Discussion on Question 20 in the morning session of the second day of 2014
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【2014-Upper】 20. Heat a certain solution in the tank using saturated steam at 135°C in the jacket. The effective liquid volume inside the tank is 6.2 m^3, the average heat transfer area is 12.5 m^2, the density of the solution is 1950 kg/m^3, and its specific heat capacity is 2.1 kJ/(kg·℃). To maintain a uniform heating temperature, a forced agitator is installed in the tank, with a total heat transfer coefficient of 1200 W/(m^2·℃). What is the time (in hours) required to heat the solution from 10°C to 90°C? 0.28 (B) 0.38 (C) 0.48 (D) 0.58 Brother zxm’s solution to this problem is as follows: I would like to discuss this with everyone: 1. Is it necessary to consider unsteady-state heat transfer here? 2. How should the method for calculating △tm be understood? @zhanghp30 @zwp997 @Higee
It cannot be considered, nor is it necessary to consider it.
Well then. . . △How should we understand the method for finding tm?
Look at Chapter 1 of the Tianjin University version; there is an empirical formula for the effective temperature difference in stirred tanks. But someone told me that the formula is a bit wrong; I don’t know where the mistake is.
This result should be fine; if derived strictly through integration, the result will be the same as well
This post was last edited by Higee on 2015-7-16 at 15:14. The calculation of the temperature difference Δtm here is different from that in steady-state heat transfer, which we are familiar with; although the forms are similar, it is purely coincidental. The specific derivation process is as follows: Assume that at time θ, the temperature of the fluid in the tank is t. After an interval of dθ, the temperature becomes t+dt. During this period, we have (mCp)c*dt = dQ = KA(T-t)dθ. Note that within this dθ interval, it is a steady-state heat transfer process with a constant temperature difference; the temperature of the hot fluid is T, while that of the cold fluid is t (as the temperature inside the tank is uniform). By separating variables and integrating, we obtain ln(T-t1)/(T-t2) = θ (#). For the entire heat transfer process, the energy balance equation gives Q = (mCp)c*(t2-t1). Using this expression to replace (mCp)c in (#), we get (t2-t1)Q = KA*——————*θ = KA△tmθ ln(T-t1)/(T-t2). PS: ① Here, Q represents the amount of heat transferred, with units of J, not the heat transfer rate, which has units of W. ② The logarithmic temperature difference here is the average of the temperature differences between the initial and final states, whereas the temperature difference mentioned in normal contexts refers to the average of the temperatures at the inlet and outlet of the heat exchanger. ③ In this example, the temperature of the hot fluid remains constant (due to steam condensation), making the situation simple; otherwise, things would be more complex. I did some calculations and gave up. Friends who are interested can give it a try
This post was last edited by tiyiss on 2015-7-17 at 16:02. Brother HIGEE is truly an amazing person~~ This afternoon I looked at the formula for the heat transfer temperature difference for the batch reactor as presented by Tianjin University; he took into account the changes in the temperature of the heating medium, which is why the formula is so complex. . . This problem involves steam condensation; in other words, Brother ZXM used this formula to solve it~ Thank you so much – you’ve opened another door for me{:3_67:}
Bro HIGEE is truly a genius~~ This afternoon I looked at the formula for the heat transfer temperature difference related to the transfer of reaction vessels as presented by Tianjin University; he took into account the changes in the temperature of the heating medium, which is why the formula is so complex. . . This problem involves steam condensation; in other words, Brother ZXM used this formula to solve it~ Thank you so much – you’ve opened another door for me{:3_67:}
A forced mixer was installed; that’s the idea. △Tm can be understood in this way: think of this jacket as a shell-and-tube heat exchanger, and that’s it; inlet temperature... outlet temperature....
Well, President Y, could you elaborate further on what the effects of a forced mixer are?
The mixer only affects the K value directly, and it has no direct computational relationship with the heat transfer temperature difference.