What are the common causes of fouling in a heat exchanger? How does fouling affe

Question

What are the common causes of fouling in a heat exchanger? How does fouling affect heat transfer and pressure drop? A double-pipe heat exchanger is constructed of a copper (k= 381) W/m middot degree C) inner tube of internal diameter Di=1.2 cm and external diameter Do = 1.6 cm and an outer tube of diameter 3.0 cm. The convection heat transfer coefficient is reported to be hi=700 W/m^2 middot degree C on the inner surface of the tube and ho= 1400 W/m^2 middot degree C on its outer surface. For a fouling coefficient h_di = 2000 W/m^2 middot degree C on the tube side and h_do = 5000 W/m^2 degree C on the shell side, determine: The thermal resistance of the heat exchanger (assume length = 1m) The overall heat transfer coefficients U_i and U_o based on the inner and outer surface areas of the tube, respectively

Explanation / Answer

Lets start with the first question.

1. Reasons for fouling in the heat exchanger;

- Scaling : This happens due to the salts (insoluble) forming some scale & causing fouling

- Particulate / Sedimentation: The name says everything. Some particulates in the solution sediment or deposit on walls of the HEx (Heat Exchanger) forming fouling

- Corrosion : Chemcials causing corrosion of the MOC (material of construction)

- Chemical : Chemical reaction resulting in some deposition causing fouling.

- Freezing: When a hot fluid gets cold to the freezing temperature causes the fouling.

Fouling mostly occurs mostly on the walls of the tubes causing the drop in the overall heat transfer coefficient. This reduces the efficiency of heat transfer operation & also increases the pressure drop for the fluid flow. The pressure drop increases as the diameter of the fluid flow is reduced.

2. The overall heat transfer co-efficient for the double pipe HEx is given by,

1/U = Do / (hi * Di) + Do * ln (Do/Di) / 2k +1/ho +hdi * Do/Di + hdo

So 1/Ui = 0.016 / (700*0.012) + 0.016 * ln (0.016/0.012) / 2(380) + 1/1400 + 2000* (0.016/0.012) + 5000

1/Ui = 7666.66 W/m2C

And for 1/Uo = 6750 W/m2C

I know the answers are not matching with the ones which you provided, but my calculation is showing the results which i published.

The thermal resistance of heat exchanger is R = L/k = 1 m / 380 W/mC = 1/380 C/m = 0.0026

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.