Determine the system\'s existing flow rate. What percentage increase in the flow
ID: 1996726 • Letter: D
Question
Determine the system's existing flow rate. What percentage increase in the flow would occur if the were completely replaced with the 16-in. pipeline? An irrigation company must transport 5.71 times 10^-2 m^3/sec of water (20 degree C) from reservoir A to reservoir B. The reservoirs are separated by 600 m and have an elevation difference of 18.4 m. Determine the diameter pipe required if the relative roughness of the pipe material is 0.36 mm. (Include minor losses.) Would the solution differ if minor losses were neglected? A 75-ft pipe must transport 2.5 cfs of 40 degree F water from a head tank to a pond. The elevation difference between the tank and the pond is 4.6 ft. Determine the size of the commercial steel pipe that is required. Assume square-edged connections and include a globe valve in the pipe. It is necessary to deliver 5 L/min of a water-glycerol solution (sp gr. = 1.1 v = 1.03 times 10^-5 m^2/sec) under a pressure head of 50 mm Hg. A glass tube is used (e = 0.003 mm). Determine the tube diameter if the tube length is 2.5 m. Assume the pressure head (50 mm Hg) is needed to overcome the friction loss in the horizontal tube; no other losses are considered. The 40-m-long, 4-in. commercial steel pipe connects reservoirs A and B as shown in Figure P4.1.6. If the pressure at point 1 is 39.3 kPa, what water is at 20 degree C, all valves are fully open, and bend losses are negligible. All the pipes in Figure P4.1.14 have a Hazen-Williams coefficient of 100. Pipe AB is 3.000 ft long. and has a diameter of 2.0 ft. Pipe BC_1 is 2, 800 with a diameter of 1.0 ft, and pipe BC_2 has a length of 3.000 and a diameter of 1.5 ft. Pipe CD is ft long with a diameter of 2.0 ft. The water surface elevation of reservoir A is 230ft (H_A) and reservoir D (H_D) is 100 ft. Determine the discharge in each pipe and the total head at points B and C if Q_B = 0 and Q_C = 0. Ignore minor losses.Explanation / Answer
Given:
length of the pipe (l) - 75ft
discharge (q) - 2.5 cu.ft/sec
length of the pipe - 75 ft
solution:
applying bernoulli's equation at the top of the water surface in the tank and outlet of the pipe
p1 / w + v21 / 2g + z1 = p2 /w + v22 / 2g + z2 + losses
where
p1 / w and p2 /w - pressure energy
v21 / 2g and v22 / 2g - kinetic energy
z1 and z2 - datum
the becomes,
0 + 0 + 4.6 = 0 + v22 / 2g +0
v22 =4.6 * 2g where g[gravity] is 32.174 ft/s2
v22 = 4.6 * 2 * 32.174 = 296
v2 = 17.20 ft/s
according to equation of continuity, discharge q=A1*V1=A2*V2 where A,V are area and velocity respectively
q=A2*V2
A2= q / v2
A2= 2.5/17.20
A2= 0.145 ft2
reuired size of the square tube = 0.38 ft (square root of 0.145)
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