Question

In a 1-shell and 2-tube heat exchanger, cold water with inlet temperature of \(20^{\circ} \mathrm{C}\) is heated by hot water supplied at the inlet at \(80^{\circ} \mathrm{C}\). The cold and hot water flow rates are \(5000 \mathrm{~kg} / \mathrm{h}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), respectively. If the shelland-tube heat exchanger has a \(U A_{s}\) value of \(11,600 \mathrm{~W} / \mathrm{K}\), determine the cold water and hot water outlet temperatures. Assume \(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short answer

Question: Determine the cold and hot water outlet temperatures in a 1-shell and 2-tube heat exchanger given the flow rates, inlet temperatures, and overall heat transfer coefficient. Answer: The cold water outlet temperature is approximately 60°C and the hot water outlet temperature is approximately 60°C.

Step-by-step solution

Convert flow rates to kg/s

Given cold and hot water flow rates are in kg/hr. Convert these flow rates to kg/s for further calculations. Divide by 3600 to convert from hours to seconds. For the cold water flow rate: \( \dot{m}_{c} = \frac{5000}{3600} \mathrm{~kg/s} \) For the hot water flow rate: \( \dot{m}_{h} = \frac{10000}{3600} \mathrm{~kg/s} \)

Calculate the heat transfer

Using the constant temperature approach, the heat transfer can be determined from the equation: \( Q = U A_{s} \Delta T_{lm} \) Where: \(Q\) - Heat transfer (W) \(U A_{s}\) - Overall heat transfer coefficient (W/K) \(\Delta T_{lm}\) - Log mean temperature difference (K) To calculate the log mean temperature difference, we need the inlet and outlet temperature differences of both cold and hot water. Express outlet temperatures by substituting T for the temperature, noting inlet and outlet temperatures: \( \Delta T_{1} = T_{hi} - T_{co} = (80-20) \mathrm{K} \) \( \Delta T_{2} = T_{ho} - T_{ci} = (80-T_{ci}) - (20) \mathrm{K} \) Log mean temperature difference is calculated as follows: \( \Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{(\Delta T_{1}/\Delta T_{2})}} \) Now, to find \(T_{ci}\) and \(T_{ho}\), we can use the energy balance equations: -Cold water energy balance: \( Q = \dot{m}_{c} c_{pc} (T_{ci} - T_{co}) \) -Hot water energy balance: \( Q = \dot{m}_{h} c_{ph} (T_{hi} - T_{ho}) \)

Combine equations and solve for outlet temperatures

Now, we can combine the energy balance equations and derive solutions for \(T_{ci}\) and \(T_{ho}\). Divide the cold water energy balance equation by the hot water energy balance: \( \frac{\dot{m}_{c} c_{pc} (T_{ci} - T_{co})}{\dot{m}_{h} c_{ph} (T_{hi} - T_{ho})} = 1 \) Now, replace \(Q\) in the heat transfer equation with the cold water energy balance equation: \( \dot{m}_{c} c_{pc} (T_{ci} - T_{co}) = U A_{s} \frac{\Delta T_{1} - \Delta T_{2}}{\ln{(\Delta T_{1}/\Delta T_{2})}} \) Substitute the ΔT1 and ΔT2 expressions calculated in Step 2 and solve for \(T_{ci}\): \( T_{ci} =20 + \frac{U A_{s}}{\dot{m}_{c} c_{pc}} \frac{(80-20) - (80-T_{ci}) - (20)}{\ln{((80-20)/(80-T_{ci}) - (20))}} \) Now substitute UA value, cold water flow rate, cold water heat capacity and solve for cold water outlet temperature: \( T_{ci} \approx 60^{\circ}\mathrm{C} \) Put the \(T_{ci}\) back into the cold water energy balance equation and solve for hot water outlet temperature: \( T_{ho} = 80 - \frac{\dot{m}_{c} c_{pc}}{\dot{m}_{h} c_{ph}}(T_{ci} - T_{co}) \) \( T_{ho} \approx 60^{\circ}\mathrm{C} \) So, the cold water outlet temperature is approximately 60°C and the hot water outlet temperature is approximately 60°C.

Key concepts

Heat Transfer Calculations

Understanding heat transfer calculations is essential when analyzing systems like shell-and-tube heat exchangers, which are commonly used in industries for cooling or heating fluids. Heat transfer in such exchangers can be described using the simple energy balance equation which states that heat lost by the hot fluid is equal to the heat gained by the cold fluid, assuming no heat losses to the surroundings.

Mathematically, this can be expressed as \( Q_{hot} = Q_{cold} \), where \( Q \) is the heat transfer rate. In practical scenarios, several variables need to be considered, such as the specific heat capacity (\( c_p \) in J/kg∙K), flow rate of fluids (\( \dot{m} \) in kg/s), and the temperature change in fluids. The heat transfer rate for a fluid can be calculated using the equation \( Q = \dot{m} c_{p} \Delta T \), where \( \Delta T \) is the change in temperature. Short and concise sentences not only help to keep calculations clear but also make it easier to pinpoint where students might have gone wrong in their own computations.

For the effectiveness in tutoring, using real-world examples where these principles are applied can be substantially helpful. Explaining concepts using familiar contexts, such as how a car radiator works (an everyday example of a heat exchanger), can deepen the understanding. Throughout the explanation, it is also beneficial to link back to the core principles to reinforce learning.

Log Mean Temperature Difference (LMTD) Method

The log mean temperature difference (LMTD) method is a crucial concept in the realm of heat exchangers, as it accounts for the variation in temperature difference between the hot and cold fluids across the length of the heat exchanger. It is especially useful when the temperature difference is not constant.

The LMTD is calculated using the formula: \[ \Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{(\Delta T_{1}/\Delta T_{2})}} \] where \( \Delta T_{1} \) is the temperature difference at one end of the heat exchanger, and \( \Delta T_{2} \) at the other end. Conceptually, this method provides an average temperature difference that can be used to calculate the heat transfer rate effectively when considering the overall heat transfer coefficient \( U \) and the area \( A_s \) through the equation: \( Q = U A_{s} \Delta T_{lm} \).

To optimize understanding, illustrating the concept with a chart showing the temperature profile of both the hot and cold fluids along the heat exchanger may help students to visualize how the temperature difference changes and why the LMTD is a necessary calculation. Highlighting the relationship between the LMTD and the overall efficiency of the heat exchanger underscores the practical significance of the method.

Energy Balance Equations

In the context of heat exchangers, energy balance equations are fundamental for determining the outlet temperatures of the fluids involved. These equations are based on the first law of thermodynamics, which dictates conservation of energy, implying that energy cannot be created or destroyed within an isolated system.

The energy balance for a fluid stream is expressed as: \( Q = \dot{m}_{c} c_{p} (T_{out} - T_{in}) \), where \( T_{out} \) and \( T_{in} \) are the outlet and inlet temperatures, respectively. For each fluid stream in the heat exchanger, one balance equation is used for the hot fluid and another for the cold fluid. These equations can be solved simultaneously to find unknown temperatures, assuming the heat transfer rate \( Q \) is the same for both fluids.

To better elucidate the energy balance concept, one can draw parallels with financial budgeting, letting students relate the conservation of energy to balancing a checkbook—nothing is lost or gained, simply redistributed. Simplified analogies like this help students to transfer the logic of everyday situations to more complex scientific principles. Also, concentrating on numerical examples and guiding students through each step reinforces their capacity for independent problem-solving.