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The inflow and outflow pipes to a turbine have diameters of \(400 \mathrm{~mm}\) and \(500 \mathrm{~mm}\), respectively, and are located at approximately the same elevation. Under a particular operating condition, the flow rate of water through the turbine is \(1 \mathrm{~m}^{3} / \mathrm{s}\) and the power extracted from the water by the turbine is \(100 \mathrm{~kW}\). Estimate the change in water pressure across the turbine. Assume water at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The change in water pressure across the turbine is 118,750 Pa.

Step by step solution

01

Understand the Problem

We need to estimate the change in water pressure across the turbine using information about the diameters of inflow and outflow pipes, flow rate, and power extracted from water.
02

Apply Continuity Equation

Inflow rate equals outflow rate due to conservation of mass. Given that the flow rate through the turbine is \(1 \ \mathrm{m}^3/\mathrm{s}\), we can find the velocity going through the inflows and outflows using \( \frac{Q}{A} \), where \( Q \) is the flow rate, and \( A \) is the cross-sectional area of the pipe.
03

Calculate Cross-sectional Areas

Cross-sectional area is given by \( A = \pi r^2 \). For the inflow, \( d = 0.4 \ \mathrm{m} \) gives \( r = 0.2 \ \mathrm{m} \). For the outflow, \( d = 0.5 \ \mathrm{m} \) gives \( r = 0.25 \ \mathrm{m} \). Calculate both areas:\[ A_{in} = \pi (0.2)^2 = 0.1257 \ \mathrm{m}^2 \]\[ A_{out} = \pi (0.25)^2 = 0.1963 \ \mathrm{m}^2 \]
04

Determine Velocities

Use the formula \( v = \frac{Q}{A} \) to find velocities:For inflow: \[ v_{in} = \frac{1}{0.1257} = 7.96 \ \mathrm{m/s} \]For outflow: \[ v_{out} = \frac{1}{0.1963} = 5.09 \ \mathrm{m/s} \]
05

Calculate Change in Kinetic Energy

The change in kinetic energy per unit volume \( \Delta KE \) across the turbine is given by:\[ \Delta KE = \frac{1}{2} \rho (v_{out}^2 - v_{in}^2) \]with density \( \rho = 1000 \ \mathrm{kg/m^3} \) for water.
06

Substitute Values to Find \(\Delta KE\)

Plugging in the velocities:\[ \Delta KE = 0.5 \times 1000 \times (5.09^2 - 7.96^2) \]\[ \Delta KE = 0.5 \times 1000 \times (25.9 - 63.4) \]\[ \Delta KE = 0.5 \times 1000 \times (-37.5) = -18750 \ \mathrm{J/m^3} \]
07

Apply Bernoulli's Equation

The power extracted \(P = 100 \ \mathrm{kW} = 100,000 \ \mathrm{J/s}\). The pressure difference can be derived considering Bernoulli's equation:\[ P = (\Delta P + \Delta KE) \times Q \]Rearrange to find \( \Delta P \):\[ 100,000 = (\Delta P - 18750) \times 1 \]\[ \Delta P = 100,000 + 18750 = 118750 \ \mathrm{Pa} \]
08

Conclusion

The change in pressure across the turbine is estimated to be \( 118750 \ \mathrm{Pa} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation
The continuity equation is a fundamental principle in fluid mechanics that ensures the conservation of mass in a fluid flow system. In simple terms, it states that for an incompressible fluid, the mass flow rate must remain constant throughout the flow path. For a pipe system, this principle can be expressed mathematically as:
  • Inflow rate = Outflow rate
This equation is particularly useful for problems involving pipes with different diameters. Given the flow rate and dimensions of the pipes, the continuity equation allows us to calculate the fluid velocity at any point in the system. By using the equation, we know that the velocity of fluid in the 400 mm diameter inflow pipe differs from that in the larger 500 mm diameter outflow pipe. The essential formula to calculate velocities is derived from focusing on constant flow rate: \[ v = \frac{Q}{A} \]Where \( v \) is the velocity, \( Q \) is the flow rate, and \( A \) is the cross-sectional area. Knowing this ensures that mass is neither created nor destroyed during the flow.
Bernoulli's Equation
Bernoulli's Equation is another key concept in fluid mechanics, capturing the energy balance along a streamline for an incompressible fluid with no friction losses. It relates the speed, pressure, and potential energy of the fluid and can be mathematically expressed as:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \ ext{constant} \]Where:
  • \( P \) is the pressure energy per unit volume
  • \( \frac{1}{2} \rho v^2 \) represents kinetic energy per unit volume
  • \( \rho gh \) is the potential energy per unit volume
By applying Bernoulli's equation in this turbine problem, we assess how kinetic and pressure energies change across the turbine. This helps us determine the pressure difference across the system, which is a crucial aspect when analyzing the performance of turbines used to extract energy from fluid flows.
Kinetic Energy
Kinetic energy in fluid flow is a measure of the energy possessed by the fluid due to its motion. It depends on both the mass of the fluid and its velocity, calculated using the formula:\[ KE = \frac{1}{2} m v^2 \]In terms of unit volume, it can also be represented as:\[ \Delta KE = \frac{1}{2} \rho (v_{out}^2 - v_{in}^2) \]In the exercise, we observe a decrease in kinetic energy as water moves from the inflow pipe to the larger outflow pipe. This results from the change in velocities, where a wider section reduces speed, decreasing kinetic energy. Understanding these variations in kinetic energy is essential when calculating energy transfer, particularly in systems like turbines where energy extraction is desired.
Pressure Measurement
Pressure measurement in fluid systems is vital for understanding how forces within the fluid affect its motion. Pressure is the force exerted by a fluid per unit area. In the context of the exercise, pressure changes across the turbine were calculated using both the change in kinetic energy and Bernoulli's equation. The formula applied was:\[ \Delta P = \frac{P}{Q} - \Delta KE \]Here:
  • \( \Delta P \) is the pressure change
  • \( P \) is the power extracted
  • \( Q \) is the volumetric flow rate
  • \( \Delta KE \) is the change in kinetic energy per unit volume
This equation shows how pressure loss or gain can be directly linked to both kinetic changes and power extracted, providing invaluable insights for designing and optimizing fluid systems like water turbines.

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Most popular questions from this chapter

A valve connects two pipes, where the upstream pipe has a diameter of \(300 \mathrm{~mm}\) and the downstream pipe has a diameter of \(450 \mathrm{~mm}\). When water flows through the valve and the velocity in the upstream pipe is \(2.5 \mathrm{~m} / \mathrm{s}\), the difference in pressure between the inflow pipe and the outflow pipe is measured as \(60 \mathrm{kPa}\). If the inflow and outflow pipes are at the same elevation, what is the head loss through the valve? What is the power loss across the valve?

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