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Question

Oil in an engine is being cooled by air in a cross- flow heat exchanger, where both fluids are unmixed. Oil (cph=2047J/kgK)\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 75C,75^{\circ} \mathrm{C}, while air (cpc=1007J/kgK)\left(c_{p c}=1007 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\right) enters the shell side at 30C30^{\circ} \mathrm{C} with a flow rate of 0.21 kg/s. The overall heat transfer coefficient of the heat exchanger is 53W/m2K53 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} and the total surface area is 1m2.1 m^2. If the correction factor is F = 0.96, determine the outlet temperatures of the oil and air.

Solution

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Answered 2 years ago
Answered 2 years ago
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Given the following in the problem;

cp,c=1007 JkgKcp,h=2047 JkgKm˙c=0.21 kgsm˙h=0.026 kgsTh,i=75CTc,i=30CU=53 Wm2KA=1 m2F=0.26\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-ε\varepsilon method, in which case, we will be applying the formulas below including;

Q˙c=Cp,c(Tc,oTc,i)Q˙h=Cp,h(Th,iTh,o)ε=Q˙Q˙maxNTU=UAsCmin\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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