7/31/2019 7D162B Pressure Drop for Double Pipe Heat Exchanger Si

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Preliminary Double Pipe Heat Exchanger Design (S.I. units)

Estimation of Heat Transfer Area Needed

Inputs Calculations

Fluid1 mass flow Overall heat transf.

rate, m1 = 11,300 kg/hr coeff. estim., U = 2448 kJ/hr-m2-K

Fluid1 temp. in, T1in = 90oC Heat Transfer Rate, Q = 1,050,900 kJ/hr

Fluid1 temp. out, T1out = 60oC Log Mean Temp

Diff, DTlm = 44.8oC

Fluid1 sp. heat, Cp1 = 3.1 kJ/kg-oC

Heat Transfer Area, A = 9.58 m2

Fluid2 temp. in, T2in = 10oC

Fluid2 mass flow

Fluid2 temp. out, T2out = 50oC rate, m2 = 6270 kg/hr

Fluid2 sp. heat, Cp2 = 4.2 kJ/kg-oC

Overall heat transf.

coeff. estim., U = 680 J/sec-m2-K

Equations used for calculations:

Q = + (m1)(Cp1)(T1in - T1out)

Q = + (m2)(Cp2)(T2in - T2out)

DTlm = [(T1in - T2out) - (T1out - T2in)]/ln[(T1in - T2out)/(T1out - T2in)]

Q = U A DTlm

7/31/2019 7D162B Pressure Drop for Double Pipe Heat Exchanger Si

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Preliminary Double Pipe Heat Exchanger Design

Determination of pipe length needed (for known heat transfer area)

Inputs Calculations

Heat Transfer Area, A = 9.58 m2

Pipe Diam. in m, D = 0.075 m

(from calculations above)

Pipe length needed, L = 41 m

pipe Diameter, Dmm = 75 mm

(in mm)

Equations used for calculations:

D = Dmm/1000

A = pDL

7/31/2019 7D162B Pressure Drop for Double Pipe Heat Exchanger Si

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Calculation of Frictional Head Loss through Pipe

for given flow rate, Q, pipe diam., D, pipe length, L,

pipe roughness, e, and fluid properties, r & m.

1. Determ. Frict. Factor, f, assuming completely turbulent flow { f = [1.14 + 2 log10(D/ )]-2

}

Inputs Calculations

Pipe Diameter, D 75 mm Pipe Diameter, D 0.075 m

Pipe Roughness, e 0.15 mm Friction Factor, f 0.02339

Pipe Length, L 41 m Cross-Sect. Area, A = 0.004418 m2

Pipe Flow Rate, Q 0.00345 m3/s Ave. Velocity, V 0.781 m/s

Fluid Density, r 910 kg/m

Reynolds number, Re 29,819

(tubeside fluid)

Fluid Viscosity, m 0.001787 N-s/m2

(tubeside fluid)

2. Check on whether the given flow is "completely turbulent flow"

(Calculate f with the transition region equation and see if it differs from the one calculated above.)

f = {-2*log10[(( /D)/3.7)+(2.51/(Re*(f1/2

))]}-2

Transistion Region Friction Factor, f: f = 0.0285

Repeat calc of f using new value of f: f = 0.0281

Repeat again if necessary: f = 0.0281

3. Calculate hL and Pf, for straight pipe flow, using the final value for f calculated in step 2

(hL = f(L/D)(V2/2g) and Pf = ghL)

Frictional Head Loss, hL 0.47 m

Frictional Pressure

Drop, DPf 4223 N/m

Frictional Pressure

Drop, DPf 4.22 kN/m2

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4. Calculate hL and Pf, for the 180o

bends

Inputs Calculations

Pipe length between No. of 180o

bends,

bends, Lsect = 4 m NB = L/Lsect = 10

Minor Loss Coefficient Head loss due to bends,

for 180o

bends, K = 1.5 hB = NBK(V2/2g) = 0.47 m

( K = 1.5 for threaded pipe or 0.2 for flanged pipe. )

Pressure Drop due to

bends, PB = 4229 N/m

Pressure Drop due to

bends in psi = 4.23 kN/m

5. Add the results from part 3 and part 4 to get total hL and Pf

Total Frictional Head Loss, hL = 0.947 m

Total Frictional Pressure Drop, Pf = 8452 N/m

Total Frictional Pressure Drop

in kN/m2= 8.45 kN/m