【Series 001 of Practical Calculations in Daily Life】Calculation of the heat load on the condenser on top of a hot water tank – Best 100 Wealth
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This post was last edited by *lihuagong on 2017-3-27 at 14:25. [Topic] Suppose there is a working condition as shown in the figure below: a 35 m3 hot water tank (with dimensions of L6400×W1700×H3200) is circulated using a pump, and the water returns to the tank through various channels (not detailed here). In any case, it is assumed that the water temperature in the system remains constant at 95°C, and the tank maintains a constant liquid level (80% of its capacity) through water addition and overflow control. The steam at the top of the tank is condensed through heat exchanger E101; the condensate is recycled, while the exhaust gas is released directly. 【Question】 How is the heat load of the can-top condenser E101 calculated? ? @arpcd @longkui1990 @tdl522 @Luo Mi @Pingsha
There is a calculation for the \"evaporation rate per unit area of an open water surface,\" with the unit being kg/(m2•h); this allows for the easy calculation of the mass that evaporates per unit area per hour, based on the water temperature and room temperature
You can take a look at the picture in the attachment; it might be helpful
On average, about 23 kg per square meter per hour evaporates
This post was last edited by zxc17835 on 2017-3-27 16:19. I was wrong
There are two questions: (1) Where did you get this image/table? It feels really good; it’s the first time I’ve seen such evaporation data. (2) Here it’s 95°C, so the evaporation rate should be around 40 kg/㎡h; you can consider it as 85°C
The key to this issue is determining the amount of evaporation; relying solely on the amount of vapor in the gas phase is likely to be inaccurate.
With this open-type evaporation, once it enters a pipe, the open environment is disrupted. Can an open design be understood as the maximum evaporation rate? An infinitely long pipe is considered an open system~
Since the liquid level in the hot water storage tank is stable and the top is also vented, I understand it to be an open-type tank. The only difference is that in this example there is no wind speed at the top of the tank, so the evaporation rate might not be that high.
Firstly, the overflow port should have a liquid seal, and its height should not be lower than the pressure drop of the heat exchanger; otherwise, the gas will escape through the overflow port. The conditions are not complete; here is a brief outline of the calculation approach. Gas composition entering the heat exchanger: calculated based on the saturated vapor pressure of water at 95 degrees Celsius, with a water vapor partial pressure of 0.845 bara. Gas flow rate entering the heat exchanger: 5 m3/h plus 20% volume of minor breathing. Heat load (without considering heat losses): flow rate × composition × enthalpy difference (2269.2 kJ/kg)
This post was last edited by arpcd on 2017-3-27 23:46. Isn’t this just a typical example of calculating consecutive large breaths? ? The flow rate of 300+50=350 m3/h is also the exhaust volume at the top of the tank; the volumetric flow rate of the wet air can be calculated based on saturated water at 95°C. . Your heat exchanger can’t be small. Don’t think that just because the liquid level in your tank remains stable it means there’s no breathing taking place; your tank is a typical example of continuous breathing. For large oil tanks, they are first filled to capacity, then emptied, filled again, and emptied once more; the volume of oil transferred over an annual cycle is used to determine the scale of such operations. Both CNPC and U.S. oil companies use this method to calculate such volumes. In your case, however, oil is continuously poured in and taken out; the annual volume handled is 350 x 8000 = 2.8 million cubic meters – what a huge oil tank! . Intermittent tank breathing is calculated, but continuous breathing isn’t? ? It hasn’t turned yet, right? . In one hour, the water in your container has to be stirred 10 times – think about that ? With constant handling like this, in order to maintain constant pressure at the top of the tank, only that much air can be allowed to enter and exit. . . Otherwise, this jar can only have two outcomes: it will either be deflated or explode. Thermal engineering professionals should be quite familiar with this type of hot water tank (an open-type hot water tank); they are commonly used in atmospheric pressure hot water boilers. When the temperature is close to that of boiling water, there are empirical data indicating that the amount of water that escapes from the top of the tank is approximately half of the circulation rate, measured in kg/h. In your case, this would be 350/2 = 175 kg/h – that’s roughly the figure. Please do the calculations carefully to confirm (it’s an estimate; you need to carry out the proper calculations yourself) The latent heat of condensation for this 175 kg of water vapor is precisely the heat load on your heat exchanger. . :lol