Process design calculations for the cooling water tank
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This post was last edited by qhhqhh on 2017-5-10 at 20:38. Some heavy oils need to be cooled before they can be removed from the facility; in such cases, water-coolers using tank water are employed (typical examples include the sludge oil at the bottom of the contact cooling towers in delayed coking units, as well as the oil-throwing water-coolers). So, how should the size and piping layout of such water-coolers be determined? I’ve read a few books and searched online as well but found nothing; I wonder if any friends on the forum have done this before?
Calculation of the heat dissipation rate Qω (kJ/s) of the cooling system: Qω = A·ge·Pe·hn/3600. Where: A is a coefficient; for diesel engines in tractors, A ranges from 0.25 to 0.35, while it is usually 0.3 for turbocharged engines. ge is the fuel consumption rate of the diesel engine, valued at 0.205 kg/kW. Pe is the effective power of the diesel engine, equal to 147 kW. hn is the lower calorific value of the fuel; for diesel, hn = 41870 kJ/kg. Therefore, Qω = 105.1460 kJ/s
Calculation of the frontal area of the radiator, FR (m2): FR = Qa/va. Here, va is the air flow velocity in front of the radiator’s frontal surface; for mining vehicles and tractors, va is taken as 8 m/s. According to the overall design requirements for tractors, the width W of the radiator core required for a 200-horsepower tractor is 670 mm. Based on the requirements regarding the frontal area of the radiator, the height of the radiator core should be: W = 0.64 m and H = 0.7768 m. By referring to the standard size tables for radiators, the standard dimensions of the radiator core are: W = 0.73 m and H = 0.74 m, with FR = 0.5402 m2
Determination of the heat dissipation surface area F′ of the radiator: F′=ΨR·Qω/(KR·△t), where ψR is a safety factor of 1.15. Considering the impact of welding defects, scale, and dust on the performance of the radiator, ΨR can be set between 1.10 and 1.15, while F′ is 39.55 m2. 9. Determination of the radiator core thickness T: T = F′ / (FR·ψ), where ψ = 800 is the volumetric compactness coefficient of the radiator core, indicating the heat dissipation area per unit volume of the core. The larger the psi value, the smaller the radiator, but the air resistance also increases. It depends on the number of fins and water tubes in the radiator, as well as its layout and shape. Generally, for 1 23, ψ is taken as 370～900 m/m. The psi value, for the selected core junction with T=0.0915m, gives a T value that should be rounded to the size of a standard radiator. The thickness T of the radiator core for tractors is generally between 60 and 100 mm. T=0.100m
Q=kSΔTm; determine the heat exchange area, which corresponds to the length of the heat exchange tubes submerged in the water tank. Q=cmΔT, which is used to calculate the volume of circulating water entering and leaving; I have indeed seen drawings of the water tanks used in coking processes. For some units, ordinary shell-and-tube heat exchangers are used for the heavy oil outlet; the fluid flows through the shell side, and a higher outlet temperature can be achieved without the need for a water tank.
Have you done the calculations, sir? Could you give me some advice?
Does anyone know why such cooling tanks are used for heavy oil? Can’t other water coolers be used?