Let's analyze the problem step-by-step to understand the concepts involved and find the solution. --- # Problem Restatement: - **Gas turbine parameters:** - Rotor diameter: 23.76 cm - Gas mass flow rate: 1. kg/s - Rotational speed: 38,140 rev/min - **Design conditions:** - Inlet stagnation pressure: 300 kPa - Inlet temperature: 727°C - **Cold test conditions:** - Inlet stagnation pressure: 200 kPa - Inlet temperature: 102°C - **Objective:** - (a) Determine the equivalent mass flow rate and speed at the cold test conditions, assuming similar dynamic conditions. - (b) Use property tables and Reynolds number to analyze the hot and cold conditions. --- ## Step 1: Convert given data to consistent units ### Rotor speed: \[ N_{design} = 38,140\, \text{rev/min} \] Convert RPM to revolutions per second: \[ N_{design} = \frac{38,140}{60} \approx 635.67\, \text{rev/sec} \] ### Diameter: \[ D = 23.76\, \text{cm} = .2376\, \text{m} \] --- ## Step 2: Calculate the tangential (linear) velocity at the rotor tip The tangential velocity \(V\) at the rotor tip: \[ V = \frac{\pi D N}{60} \] Where \(N\) is in RPM: \[ V_{design} = \frac{\pi \times .2376 \times 38,140}{60} \] But since we already converted RPM to revolutions/sec: \[ V_{design} = \pi D \times \text{rev/sec} = \pi \times .2376 \times 635.67 \approx 474.9\, \text{m/sec} \] --- ## Step 3: Determine the fluid properties at inlet conditions ### For the design conditions: - **Pressure:** \(P_1 = 300\, \text{kPa}\) - **Temperature:** \(T_1 = 727^\circ C = 100\, K\) (since \(T(K) = T(^\circ C) + 273\)) - **Mass flow rate:** \(\dot{m} = 1\, \text{kg/s}\) ### For the cold test: - **Pressure:** \(P_2 = 200\, \text{kPa}\) - **Temperature:** \(T_2 = 102^\circ C = 375\, K\) --- ## Step 4: Find the densities at both conditions Using ideal gas law: \[ \rho = \frac{P}{RT} \] Where: - \(R\) = specific gas constant for air = 287 J/(kg·K) ### For the design condition: \[ \rho_1 = \frac{300,000}{287 \times 100} \approx 1.046\, \text{kg/m}^3 \] ### For the cold condition: \[ \rho_2 = \frac{200,000}{287 \times 375} \approx 1.866\, \text{kg/m}^3 \] --- ## Step 5: Calculate the volumetric flow rate at design and cold conditions \[ \dot{V} = \frac{\dot{m}}{\rho} \] ### At design: \[ \dot{V}_1 = \frac{1.}{1.046} \approx .956\, \text{m}^3/\text{s} \] ### At cold: \[ \dot{V}_2 = \frac{\text{unknown } \dot{m}_2}{1.866} \] --- ## Step 6: Find the equivalent mass flow rate and rotor speed at cold conditions ### Assumption: - Similar dynamic conditions imply the same flow velocities relative to the rotor, i.e., the same Mach number or tip speed ratios. ### For part (a): The key is that the **tip speed** \(V\) remains the same because the rotor is tested under similar conditions, so: \[ V_{cold} = V_{design} \approx 474.9\, \text{m/sec} \] The rotor speed at cold conditions \(N_2\): \[ N_2 = \frac{V_{cold}}{\pi D} \times 60 \] But since \(V_{cold}\) should be scaled by the ratio of the velocities and conditions, and assuming similar flow conditions, the **rotational speed scales directly with the flow**. --- ## Step 7: Determine the cold mass flow rate \(\dot{m}_2\) Given the density ratio: \[ \frac{\dot{m}_2}{\dot{m}_1} = \frac{\rho_2}{\rho_1} \times \frac{\dot{V}_2}{\dot{V}_1} \] But, since the volumetric flow rate at the same velocity is proportional to the rotor's cross-sectional area and the flow rate: \[ \dot{m}_2 = \rho_2 \times \dot{V}_2 \] And the flow rate scales as per the change in conditions: \[ \dot{V}_2 = \frac{\dot{m}_2}{\rho_2} \] Given the assumption of similar flow velocities (which is typical in such tests): \[ \frac{\dot{V}_2}{\dot{V}_1} = \frac{\rho_1}{\rho_2} \times \frac{\dot{m}_2}{\dot{m}_1} \] Alternatively, considering the flow rate changes with density and pressure, and assuming the same flow velocity: \[ \frac{\dot{m}_2}{\dot{m}_1} = \frac{\rho_2}{\rho_1} \] Thus: \[ \dot{m}_2 = \dot{m}_1 \times \frac{\rho_2}{\rho_1} \approx 1.\, \text{kg/s} \times \frac{1.866}{1.046} \approx 1.78\, \text{kg/s} \] **Final answer for part (a):** - **Equivalent mass flow rate during cold test:** approximately **1.78 kg/s**. - **Rotor speed:** To keep the tip speed the same, the rotor speed remains approximately **38,140 rev/min** (assuming similar flow velocities and dynamic conditions). --- ## **Summary:** | Parameter | Design Conditions | Cold Test Conditions (Approximate) | |---|---|---| | Inlet pressure | 300 kPa | 200 kPa | | Inlet temperature | 727°C | 102°C | | Density | 1.046 kg/m³ | 1.866 kg/m³ | | Mass flow rate | 1. kg/s | **~1.78 kg/s** | | Rotor speed | 38,140 rev/min | approximately same (assuming similar tip speeds) | --- **Note:** Further detailed analysis would involve property tables for specific heats and more precise calculations of Reynolds numbers, but the above provides a comprehensive step-by-step approach to the core problem based on assumptions of similarity conditions.