%3Cp%3E%3Cspan%20class%3D%22c1%22%3EWater%20flows%20through%20a%20pipe%20as%20sh
ID: 2295219 • Letter: #
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%3Cp%3E%3Cspan%20class%3D%22c1%22%3EWater%20flows%20through%20a%20pipe%20as%20shown%20in%20the%0Afigure.%20The%20pressure%20at%20points%201%20and%202%20respectively%0Ais%20%3C%2Fspan%3E%3Cspan%20class%3D%22nobr%20c2%22%3E1.65%26nbsp%3B%3Cimg%20src%3D%0A%22http%3A%2F%2Fwww.webassign.net%2Fimages%2Fmultiply.gif%22%20%2F%3E%26nbsp%3B10%3Csup%3E5%3C%2Fsup%3E%26nbsp%3BPa%3C%2Fspan%3E%3Cspan%20class%3D%22c1%22%3E%26nbsp%3Band%26nbsp%3B%3C%2Fspan%3E%3Cspan%20class%3D%22nobr%20c2%22%3E1.35%26nbsp%3B%3Cimg%20src%3D%22http%3A%2F%2Fwww.webassign.net%2Fimages%2Fmultiply.gif%22%20%2F%3E%0A%26nbsp%3B10%3Csup%3E5%3C%2Fsup%3E%26nbsp%3BPa.%3C%2Fspan%3E%3Cspan%20class%3D%22c1%22%3E%26nbsp%3BThe%0Aradius%20of%20the%20pipe%20at%20points%201%20and%202%20respectively%0Ais%26nbsp%3B%3C%2Fspan%3E2.50%3Cspan%20class%3D%22c1%22%3E%26nbsp%3Bcm%0Aand%26nbsp%3B%3C%2Fspan%3E%3Cspan%20class%3D%0A%22nobr%20c2%22%3E1.45%26nbsp%3Bcm.%3C%2Fspan%3E%3Cspan%20class%3D%22c1%22%3E%26nbsp%3BIf%20the%0Avertical%20distance%20between%20points%201%20and%202%0Ais%26nbsp%3B%3C%2Fspan%3E%3Cspan%20class%3D%22nobr%20c2%22%3E2.75%20m%2C%3C%2Fspan%3E%3Cspan%20class%3D%0A%22c1%22%3E%26nbsp%3Bdetermine%20the%20following.%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cdiv%20class%3D%22figure%20c3%22%3E%3Cimg%20src%3D%0A%22http%3A%2F%2Fwww.webassign.net%2Fwebassignalgphys1%2F12-p-017.gif%22%20%2F%3E%3C%2Fdiv%3E%0A%3Cdiv%20class%3D%22indent%20c5%22%3E(a)%20speed%20of%20flow%20at%20point%201%3Cbr%20%2F%3E%0A%3Cspan%20class%3D%22qTextField%22%3E%3Cinput%20type%3D%22text%22%20name%3D%0A%22RN_2707451_10_0_2713856%22%20value%3D%22%22%20size%3D%2210%22%20id%3D%0A%22RN_2707451_10_0_2713856%22%20dir%3D%22ltr%22%20class%3D%0A%22c4%22%20%2F%3E%3C%2Fspan%3E%26nbsp%3Bm%2Fs%3Cbr%20%2F%3E%0A%3Cbr%20%2F%3E%0A(b)%20speed%20of%20flow%20at%20point%202%3Cbr%20%2F%3E%0A%3Cspan%20class%3D%22qTextField%22%3E%3Cinput%20type%3D%22text%22%20name%3D%0A%22RN_2707451_10_1_2713856%22%20value%3D%22%22%20size%3D%2210%22%20id%3D%0A%22RN_2707451_10_1_2713856%22%20dir%3D%22ltr%22%20class%3D%0A%22c4%22%20%2F%3E%3C%2Fspan%3E%26nbsp%3Bm%2Fs%3Cbr%20%2F%3E%0A%3Cbr%20%2F%3E%0A(c)%20volume%20flow%20rate%20of%20the%20fluid%20through%20the%20pipe%3Cbr%20%2F%3E%0A%3Cspan%20class%3D%22qTextField%22%3E%3Cinput%20type%3D%22text%22%20name%3D%0A%22RN_2707451_10_2_2713856%22%20value%3D%22%22%20size%3D%2210%22%20id%3D%0A%22RN_2707451_10_2_2713856%22%20dir%3D%22ltr%22%20class%3D%0A%22c4%22%20%2F%3E%3C%2Fspan%3E%26nbsp%3Bm%3Csup%3E3%3C%2Fsup%3E%2Fs%3C%2Fdiv%3E%0AExplanation / Answer
by bernoulli's theorem we have
P1/rho + 0.5v^2 + gh1 = P2/rho + 0.5v2^2 + gh2
P1 /rho = 1.65*10^5 /10^3 = 165
P2/rho = 1.35*10^5 /10^3 = 135
h1 = 0
h2 = 2.75 m
so 165 + 0.5v1^2 + 0 = 135 + 0.5v2^2 + 2.75
=> 0.5v1^2 + 27.25 = 0.5v2^2
=> v1^2 + 54.5 = v2^2
A1 = pi*(0.025)^2
A2 = pi*(0.0145)^2
v1*A1 = v2*A2
so v2 = v1*A1 /A2 = v1*(0.025/0.0145)^2 = 2.973*v1
so v1^2 + 54.5 = 8.839v1^2
=> v1 = 2.637 m/s
v2 = 7.84 m/s
volume flow rate = pi*(0.025)^2 * 2.637 = 5.178*10^-3 m^3/s
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