# Steam Turbine Problem Solution Below is a step-by-step solution for each part of your steam turbine problem. --- ## Given Data - **Number of stages:** 4 - **Blades per stage:** 8 (1st), 14 (2nd), 20 (3rd), 36 (4th) - **Smallest blade diameter:** .5 m - **Largest blade diameter:** 2 m - **Blockage ratio:** .5 - **Drag coefficient (\(C_d\)):** .03 - **Lift coefficient (\(C_l\)):** .6 - **Boiler pressure:** 10 bar - **Boiler outlet temperature:** \(800^\circ C\) - **Isentropic efficiency (\(\eta_{t,isen}\)):** .87 - **Condenser pressure:** 50 kPa = .5 bar - **Electrical power output (\(P_{el}\)):** 10 MW - **Generator efficiency (\(\eta_{gen}\)):** .97 --- ## a) **Find the Thermal Efficiency** ### **Step 1: Find Net Turbine Output** \[ P_{el} = P_{net} \cdot \eta_{gen} \] \[ P_{net} = \frac{P_{el}}{\eta_{gen}} = \frac{10\,000\,\text{kW}}{.97} = 10\,309\,\text{kW} \] --- ### **Step 2: Find Enthalpies from Steam Tables** #### **At Turbine Inlet (10 bar, 800°C):** - Superheated steam. - From steam tables: - \(h_1 \approx 395\,\text{kJ/kg}\) - \(s_1 \approx 8.112\,\text{kJ/kg} \cdot \text{K}\) #### **At Turbine Exit (.5 bar):** - First, find isentropic exit state (\(s_2 = s_1\)). - At .5 bar, for saturated vapor: - \(s_g = 7.593\,\text{kJ/kg} \cdot \text{K}\) - \(s_f = 1.091\,\text{kJ/kg} \cdot \text{K}\) - \(h_g = 2643\,\text{kJ/kg}\) - \(h_f = 340\,\text{kJ/kg}\) Since \(s_1 > s_g\), the steam is superheated at exit. We interpolate/extrapolate using steam tables: - At .5 bar, \(h_2s \approx 2762\,\text{kJ/kg}\) (from Mollier diagram or superheated tables with \(s = 8.112\)). #### **Actual Turbine Exit (\(h_2\))** \[ \eta_{t,isen} = \frac{h_1 - h_2}{h_1 - h_{2s}} \] \[ h_2 = h_1 - \eta_{t,isen} \cdot (h_1 - h_{2s}) \] \[ h_2 = 395 - .87 \times (395 - 2762) = 395 - .87 \times 1188 = 395 - 1034.56 = 2915.44\,\text{kJ/kg} \] --- ### **Step 3: Find Mass Flow Rate** \[ P_{net} = \dot{m} (h_1 - h_2) \] \[ \dot{m} = \frac{P_{net}}{h_1 - h_2} = \frac{10\,309\,\text{kW}}{395 - 2915.44} = \frac{10\,309}{1034.56} = 9.97\,\text{kg/s} \] --- ### **Step 4: Find Boiler Heat Input (\(Q_{in}\))** - Water enters boiler at condenser exit (\(h_3\)), which is saturated liquid at .5 bar: - \(h_3 \approx h_f@.5\,bar = 340\,\text{kJ/kg}\) \[ Q_{in} = \dot{m} (h_1 - h_3) = 9.97 \times (395 - 340) = 9.97 \times 361 = 36\,011\,\text{kW} \] --- ### **Step 5: Find Thermal Efficiency** \[ \eta_{thermal} = \frac{P_{net}}{Q_{in}} = \frac{10\,309}{36\,011} = .286 = \boxed{28.6\%} \] --- ## b) **Draw the Plant Layout and T-S Diagram** ### **Plant Layout Diagram** ``` [Boiler] -- h1 --> [Turbine] -- h2 --> [Condenser] -- h3 --> [Pump] -- h4 --> [Boiler] ``` - **Boiler:** Adds heat at high pressure (10 bar) and high temperature (800°C) - **Turbine:** Extracts work, expands steam to low pressure (.5 bar) - **Condenser:** Condenses steam to saturated liquid at .5 bar - **Pump:** Raises pressure of water to 10 bar, returns to boiler ### **T-S Diagram** - **X-axis:** Entropy (\(s\)) - **Y-axis:** Temperature (\(T\)) - Mark points: - **1:** (10 bar, 800°C), high \(T\), high \(s\) - **2:** (.5 bar, superheated), lower \(T\), same \(s\) for isentropic, slightly higher for real - **3:** (.5 bar, saturated liquid) - **4:** (10 bar, compressed liquid, slightly higher \(s\) than at point 3) **T-S diagram illustration:**  *Alt text: T-S diagram showing Rankine cycle with points 1 (turbine inlet), 2 (turbine exit), 3 (condenser exit), and 4 (pump exit).* --- ## c) **Find Pump Power Consumption** \[ P_{pump} = \dot{m} (h_4 - h_3) \] #### **Calculate \(h_4 - h_3\):** \[ h_4 - h_3 \approx v \cdot (P_1 - P_2) \] - \(v = .00101\,\text{m}^3/\text{kg}\) (specific volume at .5 bar, saturated liquid) - \((P_1 - P_2) = (10 - .5) \text{ bar} = 9.5 \times 10^5\,\text{N/m}^2\) \[ h_4 - h_3 = .00101 \times 950\,000 = 959.5\,\text{J/kg} = .96\,\text{kJ/kg} \] \[ P_{pump} = 9.97 \times .96 = 9.57\,\text{kW} \] --- ## d) **Find Torque Values for Each Turbine Stage** ### **Step 1: Find Power per Stage** - Total shaft power: \(P_{shaft} = P_{net} = 10\,309\,\text{kW}\) - Assume all stages are identical in efficiency (idealized), so power per stage is proportional to blade count (since energy extracted is proportional to number of blades and their size). #### **Estimate Power Ratio per Stage** If power extracted is proportional to blade count \(\times\) mean diameter (approximation): - Stage 1: \(8 \times .5 = 4\) - Stage 2: \(14 \times d_2\) - Stage 3: \(20 \times d_3\) - Stage 4: \(36 \times 2 = 72\) But we don't have intermediate diameters. Assume linear increase in diameter: | Stage | Blades | Diameter (m) | \(B \times D\) | |-------|--------|--------------|---------------| | 1 | 8 | .5 | 4 | | 2 | 14 | 1. | 14 | | 3 | 20 | 1.5 | 30 | | 4 | 36 | 2. | 72 | Total \(= 4+14+30+72=120\) Fractional power per stage: - Stage 1: \(4/120 = 3.33\%\) - Stage 2: \(14/120 = 11.67\%\) - Stage 3: \(30/120 = 25\%\) - Stage 4: \(72/120 = 60\%\) Now, compute power per stage: - Stage 1: \(.0333 \times 10\,309 = 343\,\text{kW}\) - Stage 2: \(.1167 \times 10\,309 = 1\,205\,\text{kW}\) - Stage 3: \(.25 \times 10\,309 = 2\,577\,\text{kW}\) - Stage 4: \(.6 \times 10\,309 = 6\,185\,\text{kW}\) --- ### **Step 2: Find Torque per Stage** \[ P = \tau \cdot \omega \implies \tau = \frac{P}{\omega} \] Let’s get \(\omega\) first (see next part e), then return to torque. --- ## e) **Find Turbine Rotational Speed** ### **Step 1: Calculate Mean Blade Velocity at Largest Stage** \[ u = \omega r \] At final stage (\(d = 2\,\text{m}\)), \(r = 1\,\text{m}\) Assume typical blade speed (for steam turbines): - \(u = 300\,\text{m/s}\) is typical for large turbines. - But let's check with lift/drag and power: From lift/drag (neglect for this estimate), use power and torque: \[ P = \tau \omega \] From largest stage: \[ \tau_4 = \frac{P_4}{\omega} \] But also, \[ \tau = F_{tan} \cdot r = (L \sin\alpha - D \cos\alpha)\cdot r \] But without blade angles, use: \[ \text{If } P = F_{tan} \cdot u \implies F_{tan} = \frac{P}{u} \] Assume \(u = 300\,\text{m/s}\): \[ \omega = \frac{u}{r} = \frac{300}{1} = 300\,\text{rad/s} \] \[ N = \frac{\omega}{2\pi} \times 60 = \frac{300}{2\pi} \times 60 = 2\,864\,\text{rpm} \] *Note: Real steam turbines (large) typically run at 300 rpm (50 Hz), so this is plausible.* --- ### **Step 2: Find Torques** \[ \tau = \frac{P}{\omega} \] Using \(\omega = 300\,\text{rad/s}\): - Stage 1: \(\tau_1 = \frac{343\,000}{300} = 1\,143\,\text{N}\cdot\text{m}\) - Stage 2: \(\tau_2 = \frac{1\,205\,000}{300} = 4\,017\,\text{N}\cdot\text{m}\) - Stage 3: \(\tau_3 = \frac{2\,577\,000}{300} = 8\,590\,\text{N}\cdot\text{m}\) - Stage 4: \(\tau_4 = \frac{6\,185\,000}{300} = 20\,617\,\text{N}\cdot\text{m}\) --- ## **Final Answers (Summary Table)** | Part | Quantity | Value | |------|------------------------------|------------------------------| | a | Thermal efficiency | \(28.6\%\) | | b | Plant layout & T-S diagram | See diagrams above | | c | Pump power | \(9.57\,\text{kW}\) | | d | Torque (N·m) per stage | 1: 1,143; 2: 4,017; 3: 8,590; 4: 20,617 | | e | Turbine speed | \(2,864\,\text{rpm}\) (\(300\,\text{rad/s}\)) | --- ### **Assumptions** - Blade power distribution is proportional to blade count \(\times\) mean diameter. - Mean blade speed is typical of steam turbines. - For more accurate torque/speed, detailed velocity triangles and steam velocities for each stage would be needed. --- **If you need more detailed calculations for any step or clarification, please specify!**