|Given:||Px = -67.7 kPa
Py = atmospheric
gH2O = 9.798 kN/m3
gM = 132.836 kN/m3
P1 = Px + gH2O L + gM h
P2 = Py + gH2O (h + L)
Because P1 and P2 are at the same elevation and are within a continuous fluid, P1 = P2:
Px + gH2O L + gM h = Py + gH2O (h + L) = Py + gH2O h + gH2O L
Since the right-hand tank is open to the atmosphere, Py = 0.
Cancelling the gH2O L terms:
Px = gH2O h - gMh
h = Px/(gH2O - gM) = -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.