Let's break down the problem step by step: ### **Given:** - Density of water, \(\rho = 1 \, \text{g/mL} = 1000 \, \text{kg/m}^3\) - Height difference, \(h = 2 \, \text{m}\) - Inlet pressure, \(P_1 = 75 \, \text{kPa} = 75000 \, \text{Pa}\) - Inlet radius, \(r_1 = 3 \, \text{cm} = 0.03 \, \text{m}\) - Exit radius, \(r_2 = 0.02 \, \text{m}\) - Gravitational acceleration, \(g = 9.8 \, \text{m/s}^2\) --- ## a) **The velocity \(V_2\):** ### **Step 1: Calculate the areas** \[ A_1 = \pi r_1^2 = \pi (0.03)^2 = 0.002827 \, \text{m}^2 \] \[ A_2 = \pi r_2^2 = \pi (0.02)^2 = 0.001257 \, \text{m}^2 \] ### **Step 2: Apply the Continuity Equation** \[ A_1 V_1 = A_2 V_2 \] \[ V_1 = \frac{A_2}{A_1} V_2 \] --- ### **Step 3: Apply Bernoulli's Equation** Assume point 1 is at the bottom and point 2 is at the top (height difference \(h\) upwards). \[ P_1 + \frac{1}{2}\rho V_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g h_2 \] Let \(h_1 = 0\), \(h_2 = 2 \, \text{m}\), so \(h_2 - h_1 = 2 \, \text{m}\). \[ P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g (2) \] If the exit is open to the atmosphere, \(P_2\) is atmospheric pressure (but since it asks for absolute, we solve in terms of \(P_2\)). --- ### **Step 4: Solve for \(V_2\)** 1. Substitute \(V_1 = \frac{A_2}{A_1} V_2\) 2. Substitute known values: \[ V_1 = \frac{0.001257}{0.002827} V_2 = 0.445 V_2 \] Now plug into Bernoulli’s equation and solve for \(V_2\): \[ P_1 + \frac{1}{2}\rho (0.445 V_2)^2 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g (2) \] If you want \(V_2\), you need either \(P_2\) or assume \(P_2\) is atmospheric. But usually, the problem will be solved as a system: #### Assume the question wants \(V_2\) in terms of \(P_2\): \[ 75000 + \frac{1}{2}(1000) (0.445 V_2)^2 = P_2 + \frac{1}{2}(1000) V_2^2 + 1000 (9.8)(2) \] \[ 75000 + 1000 \times 0.099 V_2^2 = P_2 + 500 V_2^2 + 19600 \] \[ 75000 + 99.0 V_2^2 = P_2 + 500 V_2^2 + 19600 \] \[ 75000 - 19600 + 99.0 V_2^2 = P_2 + 500 V_2^2 \] \[ 55400 + 99.0 V_2^2 = P_2 + 500 V_2^2 \] If the pipe is discharging to the atmosphere, \(P_2\) is atmospheric (let's assume \(P_{atm} = 101325\,\text{Pa}\)), then: \[ 55400 + 99.0 V_2^2 = 101325 + 500 V_2^2 \] \[ 55400 - 101325 = 500 V_2^2 - 99.0 V_2^2 \] \[ -45925 = 401 V_2^2 \] \[ V_2^2 = \frac{-45925}{401} \] \[ V_2^2 = -114.6 \] A negative value suggests an inconsistency, perhaps in the assumed pressure at the exit, or in the direction of height difference. If the flow is from low to high (uphill), the pressure at the exit may not be atmospheric. If **P2 is not atmospheric**, rearrange the last equation for \(P_2\): --- ## **b) The absolute pressure at point 2:** \[ P_2 = 55400 + 99.0 V_2^2 - 500 V_2^2 \] \[ P_2 = 55400 - 401 V_2^2 \] --- ### **Summary:** 1. **Velocity at 2:** \[ V_2 = \text{Solve using the system above, with more info about \(P_2\)} \] 2. **Absolute Pressure at 2:** \[ P_2 = 55400 - 401 V_2^2 \quad (\text{in Pa}) \] If you clarify whether the exit is to the atmosphere or if you have a value for \(P_2\), I can provide a numerical answer for \(V_2\)! --- **If you want to use atmospheric pressure for \(P_2\), you would need to check if the inlet pressure is high enough to support the height and velocity, or else the flow is not possible as described.**