Question

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.

Short answer

Answer: The rate of heat transfer (Q) and the heat transfer surface area (A) of the heat exchanger can be calculated following the steps outlined in the solution. Use the provided values and the ε-NTU method formulas to determine the values for Q and A.

Step-by-step solution

1. Define given variables

: \(c_{p,c} = 4180 \, \mathrm{J/kg \cdot K}\) (specific heat capacity of cold water) \(c_{p,h} = 4190 \, \mathrm{J/kg \cdot K}\) (specific heat capacity of hot water) \(\dot{m}_c = 0.25 \, \mathrm{kg/s} \) (cold water mass flow rate) \(\dot{m}_h = 3 \, \mathrm{kg/s}\) (hot water mass flow rate) \(T_{c,i} = 15^{\circ}\mathrm{C}\) (inlet cold water temperature) \(T_{c,o} = 45^{\circ}\mathrm{C}\) (outlet cold water temperature) \(T_{h,i} = 100^{\circ}\mathrm{C}\) (inlet hot water temperature) \(U = 950\,\mathrm{W/m^2\cdot K}\) (overall heat transfer coefficient)

2. Calculate the heat capacity rates

: \(C_c = \dot{m}_c \cdot c_{p,c}\) (cold water heat capacity rate) \(C_h = \dot{m}_h \cdot c_{p,h}\) (hot water heat capacity rate)

3. Determine the minimum and maximum heat capacity rates

: \(C_{min} = \min(C_c, C_h)\) (minimum heat capacity rate) \(C_{max} = \max(C_c, C_h)\) (maximum heat capacity rate)

4. Calculate the heat transfer rate (Q)

: \(Q = C_c \cdot(T_{c,o} - T_{c,i})\)

5. Calculate the heat exchanger effectiveness (ε)

: \(ε = \frac{Q}{C_{min} \cdot (T_{h,i} - T_{c,i})}\)

6. Calculate the ratio of heat capacity rates (C_r)

: \(C_r = \frac{C_{min}}{C_{max}}\)

7. Find the NTU value

: We need to find the appropriate relation to calculate the NTU given the ε and C_r values. For a counter-flow heat exchanger, the relation is as follows: \(ε = \frac{1 - e^{-NTU \cdot (1-C_r)}}{1 - C_r \cdot e^{-NTU \cdot (1-C_r)}}\) Solve the equation above for NTU.

8. Calculate the heat transfer surface area (A)

: \(A = \frac{NTU \cdot C_{min}}{U}\) Now, we have the values for the rate of heat transfer (Q) and the heat transfer surface area (A) of the heat exchanger.

Key concepts

Counter-Flow Heat Exchanger

A counter-flow heat exchanger is a system where two fluids flow in opposite directions, each in their own separate channels, to transfer heat from one fluid to the other. This setup provides a higher temperature gradient over the length of the exchanger, which makes heat transfer more efficient compared to parallel-flow systems.

In our exercise, cold water heats up as it flows in one direction while hot water cools down flowing in the opposite direction. This arrangement allows the cold water to continuously be exposed to the hottest water available, raising its temperature from the initial 15°C to the desired 45°C. It's particularly effective for transferring heat when the temperature difference between the hot and cold fluids varies along the length of the exchanger.

NTU Method

The Number of Transfer Units (NTU) method is a dimensional analysis for calculating the heat exchanger's effectiveness, which is its ability to transfer the maximum possible heat. Effectiveness (\( \boldsymbol{\text{ε}} \) is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate.

In the given exercise, effectiveness is necessary to determine the size and cost of the heat exchanger. By using the relevant formula for a counter-flow heat exchanger, you can determine the NTU value, which is then used to size the heat exchanger appropriately. The more transfer units, the larger the surface area required, and this relationship is pivotal to designing efficient heat exchangers.

Heat Capacity Rate

The heat capacity rate, noted as \( C \) in our calculations, is a measure of how much heat can be transferred per unit temperature. It is defined as the product of the mass flow rate (\( \boldsymbol{\text{\textbackslash{}\textbackslash{}dot{m}}} \) and the specific heat capacity (\( c_p \)) of the fluid.

When analyzing a heat exchanger, it's essential to know the hot (\( C_h \) and cold (\( C_c \) fluids' heat capacity rates, as well as the minimum (\( C_{min} \) and maximum (\( C_{max} \) values between them. These terms are a critical part of the heat transfer calculations and are used to find out the heat exchanger's effectiveness and subsequently the NTU.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient (\( U \) represents the efficiency of heat transfer across the area where the exchange takes place, including the resistance to heat flow by the tube walls and fouling. It is quantified in terms of power per unit area per unit temperature difference, typically in \( \boldsymbol{\text{W/m}^2 \text{\textbackslash{}\textbackslash{}cdot K}} \) or \( \boldsymbol{\text{BTU/hr ft}^2 \text{\textbackslash{}\textbackslash{}cdot °F}} \) in the imperial system.

In our solved problem, \( U \) of 950 \( \boldsymbol{\text{W/m}^2 \text{\textbackslash{}\textbackslash{}cdot K}} \) is used to find the required heat transfer surface area (\( A \) after calculating NTU. The coefficient is influenced by the characteristics of the fluids and the materials used, as well as their flow properties. A higher value indicates a more effective heat exchanger since it allows more heat to pass through a given area.