Given:	Px = -67.7 kPa Py = atmospheric gH2O = 9.798 kN/m3 gM = 132.836 kN/m3
Find:	h

Solution:

P₁ = P_x + g_H2O L + g_M h

P₂ = P_y + g_H2O (h + L)

Because P₁ and P₂ are at the same elevation and are within a continuous fluid, P₁ = P₂:

P_x + g_H2O L + g_M h = P_y + g_H2O (h + L) = P_y + g_H2O h + g_H2O L

Since the right-hand tank is open to the atmosphere, P_y = 0.

Cancelling the g_H2O L terms:

P_x = g_H2O h - g_Mh

h = P_x/(g_H2O - g_M) = -67.7 kPa/(9.798 kN/m³ - 132.836 kN/m³) = 0.55 m

Since h is positive, the manometer fluid is higher on the left-hand side of the U-tube than on the right.

Plugging these values into the Manometer Applet yields:

Return to Manometer Exercise.

Return to Manometer Applet.