5 dpe plate he design (syrup water)
TRANSCRIPT
Thermal design of plate HE syrup – cooling water (counter flow)
Calculation procedure
Material for courses DPE and PRO III (remember part 3).
Prepared by: Pavel Hoffman
1. Given data
Cooled solution (syrup):
Amount of incoming solution MS (kg/s)
Temperature of incoming solution tS0 (°C)
Required (or calculated) temperature of outgoing syrup tS1 (°C)
Incoming solution concentration xS0 (%)
Cooling liquid (cooling water):
Amount of incoming water MW (kg/s)
Temperature of incoming water tW0 (°C)
Given (or calculated) temperature of outgoing water tW1 (°C)
Next parameters needed for calculations:
Gap between plates s (mm)
Thickness of plates sp (mm)
Length of interplate channel L (mm)
With of interplate channel b (mm)
Equivalent diameter of channel de (m)
Flow clear area of channel fp (m2)
Heat transfer area of plate AP (m2)
Thermal conductivity of plates p (W/mK)
document.doc 1 / 59 Print date: 8 / 2007 P. Hoffman
Fouling thickness (incrustation) on side of cooling water sWi (mm)
Thermal conductivity of fouling on side of cooling water Wi (W/mK)
Fouling thickness (incrustation) on side of cooled syrup sSi (mm)
Thermal conductivity of fouling on side of cooled syrup Si (W/mK)
Max. allowable pressure loss on side of cooling water pWmax (kPa)
Max. allowable pressure loss on side of cooled syrup pSmax (kPa)
Max. allowable pressure in plates pmax (kPa)
Minimal economical temperature difference in HE tHEmin (°C)
2. Specification of physical parameters needed for calculations
Density of cooling water W (kg/m3)
Dynamic viscosity of cooling water W (Pa*s)
Kinematic viscosity of cooling water W (m2/s)
Specific heat of cooling water cW (J/kgK)
Thermal conductivity of cooling water W (W/mK)
Prandtl number of cooling water PrW (-)
Density of cooled syrup S (kg/m3)
Dynamic viscosity of cooled syrup S (Pa*s)
Kinematic viscosity of cooled syrup S (m2/s)
Specific heat of cooled syrup cS (J/kgK)
Thermal conductivity of cooled syrup S (W/mK)
Prandtl number of cooled syrup PrS (-)
(all parameters are found from tables etc. for average temperatures)
document.doc 2 / 59 Print date: 8 / 2007 P. Hoffman
Fig. 1.: Example of chart of plate HE with 2 sections
Fig.2.: Temperature profile on heat transfer surface in cooler
Fig.3.: Temperature profile in 1 section of cooler
document.doc 3 / 59 Print date: 8 / 2007 P. Hoffman
tHEmin
MS - syrup
MW - water
tS1 = ?
tS0
tW1
tW0
syrup syrup
cooling water
1.section 2.section
cooling water
cooling water
sSi; Si
sw; w
sWi; Wi
twW
q
twS
tW
tS
tW
plate fouling
twfW
W
heating of cooling waterS
cooling of syrup
cooling water(boundary layer)
fouling
twfS
tS
cooled syrup(boundary layer)
Fig. 4. Examples of plate HE design
document.doc 4 / 59 Print date: 8 / 2007 P. Hoffman
a) Inter plate channels arrangement– flow pattern
b) Plate HE with connecting plate (with necks) between two sections
document.doc 5 / 59 Print date: 8 / 2007 P. Hoffman
c) Plate HE (cooler) with press plate closed and open
d) Two types of plates with various profiles
Lb
3. Basic thermal balance of cooler
Heat transferred from syrup to cooling water (it is heated)
QHW = MW * cW * (tW1 – tW0) = k * ATreal * tL
(W; kg/s, J/kg°C, °C; W/m2K, m2, K)
Heat that is taken from cooled syrup – without heat loss it is (principle of energy
conservation in HE)
QHW = QCS
QCS = MS * cS * (tS0 – tS1) (W)
If we take into account heat loss in HE QL (W)
where QL is determined from a calculation or estimated on the basis of our pre-
vious experiments or knowledge as a part of total transferred heat (in % trans-
ferred heat – easier calculation)
QL = QCS * z / 100
QHW = QCS – QL = MS * cS * (tS0 – tS1) – QL (W)
or QHW = MS * cS * (tS0 – tS1) * (1 – z / 100) (W)
Specification of outgoing temperature of heated cooling water (if we have given
the outgoing temperature of cooled syrup)
For heat loss QL= 0
document.doc 6 / 59 Print date: 8 / 2007 P. Hoffman
tW1 = tW0 + (MS*cS*(tS0 – tS1)) / (MW*cW) (°C)
If we take into account heat loss in HE QL (W) or z (in % of transferred heat)
– owing to heat loss we have to transfer less heat from the hot syrup to the cool-
ing water (it is that tW1 will be lower than for the case without heat loss)
tW1 = tW0 + (MS*cS*(tS0 – tS1) – QL) / (MW*cW) (°C)
or tW1 = tW0 + (MS*cS*(tS0 – tS1)*(1-z / 100)) / (MW*cW) (°C)
Specification of outgoing temperature of cooled syrup (if we have given outgoing
temperature of cooling water)
For heat loss Qz = 0
tS1 = tS0 - (MW*cW*(tW1 – tW0)) / (MS*cS) (°C)
If we take into account heat loss in HE QL (W) or z (in % of transferred heat)
– owing to heat loss the hot syrup will be cooler → for needed outgoing tempera-
ture we will transfer less heat to the cooling water
tS1 = tS0 - (MW*cW*(tW1 – tW0) + QL) / (MS*cS) (°C)
or tS1 = tS0 - (MW*cW*(tW1 – tW0)*(1+z / 100)) / (MS*cS) (°C)
These outgoing temperatures we have to compare with a reality of heat transfer
(depending on given data). It is that for example we compare the calculated
temperature of cooled syrup with the given temperature of cooling water. If it is
tS1 – tW0 < tHEmin (°C)
document.doc 7 / 59 Print date: 8 / 2007 P. Hoffman
given data are unreal (see fig. 2 and 3) and it is necessary to design the cooler for
this new outgoing temperature of cooled syrup
tS1 = tW0 + tHEmin (°C)
Otherwise, with regard of temperature drops on the cooling water side, in the
fouling layer, in plate material and on the syrup side (again with fouling layer),
it would not be economical or real to transfer the needed amount of heat (or the
heat transfer area would be too large).
Then we can specify the mean logarithmic temperature difference in the HE
(counter flow) – see fig.3
(K)
4. Thermal calculation of cooler
As values of heat transfer coefficients for heating of the cooling water and
cooled syrup in HE channels depend on (among others) speeds in these channels
(an as we do not know a total heat transfer area of the HE we do not know these
speeds too) it is impossible to calculate these coefficients directly. Therefore we
must use an iterative method.
Big iteration loop
We estimate a value of coefficient of heat passage k (overall heat transfer co-
efficient) in the HE and on the base we can calculate the heat transfer area,
number of plates and liquids flow rates in one channel. If is the liquid speed too
high we design a HE with several channels connected in parallel (their number
document.doc 8 / 59 Print date: 8 / 2007 P. Hoffman
we can estimate according our practical knowledge with optimal speeds for vari-
ous liquids – from points of views of heat passage coefficient → heat transfer
area → capital cost and pressure loss → pump consumption → running costs).
Then we calculate new values of heat transfer coefficients and a new value of
the coefficient of heat passage. If there is a difference we repeat the procedure.
Simultaneously we must check pressure losses in the HE. If they are too high we
must use a solution with lower losses (it is to increase a number of parallel con-
nected channels for a liquid in question). On the contrary if are the pressure
losses too low it is good to reduce number of parallel connected channels and so
increase speed. Thus increase coefficient of heat passage and decrease needed
heat transfer area (cheaper HE). More see example about effect of speeds on a
HE size.
4.1. Estimation of coefficient of heat passage and preliminary calculation of cooler (coolers)
Estimation of coefficient of heat passage
kest = ? (W/m2K) x compare with calculated and event. change estimation
Heat transfer area of cooler (coolers) needed for syrup cooling
(m2)
Number of plates in cooler (coolers) – theoretical estimation
nPT = ATC / AP choose whole number (-)
Note 1:
document.doc 9 / 59 Print date: 8 / 2007 P. Hoffman
Exact equation for calculation valid for all types of HE is
Q = k * A * tL * F
where F is a factor with values usually in range from 0.8 to 1.0. A HE with factor
F < 0.8 is not acceptable as it is not stable in operation. For our case of a pure
counter flow is the F = 1.0. Similarly it is for pure parallel flow, evaporators and
condensers.
Fig. 5.: Examples of temperature profiles and their F factors:
document.doc 10 / 59 Print date: 8 / 2007 P. Hoffman
F = 1
Pure parallel flow in 1 or 2 passes Pure counter flow in 1 or 2 passes
Steam condensation and liquid heating in 2 passes (= as in 1)
Coolant boiling and liquid cooling in 2 passes (= as in 1)
HE with pure counter flow and 2 passes for both fluids
F factor values dependence on parameters P and R
F
F < 1
Combination of counter and parallel flows or cross flow
Evaporator
P = (tC1 – tC0) / (tH0 – tC0) = heating of cool liquid / max. possible heating
R = (tH0 – tH1) / (tC1 – tC0) = cooling of hot liquid / heating of cool liquid
Note 2:
Another variant very often used for plate HE design.
According experiences speeds of both liquids in interplate channels are in the 1st
iteration estimated. On the basis of these speeds we can specify a flow pattern. It
is numbers of parallel channels in 1 pass. From it we can specify exact speeds of
both liquids wi and for this iteration Rei, Nui and ki. Then we can specify the
heat transfer area A and from it number of passes connected in series. Next pro-
cedure is analogous to the previous case. This procedure is used in the examples
prepared in Excel.
Choice of HE arrangement
Number of parallel channels in 1 pass and number of passes in HE we
choose from a point of view of optimal speeds of liquid in channels it is that coef-
ficient of heat passage and pressure losses must be optimal. Criterion of a
proper choice (design) is a HE size (mass, total heat transfer area) of the cooler
and total pressure losses on both liquids → economic appraisal.
Choice of a number of parallel interplate channels in the cooler – syrup side
(with regard to the recommended speed)
document.doc 11 / 59 Print date: 8 / 2007 P. Hoffman
npcScalc = MS / (S * fP *wSrec) (-; kg/s, kg/m3, m2, m/s)
npcS = ? choose a whole number (-)
Choice of a number of parallel interplate channels in the cooler – cooling water
side (with regard to the recommended speed)
npcWcalc = MW / (W * fP *wWrec) (-; kg/s, kg/m3, m2, m/s)
npcW = ? choose a whole number (-)
Choice of a number of passes connected in series in HE (section) – syrup side
(one half of channels is for cooled syrup, the second one is for cooling water)
npsS = nPT / (2 * npcS) (-)
npsS = ? choose a whole number (-)
Choice of a number of passes connected in series in HE (section) – cooling water
side
npsW = nPT / (2 * npvW) (-)
npsW = ? choose a whole number (-)
A maximal difference between products npcS*npsS and npcW*npsW can be
+/- 1 (ev. max. 2) else it is impossible to set plates together according calcu-
lated values of numbers. (for = 2 in 1 pass is not the same number of parallel
channels as in others – for one liquid) – see fig. 6b.
document.doc 12 / 59 Print date: 8 / 2007 P. Hoffman
Total number of plates in the cooler (section) – again it must be a whole number
nPTreal = npsS * npcS + npsW * npcw + 1 (-)
(one end plate must close a last channel – see fig. 6a)
Real heat transfer area of one cooler (section) is
AHEreal = Ap * nPHT = Ap * (nPTreal – 2) (m2)
Both end plates close interplate channels and they do not participate on heat
transfer (see fig. 6.).
Fig.6.: Example of a flow pattern in a plate HE (plates arrangement)
a) Flow pattern: syrup 1 x 4; cooling water 2 x 2
Flow pattern on this example:
cooling water npcW = 2; npsW = 2 syrup npcS = 4; npsS = 1
= npcS*npsS – npcW*npsW
=4*1 – 2*2 = 0 O.K.
nPTreal = 1 * 4 + 2 * 2 + 1 = 9
nPTHEreal = nPTreal – 2 = 9 – 2 = 7
Cross section clear areas in 1 pass:cooling water f1pW = 2 * fp (m2)syrup f1pS = 4 * fp (m2)
b) Flow pattern: syrup 1 x 5; cooling water 2 x 2:
Flow pattern on this example:
document.doc 13 / 59 Print date: 8 / 2007 P. Hoffman
syrup
1 2 3 4 5 6 7 8
cooling water
9
f1pS = 4 * fPf1pW = 2 * fP
syrup
1 2 3 4 5 6 7 8
cooling water
9
f1pS = 5 * fpf1pW = 2 * fp
10
cooling water npcW = 2; npsW = 2 syrup npcS = 5; npsS = 1
= npcS*npsS – npcW*npsW
=5*1 – 2*2 = 1 O.K.
nPTreal = 1 * 5 + 2 * 2 + 1 = 10
nPTHEreal = nPTreal – 2 = 10 – 2 = 8
Specific heat flux in the HE (cooler) – through plates = heat transfer area
qHE = QHE / AHEreal QHE = QCW (W/m2)
4.2. Heat transfer for forced convection in interplate channel
Cross section clear area in one pass of the HE (section) – cooling water side
f1pW = npcW * fp (m2)
Cooling water speed in cannels
wW = MW / (W * f1pW) (m/s)
The speed we compare with a recommended one and ev. change a number of
parallel channels or parallel HE.
Cross section clear area in one pass of the HE (section) – cooled syrup side
f1pS = npcS * fp (m2)
document.doc 14 / 59 Print date: 8 / 2007 P. Hoffman
Cooled syrup speed in channels
wS = MS / (S * f1pS) (m/s)
The speed we compare with a recommended one and ev. change a number of
parallel channels or parallel HE.
For next calculations we will use equations from literature, e.g. criteria equa-
tions that are valid for the type of plates.
Cooling liquid side
Determination of Reynolds criterion
ReW = wW * de / W (-)
Determination of Nusselt criterion for turbulent flow
Analogous to tubular HE is for the Nusselt criterion following equation used
Nu = a * Reb * Prc * (μw / μ)0,14 = * de /
Sieder-Tate correction (μw / μ)0,14 is in practice usually neglected as it has very
small effect – especially with regard to vague effect of fouling. Value of heat
transfer coefficient or Nu depends on a character of corrugation of plates (see
fig. 4d). It follows from it that for various types of plates are in the criteria equa-
tion various constants. For this example we will use following equation that was
derived from our experiments that were done in VÚPCHT Praha on designed
plates (producer CHS Chotěboř, now Tenez Chotěboř).
NuW = 0,0303 * ReW0,809 * PrW
0,43 (-)
document.doc 15 / 59 Print date: 8 / 2007 P. Hoffman
Determination of heat transfer coefficient
W = NuW * W / de (W/m2K)
Cooled syrup side
Determination of Reynolds criterion
ReS = wS * de / S (-)
Again as in previous we use the criteria equation designed for the used type of
plates
NuS = 0,0303 * ReS0,809 * PrS
0,43 (-)
Determination of heat transfer coefficient
R = NuR * R / de (W/m2K)
4.3. Heat passage coefficient including fouling
(W/m2K)
This value we compare with the estimated one (see chap. 4.1.) and if there is a
difference we repeat these calculations till
kcalc kest
4.4. Determination of pressure loss in the HE (cooler)
document.doc 16 / 59 Print date: 8 / 2007 P. Hoffman
Pressure loss in plate HE depends among others on speed, character of corruga-
tion of plates and their arrangement. Analogous to the heat transfer we use cri-
teria equation derived from our experiments for used type of plates.
Cooling water side
EuW = 460 * ReW-0,264 (-)
Note: This equation is valid for common working conditions, it is with common
fouling layer and directions changes, inlet and outlet necks etc.
Pressure loss in one interplate channel
pW1 = EuW * W *wW2 (Pa; -, kg/m3, m/s)
Total pressure loss in the HE (or one section)
pWT = pW1 * npsW (Pa; Pa, -)
Cooled syrup side
EuS = 460 * ReS-0,264 (-)
Note: Like for previous the equation is valid for working conditions, it is with
common fouling on plates.
Pressure loss in one interplate channel
pS1 = EuS * S *wS2 (Pa; -, kg/m3, m/s)
Total pressure loss in the HE (or one section)
document.doc 17 / 59 Print date: 8 / 2007 P. Hoffman
pST = pS1 * npsS (Pa; Pa, -)
These pressure losses we compare with given maximal allowable and according
situation (for corresponding liquid) repeat this calculation with a new number of
parallel channels npci and from it calculated number of passes connected in se-
ries npsi.
Too low pressure loss in HE results in low value of heat passage coefficient and
consequently too large heat transfer area = too expensive HE. Therefore it is im-
portant to optimize capital and running costs of a designed HE.
Note:
Calculation for tubular HE
Specification of roughness of tubes
kT = ? (-)
Coefficient of friction losses for turbulent flow in tubes
(-)
Friction loss in tubes (syrup side)
(Pa; -, m, -, m, m/s, kg/m3)
Where nPS = number of passes on the syrup side in a HE
nHES = number of HE connected in series
LT = tubes lenght
dTi = internal tubes diameter
wST = syrup speed in tubes
document.doc 18 / 59 Print date: 8 / 2007 P. Hoffman
Pressure loss owing to sudden enlargement (from tubes to chamber)
(Pa; m/s, kg/m3, -)
where n is number of sudden enlargements.
Pressure loss owing to sudden contraction (from chamber to tubes)
For the case it is possible to use following equation
pLSC 0,6 * pLSE
Pressure loss in valves (approximately)
Coefficient of pressure loss in valves
(-; m, m)
where Dvalve is diameter of seat of valve and Zvalve is valve stroke above the seat.
(Pa; -, m/s, kg/m3)
where speed in valve is
Note: For valves, flap valves (butterfly valves), spherical valves, slide valves etc. it is better to calculate pressure losses according characteristics given by a manufacturer.
Total pressure loss on the syrup side
pLS = pLTS + pLSES +pLSCS + pLvalvesS (Pa)
4.5. Specification of necks diameters
For the neck diameters specification we can use following equations
document.doc 19 / 59 Print date: 8 / 2007 P. Hoffman
→
where is M (kg/s) mass flow of a fluid
(steam, vapour, inerts, solution, condensate ...)
n (-) number of necks for the fluid
wrec (m/s) reccomended speed for the fluid
(kg/m3) density of the fluid
Reccomended speeds (according my practical experiences)
These speeds are valid for common cases of evaporators and heat exchangers in
food and chemical industries.
Steam (vapour) in inlet necks 10 - 25 m/s
Condensate in outlet necks (tubes) 0,2 - 0,5 m/s
(condensate is on boundary line vapour/water) if pressure decreases owing
to pressure losses condensate starts superheated and vapour forms (in valves,
steam traps, pipelines etc.,) two-phases flow in pipeline >> volume)
Solution inlet or outlet for HEs 1 - 3 m/s
Solution outlet for evaporators 1 - 2 m/s
(boiling liquid is on boundary line vapour/liquid)
Inerts outlet 10 - 15 m/s
5. Specification of maximal performance of designed HE – syrup cooler
As the real number of plates and their configuration are little different from cal-
culated values we want to know what is the real maximal performance of this
HE. Analogical is a calculation if we want to use an existing HE for other appli-
cation etc. In our case we want to know what the lowest attainable temperature
of cooled syrup is or what the maximal attainable temperature of cooling water
is.
document.doc 20 / 59 Print date: 8 / 2007 P. Hoffman
Select number of parallel channels on side of cooled syrup
npcS = ? (-)
Select number of parallel channels on side of cooling water
npcW = ? (-)
Speed in channels – side of syrup
wS = MS / (S * fP * npcS) (m/s)
Speed in channels – side of water
wW = MW / (W * fP * npcW) (m/s)
Select number of passes connected in series – side of syrup
npsS = ? (-)
Select number of passes connected in series – side of water
npsW = ? (-)
Total number of plates in the cooler (section)
nPTreal = npsS * npcS + npsW * npcw + 1 (-)
Total real heat transfer area of one cooler (section) is
AHE = Ap * nPHT = Ap * (nPTreal – 2) (m2)
Heat transfer for forced convention in channels
document.doc 21 / 59 Print date: 8 / 2007 P. Hoffman
On the basis of our previous results we estimate outlet temperatures of syrup
and cooling water - (i –iteration) and from them physical properties. The next
procedure is similar to the previous.
tS1i = ? and tW1i = ?
- Side of cooled syrup
Remember that physical parameters depends (among others) on temperature → iteration
Specification of Reynolds and Nusselt critera
ReSi = wSi * de / Si (-)
NuSi = 0,0303 * ReSi0,809 * PrSi
0,43 (-)
Specification of heat transfer coefficient
Si = NuSi * Si / de (W/m2K)
- Side of cooling water
Specification of Reynolds and Nusselt critera
ReWi = wWi * de / Wi (-)
NuWi = 0,0303 * ReWi0,809 * PrWi
0,43 (-)
Specification of heat transfer coefficient
Wi = NuWi * Wi / de (W/m2K)
Heat passage coefficient including fouling
(W/m2K)
document.doc 22 / 59 Print date: 8 / 2007 P. Hoffman
Mean logarithmic temperature difference in the HE (counter flow)
(K)
Maximal value of heat transfered from cooled syrup in cooling water
(without heat loss)
QHWi = kcalci * ATreal * tLi (W)
New temperature of outgoing cooled syrup is (i + 1 iteration)
tS1i+1 = tS0 – QHWi / (MW * cW) (°C)
New temperature of outgoing heated cooling water is (i + 1 iteration)
tW1i+1 = tW0 + QHWi / (MW * cW) (°C)
Now we compare these temperatures with previous i –iteration. If there is a dif-
ference we repeat these calculations till are temperatures the same.
In supplement are results of several variants calculated in Excel file. Here you
can see for example an effect of a variant with 2 sections or higher pressure loss
on heat transfer area of HE.
Generalization of the procedure on other types of HE
In a case of HE where is a phase change (condensation, boiling) is the procedure
similar, it is that we firstly estimate heat passage coefficient = big iteration loop
document.doc 23 / 59 Print date: 8 / 2007 P. Hoffman
(1st iteration level). Because for condensation heat transfer coefficient depends
on a temperature difference between condensing steam and wall (that is un-
known), we must estimate it too = small iteration loop in the big one (2nd itera-
tion level). When is the calculated wall temperature on the condensing steam
side equal to the estimated one we can continue in calculations in the big itera-
tion loop. Therefore is this procedure more complicated.
Our calculations (temperatures specification) go from this standpoint: Specific
heat flux through all interfaces must be the same (from condensing steam to in-
crustation, through incrustation, through wall, through incrustation and from
incrustation to boiling liquid). In this way we can specify temperatures on single
interfaces (see fig. 2 – there are other indices).
q = s * (ts“ – tis) = is / sis *(tis – tws) =w / sw * (tws – twb) =
= ib / sib * (twb – tib) = b * (tib – tb)
In the equation is:
s – steam; is – incrustation on steam side; ws – wall on steam side;
wb – wall on boiling side; ib – incrustation on boiling side; b – boiling liquid
document.doc 24 / 59 Print date: 8 / 2007 P. Hoffman
document.doc 25 / 59 Print date: 8 / 2007 P. Hoffman
M2 - syrup
M1 - water
t22
t21
t12
t11
Temperatures in 1 section
t112
M2 - syrup
M1 - water
t221
t21
t122
t121
t222
t111
Temperatures in 2 sections
syrup
water
1st section
Connection of 2 HE sections(cooling water in parallel, syrup in series)
water (50%)
water (50 %)
syrup
2nd section
water syrup
1.section
Syrup and cooling water flow in 1 section
watersyrup
Two variants of the plate cooler design
Comparison of results of 3 variants of the syrup cooler design
Msyrob = 6705 kg/h Mcooling water = 15000 kg/h
Variant
Parameter 1 section
< p
1 section
> p
2 sections
>p
Q (kW) 191,0 190,3 205,0
k (W/m2K) 257 385 485
302
AHET (m2) 47,4 30,9 42,7
number of plates 111 73 102
mass of cooler (kg) 599 394 551
temperatures (°C)
- syrup
- cooling water
80,0 → 40,6
35,0 → 46,0
80,0 → 40,9
35,0 → 45,9
80,0 → 38,4
35,0 → 56,0
35,0 → 37,6
pressure loss (kPa)
- syrup
- cooling water
77
15
203
50
184,6
12,3
14,1
document.doc 26 / 59 Print date: 8 / 2007 P. Hoffman
Approximate economical appreciation of 3 variants
with 1 and 2 sections (both for higher p L)
Cooler with 2 sections
tSyrOut = 38,4 °C AHE = 42,7 m2 Q = 205,0 kW
Cooler with 1 section
tSyrOut = 40,9 °C AHE = 30,9 m2 Q = 190,3 kW
Data for economical appreciation
Cost of stainless steel heat transfer area (plates) CA = 8000 Kč/m2
Cost of “cool” CCOOL = 250 Kč/GJ
Costs for salaries will be the same for both variants. Costs of maintaining, deprecia-
tion etc. will be supposed as some percent of capital costs. Therefore we can neglect
them and we can calculate only with these two costs.
Syrup pre-cooled in the HE must be after-cooled in another cooler with refrigerant to
a needed temperature 10 °C with ice water from a cooling circuit – e.g. with an am-
monia compressor.
Additional costs for heat transfer area for variant with 2 sections
CCA = A * CA = (42,7 – 30,9) * 8000 = 94 400,- Kč
document.doc 27 / 59 Print date: 8 / 2007 P. Hoffman
This value represents higher capital cost.
Additional costs for maintaining and depreciation
We assume a depreciation time 12 years. Corresponding depreciation rate is c. 8
%. Cost for maintaining we assume as c. 4 % of acquisition cost. Then these ad-
ditional costs for maintaining and depreciation are
CMD = CCA * (D + M) = 94 000 * (8 + 4) / 100 = 11 330,- Kč/year
Amount of energy saved for 2 nd step of syrup cooling
QCOOL = Q2 – Q1 = 205,0 – 190,3 = 14,7 kW
Amount of saved energy (cool) per year
Line working time = 200 days / year = 200 * 24 = 4 800 h/y
QCOOLY = QCOOL *
QCOOLY = 14,7 * 4800 * 3600 / 1 000 000 = 254,0 GJ / year
Cost of saved cool per year
CCCOOLY = QCOOLY * CCOOL = 254,0 * 250 = 63 500,- Kč/year
document.doc 28 / 59 Print date: 8 / 2007 P. Hoffman
Annual profit from the variant with 2 sections
AP = CCCOOLY - CMD = 63 500 – 11 330 = 52 170,- Kč/year
Simple rate of return of the investment
SRR = CCA / AP = 94 400 / 52 170 = 1,81 years
document.doc 29 / 59 Print date: 8 / 2007 P. Hoffman
Example of centrifugal pump flow control by throttling
Design of plate HE water / syrup - example DPE - 1
Given data: Water and syrup flowrates, cooling water inlet and outlet temperatures , syrup inlet temperature
Specify: Available syrup outlet temperature, number andconfiguration of plates1st variant = t11,t12,M1,t21,M2 t22 = ?
Given dataType of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers
1 section given revise data given data
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35 35 - outlet temperature-ev. change acc. results t12 (oC) = 48 46,0 temp. calculated from HE balance t12calc (oC) = 45,8 45,9
- flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 15000 15000 - max. allowable pressure loss p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup - inlet temperature t21 (oC) = 80 80 - outlet temperature - calculated t22 (oC) = 42,3 40,9 (estimation for 1.iteration - change till be the same) 41,0 40,9 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 200 200 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
Connection (parallel or counter flow): counter fl. counter fl.
M2 - syrup
M1 - water
t22
t21
t12
t11
Temperature course in 1 section
tHEmin
Connection of 1 section of HE
syrup
water syrup
water
1.section
document.doc 30 / 59 Print date: 8 / 2007 P. Hoffman
Note: In HE inflow all cooling water and cooled hot syrup
Order: Example of DPEVarious variants of design - 1 sections
Made by: P. HoffmanDate april 2003
CONSTANTS FOR CALCULATIONS
- gap between plates s (mm) = 4 4 - channel width b (mm) = 418 418 - channel length L (mm) = 1220 1220 - ekvivalent diameter of channel de (m) = 0,0079 0,0079
- flow clear area of channel f (m2) = 0,00167 0,00167 - heat transfer area of 1 plate A (m2) = 0,435 0,435
- recommended speed - medium 1: water w1rec (m/s) = 0,25 0,25 - medium 2: syrup w2rec (m/s) = 0,25 0,25
- relations for calculation of physical parameters are from my sources (for syrup find in handbooks - for water program calculates automatically)
- minimal economical temperature difference in HE 3 3tminHE = t22-t11 tminHE (°C) =
- maximal allowable pressure in HE pmax (kPa) = 300 300
- criteria equations for calculations of heat transfer coefficient and pressure loss must be choosen according type of plates
- plate thickness sp (mm) = 1,3 1,3
- thermal conductivity of plate p (W/mK) = 15 15
- mass of 1 plate Mp (kg) = 5,4 5,4 - mass of frame Mfr (kg) = 200 200 (choose from producer data)
- heat loss in HE z (%) = 0 0 (in % transfered heat)
document.doc 31 / 59 Print date: 8 / 2007 P. Hoffman
HE CALCULATION FOR GIVEN DATA
1. Specification of physical parameters for both media
- Medium 1: water from water from - mean temperature t1 (oC) = 41,5 40,5 - specific heat c1 (kJ/kg.K) = 4,177 4,177 - density 1 (kg/m3) = 993,5 993,8 - thermal conductivity 1 (W/m.K) = 0,6292069 0,62793753 - dynamic viscosity 1 (Pa.s) = 0,000637 0,000649 - kinematic viscosity 1 (m2/s) = 6,41E-07 6,53E-07 - Prandtl number Pr1 (-) = 4,17 4,25
- Medium 2 (for 1. iteration): syrup syrup - mean temperature t2 (oC) = 60,5 60,4 - specific heat c2 (kJ/kg.K) = 2,670 2,610
(ev. change according results of iter.)
2. Heat input of the HE
Q1 = M1*c1*(t12-t11)/3600 Q1 (kW) = 226,3 191,5Q2 = 100*Q1/(100-z) Q2 (kW) = 226,3 191,5t22 = t21-(t12-t11)*M1*c1/(M2*c2)1.iteration t22 (oC) = 34,50 40,62
unreal given data!! 40,6Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 47,0 46,7 It is necessary to change given data !!!!next calculations in the column are not good - new calc. are in the right column !!
- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 57,2 60,3 - specific heat c2 (kJ/kg.K) = 2,670 2,610 - density 2 (kg/m3) = 1353,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3910 0,3850 - dynamic viscosity 2 (Pa.s) = 0,025000 0,063000 - kinematic viscosity 2 (m2/s) = 1,85E-05 4,63E-05 - Prandtl number Pr2 (-) = 170,72 427,09
document.doc 32 / 59 Print date: 8 / 2007 P. Hoffman
3. Liquids (media) speeds
calculated speed in a case of flow only through 1 interplate channel w1c = M1/(3600*1*f) w1c (m/s) = 2,51 2,51w2c = M2/(3600*2*f) w2c (m/s) = 0,82 0,82
calculated number of parallel channels in 1 pass npc1c = w1c/w1rec npc1c (-) = 10,03 10,03npc2c = w2c/w2rec npc2c (-) = 3,29 3,28
choosen number of parallel channel in 1 pass npc1 water from npc1 (-) = 6 6npc2 syrup npc2 (-) = 3 3
speeds in interplate channelsw1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,418 0,418w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,274 0,273
4. Specification of heat passage coefficient
Re1 = w1*de/1 water Re1 (-) = 5165,603 5071,93755
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 56,53 56,18
1 = Nu1*1/de 1 (W/m2*K) = 4489 4452
Re2 = w2*de/2 syrup Re2 (-) = 117,69 46,70
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,087 9,191
2 = Nu2*2/de 2 (W/m2*K) = 646 447
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 524 385
5. Specification of number of plates in HE (section)
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 7,82 15,76
document.doc 33 / 59 Print date: 8 / 2007 P. Hoffman
calculated total heat transfer area of HEATC = Q1*1000/(k*dtlog) ATC (m2) = 55,18 31,59
calculated number of plates in HE (section)nPT = ATC/A1 nPT (-) = 126,8 72,6
calculated number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 10,57 6,05
nps2c = nPT/(2*npc2) nps2c (-) = 21,14 12,10
choosen number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 5 6
nps2c = nPT/(2*npc2) nps2c (-) = 10 12
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate
nps1*npc1 = 30 36nps2*npc2 = 30 36
total number of plates in HE (section)np = nps1*npc1+nps2*npc2+1 np (-) = 61,0 73,0
Total heat transfer area oh HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 30,9
6. Specification of pressure losses
Medium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x x (-) = 1,1 1,1(x=correction according experiments Eu1 (-) = 48,19 48,42 incl. fouling)pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 8,37 8,40
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 41,8 50,4
maximal allowable pressure loss p1max = 150 150
document.doc 34 / 59 Print date: 8 / 2007 P. Hoffman
Medium 2 syrup syrup
Eu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 130,77 166,91 ( x=correction)pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 16,92
total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 133,3 203,0
maximal allowable pressure loss p2max = 200 200
maximal allowable pressure in HE (kPa) pmax = 300
If there is a high difference between calculated and allowable pressure lossesit is good to change HE arrangement (numbers of parallel channels) andrepeat the calculation.
NEW CALCULATION - HE maximal output - specification of max. heating of 2.medium t22 for choosen arrangement
- specification of max. cooling of 1.medium t12 for choosen arrangement
newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)npc1 water from npc1 (-) = 6 6npc2 syrup npc2 (-) = 3 3
speeds in passes (channels)w1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,42 0,42w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,27 0,27
newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1*npc1 = 5 6
nps2 = nPT/(2*npc2) nps2*npc2 = 10 12
document.doc 35 / 59 Print date: 8 / 2007 P. Hoffman
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate
nps1*npc1 = 30 36nps2*npc2 = 30 36
total number of plates in HEnp = nps1*npc1+nps2*npc2+1 np (-) = 61 73
Total heat transfer area oh HE (section)
AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 30,9
Specification of heat transfer passage
Re1 = w1*de/1 Re1 (-) = 5165,6 5071,9
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 56,53 56,18
1 = Nu1*1/de 1 (W/m2*K) = 4489 4452
Re2 = w2*de/2 Re2 (-) = 117,69 46,70
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,09 9,19
2 = Nu2*2/de 2 (W/m2*K) = 646 447
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 524 385
Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12) don´t iterate
unrealistic given data
estimation of t22 for i- iteration t22i (°C) = 42,50 40,86calculated value of t22 t22i+1 (°C) = 42,26 40,86estimation of t12 for i- iteration t12i (°C) = 56,66 45,94calculated value of t12 t12i+1 (°C) = 45,78 45,93
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 14,0 16,0
Max. amount of transfered heat in HEQ=k*AHE*tlog Q (kW) = 187,7 190,3
Temperature of cooled medium 2 t22i+1 (°C) = 42,3 40,9Temperature of heated medium 1 t12i+1 (°C) = 45,8 45,9
document.doc 36 / 59 Print date: 8 / 2007 P. Hoffman
Pressure losses checking
Medium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 48,19 48,42
pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 8,37 8,40
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 41,8 50,4
Medium 2 syrup syrup
Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 130,77 166,91pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) =
13,33 16,92total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 133,3 203,0
comparison of calculated and maximal allowed pressure losses (kPa)calculated max. allowedp1T = 41,8 p1max = 150 150p2T = 133,3 p2max = 200 200
maximal allowable pressure in HE (kPa)pmax = 300
If pressure losses are higher than maximal allowable we must increase numberof parallel channels npc1 or npc2 and repeat the procedure.It they are too lower we can reduce the number of parallel channels and repeat the calculations.
If results are OK it is possible to print results.
Print of results
document.doc 37 / 59 Print date: 8 / 2007 P. Hoffman
GIVEN DATA
Type of HE: CHX 1000Section: Plate cooler of syrup with cooling water from cooling towers
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 45,8 45,9 - flowrate M1 (kg/h) = 15000 15000 - max. allowable pressure lost p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup
- inlet temperature t21 (oC) = 80 80 - outlet temperature t22 (oC) = 42,3 40,9 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure lost p2max (kPa) = 200 200 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
- heat loss in HE (in % transfered heat) 0 0
Connection (parallel or counter flow): counter fl. counter fl.
Note: In HE inflow all cooling water and cooled hot syrup0
Customer:
Action: Example of DPEOrder No.: Various variants of design - 1 sections
Made by: P. HoffmanDate: april 2003
document.doc 38 / 59 Print date: 8 / 2007 P. Hoffman
CALCULATED VALUES
Medium 1 water from water fromcol. tow. col. tow.
Number of parallel channels in 1 pass npc1 (-) = 6 6Number of passes in series nps1 (-) = 5 6Liquid speed w1 (m/s) = 0,42 0,42Heat transfer coefficient 1 (W/m2*K) = 4489 4452Total pressure loss in HE (section) p1T (kPa) = 42 50
Medium 2 syrup syrup
Number of parallel channels in 1 pass npc2 (-) = 3 3Number of passes in series nps2 (-) = 10 12Liquid speed w2 (m/s) = 0,27 0,27Heat transfer coefficient 2 (W/m2*K) = 646 447Total pressure loss in HE (section) p2T (kPa) = 133 203
Heat passage coefficient k(W/m2*K) = 524 385Mean log. temperature difference tlog (oC) = 13,95 16,02Total number of plates in HE (section) np (-) = 61 73Total heat transfer area of HE (section) AHE (m2) = 25,67 30,89
Max. allowable pressure in HE (section) pmax (kPa) = 300 300
Plates (channels) arrangement npc nps medium 1 - water from cooling towers 6 * 5
------------------------------------------------------------------------------------------------ medium 2 - syrup 3 * 10
Mass of plates in HE (section) (kg) 394,2Mass of HE (section) with frame (kg) 594,2
Total mass of full HE (section) (kg) 767,1Mass of liquids in HE (section) (kg) 172,9
document.doc 39 / 59 Print date: 8 / 2007 P. Hoffman
Design of plate HE water / syrup - example DPE - 2
Given data: Water and syrup flowrates, cooling water inlet and outlet temperatures , syrup inlet temperature
Specify: Available syrup outlet temperature, number andconfiguration of plates1st variant = t11,t12,M1,t21,M2 t22 = ?
Given dataType of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers
2 sections 1.section 2.section
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35,0 35,0 - outlet temperature-ev. change acc. results t12 (oC) = 56,1 37,5 temp. calculated from HE balance t12calc (oC) = 56,0 37,6
- flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 7500 7500 - max. allowable pressure loss p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup - inlet temperature t21 (oC) = 80,0 43,0 - outlet temperature - calculated t22 (oC) = 43,1 38,5 (estimation for 1.iteration - change till be the same) 43,0 38,5 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 150 150 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
Connection (parallel or counter flow): counter fl. counter fl.
syrupsyrup
water water
voda
1.section 2.section
Connection of 2 sections in HE
t122
M2 - syrup
M1 - water
t221
t21
t121
t11
Temperature course in 2 sections
tHEmin
t222
t11
document.doc 40 / 59 Print date: 8 / 2007 P. Hoffman
Note: Both coolers sections are connected in series (syrup outlet from 1.section is inlet to 2.); both sections are cooled withthe same water - 50 % goes in 1. and 50 % in 2.section)
Order: Example of DPEVarious variants of design - 2 sections - syrup in series, cooling water in parallel
Made by: P. HoffmanDate april 2003
CONSTANTS FOR CALCULATIONS
- gap between plates s (mm) = 4 4 - channel width b (mm) = 418 418 - channel length L (mm) = 1220 1220 - ekvivalent diameter of channel de (m) = 0,0079 0,0079
- flow clear area of channel f (m2) = 0,00167 0,00167 - heat transfer area of 1 plate A (m2) = 0,435 0,435
- recommended speed - medium 1: water w1rec (m/s) = 0,25 0,25 - medium 2: syrup w2rec (m/s) = 0,25 0,25
- relations for calculation of physical parameters are from my sources (for syrup find in handbooks - for water program calculates automatically)
- minimal economical temperature difference in HEtminHE = t22-t11 tminHE (°C) = 3 3
- maximal allowable pressure in HE pmax (kPa) = 300 300
- criteria equations for calculations of heat transfer coefficient and pressure loss must be choosen according type of plates
- plate thickness sp (mm) = 1,3 1,3
- thermal conductivity of plate p (W/mK) = 15 15
- mass of 1 plate Mp (kg) = 5,4 5,4 - mass of frame Mfr (kg) = 200 200 (choose from producer data)
- heat loss in HE z (%) = 0 0 (in % transfered heat)
document.doc 41 / 59 Print date: 8 / 2007 P. Hoffman
HE CALCULATION FOR GIVEN DATA
1. Specification of physical parameters for both media
- Medium 1: water from water fromcol. tow. col. tow.
- mean temperature t1 (oC) = 45,6 36,3 - specific heat c1 (kJ/kg.K) = 4,179 4,176 - density 1 (kg/m3) = 992,0 995,3 - thermal conductivity 1 (W/m.K) = 0,6342123 0,6223785 - dynamic viscosity 1 (Pa.s) = 0,000593 0,000703 - kinematic viscosity 1 (m2/s) = 5,98E-07 7,07E-07 - Prandtl number Pr1 (-) = 3,86 4,64
- Medium 2 (for 1. iteration): syrup syrup - mean temperature t2 (oC) = 61,5 40,8 - specific heat c2 (kJ/kg.K) = 2,670 2,610
(ev. change according results of iter.)
2. Heat input of the HE
Q1 = M1*c1*(t12-t11)/3600 Q1 (kW) = 183,7 21,7Q2 = 100*Q1/(100-z) Q2 (kW) = 183,7 21,7t22 = t21-(t12-t11)*M1*c1/(M2*c2)1.iteration t22 (oC) = 43,06 38,53
43,06 38,53Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 59,0 37,8
- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 61,5 40,8 - specific heat c2 (kJ/kg.K) = 2,670 2,610 - density 2 (kg/m3) = 1353,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3910 0,3850 - dynamic viskosity 2 (Pa.s) = 0,025000 0,063000 - kinematic viskosita 2 (m2/s) = 1,85E-05 4,63E-05 - Prandtl number Pr2 (-) = 170,72 427,09
document.doc 42 / 59 Print date: 8 / 2007 P. Hoffman
3. Liquids (media) speeds
calculated speed in a case of flow only through 1 interplate channel w1c = M1/(3600*1*f) w1c (m/s) = 1,26 1,25w2c = M2/(3600*2*f) w2c (m/s) = 0,82 0,82
calculated number of parallel channels in 1 pass npc1c = w1c/w1rec npc1c (-) = 5,02 5,01npc2c = w2c/w2rec npc2c (-) = 3,29 3,28
choosen number of parallel channel in 1 pass npc1 water from npc1 (-) = 6 5npc2 syrup npc2 (-) = 3 4
speeds in interplate channelsw1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,209 0,250w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,274 0,205
4. Specification of heat passage coefficient
Re1 = w1*de/1 water Re1 (-) = 2775,4137 2807,9556
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 33,07 36,16
1 = Nu1*1/de 1 (W/m2*K) = 2647 2840
Re2 = w2*de/2 syrup Re2 (-) = 117,69 35,03
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,087 7,283
2 = Nu2*2/de 2 (W/m2*K) = 646 354
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2)
k (W/m2*K) = 485 302
5. Specification of number of plates in HE (section)
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 14,57 4,44
document.doc 43 / 59 Print date: 8 / 2007 P. Hoffman
calculated total heat transfer area of HEATC = Q1*1000/(k*dtlog) ATC (m2) = 26,01 16,24
calculated number of plates in HE (section)nPT = ATC/A1 nPT (-) = 59,8 37,3
calculated number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 4,98 3,73
nps2c = nPT/(2*npc2) nps2c (-) = 9,96 4,67
choosen number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 5 4
nps2c = nPT/(2*npc2) nps2c (-) = 10 5
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate
nps1*npc1 = 30 20nps2*npc2 = 30 20
total number of plates in HE (section)np = nps1*npc1+nps2*npc2+1 np (-) = 61,0 41,0
Total heat transfer area oh HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 17,0
in the 1.section is syrup cooled of c. 37 °C in the 2.section is syrup cooled of c. 4,6 °C S = 42,6
6. Specification of pressure losses
Medium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x x (-) = 1,1 1,1(x=correction according experiments Eu1 (-) = 56,77 56,60 incl. fouling)pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 2,47 3,53
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 12,3 14,1
document.doc 44 / 59 Print date: 8 / 2007 P. Hoffman
maximal allowable pressure loss p1max = 150 150
Medium 2 syrup syrupEu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 130,77 180,08 ( x=correction)pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 10,27
total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 133,3 51,3
maximal allowable pressure loss p2max = 150 150
maximal allowable pressure in HE pmax = 300
If there is a high difference between calculated and allowable pressure lossesit is good to change HE arrangement (numbers of parallel channels) andrepeat the calculation.
NEW CALCULATION - HE maximal output - specification of max. heating of 2.medium t22 for choosen arrangement
- specification of max. cooling of 1.medium t12 for choosen arrangement
newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)npc1 water from npc1 (-) = 6 5npc2 syrup npc2 (-) = 3 4
speeds in passes (channels)w1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,21 0,25w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,27 0,20
newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1 (-) = 5 4
nps2 = nPT/(2*npc2) nps2 (-) = 10 5
document.doc 45 / 59 Print date: 8 / 2007 P. Hoffman
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate
nps1*npc1 = 30 20nps2*npc2 = 30 20
total number of plates in HEnp = nps1*npc1+nps2*npc2+1 np (-) = 61 41
Total heat transfer area of HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 17,0
Specification of heat transfer passage
Re1 = w1*de/1 Re1 (-) = 2775,4 2808,0
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 33,07 36,16
1 = Nu1*1/de 1 (W/m2*K) = 2647 2840
Re2 = w2*de/2 Re2 (-) = 117,69 35,03
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,09 7,28
2 = Nu2*2/de 2 (W/m2*K) = 646 354
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 485 302
Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12)
estimation of t22 for i- iteration t22i (°C) = 43,20 38,42calculated value of t22 t22i+1 (°C) = 43,20 38,42estimation of t12 for i- iteration t12i (°C) = 56,01 37,56calculated value of t12 t12i+1 (°C) = 56,02 37,56
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 14,7 4,4
document.doc 46 / 59 Print date: 8 / 2007 P. Hoffman
Max. amount of transfered heat in HEQ=k*AHE*tlog Q (kW) = 183,0 22,3
Temperature of cooled medium 2 t22i+1 (°C) = 43,2 38,4Temperature of heated medium 1 t12i+1 (°C) = 56,0 37,6
Pressure losses checking
Medium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 56,77 56,60
pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 2,47 3,53
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 12,3 14,1
Medium 2 syrup syrup
Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 130,77 180,08pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 10,27
total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 133,3 51,3
comparison of calculated and maximal allowed pressure losses (kPa)calculated max. allowedp1T = 12,3 p1max = 150 150p2T = 133,3 p2max = 150 150
maximal allowable pressure in HE (kPa)pmax = 300
If pressure losses are higher than maximal allowable we must increase numberof parallel channels npc1 or npc2 and repeat the procedure.It they are too lower we can reduce the number of parallel channels and repeat the calculations.
If results are OK it is possible to print results.
document.doc 47 / 59 Print date: 8 / 2007 P. Hoffman
Print of results
GIVEN DATA t22 = ?
Type of HE: CHX 1000Section: Plate cooler of syrup with cooling water from cooling towers
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 56,0 37,6 - flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 7500 7500 - max. allowable pressure lost p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup
- inlet temperature t21 (oC) = 80 43 - outlet temperature t22 (oC) = 43,2 38,4 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure lost p2max (kPa) = 150 150 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
- heat loss in HE (in % transfered heat) 0 0 Connection (parallel or counter flow): counter fl. counter fl.
Note: Both coolers sections are connected in series (syrup outlet the same water - 50 % goes in 1. and 50 % in 2.section)
Customer:Action: Example of DPEOrder No.: Various variants of design - 2 sections - syrup in series,
cooling water in parallel
Made by: P. HoffmanDate: april 2003
document.doc 48 / 59 Print date: 8 / 2007 P. Hoffman
CALCULATED VALUES
Medium 1 water from water fromcol. tow. col. tow.
Number of parallel channels in 1 pass npc1 (-) = 6 5Number of passes in series nps1 (-) = 5 4Liquid speed w1 (m/s) = 0,21 0,25Heat transfer coefficient 1 (W/m2*K) = 2647 2840Total pressure loss in HE (section) p1T (kPa) = 12 14
Medium 2 syrup syrupNumber of parallel channels in 1 pass npc2 (-) = 3 4Number of passes in series nps2 (-) = 10 5Liquid speed w2 (m/s) = 0,27 0,20Heat transfer coefficient 2 (W/m2*K) = 646 354Total pressure loss in HE (section) p2T (kPa) = 133 51
Heat passage coefficient k(W/m2*K) = 485 302Mean log. temperature difference tlog (oC) = 14,71 4,35Total number of plates in HE (section) np (-) = 61 41Total heat transfer area of HE (section) AHE (m2) = 25,67 16,97
Max. allowable pressure in HE (section) pmax (kPa) = 300 300
Plates (channels) arrangement npc nps medium 1 - water from cooling towers 6 * 5
----------------------------------------------------------------------------------------------- medium 2 - syrup 3 * 10
Mass of plates in HE (section) (kg) 329,4 221,4Mass of HE (section) with frame (kg) 529,4 421,4
Total mass of full HE (section) (kg) 672,9 517,5Mass of liquids in HE (section) (kg) 143,5 96,1
document.doc 49 / 59 Print date: 8 / 2007 P. Hoffman
Design of plate HE water / syrup - example DPE - 3
Given data: Water and syrup flowrates, cooling water inlet and outlet temperatures , syrup inlet temperature
Specify: Available syrup outlet temperature, number andconfiguration of plates1st variant = t11,t12,M1,t21,M2 t22 = ?
Given data
Type of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers
effect of pressure losses < pL > pL
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35 35 - outlet temperature-ev. change acc. results t12 (oC) = 46,0 46,0 temp. calculated from HE balance t12calc (oC) = 46,0 46,0
- flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 15000 15000 - max. allowable pressure loss p1max (kPa) = 100 200 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup
- inlet temperature t21 (oC) = 80 80 - outlet temperature - calculated t22 (oC) = 40,6 40,6 (estimation for 1.iteration - change till be the same) 40,7 40,7 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 100 250 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
Connection of 1 section of HE
syrup
water syrup
water
1.section
M2 - syrup
M1 - water
t22
t21
t12
t11
Temperature course in 1 section
tHEmin
document.doc 50 / 59 Print date: 8 / 2007 P. Hoffman
Connection (parallel or counter flow): counter fl. counter fl.
Note: In HE inflow all cooling water and cooled hot syrup
Order: Example of DPEVarious variants of design - 1 sections
Made by: P. HoffmanDate april 2003
CONSTANTS FOR CALCULATIONS
- gap between plates s (mm) = 4 4 - channel width b (mm) = 418 418 - channel length L (mm) = 1220 1220 - ekvivalent diameter of channel de (m) = 0,0079 0,0079
- flow clear area of channel f (m2) = 0,00167 0,00167 - heat transfer area of 1 plate A (m2) = 0,435 0,435
- recommended speed - medium 1: water w1rec (m/s) = 0,20 0,60 - medium 2: syrup w2rec (m/s) = 0,20 0,60
- relations for calculation of physical parameters are from my sources (for syrup find in handbooks - for water program calculates automatically)
- minimal economical temperature difference in HEtminHE = t22-t11 tminHE (°C) = 3 3
- maximal allowable pressure in HE pmax (kPa) = 300 300
- criteria equations for calculations of heat transfer coefficient and pressure loss must be choosen according type of plates
- plate thickness sp (mm) = 1,3 1,3
- thermal conductivity of plate p (W/mK) = 15 15
- mass of 1 plate Mp (kg) = 5,4 5,4 - mass of frame Mfr (kg) = 200 200 (choose from producer data)
- heat loss in HE z (%) = 0 0 (in % transfered heat)
document.doc 51 / 59 Print date: 8 / 2007 P. Hoffman
HE CALCULATION FOR GIVEN DATA
1. Specification of physical parameters for both media
- Medium 1: water from water from - mean temperature t1 (oC) = col. tow. col. tow. - specific heat c1 (kJ/kg.K) = 40,5 40,5 - density 1 (kg/m3) = 4,177 4,177 - thermal conductivity 1 (W/m.K) = 993,8 993,8 - dynamic viscosity 1 (Pa.s) = 0,6279375 0,62793753 - kinematic viscosity 1 (m2/s) = 0,000649 0,000649 - Prandtl number Pr1 (-) = 6,53E-07 6,53E-07 - Prandtlovo číslo Pr1 (-) = 4,25 4,25
- Medium 2 (for 1. iteration): syrup syrup - mean temperature t2 (oC) = 60,4 60,4 - specific heat c2 (kJ/kg.K) = 2,610 2,610
(ev. change according results of iter.)
2. Heat input of the HE
Q1 = M1*c1*(t12-t11)/3600 Q1 (kW) = 191,5 191,5Q2 = 100*Q1/(100-z) Q2 (kW) = 191,5 191,5t22 = t21-(t12-t11)*M1*c1/(M2*c2)1.iteration t22 (oC) = 40,62 40,62
40,6 40,6Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 46,7 46,7
- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 60,3 60,3 - specific heat c2 (kJ/kg.K) = 2,610 2,610 - density 2 (kg/m3) = 1360,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3850 0,3850 - dynamic viskosity 2 (Pa.s) = 0,063000 0,063000 - kinematic viskosita 2 (m2/s) = 4,63E-05 4,63E-05 - Prandtl number Pr2 (-) = 427,09 427,09
document.doc 52 / 59 Print date: 8 / 2007 P. Hoffman
3. Liquids (media) speeds
calculated speed in a case of flow only through 1 interplate channel w1c = M1/(3600*1*f) w1c (m/s) = 2,51 2,51w2c = M2/(3600*2*f) w2c (m/s) = 0,82 0,82
calculated number of parallel channels in 1 pass npc1c = w1c/w1rec npc1c (-) = 12,54 4,18npc2c = w2c/w2rec npc2c (-) = 4,10 1,37
choosen number of parallel channel in 1 pass npc1 water from npc1 (-) = 11 4npc2 syrup npc2 (-) = 5 3
speeds in interplate channelsw1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,228 0,627w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,164 0,273
4. Určení součinitele prostupu tepla
4. Specification of heat passage coefficient
Re1 = w1*de/1 water Re1 (-) = 2766,5114 7607,90633
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 34,40 77,99
1 = Nu1*1/de 1 (W/m2*K) = 2726 6180
Re2 = w2*de/2 syrup Re2 (-) = 28,02 46,70
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 6,080 9,191
2 = Nu2*2/de 2 (W/m2*K) = 295 447
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 257 394
document.doc 53 / 59 Print date: 8 / 2007 P. Hoffman
5. Specification of number of plates in HE (section)
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 15,76 15,76
calculated total heat transfer area of HEATC = Q1*1000/(k*dtlog) ATC (m2) = 47,23 30,83
calculated number of plates in HE (section)nPT = ATC/A1 nPT (-) = 108,6 70,9
calculated number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 4,94 8,86
nps2c = nPT/(2*npc2) nps2c (-) = 10,86 11,81
choosen number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 5 9
nps2c = nPT/(2*npc2) nps2c (-) = 11 12
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate
nps1*npc1 = 55 36nps2*npc2 = 55 36
total number of plates in HE (section)np = nps1*npc1+nps2*npc2+1 np (-) = 111,0 73,0
Total heat transfer area oh HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 47,4 30,9
6. Specification of pressure losses
Médium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x x (-) = 1,1 1,1(x=correction according experiments Eu1 (-) = 56,82 43,50 incl. fouling)pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 2,93 16,99
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 14,7 152,9
document.doc 54 / 59 Print date: 8 / 2007 P. Hoffman
maximal allowable pressure loss p1max = 100 200
Medium 2 syrup syrup
Eu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 191,01 166,91 ( x=correction)
pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 6,97 16,92
total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 76,7 203,0
maximal allowable pressure loss p2max = 100 250
maximal allowable pressure in HE (kPa) pmax = 300
If there is a high difference between calculated and allowable pressure lossesit is good to change HE arrangement (numbers of parallel channels) andrepeat the calculation.
NEW CALCULATION - HE maximal output - specification of max. heating of 2.medium t22 for choosen arrangement
- specification of max. cooling of 1.medium t12 for choosen arrangement
newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)ppk1 water from ppk1 (-) = 11 4ppk2 syrup ppk2 (-) = 5 3
speeds in passes (channels)w1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,23 0,63w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,16 0,27
newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1*npc1 = 5 9
nps2 = nPT/(2*npc2) nps2*npc2 = 11 12
document.doc 55 / 59 Print date: 8 / 2007 P. Hoffman
Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate nps1*npc1 = 55 36
nps2*npc2 = 55 36
total number of plates in HEnp = nps1*npc1+nps2*npc2+1 np (-) = 111 73
Total heat transfer area oh HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 47,4 30,9
Specification of heat transfer passage
Re1 = w1*de/1 Re1 (-) = 2766,5 7607,9
Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 34,40 77,99
1 = Nu1*1/de 1 (W/m2*K) = 2726 6180
Re2 = w2*de/2 Re2 (-) = 28,02 46,70
Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 6,08 9,19
2 = Nu2*2/de 2 (W/m2*K) = 295 447
k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 257 394
Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12)
estimation of t22 for i- iteration t22i (°C) = 40,58 40,60calculated value of t22 t22i+1 (°C) = 40,58 40,59estimation of t12 for i- iteration t12i (°C) = 46,01 46,01calculated value of t12 t12i+1 (°C) = 46,01 46,01
tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 15,7 15,7
Max. amount of transfered heat in HEQ=k*AHE*tlog Q (kW) = 191,6 191,6
document.doc 56 / 59 Print date: 8 / 2007 P. Hoffman
Temperature of cooled medium 2 t22i+1 (°C) = 40,6 40,6Temperature of heated medium 1 t12i+1 (°C) = 46,0 46,0
Pressure losses checking
Medium 1 water from water fromcol. tow. col. tow.
Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 56,82 43,50
pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 2,93 16,99
total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 14,7 152,9
Medium 2 syrup syrup
Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 191,01 166,91
pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 6,97 16,92
total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 76,7 203,0
comparison of calculated and maximal allowed pressure losses (kPa)calculated max. allowedp1T = 14,7 p1max = 100 200p2T = 76,7 p2max = 100 250
maximal allowable pressure in HE (kPa)pmax = 300
If pressure losses are higher than maximal allowable we must increase numberof parallel channels npc1 or npc2 and repeat the procedure.It they are too lower we can reduce the number of parallel channels and repeat the calculations.
document.doc 57 / 59 Print date: 8 / 2007 P. Hoffman
If results are OK it is possible to print results.
Print of results
GIVEN DATA
Type of HE: CHX 1000Section: Plate cooler of syrup with cooling water from cooling towers
Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.
- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 46,0 46,0 - flowrate M1 (kg/h) = 15000 15000 - max. allowable pressure lost p1max (kPa) = 100 200 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2
Medium 2 - cooled: syrup syrup
- inlet temperature t21 (oC) = 80 80 - outlet temperature t22 (oC) = 40,6 40,6 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure lost p2max (kPa) = 100 250 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2
- heat loss in HE (in % transfered heat) 0 0
Connection (parallel or counter flow): counter fl. counter fl.
Note: In HE inflow all cooling water and cooled hot syrup0
Customer:Action: Example of DPEOrder No.: Various variants of design - 1 sections
Made by: P. HoffmanDate: april 2003
document.doc 58 / 59 Print date: 8 / 2007 P. Hoffman
CALCULATED VALUES
Medium 1 water from water fromcol. tow. col. tow.
Number of parallel channels in 1 pass npc1 (-) = 11 4Number of passes in series nps1 (-) = 5 9Liquid speed w1 (m/s) = 0,23 0,63Heat transfer coefficient 1 (W/m2*K) = 2726 6180Total pressure loss in HE (section) p1T (kPa) = 15 153
Medium 2 syrup syrup
Number of parallel channels in 1 pass npc2 (-) = 5 3Number of passes in series nps2 (-) = 11 12Liquid speed w2 (m/s) = 0,16 0,27Heat transfer coefficient 2 (W/m2*K) = 295 447Total pressure loss in HE (section) p2T (kPa) = 77 203
Heat passage coefficient k(W/m2*K) = 257 394Mean log. temperature difference tlog (oC) = 15,72 15,74Total number of plates in HE (section) np (-) = 111 73Total heat transfer area of HE (section) AHE (m2) = 47,42 30,89
Max. allowable pressure in HE (section) pmax (kPa) = 300 300
Plates (channels) arrangement npc nps medium 1 - water from cooling towers=G33411 * 5
---------------------------------------------------------------------- medium 2 - syrup 5 * 11
Mass of plates in HE (section) (kg) 599,4 394,2Mass of HE (section) with frame (kg) 799,4 594,2
Total mass of full HE (section) (kg) 1063,5 767,1Mass of liquids in HE (section) (kg) 264,1 172,9
document.doc 59 / 59 Print date: 8 / 2007 P. Hoffman