3}

Let’s calculate the total length of the tube bundle by the average diameter of the tubes.

L = F/ nπd_{ср} = 91.7/ 111·3.14·0.0225 = 11.7 m.

In accordance with the recommendations, the total length of the tube bundle should be divided into several sections of the proposed standard size, with a required margin of the heat exchange surface to be provided.

**Problem 5**

A plate heat exchanger is used to heat a 10% NaOH solution flow from 40°C to 75°C. The sodium hydroxide flow is 19000 kg/h. Water vapor condensate with a flow rate of 16000 kg/h and initial temperature of 95°C is used as a heating agent. Assume that heat transfer coefficient is 1400 W/m^{2}·deg. Please calculate basic parameters of the plate heat exchanger.

Solution: Let's find the amount of heat transferred.

Q = G_{р}с_{р} (t_{к р} – t_{н р}) = 19000/3600 · 3860 (75 – 40) = 713 028 W

From the heat balance equation, let’s determine the final temperature of the condensate.

t_{к} _{х} = (Q·3600/G_{к}с_{к}) – 95 = (713028·3600)/(16000·4190) – 95 = 56.7°C

с_{р}_{,}_{к} – heat capacity of the solution and condensate, the values are found in the reference materials.

Let’s determine average temperatures of the heat transfer media.

∆t_{н ср }= 95 - 75 = 20;

∆t_{к} _{ср} = 56.7 - 40 = 16.7

∆t_{ср} = 20 + 16.7 / 2 = 18.4°C

Let’s determine the cross-section of the channels; for calculation, let’s assume the mass velocity of the condensate as Wk = 1500 kg/m^{2}·sec.

S = G/W = 16000/3600·1500 = 0.003 m^{2}

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

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

Based on the recommendations, let’s assume the width of the spiral according to the nearest larger tabulated value B = 0.58 m.

Let’s refine the cross-section of the channel

S = B·b = 0.58·0.006 = 0.0035 m^{2}

and mass velocity of the streams

W_{р} = G_{р}/S = 19000/ 3600·0.0035 = 1508 kg/m^{3}·sec

W_{к} = G_{к}/S = 16000/ 3600·0.0035 = 1270 kg/m^{3}·sec

The heat transfer surface of the spiral heat exchanger is determined as follows.

F = Q/K∆t_{ср} = 713028/ (1400·18.4) = 27.7 m^{2}

Let’s determine the working length of the spiral

L = F/2B = 27.7/(2·0.58) = 23.8 m

Next, let’s determine a pitch of the spiral, while setting the sheet thickness as δ=5 mm.

t = b + δ = 6 + 5 = 11 mm

o calculate the number of turns of each spiral, the initial diameter of the spiral shall be assumed, based on the recommendations, as d = 200 mm.

N = (√(2L/πt)+x^{2}) – x = (√(2·23.8/3.14·0,011)+8.6^{2}) – 8.6 = 29.5

where х = 0.5 (d/t - 1) = 0.5 (200/11 – 1) = 8.6

The required diameter of the spiral is determined as follows.

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

**Problem 6**

Please determine the hydraulic resistance of a heat transfer media, created in a four-way 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 properties are as follows: flow rate G = 2.5 kg/s, velocity W = 0.240 m/s and density ρ = 776 kg/m^{3} (Reynolds criterion Re = 1573 > 50). Cooling water properties are as follows: flow rate G = 5 kg/s, velocity 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_{бс} = 15/Re ^{0.25} = 15/1573^{0.25} = 2.38

ζ_{в} = 15/Re ^{0.25} = 15/3101^{0.25} = 2.01

Let's clarify the velocity of alcohol and water in the fittings (assuming that d_{шт} = 0.3m)

W_{шт} = G_{бс}/ρ_{бс}0.785d_{шт}^{2} = 2.5/776 ·0.785·0.3^{2} = 0.05 m/s is less than 2 m/s, therefore it can be ignored.

W_{шт} = G_{в}/ρ_{в}0.785d_{шт}^{2} = 5/995 ·0.785·0.3^{2} = 0.07 m/s is less than 2 m/s, therefore it can be ignored.

Let's determine the hydraulic resistance for butyl alcohol and cooling water.

∆Р_{бс} = хζ·(l/d) · (ρ_{бс}w^{2}/2) = (4·2.38·0.9/ 0.0075)·(776·0.240^{2}/2) = 25532 Pa

∆Р_{в} = хζ·(l/d) · (ρ_{в}w^{2}/2) = (4·2.01·0.9/ 0.0075)·(995·0.175^{2}/2) = 14699 Pa.

Heat exchange equipment is designed to transfer heat energy from one environment to another, i.e., for transferring heat from hot to cold heat transfer medium. The variety of units, differed in design, purpose and heat transfer methods, allows the implementation of customer-tailored processes. Heat exchange equipment can be used either as major equipment, or auxiliary (stand-alone) equipment.

The fields of application of heat exchange equipment are as follows:

- Heat supply or removal during certain reactions;
- Process stream heating or cooling;
- Distillation;
- Adsorption and absorption;
- Melting of solids and crystallization of substances;
- Evaporation;
- Condensation.

- Block graphite heat exchangers, finned heat exchangers
- Heat exchangers
- Incinerators
- Industrial burners and combustion systems
- Plate-type heat exchangers
- Process heaters and preheaters
- Reformer furnaces
- Shell and tube heat exchangers
- Tube-in-tube heat exchangers, spriral heat exchangers
- Waste incineration and disposal plants (incinerators)

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