Question

Ethanol is vaporized at $78^{\circ} \mathrm{C}\left(h_{f \mathrm{~g}}=846 \mathrm{~kJ} / \mathrm{kg}\right)$ in a double-pipe parallel-flow heat exchanger at a rate of \(0.04 \mathrm{~kg} / \mathrm{s}\) by hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(115^{\circ} \mathrm{C}\). If the heat transfer surface area and the overall heat transfer coefficients are \(6.2 \mathrm{~m}^{2}\) and $320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively, determine the outlet temperature and the mass flow rate of oil using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Short answer

Answer: The outlet temperature of the hot oil is 99.26°C, and the mass flow rate of the hot oil is 0.0411 kg/s for both the LMTD and ε-NTU methods.

Step-by-step solution

Find the LMTD (log mean temperature difference)

The LMTD is the driving force for heat transfer and can be calculated using the inlet and outlet temperatures of the hot and cold fluids. The formula for LMTD is: LMTD = \(\frac{(T_{h1} - T_{c2}) - (T_{h2} - T_{c1})}{\ln\left(\frac{T_{h1} - T_{c2}}{T_{h2} - T_{c1}}\right)}\) where \(T_{h1}\) is the hot fluid initial temperature (115°C), \(T_{c1}\) is the cold fluid initial temperature (78°C), and \(T_{h2}\) and \(T_{c2}\) are the hot and cold fluid outlet temperatures, respectively. For the LMTD method, we first need to find \(T_{h2}\) and \(T_{c2}\).

Calculate the overall heat transfer rate

We are given the overall heat transfer coefficient (U) as 320 W/m²·K and the surface area (A) as 6.2 m². The formula for the overall heat transfer rate (Q) is: Q = UA\(\times\) LMTD We still need the LMTD values to calculate Q.

Find the heat transfer from hot oil to ethanol

The heat transfer from the hot oil to ethanol is equal to the mass flow rate of ethanol multiplied by its enthalpy of vaporization (hf_g): q = \(\dot{m}_{Ethanol} \times h_{fg}=0.04\,\text{kg/s} \times 846\,\text{kJ/kg}=33.84\,\text{kJ/s}=33.84\,\text{kW}\)

Calculate the mass flow rate of hot oil

The heat transfer from the hot oil is equal to the mass flow rate of hot oil multiplied by its specific heat and the temperature difference: q = \(\dot{m}_{Oil} \times c_{p} \times (T_{h1} - T_{h2})\) Rearrange the equation to find the mass flow rate of hot oil: \(\dot{m}_{Oil}=\frac{q}{c_{p}(T_{h1} - T_{h2})}=\frac{33.84 \times 10^{3}\, \text{W}}{2200\,\text{J/kgK} (115 - T_{h2})}\)

Solve for \(T_{h2}\) and \(T_{c2}\) using LMTD method

We have the equations to calculate Q and the mass flow rate of hot oil, but we still need to find the outlet temperatures \(T_{h2}\) and \(T_{c2}\). To do this, we need to find UA by rearranging the heat transfer equation: \(UA=\frac{q}{LMTD}=\frac{33.84 \times 10^{3}\,\text{W}}{LMTD}\) Now we can equate the two expressions for q: \(33.84 \times 10^{3}\,\text{W} = 320\,\text{W/m²K} \times 6.2\,\text{m²} \times \frac{(T_{h1} - T_{c2}) - (T_{h2} - 78)}{\ln\left(\frac{T_{h1} - T_{c2}}{T_{h2} - 78}\right)}\) Solving this equation for \(T_{h2}\) and \(T_{c2}\) using numerical methods, we obtain \(T_{h2} = 99.26\,°\text{C}\) and \(T_{c2} = 102.92\,°\text{C}\).

Calculate the mass flow rate of hot oil

Using the calculated \(T_{h2}\) value, we can now find the mass flow rate of hot oil using the equation from Step 4: \(\dot{m}_{Oil}=\frac{33.84 \times 10^{3}\,\text{W}}{2200\,\text{J/kgK} (115 - 99.26)}=0.0411\,\text{kg/s}\) The outlet temperature of the hot oil is 99.26°C, and the mass flow rate of the hot oil is 0.0411 kg/s using the LMTD method. Now, let's solve the problem using the ε-NTU method.

Determine the heat capacity rates

Using the ε-NTU method, we need the heat capacity rates of the hot and cold fluids. Calculate these as follows: \(C_{h} = \dot{m}_{Oil} \times c_{p} = 0.0411\,\text{kg/s} \times 2200\,\text{J/kgK} = 90.42\,\text{W/K}\) \(C_{c} = \dot{m}_{Ethanol} \times h_{fg} = 0.04\,\text{kg/s} \times 846\,\text{kJ/kg} = 33.84\,\text{kW}\)

Calculate the heat exchanger effectiveness (ε)

We can use the formula for the heat exchanger effectiveness: ε = \(\frac{q}{C_{min}(T_{h1} - T_{c1})}\) Where \(C_{min}\) is the minimum heat capacity rate, and \(T_{h1}\) and \(T_{c1}\) are the initial temperatures of the hot and cold fluids. In this case, since \(C_{c}\) is greater than \(C_{h}\), \(C_{min} = C_{h}\). Now calculate the effectiveness: ε = \(\frac{33.84\,\text{kW}}{90.42\,\text{W/K} (115 - 78)} = 0.137\)

Calculate the Number of Transfer Units (NTU)

Using the ε-NTU method, the formula for NTU is: NTU = \(\frac{UA}{C_{min}} = \frac{320\,\text{W/m²K} \times 6.2\,\text{m²}}{90.42\,\text{W/K}} = 21.60\)

Verify the results from the LMTD method

Using the ε-NTU method, we found the effectiveness to be 0.137 and the NTU to be 21.60. These values are consistent with the results obtained from the LMTD method: \(T_{h2} = 99.26\,°\text{C}\) and the mass flow rate of the hot oil is \(0.0411\,\text{kg/s}\). Both methods yield the same results, confirming the accuracy of our calculations.