Figure below was used to derive the equation used to determine the time it takes to empty the tank from h₁ to h₂. The equation was derived by considering that the tank is open and hence at 0 gage pressure above the water. If the tank is sealed and pressure of 5.0 psig is above the water in the tank, compute the time required to empty the tank if the original depth is 15 ft. The tank diameter is 12 ft and the orifice diameter is 6 in. dh

What pressure, p₁, is needed to produce a flowrate of 0.09 ft3/s from the tank shown in the figure below? Assume d = 0.071 ft. P1 = Mi ! psi Air P1 Gasoline diameter d→ Salt water SG = 1.1 2.0 ft 3.6 ft

Use the following given values to solve problems, unless otherwise stated. Ywater = 9790 N/m³ = 62.4 lbf/ft°; g = 9.81 m/s? = 32.2 ft/s²; 1 hp = 550 lbf-ft/s = 746 Watts Problem-1 Water is pumped from the large tank through a uniform diameter pipe (d = 3 in.) and exits as a free jet (refer figure-a). The plot of the flowrate (Q) versus pump-head (hp) is a straight-line graph that passes through points (3, 16) as shown in figure-b. If the pipe friction head loss is, hf = 7V:/2g, where V is the average fluid velocity in the pipe, find a) the equation that relates hp and Q; b) flowrate and c) If the pump is 75% efficient, what horsepower is needed to drive it? Assume steady, incompressible flow. Ignore kinetic energy correction factor (a). 12 ft Pump %3D (a) 16 3. Q, ft°ls (b) dy

Given A fluid with sg = 0.74 is flowing at 0.089 m3/s in the fabricated tube shown in the figure. The pressure in the tube before the contraction is 475 kPa. Refer to Appendix G.2. Flow 80-mm OD x 2.8-mm wall steel tube 160-mm OD x 5.5-mm wall steel tube Required Calculate the pressure in the smaller tube, to the nearest 10th. Your Answer: