Given: | P_{x} = -67.7 kPa P_{y} = atmospheric g_{H2O} = 9.798 kN/m^{3} g_{M} = 132.836 kN/m^{3} |
Find: | h |
P_{1} = P_{x} + g_{H2O} L + g_{M} h
P_{2} = P_{y} + g_{H2O} (h + L)
Because P_{1} and P_{2} are at the same elevation and are within a continuous fluid,
P_{1} = P_{2}:
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_{M}h
h = P_{x}/(g_{H2O} - g_{M}) = -67.7 kPa/(9.798 kN/m^{3} - 132.836 kN/m^{3}) = 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.