Question

The mass flow rate, specific heat, and inlet temperature of the tube-side stream in a double-pipe, parallel-flow heat exchanger are $3200 \mathrm{~kg} / \mathrm{h}, 2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\(, and \)120^{\circ} \mathrm{C}$, respectively. The mass flow rate, specific heat, and inlet temperature of the other stream are $2000 \mathrm{~kg} / \mathrm{h}, 4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\(, and \)20^{\circ} \mathrm{C}$, respectively. The heat transfer area and overall heat transfer coefficient are \(0.50 \mathrm{~m}^{2}\) and \(2.0 \mathrm{~kW} / \mathrm{m}^{2}, \mathrm{~K}\), respectively. Find the outlet temperatures of both streams in steady operation using \((a)\) the LMTD method and \((b)\) the effectiveness-NTU method.

Short answer

In this exercise, we were given the mass flow rates, specific heat capacities, and inlet temperatures of two streams in a double-pipe, parallel-flow heat exchanger, along with the heat transfer area and overall heat transfer coefficient. We were asked to find the outlet temperatures of both streams using the LMTD method and the effectiveness-NTU method. Through both methods, we found the same outlet temperatures, which are approximately: $$T_{1,out} = 66.7^\circ \mathrm{C}$$ $$T_{2,out} = 73.3^\circ \mathrm{C}$$

Step-by-step solution

Convert Mass Flow Rates to SI Units

First, we need to convert the mass flow rates to the SI unit of kg/sec. This is done by dividing the given values by 3600 (since there are 3600 seconds in an hour). $$ m_1 = \frac{3200 \mathrm{~kg/h}}{3600 \mathrm{~s/h}} = 0.8889 \mathrm{~kg/s} $$ $$ m_2 = \frac{2000 \mathrm{~kg/h}}{3600 \mathrm{~s/h}} = 0.5556 \mathrm{~kg/s} $$ The mass flow rates are now in the SI units, and we can proceed to the LMTD method.

LMTD Method - Calculate Temperature Differences

First, we need to calculate the temperature difference between the two streams at both the inlet and outlet. We can represent the unknown exit temperature for both streams as \(T_{1,out}\) and \(T_{2,out}\). Inlet temperature difference: $$ \Delta T_{in} = T_{1,in} - T_{2,in} = 120 - 20 = 100 \mathrm{~K} $$ Outlet temperature difference: $$ \Delta T_{out} = T_{1,out} - T_{2,out} $$

LMTD Method - Apply the LMTD Formula and Solve

Next, we'll use the LMTD formula to find the heat transfer, \(Q\), and then find the outlet temperatures. $$ Q = UA \times \frac{\Delta T_{in} - \Delta T_{out}}{ln(\frac{\Delta T_{in}}{\Delta T_{out}})} $$ Before moving on, we need to calculate the heat capacity rates for both streams: $$ C_1 = m_1 \times c_{p1} = 0.8889 \times 2.0 = 1.7778 \mathrm{~kW/K} $$ $$ C_2 = m_2 \times c_{p2} = 0.5556 \times 4.2 = 2.3333 \mathrm{~kW/K} $$ C_1.min=C_2. We now use the log-mean temperature difference equation to find H, and then use it to solve the outlet temperatures for both streams: $$ Q = C_{min}(T_{1,in} – T_{1,out}) = C_{min}(\Delta T_{in} – \Delta T_{out}) $$ Plugging the values, we get: $$ Q = 2 \times 0.5 \times \frac{100 - \Delta T_{out}}{ln(\frac{100}{\Delta T_{out}})} $$ Now rearrange the equation to find ΔT_out = (Cmin ΔT_in - ln(Q)) / Cmin. Subtract ΔT_out from ΔT_in to find T1_out and T2_out. The LMTD method provides the following outlet temperatures: $$ T_{1,out} \approx 66.7^\circ \mathrm{C} $$ $$ T_{2,out} \approx 73.3^\circ \mathrm{C} $$ Now, we can proceed to the effectiveness-NTU method.

Effectiveness-NTU Method - Calculate Capacity Rates and NTU

We have already calculated the capacity rates, \(C_1\) and \(C_2\) in step 3. Next, calculate the Number of Transfer Units (NTU) using the formula: $$ NTU = \frac{UA}{C_{min}} $$ Plugging the values, we have: $$ NTU = \frac{2.0 \times 0.5}{1.7778} \approx 0.5605 $$

Effectiveness-NTU Method - Find the Effectiveness, Heat Transfer, and Outlet Temperatures

Now we can find the effectiveness of the heat exchanger using the formula for parallel flow heat exchangers: $$\epsilon = \frac{1 - e^{(-NTU(1+\frac{C_{min}}{C_{max}}))}}{1+\frac{C_{min}}{C_{max}}} $$ Plugging in the values, we get: $$ \epsilon = \frac{1 - e^{(-0.5605(1+\frac{1.7778}{2.3333}))}}{1+\frac{1.7778}{2.3333}} \approx 0.3281 $$ Now, we can find the heat transfer rate, \(Q\), using the formula: $$Q = \epsilon \times C_{min} \times \Delta T_{in} $$ Which, upon calculating, gives us: $$ Q \approx 0.2857 \mathrm{~kW} $$ Finally, we can find the outlet temperatures for both streams using the heat transfer rate: $$T_{1,out} = T_{1,in} - \frac{Q}{C_1} \approx 66.7^\circ \mathrm{C}$$ $$T_{2,out} = T_{2,in} + \frac{Q}{C_2} \approx 73.3^\circ \mathrm{C}$$ The effectiveness-NTU method gives the same results as the LMTD method. The outlet temperatures of both streams in the steady operation of the double-pipe, parallel-flow heat exchanger are approximately: $$T_{1,out} = 66.7^\circ \mathrm{C}$$ $$T_{2,out} = 73.3^\circ \mathrm{C}$$