Introduction:-
Final design is a choice of optimum between the theoretical formulation results and the practical field implementation
Designing Methods: -
There are three basic methods used for designing purpose
1. Kern Method
2. Bell Delaware Method
3. Stream Analysis Method
Kern Method: -
This is the simplest method used for designing of heat exchangers. In it design is based upon analogous equations, which are solved to determine different parameters relating to design. Main disadvantage of this method id that it is unable to determine the heat transfer take place due to leakages and baffles cuts etc.
Bell Delaware Method: -
To overcome on the disadvantage of Kern Method, Bell Delaware Method was introduce in it design is based upon experimental data available. In it different standards and codes are present based upon which design is made but limitation of this method is that data available for designing is limited and it is impossible to design a product fir very special nature.
Stream Analysis Method: -
By seeing the disadvantages of Kern, Bell Delaware Methods a new method was introduce which was Stream Analysis Method. In it design is based on different graphical analysis as a result of which we become able to design a product that can fulfill our requirement completely but this method it is of very complex nature.
There are also others methods used by Heat Transfer And Flow Streams (HRTI)
In United State and Heat Transfer And Flow Streams (HTFS) research center in United Kingdom but these are of confidential nature and ate not available in literature
General Design Procedure Used For Kern Method: -
1. Define the duty: heat transfer rate, fluid flow rates, temperatures
2. Collect together the fluid physical properties required: density, viscosity, thermal conductivity
3. Decide on the type of exchanger to be used.
4. Select a trial value for the overall coefficient, U
5. Calculate the mean temperature difference LMTD
6. Calculate the area required
7. Decide the exchanger layout
8. Calculate the individual coefficients
9. Calculate the overall coefficient and compare with the trial value. If the calculated value differs significantly from the estimated value, substitute the calculated for the estimated value and return to step 6.
10. Calculate the exchanger pressure drop: if unsatisfactory return to steps 7 or 4 or 3 in that order of preference.
11. Optimize the design: repeat steps 4 to 10 as necessary to determine the cheapest exchanger that will satisfy the duty. Usually this will be the one with the smallest area.
Design Objective: -
Design a Shell and Spiral Coil Heat Exchanger for heating cold water form 25°C to 28°C by using hot water entering at 80°C and leaving at 72°C.
General Calculations: -
Outside dia of Cu-Tube = Do = 1.27 cm
Dia of shell = Ds = 1.24 cm
Length of shell = Ls = 91.44 cm
Inside dia of tube = Di = 101049 cm
Hot Fluid Specifications: -
Inlet Temperature = TH1 = 80 °C
Outlet Temperature = TH2 = 72 °C
Density = ρH = 0.9747 g/cm3
Specific Heat = CpH = 4.193 J/g. °C
Thermal Conductivity = KH = 0.00667
Dynamic Viscosity = µH = 0.00378 g/cm.sec
Hot fluid mass flow rate = m°H = 35 g/sec
Cold Fluid Specifications: -
Inlet temperature = Tc1 = 25°C
Outlet temperature = Tc2 = 28°C
Density = ρc = 0.997 g/cm3
Specific Heat = Cpc = 4.18 J/g. °C
Thermal conductivity = Kc = 0.00607 W/cm. °C
Dynamic Viscosity = µc = 0.00891 g/cm.sec
We know that
Q = m°CpΔT
For Hot Fluid:
QH = m°H CpH ΔTH
Q H = (35)(4.193)(80-72)
QH = 1174.04 J/sec
According to first law of thermodynamics, Energy can neither be created nor destroy however; it can be converted from one form to another form.
So,
QHot= QCold
For Cold Fluid:
Qc = m°c Cpc ΔTc
1174.04 = m°c (4.18)(28-25)
mc = 93.62 g/sec
LMTD: -
ΔT1 = 52°C ΔT2 = 47°C
ΔT1-ΔT2
LMTD = ----------------------
ln(ΔT1/ΔT2)
52-47
LMTD = --------------------
ln(52/47)
=49.458°C
Inside Film Heat Transfer Coefficient (hi) Calculations: -
We Know that
m°H = ρ H qH
qH = 35/0.9747
= 35.908 cm3/sec
Now
qH = uH.AT
35.908
uH = ------------
(д/4)(1.27) 2
uH = 28.35 cm/sec
So,
ρH DT uH
Reynolds No = Re = ------------
µH
(0.9747)(1.27)(28.35)
= --------------------------------
0.00378
= 9282.886
Pr# = Pr = 2.38
Dο-Di
DL = -----------
ln (Dο/Di)
1.27-1.1049
DL = ------------------
ln(1.27/1.1049)
= 1.1855 cm
Now
Pr (Re) 0.8 (Di/DL) 0.1 {1+ 0.098/{Re (Di/DL) 2} 0.2}
Nu = --------------------------------------------------------------
{26.2 (Pr) 2/3 - 0.074}
hi = 0.4041 W/cm2.°C
Outside Film heat Transfer Coefficient (hο) Calculations: -
We know that:
m°c = ρc qc
qc = 93.62/0.997
= 93.91 cm3/sec
Now
qc= uc Ae
De = Ds-DT
= 15.24-1.27
= 13.97 cm
93.91
uc = ----------------
(д/4)(13.97) 2
uc = 0.6126 cm/sec
ρc De uc
Reynolds # = Re = ----------------
µc
(0.9997)(13.97)(0.6126)
= -------------------------------
0.00891
= 958.643
Pr # = Pr = 6.14
Now
Nu = 0.063235(Re) 0.794(Pr) 0.36
hο = 1.2306* 10-2 W/cm2.°C
Fouling Factor: -
1/hdi = 5.3 cm2.°C/W
1/hdο = 5.3 cm2.°C/W
Resistance Due to Thickness: -
Tube Wall Thickness = Xw = 0.1651 cm (16gauge)
Thermal Conductivity of “Cu” = Kw = 3.799 W/cm. K
Xw/Kw = 0.1651/3.799 = 0.04346 cm2. K/W
Overall Heat Transfer Coefficient (U): -
Base on Outside Dia:
1/Uo = 1/ho+Xw/Kw (Do/DL)+1/hi (Do/Di)+1/hdi (Do/Di)+1/hdo
Uo = 1.0467*10-2 W/cm2.°C
Now
Q = Uo A (LMTD)
Q
A = ------------------------
Uo (LMTD)
1174.04
A = -----------------------
(1.0467*10-2)(49.458)
A = 2267.9 cm2
Length of the tube = L = A/дDT
L = 2267.9/д (1.27)
L = 568.4 cm
5% margin, then total length of tube:
= 568.4*1.05
= 596.8 cm
Calculations For # of turns: -
Total length of tube = 596.8 cm
Length fro turning = 576.8 cm
Length of the shell = 91.4 cm
No. of turns = 576.8/д(9.5)
= 19.3 = 19 Turns
Turn spacing = 2.54 cm
Baffle Spacing = 15.24 cm
No. of Baffles = 5
Pressure Drop Calculations : -
Coil Side
4 f Gt 2 ( L/Di )
ΔPcoil =
2 ρH
4 m°H 4 (35)
Gt = ------------ = ---------------- = 36.503 g/see cm2
Д Di Д (1.1049) 2
0.112
0.305{ 1 + ---------------- } √(Do/DL)
{Re (Do/DL)}
f = ---------------------------------------------------------
4 { Re (Do/DL)2 }0.2
f = 0.01235
Now
4 f Gt 2 ( L/Di )
ΔPcoil = ------------------
2 ρH
4(0.01235)(36.503) 2 (596.8/1.1049)
ΔPcoil = --------------------------------------------
2(0.9747)
= 18238.49 g/cm see2
= 1.8238 KN/m2
Shell Side
ΔPshell = 8 Jf (Ds/De) (L/ lb) (ρc us2 ) / 2 ( µ/µ w)-0.14
Jf = 0.05 (Fig 13.20, C & R vol. 6)
L = 596.8 cm
lb = 15.24 cm
us = 0.6126 cm/see
ρc = 0.997 g/cm3
ΔPshell = 3.19677 g/cm see
= 3.196 * 10 –4 KN/m2
Nomenclature
Q | Rate of heat |
m° | Mass Flow rate |
LMTD | Log Mean Temperature Difference |
q | Volumetric Flow Rate |
AT | Area of Tube |
u | Velocity of Fluid |
DT | Dia of Tube |
DL | Log Mean Dia |
hi | Inside Film Heat Transfer Coefficient |
hο | Outside Film heat Transfer Coefficient |
De | Equivalent Dia |
hdi | Inside Fouling Factor |
hdο | Outside Fouling Factor |
A | Heat Transfer Area |
Subscripts | |
H | Hot Fluid |
c | Cold Fluid |
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