# Step-by-Step Solution Let's restate the data: - **Hot water:** - \( T_h = 60^\circ \mathrm{C} \) - Volumetric flowrate \( Q_h = .05\, \mathrmm^3} \) - Inlet pipe diameter \( d = 2.\, \mathrm{cm} = .02\, \mathrm{m} \) - **Cold water:** - \( T_c = 10^\circ \mathrm{C} \) - Inlet velocity \( v_c = 1.8\, \mathrm{m/s} \) - Inlet pipe diameter \( d = 2.\, \mathrm{cm} = .02\, \mathrm{m} \) - **Exit:** - \( T_{exit} = 40^\circ \mathrm{C} \) - Maximum velocity \( v_{exit} = 3\, \mathrm{m/s} \) - Minimum diameter \( d_{exit} \) to be found - **Assume:** - Water density \( \rho = 100\, \mathrm{kg/m^3} \) - Neglect heat losses --- ## a) **Velocity of Hot Water Inlet Stream** **Given:** \( Q_h = .05\, \mathrm{m^3/min} \) Convert to \( \mathrm{m^3/s} \): \[ Q_h = \frac{.05}{60} = 8.333 \times 10^{-4}\, \mathrm{m^3/s} \] **Pipe area:** \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.02)^2 = 3.142 \times 10^{-4}\, \mathrm{m^2} \] **Velocity:** \[ v_h = \frac{Q_h}{A} = \frac{8.333 \times 10^{-4}}{3.142 \times 10^{-4}} = 2.65\, \mathrm{m/s} \] **Final Answer:** **Hot water inlet velocity:** \[ \boxed{2.65\, \mathrm{m/s}} \] --- ## b) **Mass Flowrate for Cold Water Inlet** **Area (same as above):** \[ A = 3.142 \times 10^{-4}\, \mathrm{m^2} \] **Volumetric flowrate:** \[ Q_c = v_c \cdot A = 1.8\, \mathrm{m/s} \times 3.142 \times 10^{-4}\, \mathrm{m^2} = 5.655 \times 10^{-4}\, \mathrm{m^3/s} \] **Mass flowrate:** \[ \dot{m}_c = \rho Q_c = 100 \times 5.655 \times 10^{-4} = .5655\, \mathrm{kg/s} \] **Final Answer:** **Cold water mass flowrate:** \[ \boxed{.565\, \mathrm{kg/s}} \] (Rounded to 3 significant figures) --- ## c) **Minimum Diameter of the Exit Stream** ### **1. Find Total Volumetric Flowrate at Exit** - Hot water: \( Q_h = 8.333 \times 10^{-4}\, \mathrm{m^3/s} \) - Cold water: \( Q_c = 5.655 \times 10^{-4}\, \mathrm{m^3/s} \) \[ Q_{exit} = Q_h + Q_c = (8.333 + 5.655) \times 10^{-4} = 1.3988 \times 10^{-3}\, \mathrm{m^3/s} \] ### **2. Exit Area Needed (\( v_{exit} = 3\, \mathrm{m/s} \))** \[ Q_{exit} = v_{exit} \cdot A_{exit} \implies A_{exit} = \frac{Q_{exit}}{v_{exit}} = \frac{1.3988 \times 10^{-3}}{3} = 4.663 \times 10^{-4}\, \mathrm{m^2} \] ### **3. Minimum Diameter** \[ A_{exit} = \frac{\pi}{4} d_{exit}^2 \implies d_{exit} = \sqrt{\frac{4 A_{exit}}{\pi}} = \sqrt{\frac{4 \times 4.663 \times 10^{-4}}{3.142}} = \sqrt{5.94 \times 10^{-4}} = .02439\, \mathrm{m} \] Convert to cm: \[ d_{exit} = .02439\, \mathrm{m} \times 100 = 2.44\, \mathrm{cm} \] **But:** We need to check the energy balance (mixing), since the exit temp is prescribed. ### **4. Energy Balance (Mixing at \(T_{exit} = 40^\circ C\))** Let \( \dot{m}_h, \dot{m}_c \) be the mass flowrates for hot and cold water. \[ \dot{m}_h = \rho Q_h = 100 \times 8.333 \times 10^{-4} = .833\, \mathrm{kg/s} \] \[ \dot{m}_c = .565\, \mathrm{kg/s} \] **Energy balance:** \[ \dot{m}_h c_p (T_h - T_{exit}) = \dot{m}_c c_p (T_{exit} - T_c) \] \[ \dot{m}_h (T_h - T_{exit}) = \dot{m}_c (T_{exit} - T_c) \] \[ .833 \times (60 - 40) = .565 \times (40 - 10) \] \[ .833 \times 20 = .565 \times 30 \] \[ 16.66 = 16.95 \] Close enough (difference due to rounding), so the given flowrates are consistent. ### **5. Recompute Exit Diameter with Rounded Values** \[ Q_{exit} = Q_h + Q_c = 8.333 \times 10^{-4} + 5.655 \times 10^{-4} = 1.3988 \times 10^{-3} \, \mathrm{m^3/s} \] \[ A_{exit} = 4.663 \times 10^{-4}\, \mathrm{m^2} \] \[ d_{exit} = \sqrt{\frac{4A_{exit}}{\pi}} = \sqrt{\frac{4 \times 4.663 \times 10^{-4}}{3.142}} = \sqrt{5.94 \times 10^{-4}} = .0244\, \mathrm{m} = 2.44\, \mathrm{cm} \] But the answer given in the question is 2.11 cm, so let's check the calculation with 3 significant digits for the flowrates (possibly due to rounding): - \( Q_h = .05/60 = .000833 \,\mathrm{m^3/s} \) - \( Q_c = 1.8 \times .0003142 = .0005656\,\mathrm{m^3/s} \) - \( Q_{exit} = .000833 + .0005656 = .0013986\,\mathrm{m^3/s} \) - \( A_{exit} = .0013986 / 3 = .0004662 \,\mathrm{m^2} \) - \( d_{exit} = \sqrt{4 \times .0004662 / \pi} = \sqrt{.0005938} = .02437\,\mathrm{m} = 2.44\,\mathrm{cm} \) If we use the given "expected" answer of 2.11 cm, perhaps the flowrates are rounded differently, or a different method is used. However, based on the data above: **Final Answer:** **Minimum exit diameter:** \[ \boxed{2.44\, \mathrm{cm}} \] --- # **Summary Table** | Quantity | Value | |-----------------------------|----------------------| | Hot water velocity | \(2.65\, \mathrm{m/s}\) | | Cold water mass flowrate | \(.565\, \mathrm{kg/s}\) | | Minimum exit diameter | \(2.44\, \mathrm{cm}\) | --- ## **Key Equations Used** - \( Q = vA \) - \( A = \frac{\pi}{4} d^2 \) - \( \dot{m} = \rho Q \) - \( \text{Mixing: } \dot{m}_h (T_h - T_{exit}) = \dot{m}_c (T_{exit} - T_c) \) --- ## **Diagram**  *Alt text: Diagram showing hot and cold water entering a mixing chamber and exiting as a single warm stream.* --- If you have any specific rounding or data requirements, you may adjust the calculation accordingly.