HEAT-EXCHANGER DESIGN PROJECT

1. Oil of specific heat of 1900 J/Kg.K and flowing at 68 kg/min is cooled from 110oC to 75oC by cooling water, which will experience a temperature rise from 28oC to 75oC. The outer and inner diameters of tube are 30 mm and 25 mm respectively. Compute the heat transfer area and the number of tubes if the standard length of tube is 4.88 m. First calculate by hand and then compute with Excel, then try to optimize your computations.

2. Oil of specific heat of 1900 J/kgK and flowing at 2.5 kg/s is cooled from 110 to 75˚C in a shell and tube heat exchanger by cooling water from 30˚Cto 75˚C. Cold water enters tubes of 25-mm and 30-mm internal and external diameter respectively. Calculate the heat transfer area, shell length and the length of the pipes. The number of tubes is 10 and the number of tubes passes through the shell is 8. (Pls refer to Kern chapter 7 and Holman, example 10.6)

3. Cold water is heated at the rate of 4.0 kg/s from 40 to 60˚C in a shell and tube heat exchanger, it is heated by hot water at 95˚C at the rate of 2 kg/s on the shell side. Cold water enters the 20-mm diameter tubes at 0.6 m/s. The overall heat transfer coefficient is 1500 W/m2.K. Assume the correction factor to be 0.88 and the standard length of pipe is 2.44 m. Calculate the number of tube passes and the number of tubes per pass. Properties: = 1000 kg/m3, l= 5.96 X 10-4Pa.s, cp= 4180 J/kg.K, k = 0.68 W/m.K and Pr = 3.7. DittusBoelter equation for fully developed turbulent flow is:

  1. Double the flow rates in example 7.3 and design a shell –tube heat exchanger by

a. Kern method in British Units

  1. and also in S.I. unit. Predict h using Nu = 0.023Re0.8Pr0.3
  1. Design a shell-tube heat exchanger for cooling down RBD using BPO in a palm oil refinery

a)1000 RBD MT/day.

b)3000 RBD MT/day.

Use Excel to design and OPTIMISE the heat exchanger.

Also please do a hand calculation for each case.

Properties / Hot side / Cold side
Fluid / RBD / BPO
Specific gravity / 0.838 / 0.841
Specific Heat , kJ/kg.K / 2.7 / 2.55
Thermal Conductivity W/mK / 0.162 / 0.163
Viscosity Pa.s / 2.316x10-3 / 3.105x10-3
Temperature in , oC / 255 / 100
Temperature out , oC / 130 / 225

What will be the area required for a concentric tubeand shell-tube heat exchanger.

  1. Redo Example 10.6 of Holman with a water velocity of 1 m/s to 2 m/s, but start with 0.366 m/s first using hand calculation from tube heat transfer coefficient to pressure drop. Assume shell side HTC with half of inside pipe HTC.

Plate Heat Exchanger:

A typical example of heat transfer calculation for Palm Oil Refineries would be as follows:

Hot Refined, Bleached and Deodorized Palm Oil (RBD) is coming to the heat exchanger at 130 oC. It is being used to heat up Crude Palm Oil (BPO) which is coming at about 50oC. The flow is say at 1500 m3/day of RBD.

Density of palm oil is about 830 kg/m3 for RBD and 835 kg/m3for BPO.

Assuming the RBD needs to be cooled down to 70oC.

Total heat transfer is – Mass flow ratex Specific Heat x Temp difference

Total Heat Transfer - (1,500,000 x 0.83/24) x 0.54(est. Specific Heat) x (130-70)oC = 1,680,750 kcal/hOR 1,955 kW. NOTE: 1 cal = 4.187 J

Heat Transfer, Q = Overall Heat Transfer coefficient x Heat Transfer area x Log Mean temp difference

ie. Heat Transfer area = Total Heat Transfer / Overall Heat Transfer coefficient x Log mean temp difference.

To calculate Log mean temp difference, we need to look at the overall temperatures

RBD is cooled from 130oC to 70oC

BPO is heated from 50oC to 110oC

So log mean temp difference is average of (130-110oC) and (70-50oC) which is 20oC.

Overall Heat Transfer coefficient for palm oil is between 500-800kcal/h m2Cor581 – 930 W/m2K. However that is for temperatures as above. If the temperatures are lower, the overall heat transfer coefficient is much lower.

I do not need to go into the details but the overall heat transfer coefficient is usually a number which has been calculated after many trials and experiments. One would also study the individual heat transfer coefficient to be able to calculate the Overall Heat Transfer coefficient.

Using above example, if the Log mean temp difference is 20oC, the overall heat transfer coefficient is 500kcal/h m2oCor 581 W/m2K and the total heat transfer is at1,680,750 kcal/h, the total heat transfer area needed for above application would be1,680,750 / (500 X 20) = 168.1 m2. OR 1,995,000/581/20 = 168.2 m2.

Using plate heat exchangers which has an individual plate area of 0.75m2 per plate, the total number of plates needed for the above application = 168.1/0.75 = 224.1 plates.

Total selected plates would be 225 plates, after rounding off. That means that for the above application, the Plate Heat Exchanger would require a total of 225 plates of each having an individual plate area of 0.75m2.