Heat Exchanger Report
Done By: Ng Jun Da (3S315)
Damian Goh (3S404)
Liu Shuang Xing (3S419)
Conversion of Information to SI units
Mass flow rate = 30000 lb/h =
13 607.7711 Kg/h
Crude oil heated from 70 F to 136 F is equivalent to
21.11°C to 57.78°C
Product to be cooled from 295 F to 225 F is equivalent to
146.11°C to 107.22°C
Properties of crude oil
Specific Heat Capacity = 0.475 BTU/lbF = 1988.73 J/kg°C
Viscosity = 2.9 mPa.s = 0.00029 N.s/m2
Thermal Conductivity = 0.13646364152 W/m.K
Density = 51.5 lb/ft³ = 824.95086376 kg/m3
Properties of Product
Specific Heat Capacity = 0.525 BTU/lbF = 2198.07 J/kg °C
Viscosity = 5.2 mPa.s = 0.00052 N.s/m2
Thermal Conductivity = 0.11934082719 W/m.K
Density = 54.1 lb/ft³ = 866.59886853 kg/m3
Specifications of heat exchanger
Shell diameter: 23.25 in = 0.59055m
No. of tubes: 324
Type of tubes: BWG 14 arranged on a 1 inch square patch, supported by baffles
Length of tubes: 12 ft = 365.76 cm = 3.6576m
Outer diameter of tubes: 0.75 in = 1.905 cm = 0.01905m
Baffles: 25% cut 22.86 cm intervals.
Inner diameter: 0.584 in = 0.0148336m
Thickness of tube wall: 0.083 in =0.0021082m
Surface area of all tubes = 𝜋 × 0.0148336 × 3.6576 × 324 = 𝟓𝟓. 𝟐𝟐𝟓𝟐𝟒𝐦𝟐
Volume of Oil = 𝜋𝑟 2 × 𝑙 = 𝜋(0.0148336 ÷ 2)2 × (3.6576 × 324)
= 0.2047973005 m³
Mass of Oil = volume x density
=168.9477092 kg
Temperature oil increases = 57.78°∁ − 21.11°∁= 𝟑𝟔. 𝟔𝟕°∁
Heat energy to be gained by crude oil to increase by 36.67 deg C =
Specific Heat Capacity × Mass of Oil × ∆𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 =
1988.73𝐽/𝐾𝑔℃ × 168.9477092𝐾𝑔 × 36.67℃ = 𝟏𝟐𝟑𝟐𝟎𝟖𝟎𝟑. 𝟖𝟐𝑱
Calculations
Reynolds number for Oil
Density = 824.95086376 kg/m3
Velocity = 20.8974523m/s
(Substitute appropriate values)
http://www.tasonline.co.za/toolbox/pipe/velocity.htm
Diameter = 0.0148336m
Viscosity = 0.00029 N.s/m2
Reynolds number = (density x velocity x diameter)/viscosity
=8.82 x104
(Substitute appropriate values)
http://www.efunda.com/formulae/fluids/calc_reynolds.cfm#calc
Re > 4000, Therefore Movement of Oil = Turbulent Flow
Heat Transfer Coefficient
Whereby
ΔQ = heat input or heat lost, J
h = heat transfer coefficient, W/(m2K)
A = heat transfer surface area, m2
ΔT = difference in temperature between the solid surface and surrounding fluid
area, K
Δt = time period, s
ΔQ = 𝟏𝟐𝟑𝟐𝟎𝟖𝟎𝟑. 𝟖𝟐J
A = 𝟓𝟓. 𝟐𝟐𝟓𝟐𝟒𝐦𝟐
ΔT = 𝟑𝟔. 𝟔𝟕°∁
Δt =
=
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
3.6576
20.8974523
= 0.17502612 s
HTC =
12320803.82
55.22524×36.67℃×0.17502612
=34760.6312997333
Rate of Heat Transfer
q = hA(Ts − Tb)
A - surface area of heat transfer.
Ts - surface temperature
Tb - temperature of the fluid at bulk temperature.
h - constant heat transfer coefficient
h= 34760.6312997333
A= 𝟓𝟓. 𝟐𝟐𝟓𝟐𝟒𝐦𝟐
Ts-Tb = 𝟑𝟔. 𝟔𝟕°∁
q = 34760.6312997333 × 55.22524 × 36.67
= 70394086.4369273
Flow Area =Surface Area of Heat Transfer
Surface Area = 𝟓𝟓. 𝟐𝟐𝟓𝟐𝟒𝐦𝟐
Mass Flow Rate
ρ = density
v = velocity
A = flow area
= 𝟏𝟏𝟓𝟒. 𝟎𝟕𝟑 /𝒔
Efficiency
E=
q
qmax
E: efficiency
=
q
Cc (Th − Tc )
q: heat transfer rate
Cc: mass flow rate x heat capacity of cool substance
T: Temperature of hot/cool substance
q = 70394086.4369273
Cc = 1154.07 × 2198.07 = 𝟐𝟓𝟑𝟔𝟕𝟐𝟔. 𝟔𝟒𝟓
Th − Tc = 107.22 − 21.11 = 𝟖𝟔. 𝟏𝟏
E=
q
qmax
=
𝟕𝟎𝟑𝟗𝟒𝟎𝟖𝟔.𝟒𝟑𝟔𝟗𝟐𝟕𝟑
𝟐𝟓𝟑𝟔𝟕𝟐𝟔.𝟔𝟒𝟓(𝟖𝟔.𝟏𝟏)
=0.322261865832
∴ Efficiency = 32.2% (3 Sig. Fig.)
Conclusion
From the above data, we conclude that at an efficiency percentage of 32.2%,
the heat exchanger is efficient. It is not very efficient, but it can still be used for
the heat exchange.