The horizontal pipe shown in the figure (Figure 1) has a cross-sectional area A1
ID: 2245632 • Letter: T
Question
The horizontal pipe shown in the figure (Figure 1) has a cross-sectional area A1 = 40.5cm2 at the wider portions and A2 = 10.2cm2 at the constriction. Water is flowing in the pipe, and the discharge from the pipe is 6.05 times 10?3m3/s (6.05L/s). I have the correct answer for part A, FInd the flow speed at the wide portion, which is v= 1.49 m/s and part B, find the flow speed at the narrow portion, which is v=5.93m/s. I need to know the answer and how to work it for part C and D. Part C ask Find the pressure difference between these portions and part D is Find the difference in height between the mercury columns in the U-shaped tube. Change is pressure is Delta p = p*g*delta h where p is the density of mercury. I know this in basic algerbia but I don't understand it.Explanation / Answer
pressure difference= 0.5*rho*(v2^2-v1^2)=16472Pa (obtained from bernoulli's equation since heeight is same)
Difference in height=P/rho*g==16472/13600*9.8=12.4cm
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