4.4 Two adjacent parallel and horizontal rectangular conduits A and B, of cross-sectional areas \(0.2 \mathrm{~m}^{2}\) and \(0.4 \mathrm{~m}^{2}\) respectively, discharge water axially into another conduit \(\mathrm{C}\) of crosssectional area \(0.6 \mathrm{~m}^{2}\) and of sufficient length for the individual streams to become thoroughly mixed. The rates of flow through \(A\) and \(B\) are \(0.6 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) and \(0.8 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) respectively and the pressures there are \(31 \mathrm{kPa}\) and \(30 \mathrm{kPa}\) respectively. Neglecting friction at the boundaries, determine the energy lost by each entry stream (divided by mass) and the total power dissipated.

Step by step solution

01

- Define Energies per unit mass

To determine the energy lost by each entry stream (divided by mass), we need to calculate the energies per unit mass for both streams A and B at the entry points and after they are fully mixed. The energy per unit mass is given by the Bernoulli Equation, which states: \[ \frac{P}{\rho} + \frac{v^2}{2} + gh = \text{constant} \] Where: \(P\) is the pressure, \(\rho\) is the density of water (assumed to be 1000 kg/m³), \(v\) is the velocity of the flow, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(h\) is the height. For horizontal conduits, the height term \(gh\) cancels out.

02

- Calculate the velocity for each stream

Using the continuity equation \(Q = A \times v\), where \(Q\) is the flow rate, and \(A\) is the cross-sectional area, we can solve for the velocities in streams A and B before they enter conduit C:For A: \[ v_A = \frac{Q_A}{A_A} = \frac{0.6 \, \text{m}^3/\text{s}}{0.2 \, \text{m}^2} = 3 \, \text{m/s} \]For B: \[ v_B = \frac{Q_B}{A_B} = \frac{0.8 \, \text{m}^3/\text{s}}{0.4 \, \text{m}^2} = 2 \, \text{m/s} \]

03

- Calculate initial energies per unit mass

Using the velocities calculated in the previous step and the pressures given, calculate the initial energy per unit mass for streams A and B:For A: \[ E_{A_{\text{initial}}} = \frac{P_A}{\rho} + \frac{v_A^2}{2} = \frac{31,000 \, \text{Pa}}{1000 \, \text{kg/m}^3} + \frac{(3 \, \text{m/s})^2}{2} = 31 \, \text{J/kg} + 4.5 \, \text{J/kg} = 35.5 \, \text{J/kg} \]For B: \[E_{B_{\text{initial}}} = \frac{P_B}{\rho} + \frac{v_B^2}{2} = \frac{30,000 \, \text{Pa}}{1000 \, \text{kg/m}^3} + \frac{(2 \, \text{m/s})^2}{2} = 30 \, \text{J/kg} + 2 \, \text{J/kg} = 32 \, \text{J/kg} \]

04

- Calculate flow rate and velocity in conduit C

The total flow rate into conduit C is the sum of the flow rates from A and B:\[ Q_C = Q_A + Q_B = 0.6 + 0.8 = 1.4 \, \text{m}^3/\text{s} \] The velocity in conduit C is then:\[ v_C = \frac{Q_C}{A_C} = \frac{1.4 \, \text{m}^3/\text{s}}{0.6 \, \text{m}^2} \ v_C = 2.33 \, \text{m/s} \]

05

- Calculate the mixed energy per unit mass

Pressure at the exit is not given but assuming it to be the same across C and taking velocity into account:\[ E_{C} = \frac{P_C}{\rho} + \frac{v_C^2}{2} = E_{\text{exit}} \]Since the exit uniform pressure is unknown yet common also assume the density to be uniform and work for the velocity contributions.Finally,\[ E_{C} = \frac{30,500}{1000}(estimated,yet persistent) + \frac{(2.33 \text{m/s})^2}{2} \]\[ E_C \text{(resulted expression success reflects uniform distribution)}\]

06

- Determine Energy Lost & Power Dissipated

The energy lost per unit mass for streams A and B are given by:For A: \[ \text{Energy lost} = E_{A_{\text{initial}}} - E_{C} \]For B: \[ \text{Energy lost} = E_{B_{\text{initial}}} - E_{C} \]Power is calculated via combining terms acting as power:A+C received\( Power_{A} = \rho Q_A \times E_{A_{\text{initial}}} - (sum-of-initialpoint-E)\)> similarly Power_{B}

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Key Concepts

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

Bernoulli equation

The Bernoulli Equation is a fundamental principle in fluid mechanics. It states that in a streamlined flow, the energy per unit mass remains constant. The equation is written as: \[ \frac{P}{\rho} + \frac{v^2}{2} + gh = \text{constant} \]Here,

• \( P \) is the pressure
• \( \rho \) is the fluid density
• \( v \) is the velocity of the fluid
• \( g \) is the acceleration due to gravity
• \( h \) is the height

In horizontal conduits, the height term \( gh \) cancels out. This simplifies our calculations when dealing with horizontal flows. By applying the Bernoulli equation, we can find out the energy per unit mass at different points within the flow. This is essential for understanding how energy transforms as the fluid moves from one point to another.

continuity equation

The Continuity Equation is another crucial principle used in fluid mechanics. It helps us understand how the flow rate remains constant in a streamlined flow. The equation is given by: \[ Q = A \times v \]Here,

• \( Q \) is the flow rate
• \( A \) is the cross-sectional area of the conduit
• \( v \) is the velocity of the fluid

Using the continuity equation, we calculate the velocities of streams in different conduits. For example, in our exercise, we found the velocities for conduits A and B as:\[ v_A = \frac{0.6 \text{ m}^3/\text{s}}{0.2 \text{ m}^2} = 3 \text{ m/s} \]\[ v_B = \frac{0.8 \text{ m}^3/\text{s}}{0.4 \text{ m}^2} = 2 \text{ m/s} \]Understanding these velocities helps us apply the Bernoulli equation more effectively.

energy loss in fluid flow

Energy loss in fluid flow occurs when the total mechanical energy of a fluid system decreases from one point to another. This often happens due to friction, turbulence, or mixing. In our exercise, to determine the energy lost by each stream, we subtract the final energy per unit mass in conduit C from the initial energies:\[ E_{\text{lost, A}} = E_{A_{\text{initial}}} - E_C \]\[ E_{\text{lost, B}} = E_{B_{\text{initial}}} - E_C \]For instance, if the energy per unit mass after mixing in conduit C is calculated as \( 32.72 \text{ J/kg} \), then the energy losses for streams A and B can be found using their respective initial energies.

power dissipation in fluids

Power dissipation in fluids is the rate at which energy is lost from the fluid due to processes like mixing or friction. To calculate the power dissipated, we multiply the energy lost per unit mass by the flow rate and density of the fluid. For streams A and B, the power dissipation is given by:\[ Power_A = \rho Q_A \times (E_{A_{\text{initial}}} - E_C) \]\[ Power_B = \rho Q_B \times (E_{B_{\text{initial}}} - E_C) \]Adding these values gives us the total power dissipated in the system. This concept is critical in designing efficient fluid systems as excessive power dissipation can lead to energy waste and decreased system performance.