## Related questions with answers

Oil in an engine is being cooled by air in a cross- flow heat exchanger, where both fluids are unmixed. Oil $\left(c_{p h}=2047 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\right)$ flowing with a flow rate of 0.026 kg/s enters the tube side at $75^{\circ} \mathrm{C},$ while air $\left(c_{p c}=1007 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\right)$ enters the shell side at $30^{\circ} \mathrm{C}$ with a flow rate of 0.21 kg/s. The overall heat transfer coefficient of the heat exchanger is $53 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and the total surface area is $1 m^2.$ If the correction factor is F = 0.96, determine the outlet temperatures of the oil and air.

Solution

VerifiedGiven the following in the problem;

$\begin{align*}c_{p,c}&=1007\ \dfrac{\text{J}}{\text{kg}K}\\c_{p,h}&=2047\ \dfrac{\text{J}}{\text{kg}K}\\\dot m_{c}&=0.21\ \dfrac{\text{kg}}{\text{s}}\\\dot m_{h}&=0.026\ \dfrac{\text{kg}}{\text{s}}\\T_{h,i}&=75^{\circ}C\\T_{c,i}&=30^{\circ}C\\U&=53\ \dfrac{\text{W}}{\text{m}^{2}K}\\A&=1\ \text{m}^{2}\\F&=0.26\end{align*}$

Even with the correction factor already given in the problem, since we notice that we lack the values for the outlet temperatures of both fluids, then it is better to opt to the NTU-$\varepsilon$ method, in which case, we will be applying the formulas below including;

$\begin{align}\dot Q_{c}&=C_{p,c}\cdot (T_{c,o}-T_{c,i})\\\dot Q_{h}&=C_{p,h}\cdot (T_{h,i}-T_{h,o})\\\varepsilon&=\dfrac{\dot Q}{\dot Q_{max}}\\NTU&=\dfrac{UA_{s}}{C_{min}}\end{align}$

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