Water discharges from a horizontal cylindricalpipe at the rate of 400 cm^3/s. At

Question

Water discharges from a horizontal cylindricalpipe at the rate of 400 cm^3/s. At a point in the pipe where theradius is 2.12 { cm}, the absolute pressure is 1.64×105 {Pa}. What is the pipe's radius at a constriction if the pressurethere is reduced to 1.20×105 { Pa}?
Please help me, thanks a lot. Water discharges from a horizontal cylindricalpipe at the rate of 400 cm^3/s. At a point in the pipe where theradius is 2.12 { cm}, the absolute pressure is 1.64×105 {Pa}. What is the pipe's radius at a constriction if the pressurethere is reduced to 1.20×105 { Pa}?
Please help me, thanks a lot.

Explanation / Answer

My calculation is worng Modified post : --------------- Flow rate V = 400 cm ^ 3 / s Radius r = 2.12 cm Area of crossection A = r ^ 2 = 14.11 cm ^ 2 Speed at that point v = V / A                                 = 28.32 cm / s = 0.2832 m / s Absoute pressure at that point P = 1.64* 10^ 5 Pa Absoute pressure at another point P ' = 1.2* 10 ^ 5Pa from Bernoulli's theorem P - P ' = ( 1/ 2) [ v ' ^ 2- v^ 2 ] where = density of water = 1000 kg / m ^3                             0.44*10^5 = ( 1/ 2) * 1000 * [ v ' ^ 2 - 0.2832 ^ 2 ]                                               = 500 [ v ' ^ 2- 0.08020224]                                           88 = v ' ^ 2 - 0.08020224                                            v ' = 9.385 m / s                                                = 938.5 cm / s Area of cross section A ' = V / v '                                       = 0.4262 cm ^ 2 we know A ' = r ' ^ 2 radius at that point r ' = [ A ' / ]                                  = 0.368 cm          

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