Question

Cold water $\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ leading to a shower enters a thin-walled double-pipe counterflow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of $1.25 \mathrm{~kg} / \mathrm{s}\( and is heated to \)60^{\circ} \mathrm{C}$ by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of $4 \mathrm{~kg} / \mathrm{s}\(. If the overall heat transfer coefficient is \)880 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger.

Short answer

Answer: The heat transfer surface area of the heat exchanger is approximately 5.61 square meters.

Step-by-step solution

Determine the heat transfer rate for cold water

To determine the heat transfer rate for the cold water, we'll use the formula \(Q_{c} = m_{c}c_{p_{c}}ΔT_{c}\), where \(m_{c}\) is the mass flow rate of cold water, \(c_{p_{c}}\) is the specific heat of cold water and \(ΔT_{c}\) is the temperature change for cold water. Given: \(m_c = 1.25 \mathrm{~kg/s}\), \(c_{p_{c}} = 4180 \mathrm{~J/kg \cdot K}\), Temperature change \(ΔT_{c} = T_{c_{out}} - T_{c_{in}} = 60 - 15 = 45 \mathrm{~K}\). Now, plug in the values into the formula: \(Q_{c} = (1.25 \mathrm{~kg/s}) \times (4180 \mathrm{~J/kg \cdot K}) \times (45\mathrm{~K})\) \(Q_{c} = 234675 \mathrm{~W}\).

Determine the heat transfer rate for hot water

Since the heat exchanger is operating in steady state and no heat is lost to the surroundings, the heat transfer rate for the hot water is equal to the heat transfer rate for the cold water: \(Q_{h} = Q_{c} = 234675 \mathrm{~W}\).

Calculate the logarithmic mean temperature difference (LMTD)

For counterflow heat exchangers, we can calculate LMTD using the following formula: \(ΔT_{lm} = \frac{(ΔT_1 - ΔT_2)}{\ln(\frac{ΔT_1}{ΔT_2})}\) where \(\Delta T_1 = T_{h_{in}} - T_{c_{out}}\) and \(\Delta T_2 = T_{h_{out}} - T_{c_{in}}\). First, we need to find the hot water outlet temperature (\(T_{h_{out}}\)) using the heat transfer rate formula. \(Q_{h} = m_{h}c_{p_{h}}(T_{h_{in}} - T_{h_{out}})\) To find the value of \(T_{h_{out}}\), we can rearrange the equation: \(T_{h_{out}} = T_{h_{in}} - \frac{Q_{h}}{m_{h}c_{p_{h}}}\) Given: \(m_h = 4 \mathrm{~kg/s}\), \(c_{p_{h}} = 4190 \mathrm{~J/kg \cdot K}\), \(T_{h_{in}} = 100 \mathrm{~^{\circ}C}\). Now, plug in the values into the formula: \(T_{h_{out}} = 100 - \frac{234675}{4 \times 4190} \approx 71.15\mathrm{~^{\circ}C}\). Now, we can find \(\Delta T_1\) and \(\Delta T_2\): \(\Delta T_1 = T_{h_{in}} - T_{c_{out}} = 100 - 60 = 40\mathrm{~K}\) \(\Delta T_2 = T_{h_{out}} - T_{c_{in}} = 71.15 - 15 = 56.15\mathrm{~K}\) Now, plug in these temperature differences into the LMTD formula: \(ΔT_{lm} = \frac{(40 - 56.15)}{\ln(\frac{40}{56.15})} \approx 47.10\mathrm{~K}\).

Determine the heat transfer surface area

To determine the heat transfer surface area, we use the overall heat transfer coefficient relation: \(U = \frac{Q}{AΔT_{lm}}\) Given: \(U = 880 \mathrm{~W/m^2 \cdot K}\), \(Q = 234675 \mathrm{~W}\), \(ΔT_{lm} = 47.10\mathrm{~K}\). Now, rearrange the equation to solve for \(A\): \(A = \frac{Q}{UΔT_{lm}}\) Now, plug in the values into the formula: \(A = \frac{234675}{880 \times 47.10} \approx 5.61\mathrm{~m^2}\). The heat transfer surface area of the heat exchanger is approximately 5.61 square meters.