LECTURE 5 cor 1

LECTURE 5 (Fluid Mechanics and Thermodynamics) 1. Fluid Mechanics 1.1 PRIMARY THERMODYNAMIC PROPERTIES OF FLUIDS Pressure (p) - is the (compression) stress at a point in a static fluid. patm  14.7 psia  101.3 kPaa Temperature (T) – is a measure of the internal energy level of a fluid.  R   F  460  K   C  273 Density () - is its mass per unit volume. m  V Specific weight () - is its weight per unit volume. W mg   g   V V Specific gravity (SG) - is the ratio of a fluid density to a standard reference fluid, water (for liquids), and air (for gases).  gas SG gas   air

air  1.20 kg m3 at 101.3 kPa and 21 C. SG liquid 

 liquid

 water

water  1000 kg m3 Example No. 1 What is the sea level (g = 32.2 ft/sec2) specific weight (lbf/ft3) of liquids with densities of 65.5 lbm/ft3? A. 32.2 B. 42.2 C. 65.5 D. 76.7 Solution:  density in lbm ft 3     g  g  gc     65.5 lbm ft 3    32.2 ft sec 2  2   32.2 lbm  ft lbf  sec 





  65.5 lbf ft 3 (C) 1.2 VISCOSITY Viscosity (1) - is that property of a real fluid by virtue of which it offers resistance to shear force. Viscosity (2) – is the fluid resistance to flow or the property of fluid to resist shear deformation. Newton’s law of viscosity states that the shear force to be applied for a deformation rate of (dV/dy) over an area A is given by, F   AdV dy  1

where F is the applied force in N, A is area in m2, dV/dy is the velocity gradient (or rate of deformation), 1/s, perpendicular to flow direction, here assumed linear, and  is the proportionality constant defined as the dynamic or absolute viscosity of the fluid. Shear Stress = F A   dV dy  Shear Strain = dV dy  F  A  Viscosity Index – is the rate at which viscosity changes with temperature. Viscosimeter – an instrument, consisting of standard orifice, used for measuring viscosity (in SSU and SSF). Absolute (Dynamic) Viscosity – is the viscosity determined by direct measurement of shear resistance (in Poise or centiPoise.) Units are 1 reyn = 1 lb-sec/in2, 1 Poise = 1 dyne-sec/cm2 = 0.1 Pa-sec. 1 centiPoise (cP) = 0.01 Poise. Kinematic Viscosity – it the absolute viscosity of a fluid divided by the density (in Stoke or centiStoke.) Units are ft2/s, m2/s, 1 stoke = 1 cm2/sec. = 0.0001 m2/sec. 1 centiStoke (cSt) = 0.01 Stoke. Example No. 2 An oil has a kinematic viscosity of 1.25 x 10-4 m2/s and a specific gravity of 0.80. What is its dynamic (absolute) viscosity in kg/(m-s)? A. 0.08 B. 0.10 C. 0.125 D. 1.0 Solution:  = 1.25 x 10-4 m2/s SG = 0.80 =x  = SG x w x   = 0.80 x (1000 kg/m3) x (1.25 x 10-4 m2/s)  = 0.10 kg/(m-s) 1.3 REYNOLDS NUMBER Reynolds number – is a dimensionless number which is the ratio of the forces of inertia to viscous forces of the fluids. It is the primary parameter correlating the viscous behaviour of all newtonian fluids. Forces of inertia DV DV Re    Viscous Forces  v where: Re = Reynolds number, dimensionless D = inside diameter, m V = velocity, m/s  = kinematic viscosity, m2/s  = absolute viscosity, Pa-sec Example No. 3 Water is flowing in a pipe with radius of 12 inches and a velocity of 11 m/sec. The viscosity of water is 1.131 Pasec. What is the Reynolds Number? A. 2964 2

B. 5930 C. 58 D. 9189 Solution: vD Re    2 12   11 39.37   Re   5930 (B) 1.131 1.4 TYPES OF FLOW Laminar Flow – particles run parallel to each other. Laminar flow occurs if the Reynolds number is less than 2000. Turbulent Flow – particles run not in same direction. Turbulent flow occurs if the Reynolds number is greater than 4000. Fully turbulent occurs at very high Reynolds number. Transitional Flow – also termed as critical flow in which this type of flow occurs if the Reynolds number is between 2000 to 4000. 1.5 SURFACE TENSION AND CAPILLARY ACTION Surface Tension – is the membrane formed on the free surface of the fluid which is due to cohesive forces. The reason why insects were able to sit on water is due to surface tension. The amount of surface tension decreases as the temperature increases. Capillary Action – this is done through the behaviour of surface tension between the liquid and a vertical solid surface. 1.6 COMPRESSIBILITY AND BULK MODULUS Compressibility, - the measure of the change in volume of a substance when a pressure is exerted on the substance. V  Vo  P Where: V = change in volume Vo = original volume P = change in pressure Bulk modulus, EB - is defined as the ratio of the change in pressure to the rate of change of volume due to the change in pressure. It is the inverse of compressibility. 1.7 HYDROSTATIC PRESSURE Hydrostatic Pressure – is the pressure of fluid exerted on the walls of the container. Notes: a. Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid. b. Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure.

3

Pressure = Weight Density x Height p   h   gh Pressure Head, p p h   g where: p = hydrostatic pressure (gage pressure) h = height of liquid (pressure head)  = Density of liquid Example No. 4 On a sea-level standard day, a pressure gage, moored below the surface of the ocean (SG = 1.025), reads an absolute pressure of 1.4 MPa. How deep is the instrument? A. 4 m B. 129 m C. 133 m D. 140 m Solution: pabs  patm  gh  h  SG  w h





1400 kN m2 101.325kN m2  1.025 9.81 kN m3 H h  129 m Example No. 5 If the absolute pressure at the bottom of the ocean is 300 kPa, how deep is the water at this point? A. 16.66 m B. 19.66 m C. 29.66 m D. 39.66 m Solution: Pabs  Pg  Patm

300  Pg  101.325 4

Pg  198.675 kPa h

Pg S.G.  



198.675  19.66 m 1.03  9.81

1.8 MANOMETER Manometer is a device to measure pressure or mostly difference in pressure using a column of liquid to balance the pressure.

pA  pB  pA  p1   p1  p2   p2  p3   p3  pB  pA  pB   1 z A  z1    2 z1  z 2    3 z 2  z3    4 z3  zB  Example No. 6 In Fig. FE2.3, if the oil in region B has SG = 0.8 and the absolute pressure at point A is 1 atm, what is the absolute pressure at point B?

A. 5.6 kPa B. 10.9 kPa C. 106.9 kPa D. 112.2 kPa Solution: p A  SGw  w 0.05 m   SG m  w 0.08  0.04 m 

 SG o  w 0.03 m   pB  0

5

pB  pA  10.05 m   13.56 0.08  0.04 m  w  0.80.03 m  pB  pA  0.5684 m w





pB  101.325 kN m 2  9.81 kN m3 0.5684 m pB  106.90 kPa

1.9 BUOYANCY Buoyancy – the tendency of a body to float when submerged in a fluid. Two Archimedes Law of Buoyancy a. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces. b. A floating body displaces its own weight in the fluid in which it floats.

W  FB FB  V where: FB = buoyant force 6

W = weight of the body V = volume of the body submerged or volume of the liquid displaced  = density of the liquid Example No. 7 A stone weighs 105 lbs in air and 83 lb in water. Find the specific gravity of the stone. A. 2.98 B. 0.35 C. 4.77 D. 2.21 Solution: Volume (stone) = Volume (water displaced) 105  83   0.352564 ft 3 62.4 Specific Weight (stone) = 105 / 0.352564 = 297.82 lb/ft3 Specific gravity (stone) = 297.82 / 62.4 = 4.77 (c) Example No. 8 An iceberg has a SG of 0.922. When floating in sea water (SG = 1.03), its exposed volume in % is nearest to, A. 5.6 B. 7.4 C. 8.9 D. 10.5 Solution: SG iceVice  SG seaVsubmerged 0.922Vice  1.03Vsubmerged Vsubmerged  0.895Vice %exposed 

Vice  0.895Vice 100% Vice

%exp osed  10.5%

1.10 BASIC SCIENTIFIC LAWS USED IN THE ANALYSIS OF FLUID FLOW a. Law of conservation of mass: This law when applied to a control volume states that the net mass flow through the volume will equal the mass stored or removed from the volume. Under conditions of steady flow this will mean that the mass leaving the control volume should be equal to the mass entering the volume. The determination of flow velocity for a specified mass flow rate and flow area is based on the continuity equation derived on the basis of this law. b. Newton’s laws of motion: These are basic to any force analysis under various conditions of flow. The resultant force is calculated using the condition that it equals the rate of change of momentum. The reaction on surfaces are calculated on the basis of these laws. Momentum equation for flow is derived based on these laws. c. Law of conservation of energy: Considering a control volume the law can be stated as “the energy flow into the volume will equal the energy flow out of the volume under steady conditions”. This also leads to the situation that the total energy of a fluid element in a steady flow field is conserved. This is the basis for the derivation of Euler and Bernoulli equations for fluid flow. d. Thermodynamic laws: are applied in the study of flow of compressible fluids. 1.11 CONTINUITY EQUATION 7

Continuity Equation - This equation is used to calculate the area, or velocity in one dimensional varying area flow, like flow in a nozzle or venturi. m  1Q1  2Q2 1 A1V1  2 A2V2 For incompressible flow.   1  2 Q  Q1  Q2 A1V1  A2V2 Example No. 9 Oil flow through a 30 tubes condenser with a velocity of 1.75 m/s. The internal diameter of tube is 20 mm an oil density is 0.90 gm/mL. Find the volume flow in liters per second. A. 16.5 B. 11.6 C. 15.6 D. 9.4 Solution: Q  AV  m3 Q  0.0202 1.75  5.4977  10 4 4 sec 3  m  Li  1000 3 30 tubes   16.5 Li sec Q   5.4977  10 4  sec  m   1.12 VELOCITY HEAD Torricelli’s Theorem: “The velocity of a liquid which discharge under a head is equal to the velocity of a body which falls in the same head”. V2 h 2g where: h = velocity head V = velocity of the liquid 1.13 FRICTION HEAD LOSS IN PIPES Darcy-Weisbach Equation fLV 2 L V2 hf  f 2gD D 2g where: hf = velocity head, m or ft V = velocity of the liquid, m/s or ft/sec L = length of pipe, m or ft D = internal diameter, m or ft f = coefficient of friction, friction factor (Darcy) g = 9.81 m/s2 or 32.2 ft/sec2 Example No. 10 What is the expected head loss per mile of a closed circular pipe (17 in inside diameter, friction factor of 0.03) when 3300 gal/min of water flow under pressure? 8

A. 38 ft B. 3580 ft C. 0.007 ft D. 0.64 ft Solution:  1 ft   Q  3300 gpm 231 in 3 gal    12 in 



Q  7.35243 ft 3 s ec V





3

 1 min     60 sec   



4Q 4 7.35243 ft 3 s ec   4.6645 fps D 2  17 12 ft 2

hf  f

L V2 D 2g

L  1 mile  5280 ft  5280   4.66452   h f  0.03    17 12   232.2 

h f  37.78 ft 1.14 COEFFICIENT OF FRICTION, f a. Coefficient of friction for laminar flow (Re < 2300). 64 f Re b. Coefficient of friction for turbulent flow Colebrook equation, turbulent flow only (Re > 2300)  2.51  D  f  2 log    12 3.7   Re f where,  = nominal roughness of pipe or duct being used A good approximate equation for the turbulent region of the Moody chart is given by Haaland’s equation:   6.9   D 1.11   f   1.8 log       Re  3.7   

2

H. Blasius Equation (4000 < Re < 105) 0.316 f 4 Re

Example No. 11 For flow of water at a Reynolds number of 1.03 x 106 through a 5-cm-diameter pipe of roughness height 0.5 mm, the approximate Moody friction factor is A. 0.012 B. 0.018 C. 0.038 9

D. 0.049 Solution: Use Haaland approximation to Colebrook equation, Re > 2300

  6.9   D 1.11   f   1.8log       Re  3.7   

2

1.11   6.9  0.5 50    f   1.8log      6  1.03 10  3.7    f  0.038

2

1.15 HYDRAULIC DIAMETER For flow in non-circular ducts or ducts for which the flow does not fill the entire cross-section, we can define the hydraulic diameter 4A Dh  P where A = cross-sectional area of actual flow, P = wetted perimeter, i.e. the perimeter on which viscous shear acts Example No. 12 Air at atmospheric pressure and with a mean velocity of 1.0 m/s, flows inside a square section duct of side b = 4 cm. If the air temperature is 350 K, determine the Reynolds Number. (at 350 K, viscosity = 20.76 x 10 -6 m2/s.) A. Re = 1296.8 B. Re = 1926.8 C. Re = 1629.8 D. Re = 1962.8 Solution: Hydraulic Diameter for square duct 4b 2 D  b  4 cm  0.04 m 4b DV 0.041.0 Re    1926.8 (B)  20.76 10 6 Example No. 13 What is the hydraulic diameter of a rectangular air-ventilation duct whose cross section is 1 m by 25 cm? A. 25 cm B. 40 cm C. 50 cm D. 75 cm Solution: 4A Dh  P 4100 cm 25 cm  Dh   40 cm 2100  25 cm  1.16 BERNOULLI’S EQUATION 10

Bernoulli’s equation – is a general energy equation that is used for solving fluid flows. It relates elevation head, pressure head and velocity head. Some conditions of using Bernoulli’s equation: (1) No fluid friction, (2) fluid is incompressible, and (3) negligible changes in thermal energy. Bernoulli’s theorem – “ Neglecting friction, the sum of the pressure head, velocity head and elevation head of a point is equal to the sum of the pressure head, velocity head and elevation head of another point”. p1 V12 p V2   Z1  2  2  Z 2  2g  2g where: p Pressure head =  V2 2g Elevation head = Z

Velocity head =

Example No. 14 Water flows through a horizontal pipe of cross sectional area of 20 cm 2. At one section the cross sectional area is 4 cm2. The pressure difference between the two sections is 29.4 psi. How many cu. meters of water will flow out of the pipe in 1 minute? A. 1.208 m3 B. 0.0185 m3 C. 0.493 m3 D. 0.008 m3 Solution: p1 p2 V22  V12     2g Q  A1V1  A2V2 20V1  4V2 V2  5V1

V22  V12 p1  p2  2g 

 101.325 kPa    202.705 kPa p1  p2  29.4 psi  14 . 696 psi  

5V1 2  V12



2 9.81 m s 2





202.705 kN m 2 9.81 kN m 3

V1  4.11 m s 2

 1m   4.11 m s  Q  A1V1  20 cm 2    100 cm 





Q  8.22  10 3 m 3 s

In 1 min. Volume = (8.22 x 10-3)(60) = 0.4932 m3 Example No. 15 11

A perfect venturi with throat diameter of 1.8 inches is placed horizontally in a pipe with a 5 in inside diameter. Eighty (80) lbm of water flow through the pipe each second. What is the difference between the pipe and venture throat static pressure? A. 29.9 psi B. 34.8 psi C. 5020 psi D. 72.3 psi Solution: 80 lb s Q  1.28205 ft 3 s 62.4 lb ft 3 V1  V2 

4Q 41.28205   9.4024 fps D12  5 12 2

4Q 41.28205   72.5492 fps D22  1.8 122

p1  p2 V22  V12   2g

p1  p2 144  72.54922  9.40242 62.4 232.2  p1  p2  34 .82 psi

1.17 VENTURI, NOZZLE AND ORIFICE METERS Venturi, Nozzle and Orifice meters are the three obstruction type meters commonly used for the measurement of flow through pipes. Flow Rate: A2 Q 1  A2 A1 2

2gh

This equation needs a modifying coefficient as viscous effects and boundary roughness as well as the velocity of approach factor that depend on the diameter ratio have been neglected. The coefficient is defined by, Qactual  Qtheoretical  C d where Cd is the coefficient of discharge. Cd for venturi meters is in the range 0.95 to 0.98. Cd for flow nozzle is in the range 0.7 to 0.9 depending on diameter ratio and Reynolds number to some extent. For orifice, The range for coefficient of discharge is 0.6 to 0.65. 1.18 DISCHARGE MEASUREMENT USING ORIFICES Actual flow rate: Qactual  C d A0 2gh where A0 is the area of orifice and Cd is the coefficient of discharge. Qactual Cd  Qtheoretical Coefficient of velocity Cv. 12

Actual velocity of jet at vena contracta V  Theoretica l velocity 2gh The value of Cv varies from 0.95 to 0.99 for different orifices depending on their shape and size. Cv 

Coefficient of contraction Cc. Area of jet at vena contracta Ac Cc   Area of velocity A0 The value of coefficient of contraction varies from 0.61 to 0.69 depending on the shape and size of the orifice. Coefficient of discharge Cd. Actual discharge Actual area Actual velocity Cd     C c  Cv Theoretical discharge Theoretical area Theoretical velocity Average value of Cd for orifices is 0.62. Example No. 16 Calculate the discharge in liters per second through a 5 in diameter orifice under a head of 7.6 m of water. Assume coefficient of discharge of 0.65. A. 78 B. 1547 C. 77 D. 100 Solution: Q  CAV  CA 2gh





  Q  0.65 5 in 0.0254 m in 2 2 9.81 m s 2 7.6 m  4

Q  0.1006 m3 s  0.1006 L s 1.19 PERIPHERAL VELOCITY FACTOR Cd   

Peripheral Velocity  DN  Velocityof Jet 2gh

2. THERMODYNAMICS 2.1 MASS AND WEIGHT Mass (1) – a property of matter that constitutes one of the fundamental physical measurements or the amount of matter a body contains. Units of mass are in lbm, slugs, or kg. Symbol m. Mass (2) - is the absolute quantity of matter in it, an unchanging quantity for a particular mass when the speed of the mass is small compared to the speed of light (no relativistic effect). Weight (1) – the force acting on a body in a gravitational field, equal to the product of its mass and the gravitational acceleration of the field. Units of weight are in lb f or N. Symbol W. Formula W = mg. Where g = 9.81 m/s2 or 32.2 ft/s2. Weight (2) - is the force exerted by a body when its mass is accelerated in a gravitational field. 2.2 VOLUME 13

Volume – the amount of space occupied by, or contained in a body and is measured by the number of cubes a body contains. Units of volume are in ft3, gallons, liters, cm3, or m3. Symbol is V. 2.3 SPECIFIC VOLUME AND DENSITY Density () of any substance – is its mass (not weight) per unit volume. Units of density are lb m/ft3 or kg/m3. Symbol . m  V Specific Volume – is the total volume of a substance divided by the total mass of a substance. Units of specific volume are in ft3/lbm or m3/kg. Symbol is v. 1 V or v  v  m Specific Weight () of any substance – is the force of gravity on unit volume. Units of specific weight are ft3/lbf or m3/N. Symbol is . 1 V or v   g W Specific Gravity (1) – is a measure of the relative density of a substance as compared to the density of water at a standard temperature. Symbol is SG. Specific Gravity (2) – a dimensionless parameter, it is defined as the ratio of the density (or specific weight) of a substance to some standard density (or specific weight). For Liquid substances:  liquid  liquid SG   H2O at std  H2O at std For Gaseous substances: gas  gas SG   air at std  air at std Example No. 17 An iron block weighs 5 N and has a volume of 200 cm3. What is the density of the block? A. 988 kg/cu. m. B. 1255 kg/cu. m. C. 2550 kg/cu. m. D. 800 kg/cu. m. Solution: 5N W    2548.4 kg m 3 3 Vg  1m   9.81 m s 2 200 cm 3   100 cm  









Example No. 18 Suppose two liquids of different densities, 1 = 1500 kg/m3 and 2 = 500 kg/m3, were poured together inside a 100-L tank, filling it. If the resulting mixture density is 800 kg/m 3, find the respective liquid amounts, in kg? A. 35,45 14

B. 53,54 C. 71,63 D. 89,72 Solution: m1 m2 m3    V3 1  2  3 m1  m 2  m3 m1  m 2  m3   3V3

m1  m 2  800 0.10   80 m 2  80  m1

m1 m  2  0.10 1500 500 m1 80  m1   0.10 1500 500 m1  380  m1   150 m1  45 kg m 2  80  45  35 kg

Masses = 35 kg and 45 kg 2.4 STANDARD DENSITIES Density of water At approx. 4oC (39.2oF) pure water has it's highest density (weight or mass) = 1000 kg/m3 or 62.4 lb/ft3. Density of air At 70 °F (21.1 C) and 14.696 psia (101.325 kPaa), dry air has a density of 0.074887 lbm/ft3 (1.2 kg/m3). 2.5 PRESSURE Pressure – is a measure of the force exerted per unit area on the boundaries of a substance (or system). It is caused by the collisions of the molecules of the substance with the boundaries of the system. Units of pressure are psi, kg/cm2, kN/m2 or kPa. Symbol is p. Formula p = F/A. 2.6 ATMOSPHERIC PRESSURE Atmospheric pressure - is the force per unit area exerted against a surface by the weight of air above that surface in the Earth's atmosphere. The standard atmosphere (symbol: atm) - is a unit of pressure and is defined as being equal to 101.325 kPa.[1] The following units are equivalent, but only to the number of decimal places displayed: 760 mmHg (torr), 29.92 inHg, 14.696 psi. Barometric pressure - is often also referred to as atmospheric pressure. Units is normally in Bar. 1 Bar = 100 kPaa. Air pressure above sea level can be calculated as p = 101325(1 - 2.25577x10-5h)5.25588, where p = air pressure (Pa) and h = altitude above sea level (m). Also

p  po

 h    h e  o

where: p = atmospheric pressure, (measured in bars) 15

h = height (altitude), km p0 = is pressure at height h = 0 (surface pressure) = 1.0 Bar (Earth) h0 = scale height = 7 km (Earth) Or for every 1,000 feet, there is a corresponding pressure decrease of approximately 1 in Hg. 2.7 ABSOLUTE AND GAUGE PRESSURES

Absolute pressure, pabs - is measured relative to the absolute zero pressure - the pressure that would occur at absolute vacuum. All calculation involving the gas laws requires pressure (and temperature) to be in absolute units. It the sum of the gauge and atmospheric pressure. Gauge pressure - the amount by which the total absolute pressure exceeds the ambient atmospheric pressure. Formula pabs = patm + pg Vacuum pressure (negative gauge pressure) - the amount by which the total absolute pressure is less than the ambient atmospheric pressure. Formula pabs = patm – pv Pressure gauge - is often used to measure the pressure difference between a system and the surrounding atmosphere. Example No. 19 A condenser vacuum gage reads 715 mm Hg when the barometer stands 757 mm Hg. State the absolute pressure in the condenser in kN/m2. A. 5.6 kN/m2 B. 5.9 kN/m2 C. 6.5 kN/m2 D. 5.2 kN/m2 Solution: pa  patm  pvac

16

pa  757  715  42 mm Hg

 101.325 kN m 2    5.6 kN m 2 pa  42 mm Hg  760 mm Hg  

Example No. 20 With a normal barometric pressure at sea level, atmospheric pressure at an elevation of 4000 ft is nearest to, A. 26 in Hg B. 27 in Hg C. 28 in Hg D. 29 in Hg Solution: A drop of 1 in Hg per 1000 ft

 1.0 in Hg  4000 ft  p  29.92 in Hg     1000 ft  p  25.92 in Hg 2.8 TEMPERATURE Temperature – is a measure of the molecular activity of a substance. It is a relative measure of how “hot” or “cold” a substance is and can be used to predict the direction of heat transfer. It is an intensive property that is a measure of the intensity of the stored molecular energy in a system. Temperature Scales: a. Fahrenheit (F) Scale – 180 units – from 32 F to 212 F. b. Celsius (C) Scale or Centigrade Scale – 100 units – from 0 C to 100 C. Relationship: 

 

9 F  32     C  5





 5 C   F  32   9 Absolute zero - is the theoretical temperature at which entropy reaches its minimum value. The laws of thermodynamics state that absolute zero cannot be reached using only thermodynamic means. Absolute temperature - is the temperature measured relative to the absolute zero. Absolute Temperature Scales: a. Kelvin (K) Scale – the absolute temperature scale that corresponds to the Celsius scale. b. Rankine (R) Scale – the absolute temperature scale that corresponds to the Fahrenheit scale. 

Relationship: 

R   F  460



K   C  273 * 273 K  273.15 K * 460 R  459.67 R Temperature Change:

 

 

 5  5  C   K     F     R 9 9

17

 

 

9 9  F   R     C     K 5    5 Ice point – the temperature of a mixture of ice and air-saturated water at 1 atm = 0 C or 32 F. Steam (boiling) point – the temperature pure liquid water in contact with its vapour at 1 atm = 100 C or 212 F. Triple point - The temperature and pressure at which a substance can exist in equilibrium in the liquid, solid, and gaseous states. The triple point of pure water is at 0.01 degrees Celsius and 4.58 millimeters of mercury and is used to calibrate thermometers. 2.9 ENERGY Energy - is defined as the capacity of a system to perform work or produce heat. Stored Energy – otherwise known as possessed energy, it is the energy that is retrieved and stored within the system; thus, dependent upon the mass flow. Potential Energy (1) – is defined as the energy of position. Symbol is P.E. Potential Energy (2) – energy due to the elevation and position of the system mgz PE  gc where m z g

gc

= mass (lbm, kg) = height above some reference level (ft, m) = acceleration due to gravity (ft/sec2, m/s2) = gravitational constant. = 32.17 ft  lbm lbf  sec2 = 1 kg  m s2  N

Kinetic Energy (1) – is the kinetic energy of motion. Symbol is K.E. Kinetic Energy (2) – energy or stored capacity for performing work; possessed by a moving body, by virtue of its momentum. KE 

mv 2 2gc

where

m v g

= mass (lbm, kg) = velocity (ft/s, m/s) = acceleration due to gravity (ft/sec2, m/s2)

gc

= gravitational constant. = 32.17 ft  lbm lbf  sec2 = 1 kg  m s2  N

Joule’s Constant “J” – 778 ft-lbf/ Btu Internal Energy (1) – is a microscopic forms of energy including those due to the rotation, vibration, translation, and interactions among the molecules of a substance. Internal Energy (2) – heat energy due to the movement of the molecules within the substance brought about its temperature. 18

Internal Energy (3) – energy stored within a body or substance by virtue of the activity and configuration of its molecules and the vibration of the atoms within the molecules. Specific Internal Energy – is the substance internal energy per unit mass. Unit is Btu/lbm or kJ/kg. Symbol is u. P-V Energy – is also called flow energy or flow work. Specific P-V Energy – is the substance P-V energy per unit mass. Enthalpy (1) – is the amount of energy possessed by a thermodynamic system for transfer between itself and its environment. It is equal to H  U  PV . Enthalpy (2) – the sum of the internal energy of a body and the product of pressure and specific volume. Unit is Btu/lbm or kJ/kg. Symbol is h. Specific Enthalpy – is defines as h  u  Pv . where u = specific internal energy P = pressure v = specific volume Chemical Energy – stored energy that is released or absorbed during chemical reactions. Nuclear Energy – energy due to the cohesive forces of the protons and neutrons within the atoms. Example No. 21 A high velocity flow of gas at 800 ft/sec possesses kinetic energy nearest to which of the following? A. 1.03 Btu/lb B. 9.95 Btu/lb C. 4.10 Btu/lb D. 12.8 Btu/lb Solution: W 1 1  2 KE  V 2   800  9937.88 ft  lb 2g 2  32.2  9937.88 ft  lb lb KE   12.77 Btu lb (D) 778.16 ft  lb Btu 2.10 HEAT AND WORK Transitory Energy – otherwise known as energy in transit or in motion; energy that loses its identity once it is absorbed or rejected within the system; independent of mass flow stream. Heat (1) - is the transfer of energy that occurs at the molecular level as a result of a temperature difference. Symbol Q. Unit is Btu, Btu/hr, kJ or kW. Heat (2) – energy in transition between a system and its surroundings because of a difference in temperature. Q is positive (+) when heat is added to the body Q is negative (-) when heat is rejected by the body Q Heat transferred per unit mass = q  m

19

Work – is defined for mechanical system as the action of a force on an object through a distance. Symbol is W. Unit is ft-lb, kJ. W  Fd Where F = force (lbf, N) d = displacement (ft, m ) 

Work is a process done by or on a system.

Power – is the rate of doing work. Unit is hp or kW. Work Power  Time 2.11 SPECIFIC HEAT Heat Capacity – is the ratio of the heat (Q) added to or removed from a substance to the change in temperature ( T ). Specific Heat (1) – is the heat capacity of a substance per unit mass. Unit is Btu/lbm-F or Btu/kg-C. Specific Heat (2) – is the heat required to raise the temperature of unit mass of a substance by a unit temperature. Specific heat at constant pressure – is the change of enthalpy for a unit mass between two equilibrium states at the same pressure per degree change of temperature. Q Cp  T Q cp  m T q cp  T Specific heat at constant volume – is the change of internal energy for a unit mass per degree change of temperature when the end states are equilibrium states of the same volume. Q Cv  T Q cv  m T q cv  T Specific heat ratio – c p h k  cv u 2.12 ENTROPY Entropy (1) – is a measure of inability to do work for a given heat transferred. Entropy (2) – is a measure of randomness of the molecules of a substance or measures the fraction of the total energy of a system that is not available for doing work. Entropy (3) – a property used to measure the state of disorder of a substance; a function of both heat and temperature. Entropy production – is the increase in entropy. 20

S 

Q Tabs

s 

q Tabs

S = the change in entropy of a system during some process (Btu/R or kJ/K). Q = the amount of heat transferred to or from the system during the process (Btu or kJ)

Tabs = the absolute temperature at which the heat was transferred (R or K). s = the change in specific entropy of a system during some process (Btu/lbm-oR or kJ/kg.K). q = the amount of heat transferred to or from the system during the process (Btu/lbm or kJ/kg).

2.13 LAWS OF THERMODYNAMICS a. ZEROTH LAW OF THERMODYNAMICS – TEMPERATURE (THERMAL EQUILIBRIUM) - states that when each of two systems is in equilibrium with a third, the first two systems must be in equilibrium with each other. This shared property of equilibrium is the temperature. - states that when two bodies have equality of temperature with a third body, they in turn have equality of temperature with each other and the three bodies are said to be in thermal equilibrium. The third body is usually a thermometer. b. FIRST LAW OF THERMODYNAMICS – LAW ON CONSERVATION OF ENERGY - states that, because energy cannot be created or destroyed [setting aside the later ramifications of the equivalence of mass and energy (Nuclear Energy)] – the amount of heat transferred into a system plus the amount of work done on the system must result in a corresponding increase of internal energy in the system. Heat and work are mechanisms by which systems exchange energy with one another. - states that during any cycle a system undergoes, the cyclic integral of heat is proportional to the cyclic integral of work or for any system, total energy entering = total energy leaving. c. SECOND LAW OF THERMODYMANICS – ENTROPY - states that the entropy – that is, the disorder – of an isolated system can never decrease. Thus, when an isolated system achieved a configuration of maximum entropy, it can no longer undergo change: It has reached equilibrium. Significant statements: - Clausius: It is impossible for a self-acting machine unaided by an external agency to move heat from one body to another at a higher temperature. - Kelvin-Planck: It is impossible to construct a heat engine which, while operating in a cycle produces no effects except to do work and exchange heat with a single reservoir. - All spontaneous processes result in a more probable state - The entropy of an isolated system never decreases. - No actual or ideal heat engine operating in cycles can convert into work all the heat supplied to the working substance. - Caratheodory: In the vicinity of any particular state 2 of a system, there exist neighboring states 1 that are inaccessible via an adiabatic change from state 2. 21

d. THIRD LAW OF THERMODYNAMICS - ABSOLUTE TEMPERATURE - states that absolute zero cannot be attained by any procedure in a finite number of steps. Absolute zero can be approached arbitrarily closely, but it can never be reached. - The entropy of a substance of absolute temperature is zero. 2.14 CONSERVATION OF MASS The law of conservation of mass states that the mass is indestructible. Steady State – is that circumstance in which there is no accumulation of mass or energy within the control volume, and the properties at any point within the system are independent of time. Continuity Equation of Steady Flow A A m  1 A11  2 A22  1 1  2 2 v1 v2 Usual English units are:  fps,  lb/ft3,  ft3/lb, A ft2, m lb/sec.

Example No. 22 A fluid moves in a steady-flow manner between two sections in the same flow line. At section (1): A1 = 0.10 m2, V1 = 6 m/s, v1 = 0.33 m3/kg. At section (2): A2 = 0.20 m2, 2 = 0.27 kg/m3. Calculate for the velocity of flow at section (2). A. 33.67 B. 37.63 C. 41.59 D. 45.55 Solution: 1 A1V1   2 A2V2

A1V1   2 A2V2 1 0.10 6  0.27 0.20 V2  0.33

V2  33 .67 m s

2.15 HELMHOLTZ FUNCTION Helmholtz function - a thermodynamic property of a system equal to the difference between its internal energy and the product of its temperature and its entropy. Symbol A. Formula A = u – Ts.

2.16 GIBBS FUNCTION Gibbs function - a thermodynamic property of a system equal to the difference between its enthalpy and the product of its temperature and its entropy. It is usually measured in joules. Symbol G. Formula G = h – Ts. Example No. 23 Water is being heated by the exhaust gases from a gas turbine. The gases leave the gas turbine at 648 C and may be cooled to 148 C. The water enters the heater at 93 C. The rate of gas flow is 25 kg/s and the water flow is 22

31.5 kg/s. Assume that the mean specific heat of the gas and water are respectively 1.088 and 4.27 kJ/kg-C. What is the available energy removed from the hot gases in kw? Take available sink temperature as 311 K. A. 8345.6 B. 4862.5 C. 6041.6 D. 6977.9 Solution: Ag  Qg  To Sg

Ag  mc p Ta  Tb   To mc p ln

Ta Tb

Ta  648  273  921 K Ta  148  273  421 K Ta  311 K Ag  251.088921  421  311251.088ln

921 421

Ag  13,600  6,622.1  6,977.9 kW 2.17 BOYLE’S LAW Boyle’s Law (1) – states that if the temperature of a given quantity of gas is held constant, the volume of gas varies inversely with the absolute pressure during a quasistatic change of state. Boyle’s Law (2) – states that the volume of a gas varies inversely with its absolute pressure during change of state if the temperature is held constant. p1V1  p2V2 Example No. 24 An ideal gas is contained in a vessel of unknown volume at a pressure of 1 atmosphere. The gas is released and allowed to expand into a previously evacuated bulb whose volume is 0.500 liter. Once equilibrium has been reached, the temperature remains the same while the pressure is recorded as 500 millimeters of mercury. What is the unknown volume, V, of the first bulb? A. 1.069 liter B. 0.853 liter C. 0.961 liter D. 1.077 liter Solution: 1 atm = 760 mm Hg p1V1  p2V2 760 V1  500 V1  0.50 L  V1  0.961 L

2.18 CHARLES’ LAW Charles’ Law (1) – the volume of a gas varies directly as the absolute temperature during a change of state if the pressure of the gas is held constant. V1 V2  T1 T2 23

Charles’ Law (2) – the pressure of a gas varies directly as the absolute temperature during a change of state if the volume of the gas is held constant in. p1 p2  T1 T2 Example No. 25 A closed vessel contains air at a pressure of 200 kN/m2 gauge and a temperature of 32 C. The air is heated to 60 C with the atmospheric pressure of 750 mm Hg. What is the final gauge pressure? A. 337.54 B. 127.54 C. 227.54 D. 427.54 Solution: Patm  750 mmHg  100 kPa

p1 p2  T1 T2 p2 200  100  32  273 60  273

p2  327 .54 kPa abs  227 .54 kPag (C)

2.19 IDEAL GAS LAW Ideal Gas Law - The equation of state of an ideal gas which is a good approximation to real gases at sufficiently high temperatures and low pressures; that is, PV = RT, where P is the pressure, V is the volume per mole of gas, T is the temperature, and R is the gas constant. pV  mRT

p1V1 p2V2  T1 T2 where p = pressure, V = volume, m = mass, R = ideal gas constant, and T = absolute temperature. Example No. 26 A volume of 400 cc of air is measured at a pressure of 740 mm Hg abs and a temperature of 18 C. What will be the volume at 760 mm Hg abs and 0 C. A. 376 cc B. 326 cc C. 356 cc D. 366 cc Solution: p2V2 p1V1  T2 T1

760 V2 740 400 

 0  273 18  273 V2  365 .4 cc

2.20 BASIC PROPERTIES OF IDEAL GAS 24

8.3143 kJ kg  K M c p  cv  R

R

cp cv

or

R

1545 ft  lb lb  R M

k

kR k 1 R cv  k 1 where: R = gas constant M = molecular weight cp = specific heat at constant pressure cv = specific heat at constant volume k = specific heat ratio cp 

Example No. 27 A 0.90 m3 tank contains 6.5 kg of an ideal gas. The gas has a molecular weight of 44 and is at 21 C. What is the pressure of the gas? A. 201.3 kPa B. 301.3 kPa C. 401.3 kPa D. 501.3 kPa Solution: R 8314.3 J kmol  K R   189 J kg  K M 44 kg kmol T  21  273  294 K mRT 6.5189 294  p   401,300 Pa  401.3 kPa V 0 .9 2.21 PROPERTIES OF AIR M  28 .97 kg air mole air

k  1 .4 R  53 .3 ft  lb lb  R  0.287 kJ kg  K

c p  0.24 Btu lb  R  0.24 kcal kg  C  1.0 kJ kg  C cv  0.171 Btu lb  R  0.171 kcal kg  C  0.716 kJ kg  C

2.22 AVOGADRO’S LAW AND NUMBER Avogadro’s Law – states that “equal volumes of all ideal gases at a particular pressure and temperature contains the same number of molecules”. Avogadro’s number, NA – is the number of molecules in a mole of any substance and is equal to 6.0225 x 1023 gmole-1. 2.23 JOULE’S LAW 25

Joule’s Law – states that the change of internal energy of an ideal gas is a function of only the temperature change. 2.24 DALTON’S LAW OF PARTIAL PRESSURE Dalton’s Law of Partial Pressure – states that the total pressure pm of a mixture of gases is the sum of the pressures that each gas would exert were it to occupy the vessel alone at the volume V m and temperature tm of the mixture. pm  pa  pb  pc     pi

Tm  Ta  Tb  Tc

i

Vm  Va  Vb  Vc 

pi  X i pm  Xi  1 i

Example No. 28 A mixture is formed at 689.48 kPaa, 37.8 C by bringing together these gases – each volume before mixing measured at 689.48 kPaa, 37.8 C: 3 mol CO2, 2 mol N2, 4.5 mol O2. Find the partial pressure of the CO2 after mixing. A. 73 kPaa B. 327 kPaa C. 145 kPaa D. 218 kPaa Solution: 3 689.48  217.73 kPaa pCO2  3  2  4.5 2.25 AMAGAT’S LAW OF PARTIAL VOLUMES Amagat's Law of Partial Volumes - states that the volume Vm of a gas mixture is equal to the sum of volumes Vi of the K component gases, if the temperature T and the pressure p remain the same. Vm  Va  Vb  Vc     Vi

Tm  Ta  Tb  Tc

i

pm  pa  pb  pc 

Vi  X iVm  Xi  1 i

2.26 THERMODYNAMIC SYSTEMS AND PROCESSES Thermodynamic Process – is the path of the succession of states through which the system passes. Cyclic Process or Cycle – is a process where a system in a given initial state goes through a number of different changes in state (going through various processes) and finally returns to its initial values. Reversible Process – is defined as a process that, once having taken place, can be reversed, and in so doing leaves no change in either the system or surroundings. Irreversible Process – is a process that cannot return both the system and the surroundings to their original conditions. That is, the system and the surroundings would not return to their original conditions if the process was reversed. Adiabatic Process – is one in which there is no heat transfer into or out of the system. The system can be considered to be perfectly insulated. 26

Isentropic Process – is one in which the entropy of the fluid remains constant. This will be true if the process the system goes through is reversible and adiabatic. An isentropic process can also be called a constant entropy process. Polytropic Process – is a process when a gas undergoes a reversible process in which there is heat transfer, the process frequently takes place in such a manner that a plot of the Log P (pressure) vs Log V (volume) is a straight line. Or stated in equation form PV n = a constant. Throttling Process – is defined as a process in which there is no change in enthalpy from state one to state two, h1  h2 ; no work is done, W  0 ; and the process is adiabatic, Q  0 . Isobaric Process – is an internally reversible (quasistatic, if nonflow) process of a pure substance during which the pressure remains constant. Isometric Process (Isochoric Process) – a constant volume process that is internally reversible (quasi-static if nonflow), involving a pure substance. Isothermal Process – is an internally reversible (quasistatic, if nonflow) constant temperature process of a pure substance. 2.27 EQUATIONS FOR THERMODYNAMIC PROCESSES dQ Entropy = S   T Non-Flow Equation Q  U  Wnf Steady-Flow Equation, K  0 P  0 Q  H  Wsf Steady-Flow Equation, K  0 P  0 Q  H  K  P  Wsf where:  U = internal energy H = enthalpy K = kinetic energy P = potential energy

Wnf = non-flow work Wsf = steady flow work Internal Energy U  mcv T Enthalpy H  mc p T Non-Flow Work Wnf   pdV Steady Flow Work, K  0 P  0 Wsf   Vdp Steady-Flow Equation, K  0 P  0 K  P  Wsf   Vdp 27

Note: k 

H U

Example No. 29 In compressing 20 kg/min of air, its enthalpy was increased by 140 kJ/kg. If the power input is 50 kW, the heat transfer in kW is, A. 3.33 B. -3.33 C. 6.67 D. -6.67 Solution: Q  H  W H  140 kJ kg 20 kg min 1 min 60 sec 

H  46.67 kW W  50 kW Q  46.67  50  3.33 kW Example No. 30 What is the temperature rise in water dropping at a height of 200 ft in F? A. 0.257 B. 1.33 C. 2.25 D. 6.67 Solution: WH Wc p t  J H t  Jc p

t 

200 ft 778.16 ft  lb Btu1.0 Btu lb  F 

t  0.257 F Example No. 31 A non-flow system contains 1 lb of an ideal gas (cp = 0.24, cv = 0.17, both in Btu/lb-R). The gas temperature is increased by 10 F while 5 Btu of work are done by the gas. What is the heat transfer in Btu? A. -3.3 B. -2.6 C. +6.7 D. +7.4 Solution: Q  U  W U  mc v T   10.17 10   1.7 Btu

W  5 Btu Q  1.7  5  6.7 Btu

28

2.28 ISOBARIC PROCESS Isobaric Process – is an internally reversible (quasistatic, if nonflow) process of a pure substance during which the pressure remains constant.

p = Constant

V2 T2  V1 T1

H  mcp T2  T1  U  mcv T2  T1  Non-flow: Q  U  Wnf

Wnf   pdV  pV2  V1  Q  H Entropy: S  

S  mc p ln

dQ T

T2 V  mc p ln 1 T1 V2

Steady Flow: Q  H  Wsf

Wsf   Vdp  0 Q  H Entropy: S  

S  mc p ln

dQ T

T2 V  mc p ln 1 T1 V2

Specific Heat Ratio: H Q k  U U 29

2.29 ISOMETRIC PROCESS (ISOCHORIC PROCESS) Isometric Process (Isochoric Process) – a constant volume process that is internally reversible (quasi-static if nonflow), involving a pure substance.

V = constant

p2 T2  p1 T1

H  mcp T2  T1  U  mcv T2  T1  Non-flow: Q  U  Wnf

Wnf   pdV  0 Q  U Entropy: S  

S  mcv ln

dQ T

T2 p  mcv ln 2 T1 p1

Steady Flow: Q  H  Wsf

Wsf    Vdp  V p2  p1  Q  U Specific Heat Ratio: H H k  U Q Example No. 32 One kg of hydrogen are cooled from 450 C to 320 C in a constant volume process. The specific heat at constant volume, cv, is 10.2 kJ/kg-K. How much heat is removed? A. 1136 kJ B. 1326 kJ C. 1623 kJ 30

D. 2136 kJ Solution:

Q  mc v T2  T1  Q  1.0 10 .2 450  320   1326 kJ

2.30 ISOTHERMAL PROCESS Isothermal Process – is an internally reversible (quasistatic, if nonflow) constant temperature process of a pure substance.

pV = constant, T = constant

H  mcp T2  T1  U  mcv T2  T1  Non-Flow Equation: Q  U  Wnf

p1V1  p2V2  pV  C  mRT Wnf  pV ln

V2 p  pV ln 1 V1 p2

Wnf  mRT ln

V2 p  mRT ln 1 V1 p2

Q  Wnf Steady Flow: Q  H  Wsf

Wsf  pV ln

V2 p  pV ln 1 V1 p2

Wsf  mRT ln

V2 p  mRT ln 1 V1 p2

Q  Wsf Entropy:

S  mR ln

V2 V1 31

S  mR ln S 

p1 p2

pV V2 pV p1 ln  ln T V1 T p2

Example No. 33 Five kmols of air initially at one atmosphere and 299 K are compressed isothermally to 8 atmospheres. How much total heat is removed during the compression? A. 25,846.3 kJ B. 28,922.9 kJ C. 39,345.6 kJ D. 44,680.9 kJ Solution: p  Q  nRT ln 1   p2 

 1 atm  kJ    Q  5 kmol  8.314 299 K ln  kmol  K    8 atm  Q  25,846.27 kJ (A) 2.31 ISENTROPIC PROCESS Isentropic Process – is a reversible adiabatic process with constant entropy.

pV k = constant, S  0 , pV  mRT , Q  0 k

p2  V1   T2   V1       p1  V2   T1   V2  , H  mcp T2  T1 

k 1

T Or  2  T1 ,

  p2      p1

  

k 1 k

U  mcv T2  T1  Non-flow: Q  U  Wnf 1 k  p V  p 1k  p1V1  V2     1  1 1  1   1 Wnf  1  k  V1   1  k  p2      

32

mRT1 Wnf  1k

Wnf 

mRT1 1k

 V 1k  mRT 1  2   1   V1   1k  

 V  k 1  mRT 1  1   1   V2   1k  

k 1    p2  k   1  p   1  

 T2  mRT2  T1  p2V2  p1V1    1  1k 1k  T1 

But, R k 1 Wnf  mcv T2  T1   U cv 

Steady Flow: Q  H  Wsf k 1    p2  k  kmRT1  1   p  1k  1   kmR T2  T1  Wsf  1k But kR cp  k 1 Wsf  mcp T2  T1   H

kp V Wsf  1 1 1k

k 1    p2  k  kmRT1  1   p  1k  1  

 V  k 1  kmRT  T  1 2  1   1   1      V2   1  k  T1   

Note: c p H   Vdp Wsf k    cv U  pdV Wnf Example No. 34 Air is compressed in a diesel engine from an initial pressure of 13 psia and a temperature of 120 F to one-twelfth of its initial volume. Calculate the final temperature assuming the compression to be adiabatic. A. 1110 F B. 980 F C. 987 F D. 1560 F Solution: T2  V1    T1  V2 

k 1

1.4 1

  t 2  460  V1    120  460  1 V   1  12  t 2  1107 .11 F

Example No. 35 Nitrogen is expanded isentropically. The temperature change from 620 F to 60 F. If the pressure ratio, p1/p2 = 13, what is the isentropic index. 33

A. 1.26 B. 1.2 C. 1.4 D. 1.46 Solution: k 1

T1  p1  k   T2  p2  k 1 620  460  13 k 60  460 2.077  13 ln2.077  

k 1 k

k 1 ln13 k

k 1  0.285 k k  1.4

2.32 POLYTROPIC PROCESS Polytropic Process – is an internally reversible process during which pVn = C where n is any constant.

pV n = constant, pV  mRT n 1

T2  p2    T1  p1  , H  mcp T2  T1  T2  V1    T1  V2 

n 1 n

U  mcv T2  T1  Non-flow Work: Q  U  Wnf 1 n  p V  p 1n  p1V1  V2     1  1 1  1   1 Wnf  1  n  V1   1  n  p2      

34

mRT1 Wnf  1n

Wnf 

mRT1 1n

 V 1n  mRT 1  2   1   V1   1n  

 V  n1  mRT 1  1   1   V2   1n  

n 1    p2  n   1  p   1  

 T2  mRT2  T1  p2V2  p1V1    1  1n 1n  T1 

But, R k 1 mR T2  T1  p2V2  p1V1 U   k 1 k 1

cv 

n  k p2V2  p1V1  1  1   1  1 Q   p2V2  p1V1    p2V2  p1V1   k  1n  1  k 1 1  n   k 1 n 1  Steady Flow: Q  H  Wsf

np V Wsf  1 1 1n Wsf 

n 1    p2  n  nmRT1  1   p  1n  1  

nmR T2  T1  1n

n 1    p2  n  nmRT1  1   p  1n  1  

But cp 

kR k 1

H  mc p T2  T1  

kmR T2  T1  k p2V2  p1V1   k 1 k 1

1  n   1  k Q   p2V2  p1V1    p2V2  p1V1   k 1 1  n   k 1 n 1  n  k p2V2  p1V1  Q k  1n  1 Entropy:

 n  k  T2 S  mcv   ln  n  1  T1 Or V  S  mcv n  k ln 1   V2   n  k   p2 S  mc v   ln  n  1   p1

  

n 1 n

 n  k   p2  mc v   ln  n   p1

  

Note: Steady flow work:



Wsf  K  P   Vdp

35

 V  n1  nmRT  T  1 2  1   1   1      V2   1  n  T1   

2.33 CURVES FOR DIFFERENT VALUES OF n.

2.34 TABLE OF IDEAL GAS FORMULAS Ideal Gas Formulas For constant mass systems undergoing internally reversible processes Process

Isometric



V C

p, V, T relations

Isobaric pC

Isentropic

T C

S C

pV n  C

p1V 1k  p 2V 2k

p1V 1n  p 2V 2n

T2 p  2 T1 p1

T2 V2  T1 V1

 pdV

0

pV2  V1 

p1V1 ln

V2 V1

2

V  p2  p1 

0

p1V1 ln

V2 V1

2

1



 Vdp 1

U 2  U1



m cv dT

mcv T2  T1 

mcv T2  T1 



p1V1  p2V2

0





k 1

T2  p2    T1  p1  p 2V2  p1V1 1 k

k  p 2V2  p1V1  1 k



m c p dT

mcv T2  T1 

mc p T2  T1 

V p1V1 ln 2 V1

n



0

1

Specific heat, c

cv

cp



36

T2  V1    T1  V2 

n 1

n 1 n

T2  p2    T1  p1  p 2V2  p1V1 1 n

n p2V2  p1V1  1 n



m cv dT

m cv dT

mcv T2  T1 

mcv T2  T1 

0

m cn dT

m Tds

m cv dT

Q

T2  V1    T1  V2 

k 1 k



m cv dT

Polytropic

Isothermal



mcn T2  T1 

k

 to  

0

k n c n  cv    1 n  k  C 



H 2  H1

m c p dT

mc p T2  T1 

mc p T2  T1 

cv dT T T mc v ln 2 T1 m

S 2  S1



m c p dT



m



c p dT

Q T

T

mc p ln m





0

T2 T1

mR ln



m c p dT

m c p dT

mc p T2  T1 

mc p T2  T1 

0

cn dT T T mc n ln 2 T1 m

V2 V1



c p dT cv dT V p  mR ln 2 , m  mR ln 2 T V1 T p1



Example No. 36 After a series of state changes, the pressure and volume of 2.268 kg of nitrogen are each doubled. What is S? cv =0.7442 kJ/kg-K, cp = 1.0414 kJ/kg-K. A. 2.8 kJ/kg-K B. 1.24 kJ/K C. 2.8 kJ/K D. 1.24 kJ/kg-K Solution: T V S  mcv ln 2  mR ln 2 T1 V1

S  mc p ln

V2 p  mcv ln 2 V1 p1

 2V 2p  S  2.2681.0414ln 1  0.7442ln 1   2.807 kJ K V1 p1   2.35 STAGNATION PROPERTIES Stagnation Properties – are those thermodynamic properties that a moving stream of compressible fluid would have if it were brought to rest isentropically (no outside work, the kinetic energy brings about the compression). Stagnation enthalpy

h0  h 

v2 2gc J

Stagnation temperature T0  T 

v2 2gc Jc p

T0  p0    T  p 

k 1 k

1

v2 2gc Jc pT

2.36 MACH NUMBER Mach Number – is the ratio of the actual speed divided by the local speed of sound a in the fluid.  M M1 M1 a [SUBSONIC] [SUPERSONIC] Acoustic speed: 1

a  g c kRT 2

37

2.37 MOLLIER DIAGRAM Mollier Diagram (h-s) – is a chart on which enthalpy is the ordinate and entropy the abscissa.

2.38 TEMPERATURE-ENTROPY DIAGRAM

38

2.39 ph-Chart

2.40 MIXTURES Mixture – substance made up of liquid and vapor portion or two-phase liquid-vapour system. x= quality or dryness factor or vapour content y = 1 – x = moisture content or wetness Properties of mixtures v  v f  xv fg 39

u  u f  xu fg h  h f  xhfg s  s f  xs fg 2.41 PROCESSES INVOLVING PURE SUBSTANCES a. Isobaric or constant pressure process: p1 = p2 b. Isothermal or constant temperature process: T1 = T2 Evaporation and condensation processes occur at constant pressure and constant temperature. c. Isometric or constant volume process: V1 = V2 For constant mass: v1 = v2 If the final state is a mixture: v1 = (vf + xvfg)2 d. Isentropic or constant entropy process: s1 = s2 Isentropic process is reversible (no friction loss) and adiabatic (no heat loss, that is, completely insulated system). e. Throttling or isenthalpic (constant enthalpy) process: h1 = h2 If the final state is a mixture: h1 = (hf + xhfg)2 If the initial state is a mixture, such as in steam calorimeter: (hf + xhfg)1 = h2 Example No. 37 After expanding 2.5 L of superheated steam at 2.5 MPaa and 400 C, its pressure was decreased to 0.01 MPaa. If its dryness fraction is 90%, what is the final volume of the steam in L? @ 2.5 MPaa and 400 C, v = 125.2 x 10-3 m3/kg @ 0.01 MPaa. vf = 1.0102 x 10-3 m3/kg and vg = 14,674 x 10-3 m3/kg. A. 164 B. 264 C. 364 D. 464 Solution:





v 2  v f  x vg  v f  v 2  1.01022103  0.90 14,674 103  1.01022103 v 2  13,206 103 m3 kg m

V1 V2  v1 v 2

2.5 L V2  3 125.2 10 13,206 103 V2  264 L

Example No. 38 Determine the heat transferred to the cooling fluid in a condenser operating under steady flow conditions with steam entering with an enthalpy of 2300 kJ/kg and a velocity of 350 m/s. The condensate leaves with an enthalpy of 160 kJ/kg and velocity of 70 m/s. A. -2199 kJ/kg B. -1922 kJ/kg C. 2190 kJ/kg D. 2910 kJ/kg 40

Solution:

Q  h2  h1 

V22  V12 21000

Q  160  2300 

70 2  350 2 21000 

2.42 THE CARNOT CYCLE

T1  T2  T4  T3 and S1  S 4  S2  S3 Q A  T1 S1  S 4 

QR  T2 S2  S 3   T2 S1  S 4 

W  Q A  QR  T1 S1  S 4   T2 S1  S 4 

e

W QA  QR T1 S1  S4   T2 S1  S4    QA QA T1 S1  S4 

e

T1  T2 TH  TL  T1 TH

QA QR  TH TL where e = Carnot cycle efficiency T1 = TH = highest absolute temperature T2 = TL = lowest absolute temperature Example No. 39 Carnot engine receives 130 Btu of heat from a hot reservoir at 700 F and rejects 49 Btu of heat. Calculate the temperature of the cold reservoir. A. -21.9 F B. -24.2 F C. -20.8 F D. -22.7 F Solution:

TH  700  460  1160 R e

W QA  QR TH  TL   QA QA TH

41

130  49 1160  TL  130 1160 TL  437 .23 R  22 .77 F

2.43 STIRLING AND ERICSSON CYCLE Ideal Stirling Cycle – is composed of two isothermal and two isometric processes, the regeneration occurring at constant volume. Ideal Ericsson Cycle – consists of two isothermal and two isobaric processes, with the regeneration occurring during constant pressure. 2.44 BASIC WORKING CYCLES FOR VARIOUS APPLICATIONS Application Steam Power Plant Gasoline Engine (Spark-Ignition) Diesel Engine (Combustion-Ignition) Gas Turbine Refrigeration System

Basic Working Cycle Rankine Cycle Otto Cycle Diesel Cycle Brayton Cycle Refrigeration Cycle

Example No. 40 One kilogram of air at a pressure and temperature of 1 bar and 15 C initially, undergoes the following processes in a cycle: isothermal compression to 2 bar; polytropic compression from 2 bar to 4 bar; isentropic expansion from 4 bar to initial condition. What is the cycle work in kJ/kg? A. 10.3 B. 13.0 C. 57.3 D. 70.3 Solution:

p1  1 bar  100 kPa T1  15  273  288 K p2  2 bar  200 kPa p3  4 bar  400 kPa

m  1 kg

1-2, T  C

T2  T1  288 K p2  2 bar  200 kPa

2-3, pV n  C 42

p2  200 kPa T2  288 K p3  400 kPa

p  T3  T2  3   p2 

n 1 n

3-1, pV k  C p3  200 kPa T1  288 K p1  100 kPa

p  T3  T1  3   p1 

k 1 k

Then

p  T3  T2  3   p2   p3     p2 

n 1 n

 400     200 

n 1 n

p   T1  3   p1 

p    3   p1 

n 1 n

k 1 k

k 1 k

 400     100 

1.4 1 1.4

n  2.333

V  W12  mRT1 ln 1   V2  n 1    nmRT2  p2  n    1 W23   1  n  p3    k 1    kmRT1  p3  k    1 W31   1  k  p1    m  1 kg

V  p  W12  mRT1 ln 1   mRT1 ln 2   V2   p1   200  W12  10.287 288 ln   57.3 kJ  100  n 1    nmRT2  p2  n    1 W23   1  n  p3    43

W23 W2  3

W31

W31

2.3331    2.33310.287 288   200  2.333   1    400   1  2.333    70 .3 kJ

k 1      kmRT1  p3 k    1    1  k  p1    1.4 1    1.4 10.287 288   400  1.4   1    100   1  1. 4  

W3 1  140 .6 kJ

Wcyc  W12  W23  W31  57.3  70.3  140.6 kJ Wcyc  13 kJ

-

End -

44