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 raising temperature from
21.11°C to 57.78°C
Product to be cooled from 295 F to 225 F is equivalent to lowering temperature from
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
cross section area of tube = 𝜋𝑟 2
=𝜋×(
0.0148336 2
)
2
=0.000172813𝑚2
Velocity =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑜𝑖𝑙 × 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑢𝑏𝑒 × 3600
Velocity = 𝟐𝟔. 𝟓𝟏𝟑𝟎𝟗𝟓𝟕m/s
Diameter = 0.0148336m
Viscosity = 0.00029 N.s/m2
Reynolds number = (density x velocity x diameter)/viscosity
=8.82 x105
(Substitute approprite 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 =𝟏𝟒𝟕. 𝟏𝟏 − 𝟐𝟏. 𝟏𝟏 = 𝟏𝟐𝟓°∁
Δt =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
3.6576×324
=
26.5130957
= 44.69587544s
HTC =
12320803.82
55.22524×125℃×44.68587544
=𝟑𝟗. 𝟗𝟑𝟐𝟐𝟔𝟕𝟑 𝑾/𝒎𝟐 ℃
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= 𝟑𝟗. 𝟗𝟑𝟐𝟐𝟔𝟕𝟑 𝑾/𝒎𝟐 ℃
A= 𝟓𝟓. 𝟐𝟐𝟓𝟐𝟒𝐦𝟐
Ts-Tb = 𝟏𝟐𝟓°∁
q = 39.9322673 × 55.22524 × 125
= 275658.6307W
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 = 275658.6307W
Mass flow Rate = 13 607.7711 Kg/h = 3.779936 Kg/s
Cc = 3.779936 x 2198.07 = 8308.564839W℃
Th − Tc = 146.11 − 21.11 = 𝟏𝟐𝟓℃
E=
q
qmax
=
𝟐𝟕𝟓𝟔𝟓𝟖.𝟔𝟑𝟎𝟕
𝟖𝟑𝟎𝟖.𝟓𝟔𝟒𝟖𝟑𝟗(𝟏𝟐𝟓)
=0.265421175
∴ Efficiency = 26.5% (3 Sig. Fig.)
Conclusion
From the above data, we conclude that at an efficiency percentage of
26.5%, the heat exchanger is not efficient. Thus we need to redesign a more
effective heat exchange
Ways to improve Heat Exchanger
1. Increase the number or length of tubes
2. Increase the mass flow rate
3. Use a product of higher specific heat capacity
4. Increase the temperature difference between the
product and the oil