Can increasing flow rate reduce the heat exchange area?
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Hello, esteemed seniors. I have a question and would appreciate your guidance. If it’s posted in the wrong forum, please let the moderator know. When using the formula Q=ATK to calculate area, can increasing the hot water flow rate reduce the heat exchange area? The question is: Given that the hot side consists of hot water with an inlet flow of 84 and an outlet flow of 50; the cold side consists of crude oil containing 30% water, with an inlet flow of 46 and an outlet flow of 80; and the heat capacity is 2500 KW, what is the heat exchange area required? Solution: Q = 2500 / 4.1868 × 3600 = 2,149,613 kcl/hr. The temperature difference t = 4°C; taking 350 Kal/m²·°C·hr, we get A = Q / (KT) = 1,535 m². The water flow rate is calculated as M = Q / (1000 × (84 – 50)) = 63.22 m³/hr. Now it seems that the heat exchange area is too large; therefore, the water flow rate is intended to be reduced to 100 m³/hr. Will this reduce the required area?
If the operating parameters of the cold material are fixed, simply increasing the water volume mainly affects the logarithmic temperature difference and the heat transfer coefficient; this results in an increase in the water outlet temperature, thereby increasing the logarithmic temperature difference. Generally, the heat transfer coefficient decreases, and at the same time the heat exchange area also decreases to a certain extent. But it’s impossible to have a reduction on the order of magnitude you mentioned. Especially when the water volume increases to a sufficient level such that the water outlet temperature exceeds 80 degrees, the heat exchange area decreases significantly compared to the current design. However, taking into account factors such as one-time equipment investment and operating costs, this idea is unreasonable.
I did the calculations and found that the value isn’t smaller at all – in fact it’s larger. Calculation: Q = 2500/4.1868 × 3600 = 2,149,613 kcl/hr for 5 tons, N; x” J% Q_water = M_water × 1000 × (84 – 50) = 100 × 1000 × 34 = 3,400,000 kcl/hr for 5 tons, N; x” J%. The temperature difference t = 4°C × x$ N – i$ ]0 `# f! {( W K; take 350 Kal/m2·°C·hr: Y. w; j8 \$ `" n# a/ R/ ]- u* l. A = Q/KT = 3,400,000 / 3(50 × 4) = 2,428 M2+. Is this calculation correct?
The temperature range difference is large; use a plate heat exchanger. With a shell-and-tube heat exchanger, the area will definitely be large; it is recommended to use a single-pass, elongated type. As the water volume increases while the heat transfer rate remains constant, the logarithmic mean temperature difference increases, and the heat transfer area decreases.
Please ask: if you want to reduce the heat exchange area, how do you do that? Thank you
Plate heat exchangers don’t seem to allow such large temperature differences either! Increasing the flow rate reduces the logarithmic mean temperature difference. If the overall size of the heat exchanger remains constant and the flow velocity increases, then the heat transfer coefficient on the side with the higher flow rate will increase slightly. However, this seems to have little effect on heat exchange between water and water. Moreover, increasing the flow velocity can either prevent the formation of fouling or it can lead to an increase in pressure drop It is estimated that the heat exchange area may increase
If your flow rate increases, the first thing to consider is increasing the capacity of the heat exchanger. An increase in flow velocity may lead to an improvement in the heat transfer coefficient, but not by that much. How high the heat transfer coefficient will go still requires careful calculation; it might be 380 or 400, but it’s not certain. It seems that the increase in your heat output should be greater than what results from an increase in the heat transfer coefficient; therefore, your calculations are likely insufficient.
Dear seniors, I have a question and would appreciate your guidance. The details are as follows: On the hot side, there is hot water, with an inlet flow of 84 and an outlet flow of 50. On the cold side, there is crude oil containing 30% water, with an inlet flow of 46 and an outlet flow of 80. The heat capacity is 2500 KW. What is the required heat exchange area? Solution: Q = 2500 / 4.1868 × 3600 = 2,149,613 kcl/hr. The temperature difference t = 4°C; taking 350 Kal/m²·°C·hr, we get A = Q / (KT) = 1,535 m². The amount of water required is M = Q / (1000 × (84 – 50)) = 63.22 m³/hr. It seems that the heat exchange area is too large; how can this area be reduced?
To reduce the heat exchange area, it is recommended that you use a plate heat exchanger. Furthermore, if the flow rate of hot water is increased, with the heat load remaining unchanged, the outlet temperature of the hot water will rise, the logarithmic mean temperature difference will also increase, and the calculated area will decrease accordingly. However, while reducing the heat exchange area, it is necessary to consider whether the actual pressure drop meets the allowable pressure drop requirement.
Can I calculate this in this situation? Given: Hot side: hot water, flow rate 100 M3/hr, specific heat 1 W-K; density 3 g/cm3, inlet temperature 84°C. Cold side: crude oil with 30% water content, flow rate 95 M3/hr, inlet temperature 46°C, outlet temperature 80°C. Heat capacity: 2500 KW. What is the heat exchange area required? 5 s$ W R& }' w- O) X Solution: Q=2500/4.1868×3600=2149613 kcl/hr. Taking 350 Kal/m2·c·hr as the value for R*, M=Q/(1000×(84-t))=100 M3/hr. The temperature at the hot water outlet is t=62.5 degrees; the temperature difference is (84-62.5)-(80-46)=8.82 degrees. A=Q/KT=696 m3. Please help me, experts – is this correct?
A pressure drop of 0.1 MPa is allowed at this time; how can the pressure drop be calculated?