ES312 Energy Transfer Fundamentals Unit A: Fundamental Concepts ROAD MAP... A-1: Introduction to Thermodynamics A-2: Engineering Properties Unit A-2: List of Subjects Basic Properties and Temperature Pressure Hydrostatic Pressure Variation Manometer Barometer
PAGE 1 of 11 Basic Properties and Temperature A constant-volume gas thermometer reads -273.15 C at absolute zero pressure BASIC PROPERTIES OF THERMODYNAMICS Density: mass per unit volume, [ (rho)], in (kg/m 3 ) or (slug/ft 3 ) Specific Volume: volume per unit mass, [v], in (m 3 /kg) or (ft 3 /slug) Specific Weight: weight per unit volume, [ (gamma)], in (N/m 3 ) or (lb/ft 3 ) Note that: g (g is the gravitational acceleration) SPECIFIC GRAVITY (SG) OF FLUIDS Specific Gravity (SG): ratio of density (or specific weight) of a given fluid to the density (or specific weight) of water Note that: SG fluid water fluid water TEMPERATURE Comparison of temperature scales If the 3rd body of zeroth law of thermodynamics is replaced with a thermometer... then, 2 bodies are in thermal equilibrium, if both have the same temperature even if they are not in contact Celsius (C): ice point = 0 C and steam point = 100 C Fahrenheit (F): ice point = 32 F and steam point = 212 F Absolute Scales: Kelvin (K) = C + 273.15 / Rankine (R) = F + 459.67 T o R 1.8 T K T o F 1.8 T o C 32 Temperature Scale Conversion: /
PAGE 2 of 11 EXERCISE A-2-1 (Do-It-Yourself) During a heating process, the temperature of a system rises 10 C. Express this rise in temperature in K, F, and R. Solution The temperature rise of a system is to be expressed in different units. Analysis Temperature change in Kelvin and Celsius scales are identical: o TK T C 10 K Temperature change in Rankine and Kelvin are related by 1.8: o o o T R 1.8 T K 18 R Temperature change in Fahrenheit and Rankine are identical: T o o F TR 18 F
PAGE 3 of 11 Pressure Basic pressure gages Absolute, gage, and vacuum pressures PRESSURE Pressure can be defined as external force acting on a surface per unit area Units of pressure (SI): N/m 2 (Pascal, or Pa ) and 1,000 Pa = 1 kpa Units of pressure (US Customary): lb/in 2 (or psi ) and lb/ft 2 (or psf ) Note that: 1 psi = 144 psf (because: 1 ft = 12 inch) Also note that: psi and psf in absolute/gage pressures can be expressed by psia/psig, psfa/psfg Atmospheric pressure: [patm] the pressure of atmosphere, which is not constant value (depends on the altitude) Note: standard sea-level value is 14.7 psi (lb/in 2 ) DEFINITIONS OF PRESSURES Absolute pressure: [pabs] the pressure measured relative to absolute zero pressure ( absolute vacuum, which is theoretical perfect vacuum) Gauge pressure: [pgage] how much higher pressure, relative to the local atmospheric pressure p p p (patm): gage abs atm Vacuum pressure: [pvac] how much lower pressure, relative to the local atmospheric pressure (patm): pvac patm pabs Note: negative gage pressure means vacuum pressure
PAGE 4 of 11 Hydrostatic Pressure Variation z (height) h (depth) The pressure of a fluid at rest increases with depth In a room filled with a gas, the variation of pressure with height is negligible Pressure in a liquid at rest increases linearly with distance from the free surface HYDROSTATIC PRESSURE The Pascal s law: states that the pressure applied to a confined liquid increases the pressure throughout by the same amount Also the pressure within a liquid at rest increases with depth (hydrostatic pressure variation) HYDROSTATIC EQUATION The hydrostatic pressure variation can be defined as: 1 2 p p p g dz g dz 2 1 2 1 HYDROSTATIC PRESSURE VARIATION WITHIN A LIQUID If the gravity (g) is assumed to be constant and also the density of liquid is assumed to be constant ( incompressible ), this can be simplified as: 2 2 p p2 p1 g dz g dz g z1 z2 g z 1 1 Note that z is the height of liquid (direction is against the gravity). In terms of the depth (h) of liquid (opposite direction to the height, measured from the free surface of liquid), the same equation can be given as: 2 2 p p2 p1 g dh g dh g h2 h1 g h 1 1
PAGE 5 of 11 Manometer The basic manometer Measuring the pressure drop across a flow section of a flow device by using differential manometer MANOMETERS Manometers: pressure measurement device, especially useful for pressures of small magnitude or small pressure differences Manometers are commonly used for airspeed measurements, combined with Pitot-static probes U-TUBE MANOMETERS A basic U-tube manometer includes a tube filled with a manometer (or gage ) fluid with one end of the tube being open to the local atmosphere Note that: p1 p2 and p2 patm gh (h is called manometer reading ) A differential U-tube manometer includes a tube filled with a manometer (or gage ) fluid that inter-connects two sections of flowing fluid, measuring the pressure drop ( p ); this pressure drop can be related to the volume flow rate (flow rate measurement device: flow meters)
PAGE 6 of 11 EXERCISE A-2-2 (Do-It-Yourself) A manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.85, and the manometer column height is 55 cm, as shown in the figure. If the local atmospheric pressure is 96 kpa, determine the absolute pressure (in kpa ) within the tank. Solution The reading of a manometer attached to a tank and the atmospheric pressure are given. The absolute pressure in the tank is to be determined. Assumptions The fluid in the tank is gas (the density is much lower than the density of manometer fluid). Properties The specific gravity (SG) of the manometer fluid is 0.85. The standard density of water is 1,000 kg/m 3. Analysis 3 3 The density of the manometer fluid is: SG water 0.851, 000 kg/m 850 kg/m Then, the absolute pressure (at the bottom of the tank) is: pabs patm gh 3 3 2 3 9610 Pa 850 kg/m 9.81 m/s 0.55 m 100.6 10 Pa (100.6 kpa)
PAGE 7 of 11 EXERCISE A-2-3 (Do-It-Yourself) The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer, as shown in the figure. The tank is located on a mountain at an altitude of 1,400 m, where the atmospheric pressure is 85.6 kpa. Determine the air pressure in the tank (in kpa absolute), if h 1 = 0.1 m, h 2 = 0.2 m, and h 3 = 0.35 m. Take the densities of water, oil, and mercury to be 1,000 kg/m 3, 850 kg/m 3, and 13,600 kg/m 3, respectively. Solution The pressure in a pressurized water tank is measured by a multi-fluid manometer. The air pressure in the tank is to be determined. Assumption The air pressure in the tank is uniform (its variation with respect to height is negligible), and thus the pressure at the air-water interface can be determined. Properties The densities of water, oil, and mercury are: 1,000 kg/m 3, 850 kg/m 3, and 13,600 kg/m 3, respectively. Analysis Starting with the pressure at point 1 (air-water interface), moving along the tube by adding or subtracting the hydrostatic pressure variation ( gh ) until reaching the point 2 (pressure is p atm ) yields: p gh gh gh p 1 water 1 oil 2 mercury 3 atm Solving for the pressure at point 1: p1 patm watergh1 oilgh2 mercury gh3 patm g mercuryh3 waterh1 oilh2 85.610 3 Pa 9.81 m/s 2 13, 600 kg/m 3 0.35 m 1, 000 kg/m 3 0.1 m 3 3 850 kg/m 0.2 m 130 10 Pa (130 kpa, absolute)
PAGE 8 of 11 A manometer is used to measure the air pressure in a tank. The manometer fluid has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank (in psia ) for the cases of the fluid level of the right arm of manometer (a) higher and (b) lower fluid level than the fluid level of the left arm (left arm is the one attached to the tank). Assumption The manometer fluid (liquid) is incompressible. Properties The specific gravity (SG) of the fluid is 1.25. The density of water (32 F) is 62.4 lbm/ft 3. Analysis The density of the fluid can be given by: 3 3 SG water 1.2562.4 lbm/ft 78 lbm/ft The pressure difference ( p ) corresponding to a manometer reading of 28 inch is: 2 3 1 slug 2 28 lb 1 ft p gh 78 lbm/ft 32.2 ft/s ft 181.44 1.26 psi 2 32.2 lbm 12 ft 12 in (a) The fluid level is higher: pabs patm pgage 12.7 psi 1.26 psi 13.96 psia (b) The fluid level is lower (means: vacuum ): pabs patm pvac 12.7 psi 1.26 psi 11.44 psia
PAGE 9 of 11 Barometer At high altitudes, a car engine generates less power. Also a person gets less oxygen The basic barometer BAROMETER The basic barometer: is a simple device (vertical glass tube, filled with a liquid) designed to measure atmospheric pressure At standard (sea-level) condition, the pressure is defined to be 1 bar Applying the hydrostatic pressure variation for barometer: patm gh (h is the barometer reading ) Note that: the barometer reading depends on the type of fluid, therefore the type of fluid must always be specified for barometer reading STANDARD ATMOSPHERIC PRESSURE (1 ATM) Barometer measures the standard 1 atm as: 760 mmhg at 0 C (or 10.332 mh2o at 4 C) under the gravity of g = 9.807 m/s 2 SI Unit: 1 atm = 101 kpa 1 bar = 100,000 Pa (100 kpa) = approximately 1 atm US Customary Unit: 1 atm = 14.7 psi = 2,116.8 psf
PAGE 10 of 11 EXERCISE A-2-4 (Do-It-Yourself) Schematic of a vertical pistoncylinder device and the free-body diagram of the piston The piston of a vertical piston-cylinder device is containing a gas (60 kg), and has a cross-sectional area of 0.04 m 2, as shown in the figure. The local atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s 2. (a) Determine the pressure (in bar ) inside the cylinder. (b) If some heat is transferred to the gas and its volume is increased, do you expect the pressure inside the cylinder change? Explain why. Solution A gas is contained in a vertical cylinder with a heavy piston. The pressure inside the cylinder and the effect of volume change on pressure are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis (a) The gas pressure in the piston-cylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the free-body diagram of the piston (figure) and balancing the vertical forces yield: pa patm AW patm A mg Solving for the pressure inside the cylinder: 2 60 kg9.81 m/s mg 1 bar p patm 0.97 bar 2 2 A 0.04 m 10,000 N/m 1.12 bar (b) The volume change will have no effect on the free-body diagram in part (a), and therefore the pressure inside the cylinder will remain the same. Note: if the gas behaves as an idea gas, the absolute temperature doubles when the volume is doubled at constant pressure.
PAGE 11 of 11 The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed (in m ). Assume an average air density of 1.2 kg/m 3. Assumptions The variation of air density and the gravity with respect to altitude is negligible. Properties 3 The density of air is given: 1.2 kg/m Analysis Pressure difference between bottom and top of the mountain can be determined by the hydrostatic pressure variation, assuming that the sir density is constant: p pbottom ptop ghair 2 100,000 N/m 3 2 0.93 bar 0.78 bar 1.2 kg/m 9.81 m/s h 1 bar This yields, h 1,274 m