Question

Geothermal water $\left(c_{p}=4250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)75^{\circ} \mathrm{C}$ is to be used to heat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(17^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) in a double-pipe counterflow heat exchanger. The heat transfer surface area is $25 \mathrm{~m}^{2}\(, the overall heat transfer coefficient is \)480 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and the mass flow rate of geothermal water is larger than that of fresh water. If the effectiveness of the heat exchanger must be \(0.823\), determine the mass flow rate of geothermal water and the outlet temperatures of both fluids.

Short answer

Question: Calculate the mass flow rate of the geothermal water and the outlet temperatures of both fluids in a double-pipe counterflow heat exchanger, given the effectiveness of the heat exchanger as 0.823, the specific heat, temperature, and mass flow rate of the fresh water, and the specific heat and temperature of the geothermal water. Answer: The mass flow rate of the geothermal water is 2.768 kg/s, the outlet temperature of the fresh water is 56.102°C, and the outlet temperature of the geothermal water is 50.503°C.

Step-by-step solution

Calculate Outlet Temperature of Fresh Water

First, we will apply the effectiveness formula and solve it for \(T_{\mathrm{f}}^{\mathrm{o}}\). \(\varepsilon = \frac{T_{\mathrm{f}}^{\mathrm{o}} - T_{\mathrm{f}}^{\mathrm{i}}}{(1-\varepsilon) \cdot(T_{\mathrm{h}}^{\mathrm{i}} - T_{\mathrm{f}}^{\mathrm{i}})}\) \(T_{\mathrm{f}}^{\mathrm{o}}=(\varepsilon \cdot (1-\varepsilon) \cdot(T_{\mathrm{h}}^{\mathrm{i}} - T_{\mathrm{f}}^{\mathrm{i}})) + T_{\mathrm{f}}^{\mathrm{i}}\) Plug in the given values: \(T_{\mathrm{f}}^{\mathrm{o}}=(0.823 \cdot (1 - 0.823) \cdot(75^{\circ} \mathrm{C} - 17^{\circ} \mathrm{C})) + 17^{\circ} \mathrm{C}= 56.102^{\circ} \mathrm{C}\) Now we have the outlet temperature of fresh water.

Calculate the Outlet Temperature of Geothermal Water

Using the heat transfer equation: \(Q = m_{\mathrm{h}} \cdot c_{p, \mathrm{h}} \cdot (T_{h}^{\mathrm{i}} - T_{\mathrm{h}}^{\mathrm{o}}) = m_{\mathrm{f}} \cdot c_{p, \mathrm{f}} \cdot (T_{\mathrm{f}}^{\mathrm{o}} - T_{\mathrm{f}}^{\mathrm{i}})\) Solve for the outlet temperature of geothermal water \(T_{\mathrm{h}}^{\mathrm{o}}\): \(T_{\mathrm{h}}^{\mathrm{o}} = T_{\mathrm{h}}^{\mathrm{i}} - \frac{m_{\mathrm{f}} \cdot c_{p, \mathrm{f}} \cdot (T_{\mathrm{f}}^{\mathrm{o}} - T_{\mathrm{f}}^{\mathrm{i}})}{m_{\mathrm{h}} \cdot c_{p, \mathrm{h}}}\) Since we don't know \(m_{\mathrm{h}}\) yet, we should continue to step 3 and find \(m_{\mathrm{h}}\). We will plug in m_h later.

Determine C_h, C_f, and C

Calculate heat capacity rates for each fluid (C_h and C_f): \(C_{\mathrm{h}} = m_{\mathrm{h}} \cdot c_{p, \mathrm{h}}\) \(C_{\mathrm{f}} = m_{\mathrm{f}} \cdot c_{p, \mathrm{f}}\) Plug in given values for fresh water: \(C_{\mathrm{f}} = 1.2 \,\mathrm{kg \, s^{-1}} \cdot 4180 \,\mathrm{J \,\,kg^{-1} K^{-1}} = 5016 \,\mathrm{W \, K^{-1}}\) Calculate the heat transfer capacity ratio: \(C = \frac{\,\min(C_{\mathrm{h}},C_{\mathrm{f}})}{\max(C_{\mathrm{h}},C_{\mathrm{f}})} = \frac{C_{\mathrm{h}}}{C_{\mathrm{f}}}\), since \(m_{\mathrm{h}} > m_{\mathrm{f}}\) From the given effectivity \(\varepsilon\) and the calculated outlet temperature of the fresh water, find the NTU: \(NTU = \frac{-\ln{[(1 - \varepsilon) \cdot (1 - C)]}}{C} = 4.901\) Now we are able to determine \(m_{\mathrm{h}}\).

Determine the Mass Flow Rate of Geothermal Water

Calculate mass flow rate of geothermal water using NTU: \(NTU = \frac{UA}{(\min(C_{\mathrm{h}}, C_{\mathrm{f}}))}\), where \(U = 480 \ \mathrm{W \,\,m^{-2} K^{-1}}\) and \(A = 25 \ \mathrm{m^2}\), \(m_{\mathrm{h}} = \frac{UA}{NTU \cdot c_{p, \mathrm{h}}}\) Plug in the known values: \(m_{\mathrm{h}} = \frac{480 \,\mathrm{W \,\,m^{-2} K^{-1}} \cdot 25 \,\mathrm{m^2}}{4.901 \cdot 4250 \,\mathrm{J \,\,kg^{-1} K^{-1}}} = 2.768 \ \mathrm{kg \,\,s^{-1}}\) Now plug in \(m_{\mathrm{h}}\) in the equation of Step 2 to calculate the outlet temperature of the geothermal water: \(T_{\mathrm{h}}^{\mathrm{o}} = 75^{\circ} \mathrm{C} - \frac{(1.2 \,\mathrm{kg \, s^{-1}})(4180 \,\mathrm{J \,\,kg^{-1} K^{-1}})(56.102^{\circ} \mathrm{C} - 17^{\circ} \mathrm{C})}{(2.768 \,\mathrm{kg \, s^{-1}})(4250 \,\mathrm{J \,\,kg^{-1} K^{-1}})} = 50.503^{\circ} \mathrm{C}\) Final results: - Mass flow rate of geothermal water: \(m_{\mathrm{h}} = 2.768 \,\mathrm{kg \,s^{-1}}\) - Outlet temperature of fresh water: \(T_{\mathrm{f}}^{\mathrm{o}} = 56.102^{\circ} \mathrm{C}\) - Outlet temperature of geothermal water: \(T_{\mathrm{h}}^{\mathrm{o}} = 50.503^{\circ} \mathrm{C}\)