1B5 :  Relationships between Different Types of Pressures  5 pts 

Fill in the blank values in the table below. Assume P_{atm} = 100 kPa and the density of liquid mercury (Hg) is 13,600 kg/m^{3}.  


Read:  This problem requires
an understanding of the relationship between absolute and gage
pressure. It will also require the effective use of unit conversions. 

Given:  P_{atm}  100  kPa  r_{H2O}  1000  kg/m^{3}  
g_{c}  1  kgm/Ns^{2}  r_{Hg}  13600  kg/m^{3}  
a.)  P_{gage}  17  kPa  c.)  P_{abs}  55  mmHg  
b.)  P_{abs}  225  kPa  d.)  P_{gage}  32  m H_{2}O  
Assumptions:  1 Assume:  g  9.8066  m/s^{2}  
Find:  P_{gauge}  ???  kPa  P_{abs}  ???  mmHg  
P_{abs}  ???  kPa  P_{gauge}  ???  m H_{2}O  
Equations / Data / Solve:  
There are two key relationships in the solution to this problem.  
The first is the relationship between absolute and gage pressure :  

Eqn 1  
The second relationship is required in order to make sense of the units for pressure in the last two columns of the table in the problem statement. The 2nd relationship is the Manometer Equation.  

Eqn 2  
The reason we use the Manometer Equation is that when a pressure unit involves a length of a given fluid, as in the last two columns of the table given in this problem, it really means that this is the height that an openended manometer (for gage pressure) or a closed end manometer (for absolute pressure) would read if the given fluid were used as the manometer fluid.  
Now, let's see how we use these 2 equations to complete the table.  
Part a.)  In order to
fill in the 2nd column, we must solve Eqn 1 for the absolute pressure : 

Eqn 3  
Therefore :  P_{abs}  117  kPa  
To complete column 3, we must convert the units from our result for P_{abs} using Eqn 2.  
In this case, P_{in} = P_{abs} and P_{out} = 0 (because it is a closedend manometer).  
Actually, P_{out} should be equal to the vapor pressure of the manometer fluid, but that is a topic for chapter 2.  
Therefore, Eqn 2 becomes : 

Eqn 4  
In this case, the answer for column 3 is actually the value of h, so we need to solve Eqn 4 for h :  

Eqn 5  
Be sure to convert kPa to Pa=N/m^{2} when plugging values into Eqn 5.  
P_{abs} = h =  0.877  m Hg  P_{abs}  877  mm Hg  
Column 4 requires a gage pressure, so the openend form of the Manometer Equation is used :  
In this case, P_{in} = P_{abs} and P_{out} = P_{atm} (because it is a closedend manometer).  

Eqn 6  
Next, we solve Eqn 6 for h and make use of Eqn 1 if we want to use the given value of the gage pressure.  

Eqn 7  
P_{gage} = h  1.7335  m H_{2}O  P_{gage}  1.734  m H_{2}O  
Parts bd)  The solution of the remaining parts of this problem involve the algebraic manipulation of Eqns 1, 3, 5 and 7, but does not involve any additional concepts, techniques or data. The final answers to parts b through d are provided in the table below.  
Answers:  P_{gage }(kPa)  P_{abs }(kPa)  P_{abs }(mmHg)  P_{gage }(m H_{2}O)  
a.)  17  117  877  1.73  
b.)  125  225  1690  22.9  
c.)  92.7  7.34  55  0.909  
d.)  314  414  3100  32  
Notice that I chose to use 3 significant figures in my answers.  
This is somewhat arbitrary since the problem statement does not make it very clear how many significant figures exist in the given information. When in doubt, 3 significant figures is a reasonable choice.  