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• # Finned-Tube Heat Exchanger Design

Discussion in 'Calculations' started by optymista93, Aug 11, 2021.

1. ### optymista93Member

Joined:
Aug 2021
Posts:
8
0
Hi,

I'm a graduate Mechanical Engineer and currently working on design of finned-tube heat exchanger. (Liquid cooling exhaust gases from diesel engine.)

I found some pre-planning calculations that my previous co-worker did. After reading about heat exchange, I re-arranged and defined some more variables. Since I didn't really learn much about heat exchange at the curses I had on university, I want to ask for explanation of some of the equations. Additionally, it would be great if anyone can confirm that equations used are in fact correct.

Input values are marked with orange color.

Constants for calculation

Density of diesel fuel: ρD = 820 kg/m3

Atmospheric pressure: patm = 1atm = 101,3 kPa

Air Gas Constant: Fin Tube Data (Constants from manufacturer, thus not an input)
Outside Surface Area of bare pipe: Surface Area of one fin, both sides: Pipe exterior surface not covered by fins: Pipe Dimensions
PipeOD = 26,7mm
PipeID = 21,4mm

Engine Data

Temperature Engine Data: TE_Data = 25°C
Engine Speed: n
eng = 2100 rev/min
Engine Power: P
eng = 243 kW
Specific Fuel Consumption: M
SC = 0,220 kg/kW*hr
Combustion Volume Flow: V
Com_air=21,11679 m3/min
Exhaust Volume Flow: V
Exh = 51,8 m3/min
Exhaust Temperature: T
Exh = 430°C

Mass Calculation

Mass Fuel Consumption: M
Diesel = Peng*MSC
Volume Fuel Consumption: VDiesel = Mdiesel / ρD
Combustion air mass flow: MCom_air = (patm*VCom_air)/(Rair*TE_Data)
Total Mass Flow Engine in: Min = Mdiesel + MCom_air
Exhaust Mass Flow: MExh = (patm*VExh)/(Rair*TExh)
Ratio inlet mass/ outlet mass: Ratio
in_out = Min/MExh

Design Limits

Ambient Design Temperature: Tamb = 40°C
Temperature Exhaust Gas Cooler Outlet: T
E_out = 130°C

Temperature Exhaust Corrected for Ambient Design: TE_in = TExh+(Tamb-TE_Data)
Mean Temperature Differance Exhaust: T
E_mean = (TE_in + TE_out)/2

Exhaust Constants

Specific Heat Exhaust @ Mean Temperature Exhaust:
C
pE_mean = Dynamic Viscosity Exhaust @ Mean Temperature Exhaust:
μE_Mean = Thermal Conducitity Exhaust:

k = Faul Factor: FaulFactor = 0,01*[(hr*ft2*R°)/BTU]
C = 0,005*[(hr*ft2*R°)/BTU]

Heat Rejection from Exhaust

∆Temperature Exhaust: ∆TE = TE_in - TE_out
Heat Rejection From Exhaust: QE = CpE_mean * MExh * ∆TE

Coolant Calculation

Coolant Flow Rate: Vcoolant = 250 L/min
Coolant Temperature Inlet: TC_in = 65°C
Water/Glycol Mixture (% Glycol): Mix = 40%

Density of Coolant Fluid: ρcoolant = (1,1*Mix+971,8)
Specific Heat of Coolant Fuild: Cpcoolant = (MF_coolant-0,0128*Mix)*10^3
Mass Flow Coolant: M
F_coolant = Vcoolant * ρcoolant
∆Temperature Coolant Fluid: ∆Tcoolant = QE/(Cpcoolant*MF_coolant)
Coolant Temperature Outlet: TC_out = TC_in + QE/(Cpcoolant*MF_coolant)

Fin Tube Calculation

Number of Tubes First Row: Firstrow = 8
Number of Tubes Second Row: Secondrow = Firstrow-1
Length of Fin Tube: Lfin_tube = 220mm
Number of fins per meter: nf = 118 fins
Fin Height: hf = 9,5mm
Fin Thickness: tf = 1,4 mm

Cross Section Area Calculation

Wfin = (nf*hf*2*tf)/m
Tube footprint area: Atube_fp = Firstrow*(PipeOD+Wfin)*Lfin_tube
Free flow Area: AFF = Aduct - Atube_fp

Temperature Fin Surface Calculation

Average Temperature Inside: Ti = (TC_in+TC_out)/2
Mean Temperature Differance Exhaust: TE_mean = (TE_in + TE_out)/2
Average Temperature Fin: TS = Ti+0,3*(TE_mean-Ti)

Reynold's Number Calculation

Mass Flow unit Area: Gn = MExh/AFF
Renynold's Number: Ren = (Gn*PipeOD)/μE_Mean

Heat Transfer Calculation

Convection Coefficient, hc: Convection Coefficient, ha = 1/(hc^(-1)+FaulFactor)

Fin Efficiency, E = 0,986-0,0068*[(hr*ft2*R°)/BTU]*ha
Final outside convection heat transfer coefficient (metric), hf = (ha*(E*Afo+Apo))/AO
Total design heat transfer coefficient(metric), u = 1/(hf^(-1)+C)

Log Mean Temperature Difference, ∆TLMTD = ([(T_(E_in )-T_(c_out ) )-(T_(E_out )-T_(C_in ) )])/(ln⁡[((T_(E_in )-T_(C_out ) ))/(〖(T〗_(E_out )-T_(C_in )))])

Area Calculation

Total Area Required, Atot = QE/∆TLMTD*u
Total Length of Fin Tube, LFT = Atot/AO
Length of 1 Array, Larray = (Firstrow+Secondrow)*Lfin_tube
Number of Arrays, narrays = LFT/Larray

Efficient fin tubes, according to design = E*(8*Secondrow+6*Firstrow)
Above equation depends on the design.

Pressure Drop Fin Tube Array

Number of rows, Nr = round(narrays)*2
Density at bulk temperature, ρb = patm/(Rair*TE_Mean)
Density of Inlet Exhuast, ρ1 = patm/(Rair*TE_in)
Density of Outlet Exhuast, ρ2 = patm/(Rair*TE_out)

Calculation factor, 1: f1 = 0,083+9,44*Ren^(-0,45)

Calculation factor, 2: f2 = [(1+(A_FF/Aduct )^2)/(4∗Nr )]∗ρb∗[(1/ρ2 )-(1/ρ1 )]

Pressure drop over fin tube: ∆Pfin_tube = (2*(f1+f2)*Gn^2*Nr)/ρb

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3. ### optymista93Member

Joined:
Aug 2021
Posts:
8
0
I didn't fit my questions above, becuase of "post sign limit". Here they come.

Questions

1.
Calculating fault factor, FaulFactor = 0,01*[(hr*ft2*R°)/BTU], we also define C, which is half in value of FaulFactor. What is it?
2. In the equation for Density of Coolant Fluid, ρcoolant = (1,1*Mix+971,8). What is the value 1,1?.
I assume that 971,8 is the density of water at 80deg Celsius. Is the value constant or should be determined as variable of temperature?

3. In the equation for Specific heat of Coolant Fluid: Cpcoolant = (MF_coolant-0,0128*Mix)*10^3, which originally was (4,4-0,0128*Mix)*10^3, I changed 4,4 with variable MF_coolant. Is that correct? Where does 0,0128 come from? I couldn't navigate this equation online. Could anyone guide me to properly understand it?

4. In the equation for cross sectional area calculation, Aduct = Firstrow*47mm*Lfin_tube
What could 47mm be? I would like to change it with a variable as well.

5. Heat transfer calculation.
• I couldn't find any equation that is similar to the one used for convection coefficient, hc = 0,2 * CpE_Mean * Gn * Ren^(-0,35) * (TE_Mean/TS)^0,25 *(k/(CpE_Mean*μE_Mean)^0,67
• Calculating fin efficiency, E = 0,986-0,0068*[(hr*ft2*R°)/BTU]*ha, are 0,986 and 0,0068 constants or variables?
6. Pressure drop calculation.
• Calculation factors 1 and 2. I assume these are friction factors?
• Are values 0.083, 9.44 and -0.45 constants? f1 = 0,083+9,44*Ren^(-0,45)
7. The value of Reynold's number is 13000+. Does this mean that flow is turbulent? If it is, shouldn't calculations check for turbulent flow and use different formulas depending on the state of flow?

Thank You so much for all the inputs!