Question

Air at \(18^{\circ} \mathrm{C}\left(c_{p}=1006 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated to \(58^{\circ} \mathrm{C}\) by hot oil at \(80^{\circ} \mathrm{C}\left(c_{p}=2150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a cross-flow heat exchanger with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{K}\) and the mass flow rate of air is twice that of oil. Determine \((a)\) the effectiveness of the heat exchanger, \((b)\) the mass flow rate of air, and \((c)\) the rate of heat transfer.

Short answer

(a) The effectiveness of the heat exchanger is calculated using the formula \(ε = 1-e^{NTU^{-1}[1-e^{-(1+C_r)NTU}]}\) with the given NTU and \(C_r\) values. (b) The mass flow rate of air can be determined using the rate of heat transfer and given temperatures in the formula: \(m_{air} = \frac{q}{c_{p_{air}} \cdot (T_{air,out} - T_{air,in})}\). (c) The rate of heat transfer can be calculated using the formula \(q = ε \cdot C_{min} \cdot (T_{oil,in} - T_{air,in})\), substituting the calculated effectiveness and known temperature values.

Step-by-step solution

(Step 1: Determine the properties of air and oil)

Given, the specific heat capacities of air and oil are \(c_{p_{air}} = 1006 J/(kg \cdot K)\) and \(c_{p_{oil}} = 2150 J/(kg \cdot K)\). Given temperature of air at the inlet, \(T_{air,in} = 18^\circ C\) and the outlet, \(T_{air,out} = 58^\circ C\). Given temperature of oil at the inlet, \(T_{oil,in} = 80^\circ C\).

(Step 2: Calculate the heat capacities of air and oil)

Let's denote the mass flow rates of air and oil as \(m_{air}\) and \(m_{oil}\), respectively. Since the mass flow rate of air is twice that of oil, we have \(m_{air} = 2m_{oil}\). Heat capacities of air and oil can be calculated as \(C_{air} = m_{air} \cdot c_{p_{air}}\) and \(C_{oil} = m_{oil} \cdot c_{p_{oil}}\). (Note: Heat capacities are in \(W/K\))

(Step 3: Calculate the capacity ratios and minimum heat capacity)

Now calculate the capacity ratios, \(C_r = \frac{C_{air}}{C_{oil}}\). Using the given mass flow rate formula, the capacity ratios can be calculated as \(C_r = \frac{2m_{oil} \cdot c_{p_{air}}}{m_{oil} \cdot c_{p_{oil}}}=\frac{2c_{p_{air}}}{c_{p_{oil}}}=0.935\). Since \(C_{air} < C_{oil}\), the minimum heat capacity is given by \(C_{min} = C_{air}\).

(Step 4: Calculate the NTU of the heat exchanger)

Given the product of heat transfer surface area and overall heat transfer coefficient is \(UA = 750W/K\). We can find the NTU (Number of Transfer Units) as, NTU = \(\frac{UA}{C_{min}} = \frac{750}{m_{air} \cdot c_{p_{air}}}\).

(Step 5: Calculate the effectiveness of the heat exchanger)

In a cross-flow heat exchanger with air mixed and oil unmixed, the effectiveness \(ε\) can be found using the following formula: \(ε = 1-e^{NTU^{-1}[1-e^{-(1+C_r)NTU}]}\). Plug in the calculated NTU and \(C_r\) values into the formula to find the effectiveness of the heat exchanger.

(Step 6: Calculate the rate of heat transfer)

The rate of heat transfer \(q\) can be found using the following formula: \(q = ε \cdot C_{min} \cdot (T_{oil,in} - T_{air,in})\). Substitute the calculated effectiveness and known temperature values into the equation to find the rate of heat transfer.

(Step 7: Calculate the mass flow rate of air)

Now, rewrite the formula for rate of heat transfer as \(q = m_{air} \cdot c_{p_{air}} \cdot (T_{air,out} - T_{air,in})\). We can solve for \(m_{air}\) using the calculated rate of heat transfer and given temperatures: \(m_{air} = \frac{q}{c_{p_{air}} \cdot (T_{air,out} - T_{air,in})}\). Now you have determined: (a) the effectiveness of the heat exchanger (b) the mass flow rate of air (c) the rate of heat transfer

Key concepts

Cross-flow Heat Exchanger

A cross-flow heat exchanger is a type of heat exchanger where the two fluids flow perpendicular to each other. In our textbook exercise, air acts as one fluid, and oil acts as another. One of the key characteristics of this exercise's cross-flow heat exchanger is that the air and oil do not mix. Specifically, the air is mixed, meaning that temperatures within the air stream equalize, and the oil is unmixed, indicating different temperatures could exist within the oil stream at the outlet.

Understanding how these two fluids interact in a cross-flow pattern is essential for predicting the heat exchanger's performance. For students tackling this problem, it's crucial to recognize that the layout of the exchanger affects heat transfer efficiency due to the temperature gradients between the fluids, with more interaction generally leading to better heat transfer. The scenario of air being heated by oil is typical in industrial processes, where maintaining a specific temperature is crucial for operational efficiency.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, represented as 'U', indicates the heat transfer rate per unit area and per unit temperature difference. This coefficient is a fundamental aspect of the formula used in our exercise to determine the heat exchanger’s effectiveness. It essentially amalgamates different forms of heat transfer — conduction, convection, and radiation — into one coefficient.

The value of 'U' given in the problem (in Watts per Kelvin) is paired with the heat transfer surface area to produce a term 'UA', which represents the heat exchanger's potential to transfer heat. It's pivotal for students to understand that a higher 'U' value typically means a more efficient heat transfer process, assuming the surface area remains constant. When dealing with heat exchanger problems like the one in our exercise, students must often solve for one part of the 'UA' product when given the other and the desired rate of heat transfer.

Number of Transfer Units (NTU)

The Number of Transfer Units (NTU) is a dimensionless parameter that indicates the size of a heat exchanger relative to its ability to transfer heat. In the given exercise, the NTU is calculated by dividing 'UA' by the minimum heat capacity rate, which in this case is for air. The resulting number gives an indication of how capable the heat exchanger is of altering the temperature of the fluid with the lowest heat capacity rate.

To delve a bit deeper, the NTU method allows for the prediction of the heat exchanger's effectiveness, an all-important metric denoting how well the heat exchanger performs compared to an 'ideal' one. Notably, the formula used in our exercise adjusts the basic NTU concept for the specifics of the cross-flow arrangement with one fluid mixed and the other unmixed. This specificity is essential when using NTU to calculate effectiveness, as different flow arrangements impact heat transfer characteristics. For a student seeking to understand and solve problems related to heat exchangers, mastering the concept of NTU and its relation to effectiveness is of the utmost importance.