1E3 :  Pressure in a Tank Using a Complex Manometer  5 pts 

The pressure gauge on the air in the tank shown below reads 87 kPa. Determine the manometer reading, h_{2}, in cm.  


Data: h_{1} = 25 cm and h_{3} = 65 cm SG_{mercury} = 13.6, SG_{oil} = 0.75, ρ_{water} = 1000 kg/m^{3} 

Read:  The density of the air is so much lover than the density of the liquids in this problem that the weight of the air can be considered negligible in the force balances we will write in this problem.  
Given:  P_{gauge}  87  kPa  
h_{oil}  0.65  m  SG_{oil}  0.7  
h_{water}  0.25  m  SG_{m}  13.6  
Find:  h_{2}  ???  cm  
Assumptions:  1 The fluids in the system are completely static.  
2 The densities of the liquids are uniform and constant.  
3 The reference density of water used to determine specific gravity is:  r_{water}  1000  kg/m^{3}  
4 The acceleration of gravity is:  g  9.8066  m/s^{2}  
g_{c}  1  kgm/Ns^{2}  
Equations / Data / Solve:  
Begin by writing the Manometer Equation for each interval between points A and F on the diagram.  

Eqn 1 

Eqn 3  

Eqn 2 

Eqn 4  

Eqn 5  
If we add all 5 of these equations together we obtain :  

Eqn 6  
The only unknown in this equation is h. So, the next step is to solve the equation for h.  

Eqn 7  
Also, because P_{f} = P_{atm} and the definition of gauge pressure we can use: 

Eqn 8  
P_{a}P_{f} =  87000  N/m^{2}  
All we need to do is convert specific gravity into density and we are ready to plug values into Eqn 7.  
The definition of specific gravity is : 

Eqn 9  
This helps us simplify Eqn 7 to : 

Eqn 10  

Eqn 11  
From Eqn 9 : 

Eqn 12  
r_{m}  13600  kg/m^{3}  
Plugging values into Eqn 11 yields :  h  0.652  m  
65.2  cm  
Answers:  h  65.2  cm 