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Monday, March 11, 2013

Bourdon and diaphragm gages

1. Bourdon gauge

The most widely used in industry for pressure and vacuum measurements (from 20 kPa to 1000 MPa) is a pressure gauge with sensitive element made of a metallic (various stainless-steel alloys, phosphor bronze, brass, beryllium copper, Monel, etc.) Bourdon tube 1 (see Figure 1). The tube was named after its inventor, E. Bourdon, who patented his invention in 1852. This tube has an elliptical or oval cross-section A-A and has the shape of a bended tube. When the pressure inside the tube 1 increases, its cross-section dimension b1 also increases by the value of ∆b1, whereas the cross-section dimension a1 reduces its length by the value of ∆a1. Therefore, the tube tends to straighten (if pressure has increased) or twist (if pressure has decreased, for example, during vacuum measurements), and the tip 2 of the tube moves linearly with applied pressure. The movement of the tip is transmitted to the pointer 3 through a mechanism 4. The tube tends to return to its original shape (the pointer returns to the starting position) after pressure is removed. A relationship between the value of the tip movement ∆x and the measured pressure is linear, so the scale of this pressure gauge is uniform.

Figure 1. Bourdon tube pressure gauge.

Some degree of hysteresis still exists during operation of these pressure gages, because metals cannot fully restore their initial elastic properties. If we have two Bourdon tubes made of the same metal, the tube with a bigger radius and a smaller thickness of the wall will have higher sensitivity. An accuracy of a typical Bourdon-tube pressure gauge is equal to ±1%, whereas a specially designed gauge may have better accuracy which varies from ±0.25 to ±0.5%.

2. Diaphragm gauge

Another type of pressure gauge, which utilises elastic-element properties, is a diaphragm pressure gauge. These gages are used when very small pressures (from 125 Pa to 25 kPa) are to be sensed. Fig. 2 shows a sensitive element for this type of pressure gauge.

Figure 2. Sensitive element of a diaphragm pressure gauge.

A flexible disc 1 made of trumpet brass, or phosphor bronze, or beryllium copper, or titanium, or tantalum, etc., is used to convert the measuring pressure to the deflection of the diaphragm. Deflection vs pressure characteristic should be close to linear as much as possible. In reality for a flat diaphragm this characteristic is non-linear. So, flat membranes are not used as sensitive elements. To linearise this relationship special diaphragms with concentric corrugations 6 are designed. Linearisation of a static characteristic of the membrane can be achieved by using a flat spring 2, which is connected, to the diaphragm through the mechanism 3. The movement of the mechanism 3 is transmitted by the link 4 to a pointer of the gauge. The measuring pressure is supplied to the pressure chamber 5 and causes the diaphragm to move upwards until the force developed by this pressure on the diaphragm is balanced by the force acted from the spring. To increase the sensitivity of this type of pressure gauge, we may increase the diameter of the diaphragm, to lengthen the spring, to change the material of the diaphragm and the spring to more elastic, to increase the depth and the number of corrugations of the diaphragm.

When pressure is applied to both sides of the membrane, then the resultant reading is proportional to the differential pressure. The space above the diaphragm is connected to atmosphere, so the diaphragm separates a measured media from the environment. In other words, it serves as a fluid or gas barrier or as a seal assembly, thus preventing contact of corrosive and aggressive fluids with pressure elements.

Accuracy of diaphragm pressure gages varies from ±1.0 to ±1.5% of the span.

3. Bellows pressure gauge

These pressure sensitive elements are usually made of stainless steel; phosphor bronze, brass and are used for pressure measurements for pressures up to 6 MPa. Bellows sensors have large displacement sensitivity. Figure 3 shows this type of sensor.

Figure 3. Bellows pressure sensitive element.

The effective area of a bellows can be calculated using the following formula:

When pressure is applied to the internal surface of a bellows the force is developed according the formula:

Let’s now develop a differential equation for bellows pressure sensor.

At steady state condition when          t=0  a displacement of the sealed end of the bellows is equal to:
.                            ,                       .                                 

These are initial conditions. The initial force is balanced by the spring force according to:


Let pressure has suddenly increased by the value of P.

Resultant force = mass * acceleration

or  ,                                                          


By subtracting (41) from (43) we get equation with variables in deviation form:



Finally, we get a second order linear differential equation with variables in deviation form:


After introduction of the following parameters:

undamped natural frequency

 damping ratio   

 and a steady-state gain      


the equation (47) can be re-written as follows:


Now we apply the Laplace transform:

                        .                                   (52)

Using initial conditions (38) and (39) we can get:



Now we can get a transfer function for bellows:


Aef                   - effective area of bellows, m2;
Fef                    - effective force due to pressure, N;
k                      - bellows stiffness (or spring factor), N/m;
m                     - mass of bellows, kg;
R1                    - small characteristic radius of bellows,m;
R2                    - big characteristic radius of bellows, m;
t                       - time, s;
x                      - displacement of the sealed end of bellows, m;
λ                      - constant (friction coefficient) for bellows, (N*s)/m.

1. At z= 1 we have a critically damped response.


2. At z< 1 we have underdamped response.


3. At z> 1 we have overdamped response.


A numerical example will help us to better understand various types of responses.
 Aef = 10, cm2 = 0.001,m2;
P’ = 6, kPa = 6000, Pa;
k=1000, N/m;
m= 100, g = 0.1, kg;
λ=10, (N*s)/m

Using the above data we can evaluate:

;                                   .

Figure 4. shows various types of responses of bellows pressure sensor.

Figure 4. Dynamic responses of bellows pressure sensor.

Article Source:: Dr. Alexander Badalyan, University of South Australia


Thursday, March 7, 2013

“U”-tube liquid filled manometers

These manometers are used for measurement of gauge pressures (up to 0.1 MPa), vacuumetric pressures (down to 0.1 MPa below atmospheric pressure) and for differential pressures of liquids and gases. The principle is based on the static balance between the measured pressure and the head of the liquid column. Fig. 1 shows the schematic of this manometer. A glass tube 1, bended to the “U”-shape, is filled by one half of its volume with liquid 2 (water, mercury). This tube is placed vertically, and a scale 3 (usually in mm) is attached along its height. Pressures P1 and P2 are supplied to legs of the tube, and levels of liquid in the legs change their position.When static balance between a measuring pressure and the head of the liquid column is reached, this pressure can be evaluated according to the equation:



P1 and  P2        - pressures supplied to the legs of the manometer,

 Pa;h1 and h2          - deviations of liquid levels from the zero point of the scale in two legs of themanometer,
 m;H= h1 + h2        - total length of the liquid column corresponded to the measuring differentialpressure, 
m;ρ                      - density of liquid filled the “U”-tube, 
kg/m3;gloc             - local gravitational acceleration, m/s2.


Figure 1. Liquid filled “U”-tube manometer.

We need always make two readings of the liquid level, namely, in each leg of the tube, because in reality due to non-uniformity of the tube diameter along its length, values of h1  and h2 are not equal. As the result of such reading the error introduced during pressure measurement will be reduced. When this type of manometer is used for pressure measurements three cases may take place:

1). P1 is above atmospheric pressure, P2 = Patm. In this case the manometer measures the difference between absolute and atmospheric pressures: P1 = Pg = ρgloc(h1+h2)

2). P is below atmospheric pressure, P1 = Patm. In this case manometer measures the difference between atmospheric and absolute pressures: P2 = Pvac = ρgloc(h1+h2).

3). In this case the equation (3) refers to measurements of differential pressures.

Since the gravitational acceleration is used for the evaluation of pressure, then, when using “U”-tube manometers, it is necessary to introduce correction which takes into account the difference between gravitational acceleration in the place where this manometer was calibrated from that where it is used.Another source of the error is the deviation of liquid temperature in the tube from that temperature when this manometer was calibrated. Due to thermal expansion of the liquid in the tube the volume of liquid will change and this inevitably introduces an error.But the most common mistake is made by not correct reading the scale in respect to the meniscus of liquid in legs of the tube. Fig. 2 gives examples how operator should make readings when using “U”-tube manometer with various liquids. We should always read a surface of the meniscus in its centre. In the case with water - in the bottom, and in the case with mercury - in the top of the meniscus. But in everyday industrial measurements the first two corrections (gravitational and thermal) are not always used, whereas the last one (the meniscus correction) must always be taken into account.

Figure 2. Correct reading of the “U”-tube manometer.

When one measures low pressures several modifications of “U”-tube manometer are used, namely, well or reservoir manometer, inclined manometer, absolute pressure gauge.Let’s develop a differential equation for this manometer.
 (4)   (5)  (6)  (7)    (8)

Hagen-Poiseuille equation applies for a laminar flow of liquid in the tube according

 (9) (10)                                                                                         
Let ∆t->0, then:                                 ,                                                           

Substitute (5)-(8) and (11) into (4):  (12)                        .                               

 Divide all terms of (12) by(2ρgA) and get a second-order differential equation:

In this equation:
a                      - acceleration of liquid in the tube, 
m/s2;A                     - cross-sectional area of the tube, 
m2;D                     - internal diameter of the tube, 
m;FP                    - displacement pressure force,
 N;Ff                     - frictional force for the laminar flow, 
N;Fg                    - gravitational restoring force,
 N;g                      - gravitational acceleration, 
m/s2;h                    - variation of the height of the liquid column in one leg of the manometer,
 m;L                      - total length of the liquid column in the tube, 
m;m                     - mass of liquid in the tube,
 kg;∆P                   - difference of pressures supplied to both legs of the manometer, 
Pa;Pg                  - pressure developed by the gravitational force acting on the liquid columnin the tube, in other words this is liquid pressure head,
 Pa;P1, P2               - pressures supplied to both legs of the manometer, 
Pa;∆Q                   - variation of the volumetric flowrate of the liquid in the tube duringdisplacement of liquid, 
m3/s;∆V                   - variation of the liquid volume displaced,
 m3;∆t                    - time during which ∆V  occurred, 
s;ρ                  - density of the liquid in the tube (density of gas above liquid is negligible), 
kg/m3η                      - dynamic viscosity of the liquid in the tube  Pa x s.

 (14)  (15)    (16)

τ                       - characteristic time of the system, s;
ζ                      - damping factor, dimensionless value;
Kp                    - steady state, or static, or simply gain of the system,  m2 * s2 /kg.

Substitute (14)-(16) into (13) and use variables in deviation form:(17)                                                                                                

Apply Laplace transform to (17):
(18) (19)                                                                                   (20)  (21)  

1. At ζ = 1 we have a critically damped response.(22)        (23)     

 Use inverse Laplace transform:(24)(25)                       
         Let,∆P’ = 2000, Pa - value of step change in measuring pressure of the manometer;L=1,m - length of mercury in the tube of the manometer;Ρ = 13533.61, kg/m3 - density of the liquid (mercury) in the manometer;g = 9.80665, m/s2 - standard gravitational acceleration.

Then,       τ = 0.2258, s,   and      Kp = 3.767*10-6, m2*s2/kgFinally, we’ve got an expression for a critically damped response:

Figure 3. Critically damped response for a “U”-tube manometer.

2. At ζ < 1 we have an underdamped response.(27)

radian frequency of oscillations:                     ,                                              

(28)  (29)

period of oscillation:     

Figure 4. Underdamped response of a “U”-tube manometer

For ζ = 0 a second-order system is free of any damping, and it will oscillate continuously with a constant amplitude and a natural frequency,natural frequency of oscillations:           


natural period of oscillations:                                                            (32)

3. At ζ > 1 we have an overdamped response. (33)           (34), and    (35) 


Figure 5. Overdamped response of a “U”-tube manometer.

Article Source:: Dr. Alexander Badalyan, University of South Australia


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