Question

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) to cool the engine-coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

Short answer

Answer: The outlet temperature of the air is 50°C, and the rate of heat transfer between the two fluids is 200 kW.

Step-by-step solution

Determine the product of the heat capacity rates of two fluids

We need to find the product of the two heat capacity rates (for air \(C_{min}\) and for coolant \(C_{max}\)) to determine the outlet temperatures and the heat transfer between the two fluids. The heat capacity rates for air and coolant can be calculated as follows: \(C_{air} = \dot{m}_{air} \cdot c_{p, air}\) \(C_{coolant} = \dot{m}_{coolant} \cdot c_{p, coolant}\)

Calculate the effective capacity rates

First, we need to determine the heat capacity rate for air and the coolant: \(C_{air} = (10 \frac{kg}{s})(1.00 \frac{kJ}{kg \cdot K}) = 10 \frac{kW}{K}\) \(C_{coolant} = (5 \frac{kg}{s})(4.00 \frac{kJ}{kg \cdot K}) = 20 \frac{kW}{K}\) Since \(C_{air} < C_{coolant}\), we have \(C_{min} = C_{air}\) and \(C_{max} = C_{coolant}\). Now we can calculate the outlet temperature of the air:

Apply the effectiveness formula

The effectiveness is given by: \(Effectiveness = \frac{T_{air,out} - T_{air,in}}{(T_{coolant,in} - T_{air,in})}\) We can rearrange the formula and solve for \(T_{air,out}\): \(T_{air,out} = Effectiveness \cdot (T_{coolant,in} - T_{air,in}) + T_{air,in}\) Plug in the given values and temperatures: \(T_{air,out} = 0.4 \cdot (80 - 30) + 30 = (0.4 \cdot 50) + 30 = 20 + 30 = 50 ^{\circ}C\) So the outlet temperature of the air is \(50^{\circ}C\).

Calculate the rate of heat transfer

To find the rate of heat transfer \(Q\), we can use the formula: \(Q = C_{min} \cdot Effectiveness \cdot (T_{coolant,in} - T_{air,in})\) Plug in the values we have: \(Q = (10 \frac{kW}{K})(0.4) \cdot (50^{\circ}C) = 4 \cdot 50 = 200 \, kW\) So the rate of heat transfer between the two fluids is \(200 \, kW\). In conclusion, the outlet temperature of the air is \(50^{\circ}C\) and the rate of heat transfer between the two fluids is \(200 \, kW\).

Key concepts

Cross-Flow Heat Exchanger

Cross-flow heat exchangers are a type of heat exchanger where two fluids cross each other perpendicularly. Imagine an automobile radiator where a cool air stream flows across the hot engine coolant, cooling it down. This is the essence of a cross-flow heat exchanger. In these applications, maximizing surface area contact between the fluids is key. The air and coolant have distinct flow paths—they don't mix, but heat is transferred between them through the radiator walls. This setup is efficient for cooling, especially in automotive contexts where space and weight are constraints. The performance of a cross-flow heat exchanger is often evaluated in terms of its heat transfer capability, typically using metrics like the UA value (overall heat transfer coefficient times the heat exchanger's surface area). The given problem describes such a system, where air and coolant flow rates impact the effectiveness of the heat exchange process. Notably, the two streams crossing at right angles contribute to the complex calculations necessary for determining outlet temperatures and other performance metrics.

Heat Capacity Rate

Heat capacity rate is a crucial concept when analyzing heat exchangers. It is essentially the product of the mass flow rate of the fluid and its specific heat capacity. In simple terms, it tells us how much heat a fluid can carry per unit degree temperature change. This is expressed mathematically as \( C = \dot{m} \cdot c_p \), where \( \dot{m} \) is the mass flow rate and \( c_p \) is the specific heat capacity of the fluid.

For the air in the problem, the heat capacity rate is calculated as \( C_{air} = 10 \frac{\text{kg}}{\text{s}} \times 1.00 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} = 10 \frac{\text{kW}}{\text{K}} \). For the engine coolant, it's \( C_{coolant} = 5 \frac{\text{kg}}{\text{s}} \times 4.00 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} = 20 \frac{\text{kW}}{\text{K}} \).

Identifying \( C_{min} \) and \( C_{max} \) is important for the heat exchange process as it determines the capacity limits. In the problem, \( C_{min} \) belongs to air since it has the lower capacity rate, and \( C_{max} \) is for the coolant. Understanding the heat capacity rate helps assess how efficiently each fluid can carry heat, influencing the effectiveness calculations of the entire system.

Effectiveness-NTU Method

The Effectiveness-NTU (Number of Transfer Units) Method is useful for determining heat exchanger performance, especially when outlet temperatures are unknown. Effectiveness, denoted as \( \varepsilon \), is defined as the ratio of actual heat transfer to the maximum possible heat transfer if the exchanger was ideal. It can be calculated using the formula: \( \varepsilon = \frac{Q}{Q_{max}} \). However, in practical applications like the given radiator problem, it is used as \( \varepsilon = \frac{T_{out} - T_{in}}{T_{max} - T_{in}} \). Here, \( T_{out} \) is the outlet temperature, \( T_{in} \) is the initial temperature, and \( T_{max} \) is the maximum temperature difference.

In this problem, effectiveness is already provided as 0.4. By using this in the effectiveness formula, one can solve for the unknown outlet temperature, \( T_{air,out} = \varepsilon \cdot (T_{coolant,in} - T_{air,in}) + T_{air,in} \), giving a result of \( 50^{\circ}C \).
This value assists in calculating the heat transferred, using \( Q = C_{min} \cdot \varepsilon \cdot (T_{coolant,in} - T_{air,in}) \), resulting in a heat transfer rate of \( 200 \text{kW} \). This method is favored for its simplicity and effectiveness in scenarios where detailed specifications might not be fully known.