Solutions to Homework Assignment #2
Problem-1: Solution
Solution Fresh and seawater flowing in parallel horizontal pipelines are connected to
each other by a double U-tube manometer. The pressure difference between the two
pipelines is to be determined.
Assumptions 1 All the liquids are
incompressible. 2 The effect of air
column on pressure is negligible.
Properties The densities of seawater and
mercury are given to be
ρ
sea = 1035 kg/m3
and
ρ
Hg = 13,600 kg/m3. We take the
density of water to be
ρ
w =1000 kg/m3.
Analysis Starting with the pressure in
the fresh water pipe (point 1) and moving
along the tube by adding (as we go down)
or subtracting (as we go up) the
gh
ρ
terms
until we reach the sea water pipe (point 2),
and setting the result equal to P2 gives
2seaseaairairHgHgw1
PghghghghP
w
=+−−+
ρρρρ
Rearranging and neglecting the effect of air
column on pressure,
)(
seaseawHgHgseaseaHgHgw21
hhhgghghghPP
ww
ρρρρρρ
−−=−+−=−
Substituting,
kPa 5.39==
⋅
−−
=−
2
2
33
32
21
kN/m 39.5
m/skg 1000
kN 1
m)] 3.0)(kg/m (1035m) 5.0)(kg/m (1000
m) 1.0)(kg/m )[(13,600m/s (9.81PP
Therefore, the pressure in the fresh water pipe is 5.39 kPa higher than the pressure in the
sea water pipe.
Discussion A 0.70-m high air column with a density of 1.2 kg/m3 corresponds to a
pressure difference of 0.008 kPa. Therefore, its effect on the pressure difference between
the two pipes is negligible.
Fresh
water
hse
h
air
Sea
water
Mercury
Air
h
Hg
hw
