Question

Cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(14^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\) where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(65^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\). Determine the maximum outlet temperature of the cold water and the effectiveness of this heat exchanger.

Short answer

Answer: The maximum outlet temperature of the cold water is approximately 40.1°C, and the effectiveness of the heat exchanger is approximately 0.784.

Step-by-step solution

Define the given parameters

In this exercise, the known values are: - inlet temperature of cold water \((T_{c,1}) = 14^{\circ}C\) - inlet temperature of hot air \((T_{h,1}) = 65^{\circ}C\) - outlet temperature of hot air \((T_{h,2}) = 25^{\circ}C\) - mass flow rate of cold water \((m_c) = 0.35\text{ kg/s}\) - mass flow rate of hot air \((m_h) = 0.8\text{ kg/s}\) - specific heat capacity of cold water \((c_{p,c}) = 4.18\text{ kJ/(kg K)}\) - specific heat capacity of hot air \((c_{p,h}) = 1.0\text{ kJ/(kg K)}\)

Calculate the heat transfer rate

Applying the energy conservation principle, the heat transfer from the hot fluid to the cold fluid can be calculated using the formula: \(Q = m_hc_{p,h}(T_{h,1} - T_{h,2})\) Substitute the known values into the formula: \(Q = (0.8\,\mathrm{kg/s})(1.0\,\mathrm{kJ/(kg\,K)})(65^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C})\) Solve the equation for Q: \(Q = 32\,\text{kJ/s}\)

Calculate the outlet temperature

Now, we will calculate the outlet temperature of the cold water \((T_{c,2})\) using the energy conservation equation for the cold side: \(Q = m_cc_{p,c}(T_{c,2} - T_{c,1})\) Rearrange the equation for \(T_{c,2}\): \(T_{c,2} = \frac{Q}{m_cc_{p,c}} + T_{c,1}\) Substitute the known values and solve for \(T_{c,2}\): \(T_{c,2} = \frac{32\,\mathrm{kJ/s}}{(0.35\,\mathrm{kg/s})(4.18\,\mathrm{kJ/(kg\,K)})} + 14^{\circ}\mathrm{C}\) \(T_{c,2} \approx 40.1^{\circ}\mathrm{C}\) So, the maximum outlet temperature of the cold water is approximately \(40.1^{\circ}\mathrm{C}\).

Calculate the effectiveness of the heat exchanger

The effectiveness of the heat exchanger can be calculated using the formula: \(\epsilon = \frac{Q}{Q_\mathrm{max}}\) Where, \(Q_\mathrm{max}\) is the maximum possible heat transfer, which can be calculated using the minimum of the two products of mass flow rate and specific heat capacities. In this case, the hot air has the smaller product, so we will use that for the \(Q_\mathrm{max}\) calculation: \(Q_\mathrm{max} = m_hc_{p,h}(T_{h,1} - T_{c,1})\) Substitute the known values into the formula: \(Q_\mathrm{max} = (0.8\,\mathrm{kg/s})(1.0\,\mathrm{kJ/(kg\,K)})(65^{\circ}\mathrm{C} - 14^{\circ}\mathrm{C})\) Solve the equation for \(Q_\mathrm{max}\): \(Q_\mathrm{max} = 40.8\,\mathrm{kJ/s}\) Now, we can find the effectiveness: \(\epsilon = \frac{32\,\mathrm{kJ/s}}{40.8\,\mathrm{kJ/s}}\) \(\epsilon \approx 0.784\) So, the effectiveness of this heat exchanger is approximately \(0.784\).

Key concepts

Heat Transfer Rate

The heat transfer rate in a heat exchanger is a measurement of the thermal energy being exchanged per unit time between two fluids at different temperatures. It's a crucial aspect of heat exchanger design and operation because it quantifies how effective a heat exchanger is at transferring heat. For a student trying to understand heat exchangers, imagine putting your cold hands near a cup of hot tea, feeling the warmth—it's a kind of heat transfer.

In the context of our exercise, we used the formula:
\[Q = m_hc_{p,h}(T_{h,1} - T_{h,2})\]
This formula represents the heat energy lost by the hot air as it cools down from its inlet temperature to its outlet temperature. Here, \(m_h\) is the mass flow rate of hot air, \(c_{p,h}\) is the specific heat capacity of hot air, and \(T_{h,1} - T_{h,2}\) is the change in temperature of the hot air. The result is the rate at which heat is transferred to the cold water, given in kilojoules per second (kJ/s).

Outlet Temperature Calculation

Calculating the outlet temperature of a fluid in a heat exchanger is critical for determining how effective the heat exchanger is. This calculation helps to understand how much the fluid warms up or cools down as it passes through the heat exchanger. We do this by equating heat gained or lost by one fluid to the heat gained or lost by another, maintaining the principle of energy conservation.

To find the outlet temperature of the cold water on the aforementioned exercise, we rearranged the conservation of energy equation for the cold water side:
\[T_{c,2} = \frac{Q}{m_cc_{p,c}} + T_{c,1}\]
After plugging in the known values and solving for \(T_{c,2}\), we determined the maximum temperature to which the cold water can be heated. This step is vital for designing heat exchangers and predicting their operation in real-world scenarios.

Energy Conservation in Heat Exchangers

Energy conservation underpins all heat exchanger calculations. It's a fundamental law of physics that in any process, energy is neither created nor destroyed. For heat exchangers, this means the heat lost by the hot fluid must equal the heat gained by the cold fluid, excluding any inefficiencies or external loss.

In our exercise, energy conservation was applied to find the rate of heat transfer \(Q\) and ultimately the outlet temperature of the cold water. The effectiveness calculation also relies on the principle of energy conservation, as it compares the actual heat transfer to the maximum possible heat transfer if all the available thermal energy from the hot fluid were transferred to the cold fluid.

Specific Heat Capacity

Understanding specific heat capacity is essential for students studying heat exchangers. It's the quantity of heat needed to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). Think of it like a measure of thermal inertia, how reluctant a material is to change temperature.

In the given problem, we had the specific heat capacities for both the cold water and hot air \(\(c_{p,c} = 4.18\, \mathrm{kJ/(kg\,K)}\)\) and \(\(c_{p,h} = 1.0\, \mathrm{kJ/(kg\,K)}\)\) respectively. These values were crucial in our calculations because they determined how much energy was needed to heat or cool the respective fluids. Different substances have different specific heat capacities, so the choice of working fluids can greatly affect the design and operation of heat exchangers.