Question

Steam at 160 bar, \(480^{\circ} \mathrm{C}\), enters a turbine operating at steady state with a volumetric flow rate of \(800 \mathrm{~m}^{3} / \mathrm{min}\). Eighteen percent of the entering mass flow exits at 5 bar, \(240^{\circ} \mathrm{C}\), with a velocity of \(25 \mathrm{~m} / \mathrm{s}\). The rest exits at another location with a pressure of \(0.06\) bar, a quality of \(94 \%\), and a velocity of \(400 \mathrm{~m} / \mathrm{s}\). Determine the diameters of each exit duct, in \(\mathrm{m}\).

Short answer

The diameters of the exit ducts are determined by calculating mass flow rates and applying fluid mechanics principles.

Step-by-step solution

Determine the Mass Flow Rate at the Inlet

Use the ideal gas law and steam tables to find the specific volume of the steam. With the volumetric flow rate given, calculate the mass flow rate entering the turbine.

Calculate the Mass Flow Rates at Each Exit

Using the information that 18% of the entering mass flow exits at one location and the rest at another, determine the mass flow rates at each of the exit points.

Apply Continuity Equation at Each Exit

Use the relationship \( \text{mass flow rate} = \text{density} \times \text{velocity} \times \text{cross-sectional area} \) to find the exit areas for both exit points.

Use the Relationship Between Area and Diameter

Since the exit ducts are circular, use the formula for the area of a circle \(A = \frac{\pi d^2}{4}\) to solve for the diameter of each exit duct.

Convert and Compile Final Results

Ensure all units are consistent for calculation, solve for the diameters of the ducts, and compile the final results.

Key concepts

headline of the respective core concept

The **mass flow rate** quantifies how much mass moves through a given point in a system per unit time. In our turbine problem, it represents the mass of steam flowing into and out of the turbine. To calculate it, we first need to understand the conditions at the turbine's inlet. We use the ideal gas law and steam tables to find the specific volume and then relate this to the volumetric flow rate provided. Using the formula:

\( \dot{m} = \frac{\dot{V}}{v} \)

where:

• \( \dot{m} \) is the mass flow rate (kg/min),
• \( \dot{V} \) is the volumetric flow rate (m³/min),
• and \( v \) is the specific volume (m³/kg).

By plugging in the given volumetric flow rate and the specific volume obtained from the tables, we can easily compute the mass flow rate entering the turbine. This allows us to further determine how this mass is split at the two exit points.

headline of the respective core concept

The **continuity equation** is a fundamental principle in fluid dynamics that helps us ensure mass conservation within a system. For our turbine, this implies that the mass flow rate entering the turbine must equal the total mass flow rate exiting the turbine. Mathematically, we write:

\( \dot{m}_{\text{in}} = \dot{m}_{\text{out1}} + \dot{m}_{\text{out2}} \)

Here, \( \dot{m}_{\text{in}} \) is the total mass flow rate entering the turbine. Given that 18% exits at the first point, we have:

• \( \dot{m}_{\text{out1}} = 0.18 \cdot \dot{m}_{\text{in}} \)
• and the remaining 82% exits at the second point: \( \dot{m}_{\text{out2}} = 0.82 \cdot \dot{m}_{\text{in}} \)

By calculating the individual mass flow rates for the given percentages, we can then apply these values to determine the exit areas, leveraging their individual velocities and specific volumes.

headline of the respective core concept

The **exit duct diameter** is crucial for understanding the physical dimensions of the outlet pipes for the turbine. To find it, we first need to know the cross-sectional area each duct must have to accommodate the determined mass flow rates. Using the relationship:

\( A = \frac{\dot{m}}{\rho \cdot v} \)

where:

• \( A \) is the cross-sectional area (m²),
• \( \dot{m} \) is the mass flow rate (kg/s),
• \( \rho \) is the density (kg/m³),
• and \( v \) is the velocity (m/s).

Once the areas are calculated, we can find the diameter using the formula for the area of a circle:

\( A = \frac{\pi d^2}{4} \)

Rearranging for \( d \), we get:

\( d = \sqrt\frac{4A}{\pi} \)

Ensuring that all units are consistent, we can use these calculations to finally determine the exact diameters of the exit ducts, providing the necessary measurements for practical implementation.