Let's carefully analyze and solve this problem step-by-step. --- ## **Given Data:** - Water height at inlet: \( h_{in} = 10\, \text{ft} \) - Water height at exit: \( h_{exit} = 19\, \text{ft} \) - Initial water height in the pipe: \( h = 10\, \text{ft} \) - Loss coefficient for the valve: \( K_L = 50 \) - Pipe area: \( A_{pipe} \) - Nozzle area: \( A_{nozzle} \) - The area of the pipe is **5 times** the area of the nozzle: \[ A_{pipe} = 5 \times A_{nozzle} \] --- ## **Objective:** Determine the **valve loss coefficient \(K_v\)** needed for the flow at \( h = 10\, \text{ft} \) and \( h = 19\, \text{ft} \). --- ## **Step 1: Understand the flow dynamics** Water flows from the pool into the pipe, passes through the control valve, and exits through the nozzle. The pressure difference drives the flow, countered by head losses in the valve. --- ## **Step 2: Convert heights to pressure heads** In fluid mechanics, pressure head \( H \) relates to velocity head through Bernoulli’s principle: \[ H = \frac{v^2}{2g} \] But since the problem involves head differences, we analyze flow using and head loss. --- ## **Step 3: Write the Bernoulli equation with head losses** The general form: \[ h_{in} + \text{(pressure head)} = h_{exit} + \text{(pressure head)} + \text{head losses} \] Since the water flows high to low, the head difference is: \[ \Delta h = h_{exit} - h_{in} = 19\, \text{ft} - 10\, \text{ft} = 9\, \text{ft} \] --- ## **Step 4: Relate flow velocity to head** Flow velocity at the nozzle: \[ v = \frac{Q}{A} \] where \( Q \) is the volumetric flow rate. Rearranged: \[ Q = A \times v \] --- ## **Step 5: Express head loss through the valve** The head loss across the valve: \[ h_{loss} = \frac{K \times v^2}{2g} \] where \( K \) is the loss coefficient for the valve. --- ## **Step 6: Connect pressure head, velocity, and head loss** Applying Bernoulli's equation between the inlet and exit points: \h_{in} - h_{exit} = h_{loss} \] or: \[ \Delta h = \frac{K \times v^2}{2g} \] --- ## **Step 7: Use the relation to find flow velocity** Rearranged: \[ v = \sqrt{ \frac{2g \times \Delta h}{K} } \] --- ## **Step 8: Find the velocity for the initial case \( h = 10\, \text{ft} \)** Assuming the valve loss coefficient \( \) is proportional to the valve's physical characteristics and is related to \(K_L\). In many flow problems, the **valve head loss coefficient \(K_v\)** relates to the flow rate as: \[ h_{loss} = \frac{Q^2}{C_v^2} \] But here, since the problem states the valve has a loss coefficient \(K_L\), we can relate the head loss as: \[ h_{loss} = \frac{K_L \times v^2}{2g} \] and the total head difference: \[ \Delta h = 9\, \text{ft} \] --- ## **Step 9: Find the flow velocity at \( h=10\, \text{ft} \)** \[ v_{10} = \sqrt{ \frac{2g \times 9}{K_L} } \] Using \( g \approx 32.2\, \text{ft/sec}^2 \): \[ v_{10} = \sqrt{ \frac{2 \times 32.2 \times 9}{50} } = \sqrt{ \frac{579.6}{50} } = \sqrt{11.592} \approx 3.41\, \text{ft/sec} \] --- ## **Step 10: Find the flow velocity at \( h = 19\, \text{ft} \)** Similarly, \[ v_{19} = \sqrt{ \frac{2g \times (19 - 10)}{K_L} } = \sqrt{ \frac{2 \times 32.2 \times 9}{50} } \] which is the same as above because the head difference is the same, so: \[ v_{19} \approx 3.41\, \text{ft/sec} \] --- ## **Step 11: Find the flow rate \( Q \)** Using area \(A_{nozzle}\): \[ Q = A_{nozzle} \times v \] But since the **area of the pipe is 5 times the area of the nozzle**: \[ A_{pipe} = 5 \times A_{nozzle} \] The flow rate: \[ Q = A_{pipe} \times v_{pipe} \] And the velocity in the pipe: \[ v_{pipe} = \frac{Q}{A_{pipe}} = \frac{Q}{5A_{nozzle}} \] Similarly, the flow velocity through the nozzle is: \[ v_{nozzle} = \frac{Q}{A_{nozzle}} = 5 \times v_{pipe} \] --- ## **Step 12: Final calculation for the valve loss coefficient \(K_v\)** The *valve loss coefficient* \(K_v\) (sometimes denoted as \(K\) in other contexts) relates to the flow: \[ h_{loss} = \frac{K_v \times v_{nozzle}^2}{2g} \] We already have the head difference \( \Delta h = 9\, \text{ft} \). Rearranged: \[ K_v = \frac{2g \times h_{loss}}{v_{nozzle}^2} \] Given that \( h_{loss} = 9\, \text{ft} \): \[ K_v = \frac{2 \times 32.2 \times 9}{(5 \times v_{pipe})^2} \] But recall that: \[ v_{pipe} = 3.41\, \text{ft/sec} \] So, \[ v_{nozzle} = 5 \times 3.41 = 17.05\, \text{ft/sec} \] Plugging into the formula: \[ K_v = \frac{2 \times 32.2 \times 9}{(17.05)^2} = \frac{579.6}{290.8} \approx 1.99 \] --- ## **Final Answer:** \[ \boxed{ \text{The valve loss coefficient } K_v \approx 2. } \] This is the coefficient needed to produce the observed head difference with the given flow conditions. --- ## **Summary:** - The head difference of 9 ft relates to velocities through the head loss equation. - Using the area ratio, we find the velocity at the nozzle. - Calculating the valve loss coefficient \(K_v\) gives approximately **2.** to produce the required flow and head difference. Let me know if you'd like further clarification!