Question

Consider a condenser unit (shell and tube heat exchanger) of an HVAC facility where saturated refrigerant \(\mathrm{R}-134 \mathrm{a}\) at a saturation pressure of \(1318.6 \mathrm{kPa}\) and at a rate of $2.5 \mathrm{~kg} / \mathrm{s}$ flows through thin-walled copper tubes. The refrigerant enters the condenser as saturated vapor and we wish to have a saturated liquid refrigerant at the exit. The cooling of refrigerant is carried out by cold water that enters the heat exchanger at \(10^{\circ} \mathrm{C}\) and exits at \(40^{\circ} \mathrm{C}\). Assuming the initial overall heat transfer coefficient of the heat exchanger to be $3500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the surface area of the heat exchanger and the mass flow rate of cooling water for complete condensation of the refrigerant. In practice, over a long period of time, fouling occurs inside the heat exchanger that reduces its overall heat transfer coefficient and causes the mass flow rate of cooling water to increase. Increase in the mass flow rate of cooling water will require additional pumping power, making the heat exchange process uneconomical. To prevent the condenser unit from underperforming, assume that fouling has occurred inside the heat exchanger and has reduced its overall heat transfer coefficient by 20 percent. For the same inlet temperature and flow rate of refrigerant, determine the new flow rate of cooling water to ensure complete condensation of the refrigerant at the heat exchanger exit.

Short answer

Question: Calculate the initial mass flow rate of cooling water and the new mass flow rate of cooling water after fouling has reduced the overall heat transfer coefficient by 20 percent. Answer: The initial mass flow rate of cooling water is 3.46 kg/s. The new mass flow rate of cooling water after fouling is 3.1 kg/s.

Step-by-step solution

Find the enthalpy change of the refrigerant

The saturated refrigerant R-134a enters as vapor and leaves as liquid. So, we need to find the enthalpy change between these two phases. Using the saturation pressure given, we can find the saturation temperature from a R-134a thermodynamic property table or online calculator. We get the saturation temperature \(T=40^{\circ}\mathrm{C}\). From the property tables, we find that \(h_{1} = h_{g} = 264.69\,\mathrm{kJ/kg}\) \(h_{2} = h_{f} = 100.1\,\mathrm{kJ/kg}\) The enthalpy change of the refrigerant is: \(\Delta h = h_{1}-h_{2} = 264.69 - 100.1 = 164.59\,\mathrm{kJ/kg}\)

Calculate the heat transfer required to cool the refrigerant

We can calculate the heat transfer rate required to cool the refrigerant as follows: \(q = \dot{m}_{\mathrm{ref}} \times \Delta h \) where \(\dot{m}_{\mathrm{ref}}\) is the mass flow rate of the refrigerant \(q = 2.5\,\mathrm{kg/s} \times 164.59\,\mathrm{kJ/kg} \times 1000\,\mathrm{J/kJ} = 411475\,\mathrm{W}\)

Find the surface area of the heat exchanger

We are given the overall heat transfer coefficient, \(U=3500\,\mathrm{W/m^2 \cdot K}\) and we have the heat transfer rate, we can calculate the surface area of the heat exchanger. The temperature differences between the hot and cold fluids are \(\Delta T_{1}=30^{\circ}\mathrm{C}\) and \(\Delta T_{2}=10^{\circ}\mathrm{C}\). We can calculate the log mean temperature difference (LMTD) as follows: \(LMTD = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})} = \frac{30-10}{\ln(30/10)}=19.1^{\circ}\mathrm{C}\) Now we can calculate the surface area: \(A = \frac{q}{U \times LMTD} = \frac{411475\,\mathrm{W}}{3500\,\mathrm{W/m^2\cdot K} \times 19.1\,\mathrm{K}} = 6.98\,\mathrm{m^2}\)

Calculate the initial mass flow rate of cooling water

We can use the heat transfer rate to calculate the mass flow rate of the cooling water: \(\dot{m}_{\mathrm{cw}} = \frac{q}{c_{p} \times \Delta T}\) where \(c_{p}\) is the specific heat capacity of the cooling water, and \(\Delta T\) is the temperature difference of the cooling water. For water, \(c_{p} = 4.18\,\mathrm{kJ/kg \cdot K}\). The temperature difference is given as \(\Delta T=30^{\circ}\mathrm{C}\). \(\dot{m}_{\mathrm{cw}} = \frac{411475\,\mathrm{W}}{4.18\,\mathrm{kJ/kg\cdot K} \times 30\,\mathrm{K} \times 1000\,\mathrm{J/kJ}} = 3.46\,\mathrm{kg/s}\)

Determine the new heat transfer coefficient with fouling

We are told that fouling reduces the overall heat transfer coefficient by 20 percent. So, the new heat transfer coefficient is: \(U_{\mathrm{new}} = 0.8 \times U = 0.8 \times 3500\,\mathrm{W/m^2 \cdot K} = 2800\,\mathrm{W/m^2 \cdot K}\)

Calculate the new mass flow rate of cooling water

The new mass flow rate of the cooling water can be calculated using the same formula as before. We use the newfound \(U_{\mathrm{new}}\) to calculate the new heat transfer rate: \(q_{\mathrm{new}} = A \times U_{\mathrm{new}} \times LMTD = 6.98\,\mathrm{m^2} \times 2800\,\mathrm{W/m^2 \cdot K} \times 19.1\,\mathrm{K} = 372058\,\mathrm{W}\) Now, calculate the new mass flow rate of the cooling water: \(\dot{m}_{\mathrm{cw,new}} = \frac{q_{\mathrm{new}}}{c_{p} \times \Delta T} = \frac{372058\,\mathrm{W}}{4.18\,\mathrm{kJ/kg \cdot K} \times 30\,\mathrm{K} \times 1000\,\mathrm{J/kJ}} = 3.1\,\mathrm{kg/s}\)