Let's solve the problem step by step: ### **Step 1: Problem Restatement** - **Given:** - Hot water temperature, \(T_1 = 65^\circ C\) - Cold water temperature, \(T_2 = 5^\circ C\) - Pressure, \(P = 250\) kPa (for water, properties don't change much with pressure at these temps) - Mass flow rate ratio (hot/cold), \(r = \frac{\dot{m}_1}{\dot{m}_2}\), varies from 0.2 to 4 (step 0.2) - Neglect heat loss, mixing is complete. - **Find:** - Outlet temperature \(T_3\) as a function of \(r\) - Table and graph of \(T_3\) vs \(r\) - Most suitable mass ratio (range) and justification --- ### **Step 2: Energy Balance Equation** The mixing process is steady and adiabatic (no heat loss). The energy balance is: \[ \dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_3 \] Where: - \(h_1\), \(h_2\), \(h_3\) are the specific enthalpies at \(T_1\), \(T_2\), \(T_3\) Assume water's specific heat \(c_p \approx 4.18 \, \text{kJ/kg·K}\) (valid at these temperatures). Enthalpy difference: \[ h = h_{ref} + c_p (T - T_{ref}) \] But since only differences matter, \[ h_1 = c_p T_1,\quad h_2 = c_p T_2,\quad h_3 = c_p T_3 \] Let \(r = \frac{\dot{m}_1}{\dot{m}_2}\), so \(\dot{m}_1 = r \dot{m}_2\). Plug into the energy balance: \[ r \dot{m}_2 c_p T_1 + \dot{m}_2 c_p T_2 = (r\dot{m}_2 + \dot{m}_2) c_p T_3 \] \[ c_p \dot{m}_2 (r T_1 + T_2) = c_p \dot{m}_2 (r + 1) T_3 \] \(c_p \dot{m}_2\) cancels on both sides: \[ r T_1 + T_2 = (r + 1) T_3 \] \[ T_3 = \frac{r T_1 + T_2}{r+1} \] --- ### **Step 3: Substitute Values** Given \(T_1 = 65^\circ C\), \(T_2 = 5^\circ C\): \[ T_3 = \frac{r \times 65 + 5}{r+1} \] --- ### **Step 4: Compute \(T_3\) for Various \(r\)** Let’s make a table for \(r = 0.2\) to \(4.0\) (step 0.2): | r | \(T_3\) (\(^\circ C\)) | |------|--------------------| | 0.2 | \(\frac{0.2\times65+5}{1.2} = \frac{13+5}{1.2} = \frac{18}{1.2} = 15.0\) | | 0.4 | \(\frac{0.4\times65+5}{1.4} = \frac{26+5}{1.4} = \frac{31}{1.4} = 22.14\) | | 0.6 | \(\frac{0.6\times65+5}{1.6} = \frac{39+5}{1.6} = \frac{44}{1.6} = 27.5\) | | 0.8 | \(\frac{0.8\times65+5}{1.8} = \frac{52+5}{1.8} = \frac{57}{1.8} = 31.67\) | | 1.0 | \(\frac{65+5}{2} = \frac{70}{2} = 35.0\) | | 1.2 | \(\frac{78+5}{2.2} = \frac{83}{2.2} = 37.73\) | | 1.4 | \(\frac{91+5}{2.4} = \frac{96}{2.4} = 40.0\) | | 1.6 | \(\frac{104+5}{2.6} = \frac{109}{2.6} = 41.92\) | | 1.8 | \(\frac{117+5}{2.8} = \frac{122}{2.8} = 43.57\) | | 2.0 | \(\frac{130+5}{3} = \frac{135}{3} = 45.0\) | | 2.2 | \(\frac{143+5}{3.2} = \frac{148}{3.2} = 46.25\) | | 2.4 | \(\frac{156+5}{3.4} = \frac{161}{3.4} = 47.35\) | | 2.6 | \(\frac{169+5}{3.6} = \frac{174}{3.6} = 48.33\) | | 2.8 | \(\frac{182+5}{3.8} = \frac{187}{3.8} = 49.21\) | | 3.0 | \(\frac{195+5}{4.0} = \frac{200}{4.0} = 50.0\) | | 3.2 | \(\frac{208+5}{4.2} = \frac{213}{4.2} = 50.71\) | | 3.4 | \(\frac{221+5}{4.4} = \frac{226}{4.4} = 51.36\) | | 3.6 | \(\frac{234+5}{4.6} = \frac{239}{4.6} = 51.96\) | | 3.8 | \(\frac{247+5}{4.8} = \frac{252}{4.8} = 52.50\) | | 4.0 | \(\frac{260+5}{5.0} = \frac{265}{5.0} = 53.0\) | --- ### **Step 5: Plot \(T_3\) vs \(r\)** - **X-axis:** Mass ratio (\(r\)) - **Y-axis:** Outlet temperature (\(T_3\), \(^\circ C\)) (You can plot this with Excel, Matlab, or Python's matplotlib.) --- ### **Step 6: Determine Most Suitable Mass Ratio** - **Practical range for shower water temperature:** Typically between **35°C and 45°C** (comfortable for most people). - From table: - **\(r = 1.0\) gives \(T_3 = 35^\circ C\)** - **\(r = 2.0\) gives \(T_3 = 45^\circ C\)** So, **suitable mass ratio range is 1.0 to 2.0**. --- ### **Step 7: Justification** - **Below \(r = 1\):** Water is too cold for a shower (\(<35^\circ C\)) - **Above \(r = 2\):** Water is too hot for a shower (\(>45^\circ C\)), may cause burns or discomfort, and wastes hot water. --- ## **Final Answers** ### **1. Formula:** \[ T_3 = \frac{r T_1 + T_2}{r + 1} \] ### **2. Table:** | r | \(T_3\) (\(^\circ C\)) | |------|--------------------| | 0.2 | 15.0 | | 0.4 | 22.14 | | 0.6 | 27.5 | | 0.8 | 31.67 | | 1.0 | 35.0 | | 1.2 | 37.73 | | 1.4 | 40.0 | | 1.6 | 41.92 | | 1.8 | 43.57 | | 2.0 | 45.0 | | ... | ... | | 4.0 | 53.0 | ### **3. Plot:** *(Please use the table above to plot \(T_3\) vs \(r\) in your preferred software.)* ### **4. Most Suitable Mass Ratio Range:** \[ \boxed{1.0 \leq r \leq 2.0} \] This range provides shower water temperature between 35°C and 45°C, comfortable and safe for daily use. --- **If you need the plot, let me know which software you want to use!**

Certainly! Let's go through the problem step-by-step, including detailed calculations, explanations, and the final answers. --- ### **Problem Restatement** - Hot water temperature: \( T_1 = 65^\circ C \) - Cold water temperature: \( T_2 = 5^\circ C \) - Pressure: \( P = 250 \text{ kPa} \) (assumed to have negligible effect on water properties at these temperatures) - The ratio of hot to cold water mass flow rates: \( r = \frac{\dot{m}_1}{\dot{m}_2} \), varies from 0.2 to 4, in steps of 0.2. - Goal: - Find the outlet temperature \( T_3 \) as a function of \( r \). - Create a table with \( r \) and \( T_3 \). - Plot \( T_3 \) vs \( r \). - Determine the most suitable range of \( r \) for comfortable shower temperature (~35°C to 45°C). --- ### **Step 1: Understand the Mixing Process** Since heat losses are negligible, the energy balance simplifies to: \[ \text{Total enthalpy in} = \text{Total enthalpy out} \] Expressed as: \[ \dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_3 \] --- ### **Step 2: Express the Mass Flow Rates** Define: \[ \dot{m}_1 = r \, \dot{m}_2 \] Total mass flow rate: \[ \dot{m}_\text{total} = \dot{m}_1 + \dot{m}_2 = r \, \dot{m}_2 + \dot{m}_2 = (r + 1) \, \dot{m}_2 \] --- ### **Step 3: Specific Enthalpy of Water** For water over the temperature range 5°C to 65°C, and assuming constant specific heat: \[ h = c_p T \] Where: - \( c_p \approx 4.18 \text{ kJ/kg·K} \) (approximate specific heat capacity of water). At temperatures \( T_1 \) and \( T_2 \): \[ h_1 = c_p T_1 = 4.18 \times 65 = 271.7 \text{ kJ/kg} \] \[ h_2 = c_p T_2 = 4.18 \times 5 = 20.9 \text{ kJ/kg} \] Similarly, \( h_3 = c_p T_3 \), where \( T_3 \) is the unknown. --- ### **Step 4: Write the Energy Balance Equation** \[ \dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_3 \] Substitute: \[ r \dot{m}_2 h_1 + \dot{m}_2 h_2 = (r + 1) \dot{m}_2 h_3 \] Divide through by \( \dot{m}_2 \): \[ r h_1 + h_2 = (r + 1) h_3 \] Solve for \( h_3 \): \[ h_3 = \frac{r h_1 + h_2}{r + 1} \] Recall: \[ h_3 = c_p T_3 \] Thus: \[ T_3 = \frac{h_3}{c_p} = \frac{r h_1 + h_2}{(r + 1) c_p} \] But since \( h_1 = c_p T_1 \) and \( h_2 = c_p T_2 \): \[ T_3 = \frac{r c_p T_1 + c_p T_2}{(r + 1) c_p} = \frac{r T_1 + T_2}{r + 1} \] **Note**: The \( c_p \) cancels out, simplifying calculations. --- ### **Step 5: Final Formula for Outlet Temperature** \[ \boxed{ T_3 = \frac{r T_1 + T_2}{r + 1} } \] --- ### **Step 6: Calculate \( T_3 \) for Given Ratios** Using the formula, plug in \( T_1 = 65^\circ C \), \( T_2 = 5^\circ C \): | \( r \) | Calculation | \( T_3 \) (\(^\circ C\)) | |---------|--------------|--------------------------| | 0.2 | \( \frac{0.2 \times 65 + 5}{0.2 + 1} = \frac{13 + 5}{1.2} = \frac{18}{1.2} = 15.0 \) | 15.0 | | 0.4 | \( \frac{0.4 \times 65 + 5}{0.4 + 1} = \frac{26 + 5}{1.4} = \frac{31}{1.4} \approx 22.14 \) | 22.14 | | 0.6 | \( \frac{39 + 5}{1.6} = \frac{44}{1.6} = 27.5 \) | 27.5 | | 0.8 | \( \frac{52 + 5}{1.8} = \frac{57}{1.8} \approx 31.67 \) | 31.67 | | 1.0 | \( \frac{65 + 5}{2} = \frac{70}{2} = 35.0 \) | 35.0 | | 1.2 | \( \frac{78 + 5}{2.2} = \frac{83}{2.2} \approx 37.73 \) | 37.73 | | 1.4 | \( \frac{91 + 5}{2.4} = \frac{96}{2.4} = 40.0 \) | 40.0 | | 1.6 | \( \frac{104 + 5}{2.6} = \frac{109}{2.6} \approx 41.92 \) | 41.92 | | 1.8 | \( \frac{117 + 5}{2.8} = \frac{122}{2.8} \approx 43.57 \) | 43.57 | | 2.0 | \( \frac{130 + 5}{3} = \frac{135}{3} = 45.0 \) | 45.0 | | 2.2 | \( \frac{143 + 5}{3.2} = \frac{148}{3.2} \approx 46.25 \) | 46.25 | | 2.4 | \( \frac{156 + 5}{3.4} = \frac{161}{3.4} \approx 47.35 \) | 47.35 | | 2.6 | \( \frac{169 + 5}{3.6} = \frac{174}{3.6} \approx 48.33 \) | 48.33 | | 2.8 | \( \frac{182 + 5}{3.8} = \frac{187}{3.8} \approx 49.21 \) | 49.21 | | 3.0 | \( \frac{195 + 5}{4} = \frac{200}{4} = 50.0 \) | 50.0 | | 3.2 | \( \frac{208 + 5}{4.2} = \frac{213}{4.2} \approx 50.71 \) | 50.71 | | 3.4 | \( \frac{221 + 5}{4.4} = \frac{226}{4.4} \approx 51.36 \) | 51.36 | | 3.6 | \( \frac{234 + 5}{4.6} = \frac{239}{4.6} \approx 51.96 \) | 51.96 | | 3.8 | \( \frac{247 + 5}{4.8} = \frac{252}{4.8} \approx 52.50 \) | 52.50 | | 4.0 | \( \frac{260 + 5}{5} = \frac{265}{5} = 53.0 \) | 53.0 | --- ### **Step 7: Plot \( T_3 \) vs \( r \)** - **X-axis:** \( r \) (from 0.2 to 4.0) - **Y-axis:** \( T_3 \) (°C) *(Use spreadsheet software, Python, MATLAB, etc., to plot.)* --- ### **Step 8: Determine Suitable Range of \( r \)** - Typical comfortable shower temperature: **35°C to 45°C**. - From the table: - At \( r = 1.0 \), \( T_3 = 35^\circ C \). - At \( r = 2.0 \), \( T_3 = 45^\circ C \). - **Conclusion:** - **Range of \( r \) from 1.0 to 2.0** yields temperatures from 35°C to 45°C, ideal for showers. - **Below 1.0:** water too cold (<35°C). - **Above 2.0:** water too hot (>45°C), risking burns or discomfort. --- ### **Final Summary:** **1. Final formula:** \[ \boxed{ T_3 = \frac{r \times 65 + 5}{r + 1} } \] **2. Calculated temperatures:** | \( r \) | \( T_3 \) (°C) | |---------|----------------| | 0.2 | 15.0 | | 0.4 | 22.14 | | 0.6 | 27.5 | | 0.8 | 31.67 | | 1.0 | 35.0 | | 1.2 | 37.73 | | 1.4 | 40.0 | | 1.6 | 41.92 | | 1.8 | 43.57 | | 2.0 | 45.0 | | 2.2 | 46.25 | | 2.4 | 47.35 | | 2.6 | 48.33 | | 2.8 | 49.21 | | 3.0 | 50.0 | | 3.2 | 50.71 | | 3.4 | 51.36 | | 3.6 | 51.96 | | 3.8 | 52.50 | | 4.0 | 53.0 | **3. Suitable \( r \) range for comfortable shower temperature:** \[ \boxed{ 1.0 \leq r \leq 2.0 } \] This ensures the water temperature is comfortable and safe. --- **Let me know if you'd like assistance with plotting or further analysis!**