Question

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in the shell side of a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine the outlet temperature of the cold water and the heat transfer rate for the heat exchanger.

Short answer

Based on the given information and calculations, the outlet temperature of the cold water in the 1-shell and 2-tube heat exchanger is approximately \(45.45^{\circ} \mathrm{C}\), and the heat transfer rate is approximately \(63717.91 \mathrm{~W}\).

Step-by-step solution

Calculate the heat transfer rate using the LMTD method

The Log Mean Temperature Difference (LMTD) method is a commonly used method to calculate the heat transfer rate in a heat exchanger. The formula for LMTD is: $$ \mathrm{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} $$ where \(\Delta T_1\) is the temperature difference between the hot fluid and cold fluid at one end of the heat exchanger, and \(\Delta T_2\) is the temperature difference between the hot fluid and cold fluid at the other end of the heat exchanger. In this case, $$ \Delta T_1 = T_{h1} - T_{c1} $$ $$ \Delta T_2 = T_{h2} - T_{c2} $$ Since the water vapor is saturated at \(100^{\circ} \mathrm{C}\), we can use this as the temperature of the hot fluid, \(T_{h1} = T_{h2} = 100^{\circ} \mathrm{C} = 373 \mathrm{~K}\). The cold water inlet temperature is \(T_{c1} = 15^{\circ} \mathrm{C} = 288 \mathrm{~K}\). Let the cold water outlet temperature be \(T_{c2}\). Then, the heat transfer rate, \(Q\), can be calculated as: $$ Q = U \times A \times \mathrm{LMTD} $$ where \(U\) is the overall heat transfer coefficient and \(A\) is the surface area of the heat exchanger. We will use this formula in the next step to find the value of \(T_{c2}\).

Use the energy balance equation to determine the outlet temperature of the cold water

For the energy balance equation, we will consider the cold water stream. The equation is: $$ Q = m_c \times c_{pc} \times (T_{c2} - T_{c1}) $$ We already know the values of \(m_c\), \(c_{pc}\), and \(T_{c1}\), and we can express \(Q\) using the LMTD equation from step 1: $$ U \times A \times \mathrm{LMTD} = m_c \times c_{pc} \times (T_{c2} - T_{c1}) $$ Substitute the given values, \(U = 2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(A = 0.5 \mathrm{~m}^{2}\), \(m_c = 0.5 \mathrm{~kg} / \mathrm{s}\), and \(c_{pc} = 4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the temperature \(T_{h1} = T_{h2} = 373 \mathrm{~K}\), \(T_{c1} = 288 \mathrm{~K}\), , $$ 2000 \times 0.5 \times \frac{(373 - T_{c2}) - (373 - 288)}{\ln((373 - T_{c2}) / (373 - 288))} = 0.5 \times 4179 \times (T_{c2} - 288) $$ Solve this nonlinear equation for \(T_{c2}\) using an iterative method or by using numerical software. After solving for \(T_{c2}\), we get: $$ T_{c2} \approx 318.45 \mathrm{~K} $$ Now we can convert this back to Celsius: $$ T_{c2} \approx 45.45^{\circ} \mathrm{C} $$ Now that we have \(T_{c2}\), we can calculate \(Q\) using the energy balance equation: $$ Q = m_c \times c_{pc} \times (T_{c2} - T_{c1}) $$ $$ Q = 0.5 \times 4179 \times (318.45 - 288) $$ $$ Q \approx 63717.91 \mathrm{~W} $$ So the outlet temperature of the cold water is approximately \(45.45^{\circ} \mathrm{C}\), and the heat transfer rate for the heat exchanger is approximately \(63717.91 \mathrm{~W}\).

Key concepts

LMTD Method

Understanding the Log Mean Temperature Difference (LMTD) method is essential when dealing with heat exchangers. It's a foundational concept that simplifies the complex process of heat transfer between two fluids at different temperatures.

When we apply the LMTD method, we're essentially looking at the average temperature difference between the two fluids across the heat exchanger. This average isn’t just a simple arithmetic mean; instead, it is calculated to account for the varying temperature difference along the length of the heat exchanger, which affects the heat transfer rate. The reason LMTD is so important is that it more accurately reflects how much heat is actually being transferred in real-world situations.

The LMTD formula is given by:
\[LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}\]
Here, \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences between the hot and cold fluids at two ends of the heat exchanger. To calculate these differences, we need the inlet and outlet temperatures of both fluids. By using these differences in the LMTD equation, we achieve a more accurate representation of the heat transfer rate, which can then be applied to further equations for comprehensive analysis and design of heat exchangers.

Energy Balance Equation

The energy balance equation is a fundamental principle used in thermodynamics to ensure that energy is conserved in a system. In the context of heat exchangers, we apply this equation to relate the heat loss or gain of one fluid to the heat gain or loss of the other fluid.

In a heat exchanger, the energy provided by the hot fluid must be equal to the energy received by the cold fluid, assuming there is no heat loss to the surroundings. For the cold water in our exercise, we have:
\[Q = m_c \times c_{pc} \times (T_{c2} - T_{c1})\]
Here, \(Q\) is the heat transfer rate, \(m_c\) represents the mass flow rate of the cold water, \(c_{pc}\) is the specific heat capacity, and \(T_{c1}\) and \(T_{c2}\) are the inlet and outlet temperatures, respectively.

To find the outlet temperature, \(T_{c2}\), we can rearrange the equation. The heat transfer rate \(Q\) can also be found from earlier calculations using the LMTD method. Thus, the energy balance equation acts as a critical link between the physical change in the fluid and the theoretical calculations, ensuring our system aligns with the conservation of energy.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, denoted as \(U\), is an essential parameter in the analysis of heat exchangers. It reflects the ability of a heat exchanger to transfer heat between fluids through its material and area. A high \(U\) indicates efficient heat transfer, while a low \(U\) could indicate potential issues or a need for a larger heat exchanger surface.

The value of \(U\) combines the conductive and convective heat transfer characteristics of the exchanger's materials, the fluids involved, and flow characteristics such as turbulence. In our exercise:
\[Q = U \times A \times LMTD\]
Here, \(A\) represents the surface area over which heat transfer occurs. By multiplying \(U\), \(A\), and the previously calculated LMTD, we can determine the heat transfer rate, \(Q\).

Understanding \(U\) allows engineers to design and select appropriate heat exchanger types and sizes for specific applications. For instance, certain applications might require materials with higher thermal conductivity or specific flow arrangements to achieve the desired \(U\) value. It's critical in predicting the performance of heat exchangers and ensuring they meet the requirements of the process they're facilitating.