Soda water heat exchange calculation

Calculate the heat load based on steam, and then determine the flow rate of cooling water. This is supersaturated steam at 0.4MPA; the calculations below are approximate for reference only: Q=3513kW, heat exchange area: 22.9 M2, BEM500-1.0-22.9-2.0/25-1

Reply to 2# jhyy19802003: How is Q=3513KW calculated? ? ? ?

What does the heat of 3513KW include? ? Since there is no phase change, should latent heat be considered? What is the Cp of superheated steam? ? ?

When a substance undergoes a phase change (a change in its state of matter), the heat absorbed or released without a change in temperature is called \"latent heat\"

0.4 MPa, 140 degrees – it’s no longer steam; it has become condensed liquid

Are you at 140 degrees or in a steam state? In that case, your pressure must be below 0.27 Mpag; please confirm.

Assume your steam pressure is 0.2 Mpag. The heat energy at 180 degrees is 674.5 kcal/kg, while it is 654.5 kcal/kg at 140 degrees. The heat released = 6000 * (674.5 – 654.5) = 120,000 kcal/h. Converting this to KW gives 139.5. Since no phase change occurs, the value of K will not be high.

The steam pressure is 0.4 MPA; the inlet temperature is 180 degrees Celsius, so it is superheated steam. The outlet temperature is 140 degrees Celsius, and the state at the steam outlet remains that of steam. The outlet pressure must be less than 0.27 MPA, so it can be assumed to be 0.26 MPA. The heat enthalpy at 0.4 MPa is approximately 2738.5 KJ/KG, while that at 0.25 MPa is 2717.2 KJ/KG. The heat released is 6000 * (2738.5 – 2717.2) = 127800 KJ/H.

Results for floor 9 vs floor 2: 35.5KW/3515KW – the difference is huge :'(

This post was last edited by zpg on 2010-11-19 at 10:49. The operating condition described by the original poster is quite common: the heater uses superheated steam for heating, and this superheated steam condenses inside the heater. The pressure drop in the superheater is generally around several dozen kPa, which is negligible compared to a steam pressure of 0.4 MPaG. In such a situation, you should first determine the state of the superheated steam. A pressure of 0.4 MPa is considered gauge pressure; the corresponding zone is Zone 2. For superheated steam, the pressure is P = 0.50132500 MPa (absolute pressure), the temperature is T = 180.00 °C, and the specific enthalpy is H = 2812.37 kJ/kg. For saturated steam at the same pressure, the zone is Zone 4: the pressure is P = 0.50132500 MPa (absolute pressure), the temperature is T = 151.94 °C, and the specific enthalpy is H = 2748.23 kJ/kg. Therefore, it can be determined that it is superheated steam. If the heat exchanger has a sufficient area, then the conditions of the steam will change as follows: superheated steam at P = 0.4 MPaG and T = 180 °C → (sensible heat cooling) to saturated steam at P = 0.4 MPaG and T = 151.94 °C → (latent heat condensation) to saturated water at P = 0.4 MPaG and T = 151.94 °C → (sensible heat cooling) to unsaturated water at P = 0.4 MPaG and T = 140 °C. Unsaturated water refers to cold water, that is, water that has not reached its boiling point at the given pressure; the plain water we drink on a daily basis is unsaturated water. In steam heaters used in actual industry, the process for superheated steam is one of cooling and condensation; the subcooling of saturated condensate does not necessarily occur in all heaters. Therefore, when calculating the area, emphasis should be placed on determining the area of the cooling section. The area in question is Zone 4; for saturated condensate, the pressure is P = 0.50132500 MPa, the temperature is T = 151.94 °C, and the specific enthalpy is H = 640.62 kJ/kg. The calculation of the heat load is straightforward – if the sensible heat associated with the supercooling of the condensate is ignored, then Q = Q1 + Q2. Here, Q1 represents the sensible heat released when superheated steam cools down to saturated steam, while Q2 represents the latent heat released during the condensation of saturated steam. In practice, this can be calculated as (enthalpy of superheated steam – enthalpy of saturated water), i.e., Q = 6000 × (2812.37 – 640.62) = 3619.6 kW. Among this, Q1 = 6000 × (2812.37 – 2748.23) = 106.9 kW, accounting for 2.95%, while Q2 = 6000 × (2748.23 – 2748.23) = 3512.7 kW, accounting for 97.05%. The medium being heated is water, with its temperature rising from 50 °C to 60 °C. The flow rate of water is given by Q/(Cp×ΔT) = (3619.6 × 3600) / (4.184 × 10) = 311438 kg/h, or 311.4 tons/h. In other words, approximately 1 ton of steam can heat around 50 tons of water over a temperature difference of 10 °C. The calculation of 3513 kW on the 2nd floor is completely correct, which also shows that latent heat accounts for the vast majority of the heat load. In the area calculation, the heat exchange area should be divided into a cooling section and a condensation section. Separate total heat transfer coefficients should be assumed to calculate the heat transfer temperature differences for each section: For the cooling section, assuming K=100 W/m2·°C and ΔT=110.7°C, then F1=Q1/(K·ΔT)=106.9×1000/(100×110.7)=9.65 m2. For the condensation section, assuming K=1500 W/m2·°C and ΔT=96.9°C, then F2=Q2/(K·ΔT)=3512.7×1000/(1000×96.9)=24.18 m2. The total area is F=F1+F2=33.83 m2, of which F1 accounts for 28.54% and F2 accounts for 71.46%. It can be seen that the area of the sensible heat section cannot be ignored; although sensible heat constitutes only 3% of the total heat load, its proportional area is much larger. In the area calculation for heaters, this section is often overlooked. The area calculated by the friend on the second floor for the condensation section is appropriate, but since the area of the cooling section with a lower heat transfer coefficient was ignored, the calculated area of 22 m2 is slightly too low. In thermal calculations, the above calculations are all very fundamental and should be mastered.