Let's break down part (a) step by step: ### **Step 1: Write the Bernoulli Equation** For a tank sealed with air pressure, the Bernoulli equation (neglecting losses and assuming velocity at the water surface is negligible) between the water surface (point 1) and the nozzle exit (point 2) is: \[ P_1 + \gamma z_1 + \frac{1}{2}\rho v_1^2 = P_2 + \gamma z_2 + \frac{1}{2}\rho v_2^2 \] - \( P_1 \) = pressure on water surface (20 psig) - \( z_1 \) = height of water surface (h) - \( v_1 \approx 0 \) (large tank, negligible surface velocity) - \( P_2 \) = atmospheric pressure (since nozzle opens to atmosphere) - \( z_2 = 0 \) (datum at nozzle) - \( v_2 \) = velocity at nozzle (unknown) #### Convert 20 psig to psia: - Atmospheric pressure \( = 14.7 \) psia - \( 20 \) psig \( = 20 + 14.7 = 34.7 \) psia **Explanation:** We set up the Bernoulli equation, recognizing the tank is sealed and the nozzle is open to the atmosphere. The tank's pressure is given as gauge, so we convert to absolute. --- ### **Step 2: Substitute Values and Solve for \( v_2 \)** - \( P_1 = 34.7 \) psia \( = 34.7 \times 144 = 4996.8 \) psf (1 psi = 144 psf) - \( P_2 = 14.7 \) psia \( = 14.7 \times 144 = 2116.8 \) psf - \( h = z_1 = 8 \) ft, \( z_2 = 0 \) - \( \gamma = 62.4 \) lb/ft³ (unit weight of water) - \( \rho = 1.94 \) slugs/ft³ (density of water) Plug into Bernoulli: \[ P_1 + \gamma h = P_2 + \frac{1}{2}\rho v_2^2 \] \[ 4996.8 + 62.4 \times 8 = 2116.8 + \frac{1}{2} (1.94) v_2^2 \] \[ 4996.8 + 499.2 = 2116.8 + 0.97 v_2^2 \] \[ 5496.0 - 2116.8 = 0.97 v_2^2 \] \[ 3379.2 = 0.97 v_2^2 \] \[ v_2^2 = \frac{3379.2}{0.97} = 3484.7 \] \[ v_2 = \sqrt{3484.7} \approx 59.04\ \text{ft/s} \] **Explanation:** We substituted all known values into Bernoulli's equation and solved for the velocity of water at the nozzle. --- ### **Step 3: Find the Volume Flow Rate \( Q \)** The diameter of the nozzle is 25 mm (\( = 0.025 \) m \( = 0.0820 \) ft). \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.0820)^2 = 0.00528\ \text{ft}^2 \] \[ Q = v_2 \times A = 59.04 \times 0.00528 = 0.311\ \text{ft}^3/\text{s} \] **Explanation:** We used the calculated velocity and the cross-sectional area of the nozzle to find the volumetric flow rate. --- ## **Final Answer (a):** **The volume flow rate of water from the nozzle is approximately \( \boxed{0.31\ \text{ft}^3/\text{s}} \).** --- Let me know if you want to proceed with part (b)!

Let's proceed with part (b), which involves calculating the pressure in the sealed tank that would result in a flow velocity of 20 ft/s from the nozzle, given the same nozzle diameter. ### **Step 1: Write Bernoulli Equation for the new velocity** Similar to part (a), the Bernoulli equation between the water surface (point 1) and the nozzle exit (point 2): \[ P_1 + \gamma z_1 = P_2 + \frac{1}{2}\rho v_2^2 \] - \( P_1 \) = pressure in the tank (unknown, to be found) - \( z_1 = 10\, \text{ft} \) - \( v_2 = 20\, \text{ft/s} \) Assuming atmospheric pressure at the nozzle exit: \[ P_2 = 14.7\, \text{psia} = 2116.8\, \text{psf} \] - \( \gamma = 62.4\, \text{lb/ft}^3 \) - \( \rho = 1.94\, \text{slugs/ft}^3 \) Express \( P_1 \) in psf: \[ P_1 = P_{tank} + \text{gauge pressure} \] But since we're solving for the absolute pressure in the tank, we will convert everything to consistent units. ### **Step 2: Calculate \( P_1 \)** Bernoulli rearranged: \[ P_1 = P_2 + \frac{1}{2} \rho v_2^2 - \gamma z_1 \] Calculating each term: - \( \frac{1}{2} \rho v_2^2 = 0.97 \times (20)^2 = 0.97 \times 400 = 388\, \text{psf} \) - \( \gamma z_1 = 62.4 \times 10 = 624\, \text{psf} \) - \( P_2 = 2116.8\, \text{psf} \) Plug in: \[ P_1 = 2116.8 + 388 - 624 = 1880.8\, \text{psf} \] This is the absolute pressure at the water surface in the tank. ### **Step 3: Convert \( P_1 \) to gauge pressure** - Absolute pressure \( P_{abs} = 1880.8\, \text{psf} \) - Atmospheric pressure \( P_{atm} = 2116.8\, \text{psf} \) Gauge pressure in the tank: \[ P_{gauge} = P_{abs} - P_{atm} = 1880.8 - 2116.8 = -236\, \text{psf} \] Negative gauge pressure indicates the tank pressure is below atmospheric, which is physically impossible unless the tank is under vacuum. **Alternatively**, the pressure in the tank (gauge) that causes flow at 20 ft/s is approximately: \[ P_{tank\, gauge} = P_{abs} - P_{atm} \approx -236\, \text{psf} \] which suggests that to get this flow rate, the tank must be under a slight vacuum (negative gauge pressure). --- ### **Final Answer (b):** **The pressure in the sealed tank that would cause a flow velocity of 20 ft/s from the nozzle is approximately a gauge pressure of \(\boxed{-236\, \text{psf}}\), or about \(\boxed{-1.63\, \text{psi}}\) (since 1 psi = 144 psf).** --- **Summary:** - The tank must be under a slight vacuum of about 1.63 psi to produce a flow of 20 ft/s from the nozzle. - This negative gauge pressure reflects the reduced pressure necessary to achieve the specified flow velocity at the given head and nozzle size.