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A heat exchanger is used to cool down a stream of $\pu{20 kg s-1}$ of water from $\pu{80 °C}$ to $\pu{60 °C}$. The cooling water enters at a rate of $\pu{12 kg s-1}$ and $\ce{20 °C}$. The overall heat transfer coefficient is $\pu{2 kW m-2 K-1}$ and the heat capacity of water is $\pu{4.183 kJ kg-1 K-1}$

Heat exchange tube

Calculate the surface area.

I am stuck on what the last step of the calculations should be. I began with calculating the outlet temperature for the cooling stream using that $q$ is constant, I set IN = OUT:

$$m_hC_pΔT_\mathrm{in} = m_cC_pΔT_\mathrm{out}\label{eq:1}\tag{1}$$

From $\eqref{eq:1}$ I got that $T_\mathrm{2out} = \pu{53 °C}$

Then I calculated the logarithmic temperature difference, which I got to be $ΔT_\mathrm{ln} = \pu{-33.075 °C}$.

After this I am confused as to how I can get the surface area of the tube.

Edit

I did a heat balance, IN - OUT = 0:

$$(m_hC_pΔT_\mathrm{in} + m_cC_pΔT_\mathrm{in}) - (m_hC_pΔT_\mathrm{out} + m_cC_pΔT_\mathrm{out}) = 0$$

simplified into:

$$m_hC_pΔT_\mathrm{in} - m_hC_pΔT_\mathrm{out} - UAΔT_\mathrm{ln} = 0$$

However when I try to solve for A I get a negative value, which isn't correct.

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  • $\begingroup$ There is no way to solve the problem with two variables. So the overall heat difference must be directly calculable and the only variable is the surface area of the heat exchanger. So "Hot in" is 80 C and "Hot out" is 60 C. "Coolant in" is 20 C and "Coolant out" is 80 C. (Edited - I had mistakenly stated "Coolant out" was 60 C.) $\endgroup$ – MaxW Feb 14 at 7:32
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The outlet temperature should be 53.3. The log-mean temperaturr difference should be positive. Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.

ADDENDUM

This addendum is for the benefit of @MaxW (the downvoter).

  1. The heat load $\dot{Q}$ of the heat exchanger is given by: $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.

  2. The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$

Solving for the exit temperature of the cold stream gives $$T_{C,out}=53.33\ C$$ 3. The log-mean temperature difference is given by: $$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$

  1. The heat transfer area A is determined from the equation: $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives: $$A=25.4\ m^2$$

Questions?

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    $\begingroup$ Why did this get downvoted? Indeed the questioner miscalculated the log-mean temperature difference. Also he just needs the heat transferred from one side to the other, so $Q=m_hC_p\left(T_{h,in}-T_{h,out}\right)=UA\Delta T_{ln}$ $\endgroup$ – user5713492 Feb 13 at 21:32
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    $\begingroup$ @user5713492 If it pisses you off so much, why don't you just cancel it with an upvote? $\endgroup$ – Chet Miller Feb 14 at 2:05
  • $\begingroup$ This is more meta than chem, but I don't vote a whole lot, and I don't think I have ever downvoted, leaving to other more frequent participants to set effective rankings. I consider it sad that some feel the need to denigrate other's contributions with rather arbitrary downvotes, the kind of culture I avoid by eschewing social media. I encourage the downvoter in this instance to justify so at least we could tell what's lacking. Or maybe I could edit your post to derive everything from first principles and perhaps get you some upvotes... $\endgroup$ – user5713492 Feb 14 at 3:03
  • $\begingroup$ No thanks. I said what I said, and I stand by it. Upvotes are not important to me. $\endgroup$ – Chet Miller Feb 14 at 3:07
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    $\begingroup$ @MaxW See my ADDENDUM $\endgroup$ – Chet Miller Feb 14 at 12:37

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