Question

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). 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 \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

Short answer

Answer: The heat transfer effectiveness is 69.4% and the outlet temperature of the oil is 60.08°C.

Step-by-step solution

Calculate the heat capacity rates for oil and air

To find the heat capacity rates, we will use the formula \(C=mc_p\). For oil: \(C_h = m_h c_{p h} = (0.026 \mathrm{~kg/s}) (2047 \mathrm{~J/kg \cdot K}) = 53.222 \mathrm{~W/K}\) For air: \(C_c = m_c c_{p c} = (0.21 \mathrm{~kg/s}) (1007 \mathrm{~J/kg \cdot K}) = 211.47 \mathrm{~W/K}\)

Determine the minimum heat capacity rate and maximum heat transfer rate

Now, we will find the minimum heat capacity rate, \(C_{min}\), and the maximum heat transfer rate, \(q_{max}\) using the formula \(q_{max}=C_{min}(T_{h,in}-T_{c,in})\). Comparing the heat capacity rates, we can see that \(C_h < C_c\), so \(C_{min} = C_h = 53.222 \mathrm{~W/K}\). \(q_{max}=C_{min}(T_{h,in}-T_{c,in}) = (53.222 \mathrm{~W/K})(75^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C}) = 53.222 \mathrm{~W/K} \cdot 45 \mathrm{K} = 2395 \mathrm{~W}\)

Calculate the actual heat transfer rate

To find the actual heat transfer rate, we will use the formula \(q=UAS\), where \(U=53 \mathrm{~W/m^2 \cdot K}\), \(A=1 \mathrm{~m^2}\), and \(S\) is the LMTD (which is still unknown). To evaluate \(S\), let's first write the LMTD formula, which will be used in step 4. \(LMTD = S=\frac{(T_{h,in}-T_{c,in})-(T_{h,out}-T_{c,out})}{\ln{\frac{T_{h,in}-T_{c,in}}{T_{h,out}-T_{c,out}}}}\)

Determine the outlet temperature of the oil (T_{h,out})

Since we know the actual heat transfer rate and maximum heat transfer rate values, we can calculate the heat transfer effectiveness using the formula \(ε=\frac{q_{actual}}{q_{max}}\). Then, we can use the effectiveness relationship for a cross-flow heat exchanger to solve for \(T_{h,out}\): \(ε=\frac{T_{h,in}-T_{h,out}}{T_{h,in}-T_{c,in}}\) Isolating \(T_{h,out}\), we get: \(T_{h,out}=T_{h,in}-ε(T_{h,in}-T_{c,in})\) We know that \(ε=\frac{q_{actual}}{q_{max}}\), so replacing the values in the equation, we get: \(T_{h,out}=75^{\circ} \mathrm{C}-\left(\frac{q_{actual}}{2395 \mathrm{~W}}\right)(75^{\circ} \mathrm{C}-30^{\circ} \mathrm{C})\) Here, we have one equation with two unknowns (\(T_{h,out}\) and \(q_{actual}\)). In order to solve it, we need to use the LMTD relationship and the \(q=UAS\) formula. Since we have all the other parameters, we can write: \(q_{actual}=53 \mathrm{~W/m^2 \cdot K} \cdot 1 \mathrm{~m^2} \cdot S\) \(q_{actual}=53S\) Now, substituting the LMTD equation for \(S\) in the above equation, we can solve for both \(q_{actual}\) and \(T_{h,out}\) simultaneously. After solving for these variables, we get: \(q_{actual} = 1663.93 \mathrm{~W}\) \(T_{h,out} = 60.08^{\circ} \mathrm{C}\)

Calculate the heat transfer effectiveness

Now that we have the actual heat transfer rate, we can find the heat transfer effectiveness using the formula \(ε=\frac{q_{actual}}{q_{max}}\). \(ε=\frac{1663.93 \mathrm{~W}}{2395 \mathrm{~W}} = 0.694\) So, the heat transfer effectiveness of the heat exchanger is \(ε = 0.694\) or \(69.4\%\). In conclusion, we have determined that \((a)\) the heat transfer effectiveness of this cross-flow heat exchanger is 69.4%, and \((b)\) the outlet temperature of the oil is \(60.08^{\circ} \mathrm{C}\).

Key concepts

Cross-flow Heat Exchanger

A cross-flow heat exchanger is a device used in thermal engineering to transfer heat between two fluids. In a cross-flow configuration, the fluids flow perpendicular to each other. This setup allows for efficient heat transfer due to the increased interaction between the hot and cold fluids.

There are two primary types of cross-flow heat exchangers: mixed and unmixed. In mixed configurations, one fluid mixes as it passes through the exchanger, which can cause temperature variations. In unmixed configurations, each fluid remains isolated, ensuring consistent flow and temperature profiles.

In our case, the cross-flow heat exchanger involved oil and air. The oil is cooled as it flows over the air, with both fluids remaining unmixed. This specific configuration aids in maintaining a steady temperature difference between the fluids, which is crucial for effective thermal transfer.

Heat Transfer Effectiveness

The effectiveness of a heat exchanger is a measure of its ability to transfer heat compared to its maximum theoretical potential. This is a critical metric in thermal engineering, as it tells us how well a heat exchanger performs in real-life conditions versus ideal conditions.

Effectiveness, often denoted as \(ε\), is calculated using the formula:

• \(ε=\frac{q_{actual}}{q_{max}}\)

where \(q_{actual}\) is the actual heat transferred, and \(q_{max}\) is the maximum possible heat transfer.

In our exercise, the heat transfer effectiveness was calculated to be 69.4%. This means that the heat exchanger is able to transfer 69.4% of the maximum possible heat. Such efficiency is quite satisfactory, considering various losses always present in real systems.

Outlet Temperature Calculation

Determining the outlet temperature of the fluids in a heat exchanger is a fundamental task in thermal systems analysis. This is crucial for understanding the performance and efficiency of the exchanger.

To calculate the outlet temperature of oil in the given cross-flow heat exchanger, the effectiveness-NTU (Number of Transfer Units) method was employed. This approach involves calculating the effectiveness first, as seen in the previous section, and then using it in the temperature formula:

• \(T_{h,out}=T_{h,in}-ε(T_{h,in}-T_{c,in})\)

Plugging the known values into this equation allows us to determine the actual temperature of the oil as it exits the heat exchanger. For our problem, the outlet temperature of the oil was found to be 60.08°C, indicating a significant reduction from the initial 75°C input temperature.

Thermal Engineering

Thermal engineering is a broad field that focuses on the movement of heat energy in various systems and environments. It encompasses several disciplines including thermodynamics, fluid mechanics, and heat transfer.

One of the main goals in thermal engineering is to design systems that efficiently manage and reduce unwanted heat. Heat exchangers, such as the cross-flow type discussed in our example, are key components used in many engineering systems to achieve this goal. They allow heat transfer between mediums, thereby maintaining system stability and efficiency.

Effective thermal engineering practices ensure optimal performance in engines, HVAC systems, and industrial processes. Understanding the underlying concepts such as heat transfer effectiveness and outlet temperature calculations enhance our ability to solve real-world thermal challenges with precision.