Question

Air flows through this tube at a rate of 1400cm3/s . Assume that air is an ideal fluid.

What is the height h of mercury in the right side of the U-tube?

Answer 1

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air flows through a tube at a rate of 1200 cm³/s.Assume tat air is an ideal fluid. what is the height (h) of mercuryin the right side of the tube?   

{Volume Flow Rate(given)} = R = 1200 cm³/s = 1200.0e-6 m³/s
{Area, Left Side Tube} = A_L = (PI/4)*D² =
          = (PI/4)*(2.0e-2 m)^2
          = 0.00031415926 m^2
{Flow Velocity, Left Side} = v_L = R/A =
          = (1200.0e-6)/(0.00031415926)
          = 3.8197186 m/sec
{Area, Right Side Tube} = A_R = (PI/4)*D² =
          = (PI/4)*(4.0e-3 m)^2
          = 0.000012566370 m^2
{Flow Velocity, Right Side} = v_R = R/A =
          = (1200.0e-6)/(0.000012566370)
          = 95.492970 m/sec

From Bernoulli's Law, we have:

     P_L - P_R = (1/2)*?_a*(v_R)^2 -(1/2)*?_a*(v_L)^2

          = (1/2)*(1.2)*(95.492970)^2 - (1/2)*(1.2)*(3.8197186)^2

          = 5462.5902 Pa

Finally, to determine mercury height, we use following
relationship:

     ?_m*g*h = P_L -P_R = 5462.5902 Pa

     ---->    h = (5462.5902)/(?_m*g)

     ---->    h = (5462.5902)/((13534)*(9.81))

     ---->   h = 0.04114 m

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