1
Guillermo Sanchez Gedeon
CE 343 Lab 3
Due: 09/14/2021
Ankit Ghanghas
Assignment 3
Exercise 1:
1. ρ Hg = 13600 kg/m3
g = 9.81 m/s2
𝑃 = #𝜌∗𝑔∗ℎ
𝑃# = 13600#(𝑘𝑔
𝑚!)∗9.81#(𝑚
𝑠“)∗0.001#(𝑚) = 133.416#𝑃𝑎
2.
𝑃 = 𝑎𝑡𝑚−#𝜌𝑔ℎ
i.
0 = 101.3∗10!#(𝑃𝑎)−1000#(#$
%!)∗9.81#(%
&“)∗ℎ#(𝑚)
hw = 10.33 m
hw = 33.9 ft
ii.
0 = 101.3∗10!#
(
𝑃𝑎
)
−13600#
8
#$
%!
9
∗9.81#
8
%
&“
9
∗ℎ#
(
𝑚
)
hHg = 0.76 m
hHg = 2.5 ft
iii. Mercury, as can be seen, is a far more practical liquid for a barometer since it
requires less distance to ascend to record a measurement. Mercury just requires
0.76 m to record the same measurement that water requires 10.33 m to record.
3. z = 0.5 m
Top of the liquid and exposed to the atmosphere:
Pressure head = 0.76m * 13600 kg/m3 * 9.81 m/s2 = 101.3 kPa
Piezometric head = 0.76 + 0.5 = 1.26 m
4. The atmospheric pressure measurement would be smaller than the real number if there is
vapor pressure present instead of an absolute vacuum. Because mercury vapor pressure is
2
lower than water vapor pressure, a mercury barometer would provide a more exact
reading than a water barometer.
