Let's solve the problem step by step. --- ##Given Data** | Parameter | Value | |-----------------------------|-------------| | Penstock length, \( L \) | 500 m | | Penstock diameter, \( D \) | 1 m | | Elevation difference, \( h \)| 50 m | | Average roughness, \( \epsilon \) | .046 mm \( = .000046 \) m | | Turbine efficiency, \( \eta_t \) | 95% = .95 | | Generator efficiency, \( \eta_g \) | 90% = .90 | | Minor loss coefficients | | --- ## **a) Maximum Power Flow Rate (\( Q \))** ### **1. Head Loss Calculation** The net head available to the turbine is: \[ H_{net} = h - h_L \] Where \( h_L \) is the head loss due to friction: \[ h_L = f \frac{L}{D} \frac{V^2}{2g} \] where: - \( f \) = Darcy-Weisbach friction factor - \( V \) = mean velocity \((V = Q/A)\) - \( A \) = cross-sectional area = \( \pi D^2/4 \) #### **Friction Factor (for turbulent flow):** Use the Colebrook-White equation, but for engineering estimation, use the Swamee-Jain equation: \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] where - \( Re = \frac{VD}{\nu} \), \( \nu \approx 1 \times 10^{-6} \, \text{m}^2/\text{s} \) for water. #### **Power Delivered to Water:** \[ P_{water} = \rho g Q H_{net} \] To maximize power delivered, set \( Q \) such that \( Q H_{net} \) is maximized. \[ P_{turbine} = \eta_t \rho g Q H_{net} \] \[ P_{generator} = \eta_t \eta_g \rho g Q H_{net} \] #### **Net Head as Function of \( Q \):** Let \( V = Q/A \), \[ h_L = f \frac{L}{D} \frac{(Q/A)^2}{2g} \] \[ H_{net} = h - f\frac{L}{D} \frac{(Q/A)^2}{2g} \] \[ P_{water} = \rho g Q \left(h - f\frac{L}{D} \frac{(Q/A)^2}{2g}\right) \] --- ### **2. Maximizing Power with Respect to \( Q \):** Set derivative of \( P_{water} \) with respect to \( Q \) to zero: \[ \frac{d}{dQ} \left[ Q \left(h - f\frac{L}{D} \frac{(Q/A)^2}{2g}\right) \right] = \] \[ h - f\frac{L}{D} \frac{3(Q/A)^2}{2g} = \] \[ h = f\frac{L}{D} \frac{3(Q/A)^2}{2g} \] \[ (Q/A)^2 = \frac{2ghD}{3fL} \] \[ Q = A \sqrt{ \frac{2ghD}{3fL} } \] --- ### **3. Area and Substitute Values** \[ A = \frac{\pi D^2}{4} = \frac{\pi (1)^2}{4} = .785 \, \mathrm{m}^2 \] ### **4. Estimate Friction Factor (\( f \))** Assume turbulent flow. - \( \epsilon/D = .000046/1 = 4.6 \times 10^{-5} \) - We'll iterate for correct \( f \), but let's estimate with \( f \approx .015 \) (for smooth pipes, turbulent flow). --- ### **5. Calculate \( Q \):** \[ Q = .785 \times \sqrt{ \frac{2 \times 9.81 \times 50 \times 1}{3 \times .015 \times 500} } \] Calculate denominator: \[ 3 \times .015 \times 500 = 22.5 \] Numerator: \[ 2 \times 9.81 \times 50 \times 1 = 981 \] So, \[ Q = .785 \times \sqrt{ \frac{981}{22.5} } \] \[ \frac{981}{22.5} = 43.6 \] \[ \sqrt{43.6} = 6.61 \] \[ Q = .785 \times 6.61 = 5.18 \ \text{m}^3/\text{s} \] --- ### **6. Check Friction Factor with Actual Reynolds Number** \[ V = Q/A = 5.18 / .785 = 6.60\, \mathrm{m/s} \] \[ Re = \frac{VD}{\nu} = \frac{6.60 \times 1}{1 \times 10^{-6}} = 6.6 \times 10^6 \] Swamee-Jain equation: \[ f = .25 \left[ \log_{10} \left( \frac{.000046}{3.7 \times 1} + \frac{5.74}{(6.6 \times 10^6)^{.9}} \right) \right]^{-2} \] Calculate inside log: - \( \frac{.000046}{3.7} = 1.24 \times 10^{-5} \) - \( (6.6 \times 10^6)^{.9} \approx 1.13 \times 10^6 \) - \( \frac{5.74}{1.13 \times 10^6} = 5.08 \times 10^{-6} \) - Total: \( 1.24 \times 10^{-5} + 5.08 \times 10^{-6} = 1.75 \times 10^{-5} \) - \( \log_{10}(1.75 \times 10^{-5}) = -4.76 \) - \( [ -4.76 ]^{-2} = .044 \) - \( f = .25 \times .044 = .011 \) Update \( f \) and recalculate: \[ 3 \times .011 \times 500 = 16.5 \] \[ \frac{981}{16.5} = 59.5 \] \[ \sqrt{59.5} = 7.72 \] \[ Q = .785 \times 7.72 = 6.06 \ \text{m}^3/\text{s} \] **Final Flow Rate:** \[ \boxed{Q = 6.1\ \text{m}^3/\text{s}} \] (rounded to two significant digits) --- ## **b) Maximum Output (MW)** \[ P_{out} = \eta_t \eta_g \rho g Q H_{net} \] First, find \( H_{net} \): \[ h_L = f \frac{L}{D} \frac{V^2}{2g} \] \[ V = Q/A = 6.06/.785 = 7.72\, \text{m/s} \] \[ h_L = .011 \times 500 \times \frac{(7.72)^2}{2 \times 9.81} \] \[ (7.72)^2 = 59.6 \] \[ \frac{59.6}{19.62} = 3.04 \] \[ .011 \times 500 = 5.5 \] \[ h_L = 5.5 \times 3.04 = 16.7\ \text{m} \] \[ H_{net} = 50 - 16.7 = 33.3\ \text{m} \] \[ \rho = 100\ \text{kg/m}^3 \] \[ g = 9.81\ \text{m/s}^2 \] \[ P_{out} = .95 \times .90 \times 100 \times 9.81 \times 6.06 \times 33.3 \] \[ = .855 \times 100 \times 9.81 \times 6.06 \times 33.3 \] Calculate stepwise: - \( 100 \times 9.81 = 981 \) - \( 981 \times 6.06 = 59,489 \) - \( 59,489 \times 33.3 = 1,980,995 \) - \( .855 \times 1,980,995 = 1,693,750 \) W \[ P_{out} = 1,693,750\ \text{W} = 1.69\ \text{MW} \] **Final Power Output:** \[ \boxed{1.7\ \text{MW}} \] --- ## **c) Reasonable Turbine RPM** For a Kaplan turbine, the **specific speed** (\( N_s \)) is used to select RPM: \[ N_s = N \sqrt{P} / H^{5/4} \] For Kaplan turbines, typical \( N_s \) is 250–850 (SI units), but let's use the **dimensionless form**: \[ N_s = \frac{N \sqrt{P}}{H^{1.25}} \] Rearrange for \( N \): \[ N = N_s \frac{H^{1.25}}{\sqrt{P}} \] - \( H = 33.3 \) m - \( P = 1,693,750 \) W Let's choose \( N_s = 500 \): \[ N = 500 \times \frac{(33.3)^{1.25}}{\sqrt{1,693,750}} \] Calculate: - \( (33.3)^{1.25} = e^{1.25 \ln(33.3)} = e^{1.25 \times 3.51} = e^{4.39} = 80.8 \) - \( \sqrt{1,693,750} = 1,301 \) \[ N = 500 \times \frac{80.8}{1,301} = 500 \times .0621 = 31 \ \text{RPM} \] **Final Reasonable RPM:** \[ \boxed{31\ \text{RPM}} \] --- ## **Summary Table** | Quantity | Value | |----------------------|----------------| | Flow Rate (\( Q \)) | \( 6.1\ \text{m}^3/\text{s} \) | | Max Output (\( P \)) | \( 1.7\ \text{MW} \) | | Turbine RPM (\( N \))| \( 31\ \text{RPM} \) | --- ### **Let me know if you want to see any calculation steps explained in more detail!**