- CUMBRES DE CREDOS
2) Depósitos higroturbosos:
4.1.2 Registros actuales
The types of pressure-sensing devices include the Bourdon gauge, diaphragm capsule, and capacitance sensor, column of fluid, manometer, barometer, sili-con diaphragm, and semisili-conductor strain gauges. The latter are used for better response and increased sensitivity. A manometer is a U-shaped device that uses a fluid having greater density than other fluids in the process unit.
Manometer operation is based on the fact that hydrostatic pressure at the same level in the same fluid must be the same in each leg. To understand how a manometer works, we must understand how to determine the hydro-static pressure caused by a mass of a column of fluid.
By definition of pressure, P FA mg
A Vg
A Ahg
A gh
= = = ρ = ρ =
ρ
Whenever we need to determine the hydrostatic pressure caused by a mass of fluid, it is simply given by
P = ρgh
Gauge pressure is the pressure relative to atmospheric pressure. Gauge pres-sure is positive for prespres-sures above atmospheric prespres-sure, and negative for pressures below it. Atmospheric pressure does add to the pressure in any fluid not enclosed in a rigid container. The total pressure, or absolute pres-sure, is thus the sum of gauge pressure and atmospheric pressure:
Pabs=Pgauge+Patm
where
Pabs is absolute pressure Pgauge is gauge pressure Patm is atmospheric pressure
Example 1.14 Fundamentals of Pressure Problem
Consider the manometer in Example Figure 1.14.1. If h = 10 in. and the manometer fluid is mercury (ρ = 13.6 g/cm3), calculate the gauge pres-sure and absolute prespres-sure (14.7 psia).
Solution
Known quantities: Manometer fluid is mercury, density of mercury, and change in mercury height, h.
Find: The gauge and absolute pressure.
Analysis: Absolute pressure is obtained by adding the gauge pressure to atmospheric pressure.
Gauge pressure = absolute pressure − atmospheric pressure ρ = 13.6 g/cm3 (or 13.6 times greater than H2O) Manometer filled with mercury.
Example 1.15 Pressure Gauges in a Tank Problem
A pressure gauge on a tank reads 25 psi. What is the absolute pressure, considering the atmospheric, Patm = 14.7 psi?
Solution
Known quantities: Pressure gauge.
Find: The gauge pressure.
Analysis: Absolute pressure is obtained by adding the gauge pressure and local atmospheric pressure.
The gauge reads the gauge pressure directly. Therefore, the absolute pressure
Pabs = 25 psig + 14.7 psi = 39.7 psia = 40 psia
Example 1.16 Absolute Pressures from Vacuum Pressure Problem
The pressure gauge on a tank reads 20 cmHg vacuum. The atmospheric pressure in that tank area is 760 mmHg. What is the absolute pressure in the tank?
Solution
Known quantities: Vacuum pressure.
Find: The absolute pressure.
Analysis: The gauge reads a vacuum gauge pressure directly. The pressure is measured below atmospheric pressure relative to the lat-ter and in the reverse direction toward zero absolute pressure, that is, Pvac = Patm − Pabs. Thus, the absolute pressure is
Pabs=Patm−Pvac=760 200 560− = mmHg
Example 1.17 Pressure Caused by a Column of Fluid Problem
The fluid used in a manometer has a specific gravity of 0.80, and the col-umn manometer height is 0.50 m. The manometer is used to measure the pressure in a pipe. The local atmospheric pressure is 100 kPa. Determine the pressure in the column.
Solution
Known quantities: Fluid specific gravity and column height are known.
Find: The pressure in the manometer column.
Analysis: The density of the fluid is obtained by multiplying its specific gravity by the density of water, which is 1000 kg/m3.
ρ ρ= s×ρH O 3= 3
2 =0.8 1000 kg/m× 800 kg/m
From Equation 1.14 and substituting known quantities,
Example 1.18 Pressure Calculations in American Engineering System Problem
What is the pressure (in psi) caused by a 10 ft column of fluid with a spe-cific gravity of 0.8 at ambient temperature? The gravitational acceleration is 32.174 ft/s2.
Solution
Known quantities: Fluid specific gravity, height, and acceleration are known.
Find: The column pressure.
Analysis: Use P = ρgh.
Density of water (reference fluid)
ρref lbm
Example 1.19 Pressure Calculations in SI Units Problem
What is the pressure (in kPa) caused by a 10 m long column filled with water at ambient temperature? Assume the gravitational acceleration to be 9.81 m/s2.
Solution
Known quantities: Fluid specific gravity, height, and acceleration are known.
Find: The column pressure.
Analysis: Use P = ρgh.
Density of water
Calculation of pressure based on the type of manometer used
Three different arrangements of manometers are shown in Figure 1.5.
These manometers can be used to measure pressure using a column of a dense liquid.
1. An open-end manometer can give the gauge pressure.
2. A differential manometer gives ΔP between two points. Note that the pressure decreases in the direction of flow.
3. A closed-end manometer gives absolute pressure. A manom-eter that has one leg sealed and the other leg open measures atmospheric pressure. It is called a barometer.
The basic variables that need to be considered in a manometer are those that measure pressure or pressure difference as shown in Figure 1.6. The line 1–2 is at the interface between the manometer fluid and the higher-pressure fluid. The hydrostatic higher-pressure on each leg is the same along the line within the same fluid. It becomes our reference point. We do a pressure balance by equating the pressures on each leg. Applying this allows us to develop the general manometer equation as
Pressure on leg Pressure on leg i e
(a) Open-end, (b) differential, and (c) sealed-end manometers.
Example 1.20 Pressure Drop across an Orifice Problem
Determine the pressure drop across the orifice meter as shown in Example Figure 1.20.1.
Solution
Known quantities: Water density and manometer fluid height differ-ence are known.
Find: The pressure drop across the orifice meter P1−P2.
Analysis: This is a differential manometer. Note that the hydrostatic pressure above the x mark is the same on both sides. Manometers are simple, inexpensive, and accurate devices used to measure fluid pressure.
P1+ρ1gd P= 2+ρfgd P1
(1) (2)
ρ1
ρ3 ρ2
P2
d2
d1
h
FIGURE 1.6
Open-end manometer.
Water P1 P2
x x¢
d = 0.022 m
ρf =1100 kg/m3 ρ1= 1000 kg/m3
EXAMPLE FIGURE 1.20.1 Pressure drop across an orifice.
We can now write the pressure difference across the orifice from the earlier equation as
P P gd1− 2= (ρf−ρ1)
Substituting the appropriate known quantities, we get
P P1 2 9 8 2 0 022 1100 1000 3 1 2
1 1
− = ⎛
⎝⎜ ⎞
⎠⎟
( ) (
−)
⎛⎝⎜⎜ ⎞
. m . ⎠
s m kg
m N
kg m/s⋅ ⎟⎟⎟
=21 6. N2 =21 6.
m Pa
Example 1.21 Pressure in a Manometer Problem
Consider the tank filled with natural gas as shown in Example Figure 1.21.1. Assume that the manometer fluid is a liquid with a spe-cific gravity of 0.87 and the elevation difference is of 0.01 m (h = 0.01 m).
Calculate the pressure inside the tank (PB).
Solution
Known quantities: Elevation difference and manometer fluid specific gravity.
Find: The pressure in the tank.
Analysis: Since the manometer used is a closed-end one, the pressure inside the tank is calculated using the following equation:
Pgauge=ρgh
PA= 0 (vacuum)
PB
h
Manometer fluid
EXAMPLE FIGURE 1.21.1
Pressure in a closed tank measured using a closed-end manometer.
Substitute known quantities:
Example 1.22 Open-End Manometers Problem
An open-end manometer is connected to a pipe in which a gas is flow-ing (Example Figure 1.22.1). The manometer fluid density is 2.00 g/cm3. A gas bubble is trapped in the left leg of the manometer as shown in Example Figure 1.22.1. What is the pressure of the flowing gas inside the pipe (in atm)? Is this pressure a gauge pressure or an absolute pres-sure? How do you know?
Solution
Known quantities: As per schematic diagram.
Find: The pressure of the flowing gas.
Analysis: At the level marked with the letter x, the pressure in the more dense fluid (density = 2.00 g/cm3) must be the same since the fluid is not moving. On the right-hand side of the manometer, this pressure must be the pressure exerted on its free surface plus the pressure from the 12.00 cm of this fluid. Since the tube is open to atmospheric pressure, the pressure at level x in the manometer fluid is
Px=ρgh P+ atm= kg× 2× +Patm
Pressure in a gas pipe measured using an open-end manometer.
Px=
Adding up all of the hydrostatic pressures and the pressure from the flowing gas in this leg gives
Px=0 0232. atm+Patm
Pxʹ=Pgas+ρgas gas gasg h +ρfluid fluid fluidg h
We can ignore the contribution of the gas bubble to the hydrostatic head, since gas densities are much smaller than liquid densities. Thus the ear-lier equation gives
The difference between gas pressure and atmospheric pressure is the gauge pressure:
Pguage=Pgas−Patm=0.0037 atm
Had the right leg of the manometer been open to the atmosphere, the manometer reading would have been a gauge pressure.
Example 1.23 Pressure in a Gas Pipe Problem
Gas is flowing in a circular pipe. A closed-end manometer is connected to the gas pipe. The manometer fluid density is 2000 kg/m3. A gas bubble has been trapped in the left leg of the manometer as shown in Example Figure 1.23.1. The closed end of the manometer is in a vacuum.
What is the pressure of the flowing gas inside the pipe (in cmHg)? Is the manometer reading the gauge or the absolute pressure of the pipe?
Solution
Known quantities: See the schematic diagram.
Find: The pressure of the flowing gas.
Analysis: At the level marked with the letter x the pressure in the more dense fluid (specific gravity = 2.00) must be the same since the fluid is not moving. On the right-hand side of the manometer, this pressure must be the pressure exerted on its free surface plus the pressure from the 12.00 cm of this fluid. Since there is a vacuum above the fluid, the pres-sure exerted on its free surface is zero. Thus, the prespres-sure at level x in the manometer fluid is given by
Px= × ×
× + =
0 12 9 81 2000 1
1 013 10 76
1 0 1 77
. . 5
. .
m m
s kg
m atm
Pa cmHg
atm c
2 3 mmHg
At level x in the left leg of the manometer, the pressure must be the same as the afore-calculated pressure (1.77 cmHg). Adding up all of the hydrostatic pressures and the pressure from the flowing gas in this leg gives
Pxʹ=Pgas+ρgasghgas+ρfluidghfluid
The contribution to the hydrostatic head from the gas bubble can be ignored since gas densities can be neglected relatively to liquid densi-ties. Thus the earlier equation gives
Vacuum Gas flow
Gas bubble
0.12 m x x¢
Density = 2 g/cm3 0.10 m
EXAMPLE FIGURE 1.23.1
Measuring pressure in a gas pipe using a closed-end manometer.
P Px= x = =P + ×
The manometer reading is the absolute pressure. We know this because the pressure exerted on the fluid surface in the right leg of the manom-eter is zero (vacuum). Thus the manommanom-eter reading is the absolute pres-sure. Had the right leg of the manometer been open to the atmosphere, the manometer reading would have been a gauge pressure.
Example 1.24 Pressure in Closed and Open Tanks Problem
Consider the equipment connected to a closed-end manometer, as shown in Example Figure 1.23.1. When the equipment is fully open to the atmo-sphere, the difference in the manometer liquid height in the closed leg of the manometer is h1. When the equipment is in operation, a pressure gauge on the equipment reads 50.0 mmHg, and the difference between the liquid level in the closed leg of the manometer and the liquid level in the leg connected to the equipment is now 30.0 cm more than it was when the equipment was open. Assume that there is a partial vacuum pressure above the liquid in the closed end of the manometer. Assume that the ambient pressure is 710 mmHg. What is the density of the liquid in the manometer?
Solution
Known quantities: See the schematic diagram.
Find: The pressure of the flowing gas.
Analysis: In this problem, the two scenarios are schematically repre-sented in Example Figure 1.24.1.
At the position marked “×,” the pressure in the manometer fluid must be the same in both legs. The pressure above the left leg is an ambient pressure (710 mmHg). The pressure on the other side is the hydrostatic head from the manometer fluid plus any pressure resulting from the gas in the closed end. Let us call this pressure Pv. Thus,
710mmHg=ρgh1+ Pv
The system will look essentially the same when the equipment is in operation, except that the manometer fluid in the right leg will be 30 cm above the fluid in the left leg. The pressure gauge on the equipment will read 50 mmHg, but remember that this is a gauge pressure. Thus the
actual pressure is 710 + 50 mmHg. Proceeding as mentioned earlier, you would get
710mmHg+50mmHg=ρg h( 1+30cm)+Pv
Subtracting the first equation from the second equation gives 50 mmHg = ρg(0.30 m)
Solving for ρ gives
ρ =
Example 1.25 Inclined Manometers Problem
The manometer shown here contains three liquids with densities ρ1 = 2000 kg/m3, ρ2 = 1000 kg/m3, and ρ3 = 1600 g/cm3. One end of the manometer is connected to a vessel filled with a gas, while the other end is open to the atmosphere (PB). The ambient pressure is 100 kPa. The inclined leg of the manometer is at an angle of 30° from the horizontal.
What is the pressure (PA) inside the vessel?
Solution
Known quantities: See schematic diagram (Example Figure 1.25.1).
Find: The pressure in the tank, PA.
Pressure inside open tank and closed tank.
Analysis: Starting at the lowest point in the manometer and going up each leg, we can balance the pressure as
PA+ρ3g( .0 18m)+ρ1g( .0 10m)=PB+ρ2g( .0 09m)+ρ1g h( )
Using this to converse units gives
PA N2 2 2 2
Pressure in a closed tank measured using an inclined manometer.
Simplifying,
PA+2.825 kPa 1.962 kPa 100 kPa 0.882 kPa 3.139 kPa+ = + + The pressure in the tank PA = 99 kPa.