Problem Statement:
50
The water in a tank is pressurized by air, and the pressure is measured by a mulitfluid manometer
as shown below.
Determine the gage pressure of air in the tank if:
0.20 m
0.30 m
0.46 m
Schematic:
Assumptions/Approximations:
101.33 kPa
Physical Laws:
[Equation 115]
Properties/Sources:
1000.00
Conversions:
850.00
1 N =
1.00
13600.00
1 kPa =
1000.00
g =
9.81
Calculations:
In the diagram above for absolute pressure,
Check units:
h
1
=
h
2
=
h
3
=
P
ATM
=
P =
ρ
*g*h
P
GAGE
= P
ABS
 P
ATM
ρ
H2O
=
kg/m
3
ρ
OIL
=
kg/m
3
kg*m/s
2
ρ
Hg
=
kg/m
3
N/m
2
m/s
2
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158.215 kPa
56.89
kPa
Reasoning/Verification/Discussion:
By using Pascal's Law and the sum of Forces rule from statics, you could determine the answer using
each reference point as a place to sum forces.
We will learn that "Forcing Functions" are required for any
P
ATM
= P
1
+
ρ
H2O
*g*h
1
+
ρ
OIL
*g*h
2

ρ
Hg
*g*h
3
(kg/m
3
)*(m/s
2
)*(m)
= kg/(m*s
2
) = kg*m/(m
2
*s
2
) = N/m
2
P
1
= P
ATM

ρ
H2O
*g*h
1

ρ
OIL
*g*h
2
+
ρ
Hg
*g*h
3
=
[Note that
ρ
*g*h is divided by 1000 for units]
P
1GAGE
= P
1
 P
ATM
=
change in energy state such as
∆
P or
∆
T.
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 Spring '09
 n/a
 Problem Statement, kPa, physical laws

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