The following figure shows a bent in a pipe with a diameter of D (cm). Pressure is measure in cross section#1 as demonstrated in the figure as P1 (kPa). In cross section#2 the diamete reduced through a nozzle to 5.0 cm and the water is discharged into atmosphere. Th elevation difference between two cross sections is 0.5 m. Assume all losses are negligib and density of the fluid, water, is 1000 kg/m³. a) Find the velocity in cross section#%; b) Calculate the horizontal and vertical forced acting on the pipe and the direction the resultant force. D = 20 cm, Pj= 450 kPa Cross section#2 0.5 m Diameter: 5 cm P1 60° Cross section#1 Diameter: D cm

The following figure shows a bent in a pipe with a diameter of D (cm). Pressure is measure in cross section#1 as demonstrated in the figure as P1 (kPa). In cross section#2 the diamete reduced through a nozzle to 5.0 cm and the water is discharged into atmosphere. Th elevation difference between two cross sections is 0.5 m. Assume all losses are negligib and density of the fluid, water, is 1000 kg/m³. a) Find the velocity in cross section#%; b) Calculate the horizontal and vertical forced acting on the pipe and the direction the resultant force. D = 20 cm, Pj= 450 kPa Cross section#2 0.5 m Diameter: 5 cm P1 60° Cross section#1 Diameter: D cm

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Find the difference in pressure head and velocity of flow of an oil through a pipe, when the difference of mercury level in a different U-tube manometer connected to the two tappings of the pitot tube as 108 mm. Take co-efficient of the pitot tube ("L" shaped measuring instrument) as 0.9 and sp. gr of oil = 0.8 Difference in pressure head (unit in mm) = Velocity of flow (unit in km/hr) =
As shown in the figure below, a cylindrical tank with a hole in the bottom and an air vent in the top is being filled with water. Given the following: Dtank = 4 ft hill = 8 ft NOG height of water in tank when tank is filled Qin = 10 gal/min %3D NOT UPL TO CHE diameter of hole Find: volumetric flowrate into tank EGG EXAM into the tank in slugs/second OT UPS XAM O NOT U EXAM (c) The rate at which the water level in the tank is mass flowrate (in slugs/second) of the water (b) The mass flowrate (in slugs/second) out of the tank UPLOAD TO CHE NOT UPLOAD when the water level reaches the fill height DO NOT UPL rising (in inches/minute) when the water level NOT UPL PLOAD TO CHEGG NOT UPLOAD TO CHEGG EXAM TO CHEGG EXAM OD TO CHEGG UP reaches the filled height то снE LOAD vent Dtank DO DO DO NOT UPLOADTO EXAM EX DO 0 DONOT UPLOA T Dhole G EXAM TO CNGO UPLOA hole mdotout
An industrial oil (s.g. = =0.85) is flowing at 160 gallons per minute (gpm) in a system with the following details: SUCTION: Standard elbow and the pipe is 7 foot-long. DISCHARGE: Street type elbow and the pipe length is 42 feet. Neglecting losses due to the filter and the hydraulic press, compute: (a) The total energy loss due to friction (suction and discharge pipes) (b) The total energy loss due to minor losses (suction and discharge) 3" Schedule 40 Pump 4" Schedule steel pipe 40 steel pipe Hydraulic Filter press Flow 4.0 ft 早 L0 ft 2.0 ft Reservoir
1. To model power production in a traditional hydroelectricity unit, we can use the below schematic figure. Water (e= 998 kg/m3, µ = 0.001 kg/m.s) flows from a large tank at elevation 60 m through a cast iron pipe with total length of L diameter 0.1 m. There are three losses associated with sharp-edge entrance, 90° reg- ular elbow and an open globe valve. Outlet of the valve is discharged to atmosphere. The velocity of the water in the pipe is 3 m/s. For minor losses use values for a flanged pipe with diameter of 4". Find the power produced in the turbine in kW 100 m and unit. Hint: If we write energy grade equation between point 1 (surface of water in tank) and point 2 (left hand side of the valve);" P V P V + 21 – hturbine - Ahtot 2g + 2 pg pg 2g where, P = P2 = Patm and Vi 0 (large tank) Sharp-odge entrance 60 m Elbow Turbine Ginhe Valhe
None
i need the answer quickly
Water at 25°C is pumped at 1800 gal/min from the lower to upper reservoir, as shown in Figure below (Zı = 30 ft and z2 = 180 ft). Pipe friction losses are approximated by hf = 35V/2g, where V is the average velocity in the pipe. The pump is 80% efficient. (a) Find the head loss due to friction, hẹ in ft. in horse power (b) Compute the pump head, h, in ft and the pump power needed, Ppump (hp). Given: Specific weight of water, ywater = 62.4 lb/ft'; 1 ft/s = 448.8 gal/min. Show all calculations. Z2 = 180 ft Z¡ = 30 ft D= 6 in Pump
2. Oil with a S.G. pf 0.82 is flowing in a pipe under the condition below (see figure). If the total head loss from point 1 to point 2 is 3 ft, find the pressure at point 2 in PSI. Note: cfs = cubic ft per sec Point 1 D₁ - 6 km P₁-65 pal 0 Q-2.08 efs Q-2.08 efs Point 2 D₂ - 9 in 10.70 ft 4.00 ft Reference datum
Fluid 2
A pump has one inlet and two outlets as shown in the figure all at the same elevation. What pump power is required if the pump is 85% efficient. Neglect pipe losses
A hydraulic pump is shown in the following Figure, flow rate is 25 L/min, the diameter of suction pipe is 25mm, and the suction port of pump is higher than the oil surface of tank, the height is h, Kinematic viscosity of oil is 30×10“m²/s, density of oil is 900kg/m³. Only the pressure loss along the straight suction pipe with length h is 75 considered, if the vacuum degree at the inlet of pump is not bigger than 4500P ( 2 = R. a = 2, g=9.8) (1) Calculate the flow velocity in the suction pipe; (2) Judge the flow state in suction pipe; (3) Calculate the pressure loss along the straight suction pipe; (4) Calculate the oil absorption height h.