6.

# Technical calculations

Industrial production requires a daily check of the plant capacity, product output yield, consumption of energy etc. by collection of various production data and calculations. Examples of some useful technical calculations are given in the following sections.

6.1.

## Evaporation and product output

The symbols used in the examples consist of three letters, the meaning of which is as follows:
First letter Second letter Third letter
E = evaporator F = feed R = rate (kg/h)
D = spray dryer E= evaporation S = total solids (%)
B = fluid bed P = product
Índice. 6.1. Evaporation, symbols used in the examples consist of three letters, the meaning of which is as follows
a) Evaporation and product rate from the evaporator:
[6,1]
[6,2]
[6,3]
[6,4]
[6,5]
b) Evaporation and product rate from the spray dryer:
[6,6]
[6,7]
[6,8]
[6,9]
[6,10]
[6,11]
c) Evaporation and product rate from the fluid bed:
[6,12]
[6,13]
[6,14]
[6,15]
[6,16]
[6,17]
d) Example:

Evaporator feed rate: EFR = 50000 kg/h
Total solids of evaporator feed: EFS = 12.2 %
Total solids of the concentrate: EPS = 48.0 %
Total solids of powder from spray dryer: DPS = 94.2 %
Total solids of powder from fluid bed: BPS = 97.2 %

Notice that DFR = EPR, BFR = DPR, DFS = EPS and BFS = DPS.
The results are given in table below..
Calculated Calculation Result Equation
EPR = DFR 50000x12.2/48 = 12708.3 [6,3] and [6,6]
EER 50000-12708.3 =  37291.7 [6,1]
DPR = BFR 12708.3x48/94.2 = 6475.6 [6,8] and [6,12]
DER 12708.3-6475.6 = 6232.7 [6,7]
BPR 6475.6x94.2/97.2 = 6275.7 [6,15]
BER 6475.6-6475.7 =  199.9 [6,13]
Índice. 6.2.
There are examples of calculations in all the following sections and the results are rounded up to the reasonable decimals. However, the calculations have been made using the exact figures.
6.2.

## Heating of atmospheric air

If an amount Aa of atmospheric air with humidity ya is to be heated from a temperature t1 to a temperature t2, the amount of heat Q necessary is calculated by the equation:
[6,18]
If Aa =30000 kg/h, ya =0.009 kg/kg, t1=15°C and t2=200°C then from equation [3,8] ca1=0.240, ca2=0.245 and from equation [3,10] cv1=0.445 and cv2=0.463 kcal/kg/°C. The result is:
[6,19]
6.3.

## Mixing of two air stream

Two quantities of air A1 and A2, with humidities y1 and y2 and temperatures t1 and t2 are to be mixed, the temperature t3 of the mixed air calculated by a simplified method is as follows:
[6,20]
Example:
A1 = 50000 kg/h, A2 = 10000 kg/h, t1 = 90°C, t2 = 20°C,
y1 = 0.0442 kg/kg and y2 = 0.007 kg/kg:
[6,21]
For a more precise calculation, heat capacities of air must be considered. From equation [3,8] values are ca1 = 0.240 kcal/kg/°C at 90°C and ca2 = 0.241 kcal/kg/°C at 20°C and calculation of the resulting air humidity y3 and enthalpy h3 is by:
[6,22]
[6,23]
The enthalpy of both components using heat capacities of water vapour from equation [3,10] at 20°C cv1 = 0.445,at 90°C cv2 = 0.451 and using equation [3,11]:
[6,24]
[6,25]
The enthalpy of the resulting air mixture is then:
[6,26]
[6,27]
Using equation [6,22] which is another form of equation [3,11] the resulting temperature t3 is:
[6,28]
The values ca3 and cv3 are calculated from equation [3,8] and equation [3,10] using the approximate mixing temperature calculated by [6,19], i.e. 78°C which are 0.241 and 0.450 kcal/ kg/°C respectively.
[6,29]
In comparison with the simplified calculation there is a difference of 1.04°C.
6.4.

## Dry air rate, water vapour rate and air density

Using figures from the previous example and equations [3,5] through [3,7] and [3,13], the dry air rates Adn, water vapour rates Avn and air densities ρn are:
[6,30]
The volume of air V is calculated as follows:
[6,31]
Thus:

V1 = A1/ρ1 = 50000/0.9484 = 52722 m3/h
V2 = A2/ρ2 = 10000/1.1997 = 8336 m3/h
V3 = A3/ρ3 = 60000/0.9804 = 61199 m3/h
6.5.

## Air velocity in ducts

Using a Pitot tube (Fig.6.1.) it is possible to measure the static pressure Ps, dynamic pressure Pd and total pressure Pt.
[6,32]
[6,33]
and,
[6,34]
where:
Pd = dynamic pressure in mm water gauge
g = gravity constant 9.81 m/s²
ρ = density of air kg/m³
v = air velocity in m/s.
6.6.

## Air flow measurements

In practice, it is not easy to measure exactly the amount of air passing through a duct, filter or spray dryer. The methods available are listed below:

a) Measuring the air velocity and duct area.

If the air velocity is measured by Pitot tube or by wind or hot wire anemometer in a duct of SA
area, then the volume of air flow V in m3/h is:
[6,35]
Knowing the air temperature t and humidity y, the air rate A in kg/h can be calculated using equations [3,12], [3,13] and [6,23]. In a similar way, for round ducts of diameter D, the equation is:
[6,36]
b) Measuring pressure drop across the cyclone:
[6,37]
where:
A = air rate in kg/h
D = cyclone diameter in m
n = number of cyclones
ρ = density of air in kg/m3
ΔP= pressure drop across the cyclone in mm WG
K = cyclone constant

The cyclone constant depends upon the cyclone design and powder loading in air. For various types of the cyclones constants lay between 200 - 1000, the exact values being proprietary manufacturer know-how. However, if the air flow has been measured by one of the described methods and at the same time the pressure drop over the cyclone measured the cyclone constant can be calculated backwards and used in later routine measurements.

c) Measuring the amount of heat necessary for air heating:

If air is heated from a temperature t1 to a temperature t2 the amount of air can be calculated from the amount of heat used and the temperature difference. In principle the same method of calculation is used for steam, oil, gas and electric air heaters.

The basic equation is:
[6,38]
where:
E = efficiency of the heater
t1 = air temperature at heater inlet
t2 = air temperature at heater outlet
ca1 = heat capacity of air at heater inlet
ca2 = heat capacity of air at heater outlet
X = the heat consumed kcal/h.

Calculation of X for various types of heaters is explained below.

c1) Measuring of the condensate from the steam heater:

X for the equation [6,30] is:
[6,39]
where:
W = amount of the condensate in kg/h
hs = enthalpy of steam at heater inlet in kcal/kg
hc = enthalpy of the condensate (which is equal to the temperature)

c2) Measuring of gas or oil consumption:

X for equation [6,30] is:
[6,40]
where:
G = amount of oil in kg/h or gas in Nm3/h
Qh = caloric value of the fuel in kcal/kg or kcal/Nm3.

c3) Measuring of consumption of electricity:

X for the equation [6,30] is:
[6,41]
d) Measuring the water evaporation rate:

The dryer is operated on water under constant t1 and t2 for at least 1 hour to obtain stable conditions. The amount of water, supplied to the dryer over a period of time is measured. The most suitable way is to measure the level difference h in a cylindrical feed tank of diameter D. The volume and weight of the evaporated water and the evaporation rate are:
[6,42]
[6,43]
[6,44]
where:
D = feed tank diameter in m
h = level difference in m
t = time of measuring in s
ρw = density of water at feed temperature tf in °C
DER = rate of evaporation in kg/h.

The drying air rate A is then:
[6,45]
where:
t1 and t2 = the air inlet and outlet temperatures,
ca1 and ca2 = heat capacities of air at t1 and t2,
cv2 = heat capacity of water vapour at t2
tf = feed temperature,
Ar = surface area of the dryer in m²,
ts = spray dryer surrounding temperature,
K = radiation constant in kcal/m²/h.
6.7.

## Barometric distribution law

The barometric pressure in an altitude of m meters above sea level is calculated:
[6,46]
where:
p0 = barometric pressure at sea level at 0°C,
M = molecular weight of air (= 0.029 kg/mol),
g = gravity constant (= 9.807 m/s2),
R = gas constant (=8.3144 J/K/mol)
T = absolute temperature in °K, and
m = altitude in m.
6.8.

## The heat balance of a spray dryer

The spray dryer in operation is a system where air and product move through under changing temperatures and humidities and as the product is concerned, also changing physical properties. Entering components are: drying gas, which is usually heated ambient air, some auxiliary air flows (as cooling air, fines transport air etc.) and the feed to be dried. The humidity of the entering air corresponds to the ambient air humidity, possibly increased somewhat by moisture generated during combustion (in case of direct gas heating) or by moisture picked up by the air on passage through the building. The feed is the milk concentrate. Exhaust air and powder leave the system. The exhaust air is made up of all entering air flows plus water formed from the evaporation and besides it contains also some traces of the dried solids (fine particles). The dry products in powder form contain practically all the feed solids, but have residual moisture. The amount of air necessary to evaporate a required amount of water from a given amount of feed can be found by calculating all the individual heat requirements necessary for evaporating the water, heating or cooling each individual component from its inlet to its outlet temperature, while compensating for heat losses. The sum of these contributions is then recalculated into the air rate on the basis that the drying air enters the system with a temperature t1 and leaves with a temperature t2.

The following example is a demonstration of the calculation of the heat requirement and drying air rate of a spray dryer as specified below by points a) through g). The source of drying air is ambient thus the inlet humidity is considered as ambient humidity ya:

a) The drying air of the inlet temperature t1 and humidity ya, and the drying air rate Ad (which has to be calculated),
b) The auxiliary cooling air for the atomizing device of temperature tc, humidity ya and rate Ac,
c) The fines transport air of temperature tt, humidity ya and rate At,
d) The feed concentrate of temperature tf, rate DFR and solids content DFS,
e) Recycled fines, collected from all cyclones, into the dryer in the amount given by ratio R to the total powder production and temperature tt (i.e. the same as of fines transport air).

The mass flows at the outlet of the system are:

f) The exhaust air consisting of all the entering air flows plus the moisture generated during drying, having final temperature t2 and humidity y2,
g) The powder consisting of the feed solids plus some residual moisture.

For the calculation, the operating conditions at the inlet and the outlet have to be first set. Product experience, knowledge of the drying installation and applied process is used to estimate the permissible air inlet temperature and feed total solids content and the required air outlet temperature for the specified powder moisture. Besides there is also some heat loss due to the radiation from the equipment surface of area SA. The radiation coefficient is K and the surrounding temperature around the dryer is estimated to be 20°C above the ambient temperature. Having fixed the operating parameters the heat balance can be calculated as follows:

Heat of evaporation:
[6,47]
Heat of product solids:
[6,48]
Heat of cooling air:
[6,49]
Heat of fines transport air:
[6,50]
Heat of fines:
[6,51]
[6,52]
The sum of heat requirements:
[6,53]
The drying air rate:
[6,54]
Example: Calculate the drying air rate and size of dryer for following duty:

Feed rate: DFR = 4000 kg/h
Feed concentration: DFS = 48 %
Feed temperature: tf = 60 °C
Product solids content: DPS = 95 %
Ambient temperature: ta = 15 °C
Ambient humidity: ya = 0.01 kg/kg
Inlet temperature: t1 = 200 °C
Outlet temperature: t2 = 80 °C
Cooling air rate: Ac = 200 kg/h
Fines transport air: At = 500 kg/h
Transport air temperature tt = 60 °C

The values for operating parameters were estimated from product experience with respect to the required quality specification. The powder temperature is estimated to be 5°C below the outlet air temperature, fines recirculation ratio R = 0.5, radiation loss coefficient K = 3.0 kcal/ m2/h and dryer surface area 300 m2. The heat capacities of air and water vapour are taken from Equation 3.8 and 3.10 (ca1=0.245, ca2 = 0.241, caa = 0.24, cat = 0.241, cv1 = 0.463, cv2 = 0.45, cva = 0.444 and cvt = 0.448kcal/kg/°C). The product is whole milk with 28% fat, thus the heat capacity of solids, using values from Table 3.2. is:

cs = (28 * 0.5 + (100 - 28) * 0.3)/100 = 0.356 kcal/kg/°C

Calculation (according to equations [6,9], [6,7] and [6,39] through [6,46]:

DPR = 4000 * 48/95 = 2021.1 kg/h
DER = 4000 - 2021.1 = 1978.9 kg/h
Qev = 1978.9 * (597.3 + 0.45 * 80 - 60) = 1134530.5 kcal/h
Qpr = 2021.1 * (80 - 5 - 60) * (0.356 * 95 / 100 + (1 - 95 / 100)) = 11768.6 kcal/h
Qco = 200 / 1.01 * ((80 * 0.241 - 15 * 0.24) + * (80 * 0.45 - 15 * 0.444)) = 3163.0 kcal/h
Qtr = 500 / 1.01 * ((80 * 0.241 - 60 * 0.241) + 0.01 * (80 * 0.45 - 60 * 0.448)) = 2431.3 kcal/h
Qfi = 2021.1 * 0.5 *(80 - 5 - 60) * (0.356 * 95 / 100 +(1 - 95 / 100)) = 5884.3 kcal/h
Qrl = 300 * 3.0 * (80 - 15 - 20) = 40500.0 kcal/h
ΣQ = 1134530.5 + 11768.6 + 3163 + 2431.3 + 5884.3 + 40500 = 1198277.7 kcal/h
Adr = 1198277.7 / (200 * .245 - 80 * .241 + 0.01 * (200 * .463 - 80 * 0.45)) = 39565 kg/h

The above calculation can be simplified by neglecting the air moisture content and powder moisture and using for heat capacities of air and of water vapour constants 0.24 and 0.46 kcal/ kg/°C respectively:

Qev = 1978.9 * (597.3 + 0.46 * 80 - 60) = 1136113.7 kcal/h
Qpr = 2021.1 * 0.356 * (80 - 5 - 60) = 10792.4 kcal/h
Qco = 200 * 0.24 * (80 - 15) = 3120.0 kcal/h
Qtr = 500 * 0.24 * (80 - 60) = 2400.0 kcal/h
Qfi = 2021.1 * 0.356 * (80 - 5 - 60) = 5396.1 kcal/h
Qrl = 300 * 3.0 / (80 - 15 - 20) = 40500.0 kcal/h
ΣQ = 1198322.2 kcal/h
Adr = 1198322.2 / (0.24 * (200 - 80)) = 41608 kg/h

This comparison demonstrates that the simplified calculation results in more than 5% higher amount of air and emphasizes the importance of calculation on the enthalpy basis. The difference is even greater if the available ambient air has high humidity values. The absolute humidity of the exhaust air is:
[6,55]
Total exhaust air = 39565.4 + 200 + 500 + 1978.9 = 42244.3 kg/h
Exhaust air density Vexair and volume Vexair are then:
[6,56]
[6,57]
The dryer should have two main cyclones of cyclone constant 380 operating at a pressure drop of 150 mm WG. The cyclone diameter according to the equation [6.29] will be:
[6,58]
Reference: Schlünder,E.U.:Dissertation Techn.Hochschule Darmstadt D 17, 1962.