Let's break down the problem step by step: --- ## **Step 1: Given Data** - **Air leak to be removed (\( \dot{m}_{air} \))**: 300 kg/h = **.0833 kg/s** - **Condenser system pressure**: 27.5 kPa - **Condenser system temperature**: 50°C - **Medium pressure steam**: 180°C, 100 kPa - **Ejector discharge pressure**: 110 kPa (where partial condensation occurs) --- ## **Step 2: Find Required Steam Flowrate** This is a typical ejector (jet pump) calculation. The required steam flow is determined by the **entrainment ratio** (air/steam), which depends on the thermodynamic properties and ejector design. ### **2.1. Determine the Required Motive Steam to Remove Air** The general formula for **mass ratio** for a single-stage ejector is: \[ \frac{\dot{m}_{air}}{\dot{m}_{steam}} = \frac{(h_s - h_{d})}{(h_{a} - h_{d})} \] Where: - \( \dot{m}_{air} \): Mass flow of air (kg/s) - \( \dot{m}_{steam} \): Mass flow of steam (kg/s) - \( h_s \): Enthalpy of motive steam at the nozzle inlet (kJ/kg) - \( h_a \): Enthalpy of air at suction conditions (kJ/kg) - \( h_d \): Enthalpy at the ejector discharge (kJ/kg) But more commonly, steam ejector design handbooks provide the **steam-to-air ratio** for typical conditions. For preliminary calculations, we can use the following empirical relationship: \[ \text{Steam required} = \text{Air load} \times \text{Steam-to-air ratio} \] For **single-stage steam ejectors** at similar conditions (from Perry's or ejector manufacturer data): - **Steam-to-air ratio** ≈ **1.2 to 1.5** (for suction pressure around 27.5 kPa, discharge pressure 110 kPa) --- ## **Step 3: Calculate Steam Flowrate** Let's use the upper end for safety: \[ \dot{m}_{steam} = \dot{m}_{air} \times \text{(steam-to-air ratio)} \] \[ \dot{m}_{steam} = .0833 \text{ kg/s} \times 1.5 = .125 \text{ kg/s} \] Convert back to kg/h: \[ .125 \text{ kg/s} \times 360 = 450 \text{ kg/h} \] --- ## **Step 4: Summary Table** | Parameter | Value | |--------------------------|-------------------| | Air leak (\( \dot{m}_{air} \)) | 300 kg/h (.0833 kg/s) | | Steam-to-air ratio | 1.5 | | Required steam (\( \dot{m}_{steam} \)) | 450 kg/h (.125 kg/s) | --- ## **Step 5: Final Answer** \[ \boxed{\text{The required flowrate of medium pressure steam is approximately } \mathbf{450 \text{ kg/h}}.} \] > If more accuracy is needed, you can use steam tables to get actual enthalpies and calculate the ratio, but this estimate is standard for engineering calculations. --- ### **Explanation** - The steam ejector uses high-pressure steam to entrain and remove air from the condenser. - The steam-to-air ratio is based on pressure and temperature conditions typical for single-stage ejectors. - Calculating enthalpy difference would refine the answer, but the steam-to-air ratio (empirical) is widely used for quick estimates. --- **If you want a detailed calculation using steam tables (enthalpy values), let me know!**