Determine surface area of heat exchange tube

Question

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.

Answer 1

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: $$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$

4. 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?