Let's solve the problem **step by step**. ### **Given Data** - Vertical distance, \( z_0 = 50 \) m - Exit velocity, \( V_0 = 1 \) m/s - Pipe diameter, \( D = 0.1 \) m (not needed for energy per kg) - Reservoir gage pressure, \( P_r = 303 \) kPa \( = 303,000 \) Pa - Water density, \( \rho = 1000 \) kg/m³ - Atmospheric pressure at outlet (gauge): 0 We are to **calculate the useful work added to the water per unit mass (\( N \cdot m / kg \))**. --- ## **Step 1: Apply the Steady-Flow Energy Equation** The **steady-flow energy equation** (Bernoulli + shaft work per unit mass): \[ \frac{P_r}{\rho} + g \cdot 0 + 0 + w_{\text{useful}} = \frac{P_0}{\rho} + g z_0 + \frac{V_0^2}{2} \] where: - \( w_{\text{useful}} \) = useful work added per unit mass - \( P_0 \) = atmospheric pressure (gauge) = 0 --- ## **Step 2: Substitute Values** \[ \frac{P_r}{\rho} + w_{\text{useful}} = g z_0 + \frac{V_0^2}{2} \] \[ w_{\text{useful}} = \left[ g z_0 + \frac{V_0^2}{2} \right] - \frac{P_r}{\rho} \] --- ## **Step 3: Plug in Numbers** ### Constants: - \( g = 9.81 \) m/s² ### Calculate Each Term: #### (a) \( g z_0 \) \[ g z_0 = 9.81 \times 50 = 490.5 \ \text{J/kg} \] #### (b) \( \frac{V_0^2}{2} \) \[ \frac{V_0^2}{2} = \frac{1^2}{2} = 0.5 \ \text{J/kg} \] #### (c) \( \frac{P_r}{\rho} \) \[ \frac{P_r}{\rho} = \frac{303,000}{1000} = 303 \ \text{J/kg} \] --- ## **Step 4: Compute the Useful Work** \[ w_{\text{useful}} = (490.5 + 0.5) - 303 \] \[ w_{\text{useful}} = 491 - 303 = 188 \ \text{J/kg} \] --- ## **Step 5: Final Answer** \[ \boxed{188 \ \text{J/kg}} \] --- ### **Summary of Steps** 1. Write steady-flow energy equation. 2. Rearrange for useful work per unit mass. 3. Substitute all known values. 4. Calculate each term. 5. Subtract reservoir pressure term from sum of elevation and kinetic energy terms. --- **The amount of useful work added to the water is \( \boxed{188 \ \text{J/kg}} \).**

**1. Introduction** Understanding the flow of water in a piping system involving elevation changes, pressure differences, and velocity variations relies on the application of fundamental principles of fluid mechanics. The steady-flow energy equation, often derived from the Bernoulli equation, is central to analyzing such systems. It relates the energy per unit mass at different points in the flow, accounting for pressure energy, potential energy due to elevation, and kinetic energy associated with velocity. When a pump adds energy to the fluid to overcome elevation and pressure head losses, the work input must be carefully quantified to determine the efficiency and performance of the system. The problem involves calculating the net useful work supplied by the pump, which is the difference between the energy imparted to the water and the energy losses due to pressure and velocity effects. This requires knowledge of pressure heads, elevation heads, velocity heads, and the relationship between pressure and energy per unit mass. Recognizing the importance of these parameters and their interplay allows for an accurate assessment of the work added per unit mass, which is critical in optimizing pump operation and system efficiency. **Explanation:** The introduction clarifies how fundamental principles such as Bernoulli’s equation relate to the problem of analyzing water flow in a pumped system. It emphasizes the significance of understanding energy transformations—pressure, elevation, and velocity—and how they influence the work input required by the pump. This foundational knowledge supports the subsequent application of energy equations and ensures clarity in interpreting the physical phenomena involved in the problem. --- **2. Presentation of Relevant Formulas Required To Solve The Question** - **Steady-Flow Energy Equation (Bernoulli Equation with Work per unit mass):** \[ \frac{P_1}{\rho} + g z_1 + \frac{V_1^2}{2} + w_{\text{add}} = \frac{P_2}{\rho} + g z_2 + \frac{V_2^2}{2} + h_{\text{loss}} \] In a simplified form, when considering the point at the reservoir (initial point) and the exit (final point): \[ w_{\text{useful}} = \left( g z_0 + \frac{V_0^2}{2} \right) - \frac{P_r}{\rho} \] where: - \( w_{\text{useful}} \) is the work added per unit mass, - \( P_r \) is the reservoir gauge pressure, - \( z_0 \) is the elevation head, - \( V_0 \) is the velocity at the outlet, - \( \rho \) is the fluid density, - \( g \) is acceleration due to gravity. - **Pressure Head:** \[ h_p = \frac{P}{\rho g} \] This relates pressure to energy per unit weight, useful for converting pressure to head. - **Velocity Head:** \[ h_v = \frac{V^2}{2g} \] Represents the kinetic energy per unit weight. **Explanation:** These formulas are essential because they allow the conversion of pressure and velocity measurements into equivalent energy units (heads), facilitating energy balance calculations. The Bernoulli equation with work term directly relates the energy supplied by the pump to the energy changes in the flow, enabling the calculation of the actual work input per unit mass. --- **3. A Detailed Step-by-Step Solution** **Step 1: Identify Known Data** - Elevation difference, \( z_0 = 50\, \text{m} \) - Exit velocity, \( V_0 = 1\, \text{m/s} \) - Reservoir gauge pressure, \( P_r = 303\, \text{kPa} = 303,000\, \text{Pa} \) - Water density, \( \rho = 1000\, \text{kg/m}^3 \) - Gravitational acceleration, \( g = 9.81\, \text{m/s}^2 \) *Explanation:* Accurate identification of given parameters is fundamental to applying the energy equations correctly. The data includes the physical parameters of the system and the initial conditions necessary for energy calculations. --- **Step 2: Convert Pressure to Specific Energy (Pressure Head)** \[ h_p = \frac{P_r}{\rho g} = \frac{303,000}{1000 \times 9.81} \approx 30.87\, \text{m} \] *Explanation:* Converting the reservoir pressure from Pascals to head (meters of water) allows for a consistent energy comparison with elevation and velocity heads. --- **Step 3: Calculate Velocity Head** \[ h_v = \frac{V_0^2}{2g} = \frac{1^2}{2 \times 9.81} \approx 0.051\, \text{m} \] *Explanation:* Velocity head represents the kinetic energy contribution per unit weight of water at the outlet, which is necessary for an accurate energy balance. --- **Step 4: Calculate Elevation Head** \[ z_0 = 50\, \text{m} \] *Explanation:* Elevation head directly influences the potential energy component in the energy balance, representing the work needed to lift water from the reservoir to the outlet level. --- **Step 5: Apply the Energy Balance Equation** \[ w_{\text{useful}} = \left( g z_0 + \frac{V_0^2}{2} \right) - \frac{P_r}{\rho} \] Substituting the computed values: \[ w_{\text{useful}} = 9.81 \times 50 + 0.051 - 303 \] \[ w_{\text{useful}} = 490.5 + 0.051 - 303 \] \[ w_{\text{useful}} \approx 490.55 - 303 = 187.55\, \text{J/kg} \] *Explanation:* The energy available to the water is the sum of elevation and velocity heads, reduced by the pressure head at the reservoir. The difference indicates the net work per unit mass that effectively contributes to lifting and accelerating the water. --- **Step 6: Final Result** \[ \boxed{w_{\text{useful}} \approx 188\, \text{J/kg}} \] *Explanation:* The calculated value represents the net useful work per unit mass added to the water by the pump, accounting for the system’s elevation, velocity, and pressure conditions. --- ### **Conclusion** The comprehensive analysis employing the steady-flow energy equation reveals that approximately **188 Joules per kilogram** of work are effectively imparted to the water, overcoming pressure, elevation, and velocity effects in the system. This quantification is critical for designing and optimizing pump systems and evaluating their efficiency.