Pressure Measurement
Absolute and Gage Pressure P abs = P gage + P atm where P abs = Absolute pressure P abs = Gage pressure P abs = atmospheric pressure
A perfect vacuum is the lowest possible pressure. Therefore, an absolute pressure will always be positive A gage pressure above atmospheric pressure is positive A gage pressure below atmospheric pressure is negative (sometimes called vacuum) Gage pressure will be indicated in the units of Pa(gage) or psig Absolute pressure will be indicated in the units od Pa(abs) or psia
The magnitude of the atmospheric pressure varies with location and with climatic conditions. The range of normal variation of atmospheric pressure near Earth s surface is approximately from 95kPa(abs) to 105kPa(abs)
Absolute pressure pressure measured relative to a perfect vacuum Atmospheric pressure pressure exerted by weight of air in the atmospheric of earth common reference pressure its magnitude varies with location and with climatic condition. Also known as barometric pressure Gage pressure the resulting pressure measured above reference pressure (atmospheric pressure)
Example Express pressure of 155kPa (gage) as an absolute pressure. The local atmospheric pressure is 98kPa(abs) P abs = P gage + P atm P abs = 155kPa gage + 98kPa abs = 253kPa(abs) Express pressure of 225kPa (abs) as a gage pressure. The local atmospheric pressure is 101kPa(abs) P gage = 225kPa abs 101kPa abs = 124kPa(gage)
Example Express a pressure of 10.9psia as a gage pressure. The local atmospheric pressure is 15.0 psia P gage = 10.9psia 15.0psia = 4.1psig Negative gage pressure is called vacuum. This can also be read 4.1psi below atmospheric pressure or 4.1psi vacuum Express a pressure of -6.2psig as an absolute pressure P abs = 6.2psig + 14.7psia = 8.5psia Atmospheric pressure is taken as 14.7psia
Relationship Between Pressure and Elevation As we go deeper in fluids, the pressure increases (e.g. diving in ocean) The pressure varies with a change in depth or elevation Elevation is the vertical distance from some reference level to a point of interest Elevation will always be measure positively in upward direction Reference level can be taken at any level It is best to choose lowest point of interest in a problem as reference level
Illustration Figure 3.3
Pressure-Elevation Relationship Pressure-elevation relationship - the change in pressure in a homogenous liquid at rest due to a change in elevation: p = γh change in pressure = sp. Weight of liquid x change in elevation
Pressure-Elevation Relationship The equation is valid only for homogenous liquid at rest Points on the same horizontal level have the same pressure The change in pressure is directly proportional to the sp. weight of liquid Pressure varies linearly with the change in elevation or depth A decrease in elevation causes an increase in pressure An increase in elevation causes a decrease in pressure The equation does not apply to gases (sp. weight of gases changes with changes of pressure)
Example Calculate the change in water pressure from the surface to a depth of 5m p = γh = 9.81kN 5m = 49.05kPa m 3 If the surface of the water is exposed to the atmosphere, the pressure there is 0 Pa(gage). Descending in water (decrease elevation) produce an increase in pressure. Therefore, at 5m the pressure is 49.05kPa(gage)
Example Calculate the change in water pressure from the surface to a depth of 15ft p = γh = 62.4lb 1ft 2 ft 3 15ft 144in 2 = 6.5lb in 2 If the surface of the water is exposed to the atmosphere, the pressure there is 0 psig. Descending in water (decrease elevation) produce an increase in pressure. Therefore, at 15ft the pressure is 6.5psig
Example A tank of oil with one side open to the atmosphere and the other side sealed with air above the oil. The oil has a specific gravity of 0.90. Calculate the gage pressure at points A, B, C, D, E and F and the air pressure in the right side of the tank.
Example Point A at this point, the oil is exposed to the atmosphere p A = 0Pa gage Point B the change in elevation between point A and B is 3.0m γ oil = sg oil γ water = 0.9 9.81kN/m 3 = 8.83kN/m 3 p A B = γ oil h = 8.83kN/m 3 3m = 26.5kN/m 2 Hence pressure at B is p B = p A + p A B = 0Pa gage + 26.5kPa = 26.5kPa(gage)
Point C the change in elevation from point A to point C is 6m p A c = γ oil h = 8.83kN/m 3 6m = 53kN/m 2 p C = p A + p A C = 0Pa gage + 53kPa = 53kPa(gage) Point D this point Is at the same level as Point B p D = p B = 26.5kPa gage Point E = point A = 0 Pa(gage)
Point F the change in elevation is 1.5m. F is higher than A p A F = γ oil h = 8.83kN 3 m 1.5m = 13.2 kn m p F = p A + p A F = 0Pa gage + 13.2kPa = 13.2kPa gage 2 Because air in the right side of the tank is exposed to the surface of the oil, where p F = 13.2kPa gage, the air pressure is also - 13.2kPa(gage) or 13.2kPa below atmospheric pressure.
Pascal s Paradox In the pressure-elevation relationship p = γh, the shape and size of the small volume of fluid does not affect the result The change in pressure depends only on the elevation (h) and type of fluid (γ) Pascal s Paradox is shown in the figure - all containers have same pressure at bottom even though the amount of fluid is different
Pascal s Paradox This phenomenon is useful especially in water distribution system Water tanks are located at higher elevation high pressure
Pressure Measurement Devices Manometers Barometers Pressure gages and transducers
Manometers Uses the relationship between a change in pressure and a change in elevation in static fluid Simplest kind U-tube manometer One end is connected to the pressure that is to be measure Other end is left open to the atmosphere Tube contained a liquid gel called the gage fluid which does not mix with the fluid whose pressure is to be measured (typical fluid water, mercury, colored light oils
Manometer Writing the equation 1. Start from one end of the manometer and express the pressure there in symbol form (p A ). If one end is open, the pressure is atmospheric pressure, taken to be zero gage pressure 2. Add terms representing changes in pressure using p = γh, proceeding from the starting point and including each column of each fluid separately. 3. When the movement from one point to another is downward, the pressure increases and the value of p is added. Conversely, when the movement from one point to the next is upward, the pressure decreases and p is subtracted.
Manometer Writing the equation 4. Continue this process until the other end point is reached, the result is an expression for the pressure at that end point. Equate this expression to the symbol for the pressure at the final point, giving a complete equation for the manometer 5. Solve the equation algebraically for the desired pressure at a given point or the difference in pressure between two points of interest. 6. Enter known data and solve for the desired pressure.
Example 1. Calculate pressure at point A
Let s start from point 1. p 1 = 0Pa(gage) At point 2, movement is downward hence the pressure increases: p 1 2 = γ mercury (0.25) p 2 = p 1 + p 1 2 = p 1 + γ mercury (0.25) At point 3, pressure is equal at point 2 due to same level p 3 = p 2
At point 4, the movement is upward hence pressure decreases. And also, the fluid is different than the previous column. p 3 4 = γ water (0.4) p 4 = p 3 γ water 0.4 = p 1 + γ mercury (0.25) γ water (0.4) Pressure at A is equal to pressure at point 4 (same level) p A = p 4 = p 1 + γ mercury (0.25) γ water (0.4)
Hence the equation for pressure at point A: p A = p 1 + γ mercury (0.25) γ water (0.4) Calculate γ mercury and substitute the known value of γ water and p 1 : γ mercury = (sg)γ water = 13.54 9.81 = 132.8kN/m 3 p A = 0 + 132.8 0.25 9.81 0.4 p A = 29.28kN/m 2 p A = 29.28kPa(gage)
Example Calculate the difference in pressure between points A and B and express it as p B -p A Solution: Let s start at A. At A, the pressure is unknown and we can write as p A At point 1: p 1 = p A + γ oil (29.5 + 4.25in) At point 2: p 2 = p 1 At point 3: p 3 = p 2 γ w (29.5in) At point 4: p 4 = p 3 γ oil (4.25in)
Point B = pressure at point 4 At point 1: p 1 = p A + γ oil (29.5 + 4.25in) At point 2: p 2 = p 1 At point 3: p 3 = p 2 γ w (29.5in) At point 4: p 4 = p 3 γ oil (4.25in) p 4 = p A + γ oil (33.75in) γ w (29.5in) γ oil (4.25in) p B = p A + γ oil (33.75in) γ w (29.5in) γ oil (4.25in) p B p A = γ oil (33.75in) γ w (29.5in) γ oil (4.25in)
Solve the common term (γ oil ) p B p A = γ oil (33.75in 4.25in) γ w (29.5in) p B p A = γ oil (29.5in) γ w (29.5in) p B p A = (29.5in)(γ oil γ w ) The difference between p B and p A is a function of the difference between the specific weights of the two fluids
Substitute known value and convert sg to γ oil : γ w = 62.4lb ft 3 Conversion feet to in = 1ft 3 1728in 3 p B p A = (29.5in)(γ oil γ w ) p B p A = (29.5in)(0.86)[(62.4)( 1ft3 1ft3 1728in3) (62.4)( 1728in 3)] p B p A = 0. 149lb/in 2
Negative signs indicate p A is greater than p B Using a gage fluid with a specific weight very close to that of the fluid in the system makes the manometer very sensitive A large displacement of the column of gage fluid is caused by a small differential pressure and this allows a very accurate reading
Quiz 1 Calculate the difference in pressure between points A and B and express it as p B -p A
well-type manometer When a pressure is applied to a well-type manometer, the fluid level in the well drops a small amount while the level in the right leg rises a larger amount in proportion to the ratio of the areas of the well and the tube
Inclined well-type manometer Same features as well-type manometer but offers a greater sensitivity This is achieved by placing the scale along the inclined tube The scale length is increases as a function of the angle of inclination of the tube, θ. h L = sin θ
Example - θ=15 o h L = sin 15o The ratio of scale length L to manometer deflection h is L h = 1 = 3.86 sin 15o The scale would be calibrated so that the deflection could be read directly
Example An inclined well-type manometer in which the distance L indicates the movement of the gage fluid level as the pressure p A is applied above the well. The gage fluid has a specific gravity of 0.87 and L=115mm. Neglecting the drop in fluid level in the well, calculate p A.
γ f = 0.87 9.81 = 8.535kN/m 3 h L = sinθ h 0.115 = sin15o h = 0.0298m p A = γ f h = 8.535 0.0298 = 0.254kN/m 2 = 0.254kPa(gage)
Example Calculate p A p A = γ w h = 62.4 1 1728 6.8 = 0.246lb/in 2
Barometers It is a device for measuring atmospheric pressure It consists of a long tube closed at one end that is initially filled completely with mercury and allowed to come to equilibrium A void is produced at the top of the tube that is very nearly a perfect vacuum, containing mercury vapour at a pressure of only 0.17 Pa at 20 o C
Barometers By starting from the nearly perfect vacuum, the equation: 0 + γ m h = p atm p atm = γ m h Because the specific weight of the mercury is approximately constant, a change in atmospheric pressure will cause a change in the height of the mercury column This height is often reported as barometric pressure to obtain true atm pressure, p atm = γ m h
Example A new broadcaster reports that the barometric pressure is 772mm of mercury. Calculate the atmospheric pressure In kpa(abs).γ m =133.3kN/m 3 p atm = γ m h = 133.3 0.772 = 102.9kN/m 3 = 102.9kPa(abs)
Example The standard atmospheric pressure is 101.325kPa. Calculate the height of a mercury column equivalent to this pressure. p atm = γ m h h = p atm /γ m = 101.325kNm 2 133.3kNm 3 = 0.76m = 760mm
Example A news broadcaster reports that the barometric pressure is 30.40in of mercury. Calculate the pressure n psia. p atm = γ m h = 848.7lb ft 3 30.40in 1ft3 1728in 3 = 14.93lb in 2 = 14.93psia
Pressure Expressed as the Height of the Column of Liquid Magnitude of the pressure reading in some fluid systems (air flow in heating ducts) is often small The readings are given in units such as inches of water (inh2o or inwc for inches of water column) rather than the conventional units of psi or Pa. To convert such units, pressure-elevation relationship must be used p = 62.4lb ft 3 1inH 2 O 1ft 3 1728in 3 = 0.0361lb in 2 = 0.0361psi
Convert to Pa 1inH 2 O = 0.0361psi 6895Pa 1psi = 249.0Pa Similarly for a mercury manometer p = 848.7lb ft 3 1inHg 1ft 3 1728in 3 = 0.491lb in 2 = 0.491psi 1.0 inhg 1.0 in of mercury 0.491 psi 1.0 mmhg 1.0 mm of mercury 0.01934 psi 1.0 mmhg 1.0 mm of mercury 133.3 Pa
Pressure Gages and Transducers Pressure gages is used when only a visual indication is needed at the site where the pressure is being measure Pressure transducer used when pressure is measure at one point and the readings are displayed at another point the sensed pressure causes an electrical signal to be generated that can be transmitted to a remote location i.e. control centre
Pressure Gages A widely used pressure measuring device is the Bourdon tube pressure gage The pressure to be measure is applied to the inside of a hollow tube with a flatten oval cross section normally formed into a segment of a circle or a spiral
Pressure Gages The increased pressure inside the tube causes the spiral to be opened somewhat The movement of the en of the tube is transmitted through a linkage that causes a pointer to rotate
The scale of the gage normally reads zero when the gage is open to the atmospheric pressure Calibrated in pascals (Pa) or other units of pressure above zero Therefore this type of gage reads gage pressure directly Some gages are capable of reading pressures below atmospheric.
Pressure Transducer and Transmitters The pressure to be measured is introduced to the port and acts on a sensing element that generates a signal proportional to the applied pressure The sensing element can be a strain gage bonded to a diaphragm that is deformed by pressure. As the strain gages sense the deformation of the diaphragm, their resistance changes
Pressure Transducer and Transmitters Passing an electrical current through the gages and connecting them in a network called a Wheatstone bridge, causes a change in electrical voltage to be produced The readout device is typically a digital voltmeter
Example of Pressure Transducer