Question

A U-tube manometer is used to measure the pressure at the stagnation point of a model in a wind tunnel. One side of the manometer goes to an orifice at the stagnation point; the other side is open to the atmosphere ( Fig. P1.24 ). If there is a difference of 3.0 cm in the mercury levels in the two tubes, what is the pressure difference in N>m2 .

Answer 1

ΔP = 3.98 10³ Pa

Step-by-step explanation:

We use that the pressure is given

P = ρ g y

where ρ is the density of mercury (ρ = 13534 kg / m³) and y the height of the column

In this case, to measure the pressure, a line is drawn at the lowest point of the mercury column, on the right side of the U-shaped tube, for this point the pressure on both sides of the tube is the same

ΔP = P_l{eft} - P_{right}

ΔP = (P_a + P) - P_a

done P is the gauge pressure on the left side and P_a is the atmospheric pressure

ΔP = ρ g h

let's calculate

ΔP = 13534 9.8 0.03

ΔP = 3.98 10³ Pa

remember that [Pa] = [N / m²]