# Determine surface area of heat exchange tube

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 $$\pu{4.183 kJ kg-1 K-1}$$

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.

• 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.)
– MaxW
Commented Feb 14, 2019 at 7:32

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.

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?

• @user5713492 I thought that I had to take the $\ce{T_{c,out} - T_{h,in}}$ but it was the other way around. I now get that the log-mean temperature is $\ce{32.9^oC}$. I hope this is correct. Commented Feb 14, 2019 at 12:37
• This is the correct value for the LMTD. Commented Feb 14, 2019 at 12:39
• I just saw that @ChesterMiller added the calculations which confirms that I now have the right answer. Thank you both for your help! Commented Feb 14, 2019 at 12:40