Question:
(Refer to Q 6.72, Fig 6.29 on page 184.)
A large tank contains with a specific gravity of . It drains freely into the air through a nozzle with a diameter "d2" of mm. The main pipe has a diameter "d1" of mm and the elevation differences are: "h1" = mm and "h2" = mm.

Calculate the following:

Part 1: The flow rate through the pipe (in L/s).
(Answer = )

Part 1: Use the General Energy (Bernoulli) Equation (no Head Losses) between the fluid surface in the tank and the exit nozzle: 0 + 0 + h1 = v^{2}/2g + 0 + 0 (datum at nozzle). Hence calculate the exit velocity ( m/s) and then the corresponding flow rate using Q = Av. Calculations are done in metres (convert "h1" and the nozzle diameter) and m^{3}/s but the answer is required in L/s..

Part 2: The pressure at "A" (in kPa, sign with no space if negative).
(Answer = )

Part 2: Note that the pressures at the fluid surface and at "A" are not the same. The fluid is stationary at the surface of the large tank but moving (v^{2}/2g term) inside the pipe. Use the General Energy (Bernoulli) Equation (no Head Losses) between the fluid surface in the tank and "A" inside the main pipe: 0 + 0 + h1 = v^{2}/2g + P_{A}/specific weight + h1 (datum at nozzle). Use Q = Av to find the velocity inside the pipe ( m/s) and the specific gravity of the fluid to find its specific weight ( kN/m^{3}).

Part 3: The pressure at "B" (in kPa, sign with no space if negative).
(Answer = )

Part 3: Use the General Energy (Bernoulli) Equation (no Head Losses) between the fluid surface in the tank and "B" inside the main pipe: 0 + 0 + 0 = v^{2}/2g + P_{B}/specific weight + h2 (datum at the fluid surface). Note that the lowest pressure in the siphon system is at its highest point "B".

Part 4: The pressure at "C" (in kPa, sign with no space if negative).
(Answer = )

Part 4: The pressures at "A" and "C" are the same since they have the same elevation and the same v^{2}/2g term. Note that v^{2}/2g is an energy term and the direction of flow is not important.

Part 5: The pressure at "D" (in kPa, sign with no space if negative).
(Answer = )

Part 5: Use the General Energy (Bernoulli) Equation (no Head Losses) between the fluid surface in the tank and "D" inside the main pipe: 0 + 0 + h1 = v^{2}/2g + P_{D}/specific weight + 0 (datum at nozzle). Alternatively, use the equation between "D" inside the pipe and the unrestrained fluid after it exits the nozzle: v_{pipe}^{2}/2g + P_{D}/specific weight + 0 = v_{nozzle}^{2}/2g + 0 + 0.

Enter the answer to Part 5 in the answer box below: