Review of Thermodynamics

ME 306 Fluid Mechanics II • Ideal gas equation of state is 푝 = 휌푅푇 Part 4 where 푅 is the gas constant. • By defining specific volume as 푣 = 1/휌 ideal gas law becomes 푝푣 = 푅푇 • For an ideal gas internal energy (푢 ) is a function of only. • Ideal gas specific heat at constant volume is defined as These presentations are prepared by 푑푢 Dr. Cüneyt Sert 푐 = 푣 푑푇 Department of Mechanical Engineering Middle East Technical University • 푐푣 is also a function of temperature, but for moderate temperature changes it can be Ankara, Turkey taken as constant. In this course we’ll take 푐푣 as constant. [email protected] • Change in internal energy between two states is (considering constant 푐푣) 푢 − 푢 = 푐 (푇 − 푇 ) Please ask for permission before using them. You are NOT allowed to modify them. 2 1 푣 2 1

4-1 4-2

Review of Ideal Gas Thermodynamics (cont’d) Review of Ideal Gas Thermodynamics (cont’d)

• Enthalpy is defined as • For air 푝 ℎ = 푢 + = 푢 + 푅푇 푐푝 − 푐푣 = 푅 휌 푘퐽 푘퐽 • For an ideal gas enthalpy is also a function of temperature only. 1.005 0.287 푘𝑔퐾 푘퐽 푘𝑔퐾 • Ideal gas specific heat at constant is defined as 0.718 푘𝑔퐾 푑ℎ 푐푝 = • 훾 푑푇 Specific heat ratio is (also shown with ) 푐푝 • 푐푝 will also be taken as constant in this course. For constant 푐푝 change in enthalpy is 푘 = 푐푣 ℎ − ℎ = 푐 (푇 − 푇 ) 2 1 푝 2 1 which has a value of 1.4 for air. • Combining the definition of 푐푣 and 푐푝 • 푑ℎ 푑푢 Combining the above relations we can also obtain 푐푝 − 푐푣 = − = 푅 푅푘 푅 푑푇 푑푇 푐 = , 푐 = 푝 푘 − 1 푣 푘 − 1

4-3 4-4 Review of Ideal Gas Thermodynamics (cont’d) and the

• change for an ideal gas is expressed with 푇푑푠 relations • Compressibility effects become important when a fluid moves with speeds comparable 푐 1 1 to the local speed of sound ( ). 푇푑푠 = 푑푢 + 푝 푑 , 푇푑푠 = 푑ℎ − 푑푝 휌 휌 • Mach number is the most important nondimensional number for compressible flows 푀푎 = 푉 / 푐 • Integrating these 푇푑푠 relations for an ideal gas • 푀푎 < 0.3 Incompressible flow (density changes are negligible) 푇2 휌1 푇2 푝2 푠2 − 푠1 = 푐푣 푙푛 + 푅 푙푛 , 푠2 − 푠1 = 푐푝 푙푛 − 푅 푙푛 • 0.3 < 푀푎< 0.9 Subsonic flow (density changes are important, shock waves 푇1 휌2 푇1 푝1 do not develop) • For an adiabatic (no heat transfer) and frictionless flow, which is known as isentropic • 0.9 < 푀푎< 1.1 flow, entropy remains constant. Transonic flow (shock waves may appear and divide the flow field into subsonic and supersonic regions) Exercise : For isentropic flow of an ideal gas with constant specific heat values, derive • 1.1 < 푀푎< 5.0 Supersonic flow (shock waves may appear, there are no the following commonly used relations, known as isentropic relations subsonic regions) 푘/(푘−1) 푘 푇2 휌2 푝2 • 푀푎 > 5.0 Hypersonic flow (very strong shock waves and property = = 푇1 휌1 푝1 changes) 4-5 4-6

Speed of Sound (푐) Speed of Sound (cont’d)

• Speed of sound is the rate of propagation of a pressure pulse (wave) of infinitesimal • To obtain a relation for the speed of sound consider the following experiment strength through a still medium (a fluid in our case). • A duct is initially full of still gas with properties 푝,휌, 푇and 푉 = 0 • It is a thermodynamic property of the fluid.

• For air at standard conditions, sound moves with a speed of 푐 = 343 m/s. 푝 푇 휌 푉 = 0 Exercise: a) What’s the speed of sound in air at 5 km and 10 km altitudes?

b) What’s the speed of sound in water at standard conditions? • The piston is pushed into the fluid with an infinitesimal velocity. • A pressure wave of infinitesimal strength will form and it’ll travel in the gas with the speed of sound 푐. • As it passes over the gas particles it will create infinitesimal property changes.

Wave front moving with speed 푐

푝 +푑푝 푝 휌 +푑휌 푐 휌 푇 +푑푇 푇 4-7 0 + 푑푉 푉 = 0 4-8 Speed of Sound (cont’d) Speed of Sound (cont’d)

• For an observer moving with the wave front with speed 푐, wave front will be stationary Exercise: Using conservation of mass and momentum on the CV of the previous slide, and the fluid on the left and the right would move with relative speeds derive the following expression for the speed of sound.

푝 + 푑푝 푝 푑푝 휌 + 푑휌 휌 푐 = Propagation of a sound wave 푑휌 푇 + 푑푇 푇 푠 is an

푐 − 푑푉 푐 Stationary wave front with respect Exercise : In deriving the speed of sound equation, we did not make use of the energy to an observer moving with it equation. Show that it can also be used and gives the same result.

• Consider a control volume enclosing the stationary wave front. The flow is one- Exercise : What is the speed of sound for a perfectly incompressible fluid. dimensional and steady. It has one inlet and one exit.

Exercise : Show that speed of sound for an ideal gas is given by 푐 − 푑푉 푐 Inlet and exit cross sectional areas 푐 = 푘푅푇 퐴 are the same ( ) 4-9 4-10

Wave Propagation in a Compressible Fluid Wave Propagation in a Compressible Fluid (cont’d)

• Consider a point source generating small pressure pulses (sound waves) at regular • Case 2 : Source moving with less than the speed of sound (푀푎 < 1) intervals. • Waves are not symmetric anymore. • Case 1 : Stationary source • An observer will hear different sound frequencies depending on his/her location. • Waves travel in all directions symmetrically. • This asymmetry is the cause of the Doppler effect. • The same sound frequency will be heard everywhere around the source.

4-11 4-12 Wave Propagation in a Compressible Fluid (cont’d) Wave Propagation in a Compressible Fluid (cont’d)

• Case 3 : Source moving the speed of sound (푀푎 = 1) • Case 4 : Source moving with more than the speed of sound (푀푎> 1) • The source moves with the same speed as the sound waves it generates. • The source travels faster than the sound it generates. • All waves concentrate on a plane passing through the moving source creating a • Mach cone divides the field into zones of action and silence. Mach wave, across which there is a significant pressure change. • Half angle of the Mach cone is called the Mach angle 휇. • Mach wave separates the field into two as zone of silence and zone of action.

Zone of Zone of Zone of 휇 Zone of action silence action silence

푉 = 푐 푉 > 푐

• First aircraft exceeding the speed of sound : http://en.wikipedia.org/wiki/Bell_X-1 4-13 4-14

Wave Propagation in a Compressible Fluid (cont’d) 1D, Isentropic, Compressible Flow

Exercise : For ‘‘Case 4’’described in the previous slide show that • Consider an internal compressible flow, such as the one in a duct of variable cross sectional area sin 휇 = 1/푀푎 Exercise : A supersonic airplane is traveling at an altitude of 4 km. The noise generated by the plane at point A reached the observer on the ground at point B after 20 s. Assuming isothermal atmosphere, determine a) Mach number of the airplane b) distance traveled by the airplane before the observer hears the noise c) velocity of the airplane • Flow and fluid properties inside this duct may change due to d) temperature of the atmosphere • Cross sectional area change 퐿 = 5 km F/A-18 breaking the sound barrier • Frictional effects A NOT the subject of ME 306. • Heat transfer effects Without these the flow is isentropic. 퐻 = 4 km • In ME 306 we’ll study these flows as 1D and consider only the effect of area change, Ground http://en.wikipedia.org i.e. assume isentropic flow. B 4-15 4-16 1D, Isentropic, Compressible Flow (cont’d) Stagnation Enthalpy

푉2 • The sum ℎ + is known as stagnation enthalpy and it is constant inside the duct. • Conservation of energy for a control volume 2 enclosing the fluid between sections 1 and 2 is 푉2 ℎ0 = ℎ + = constant along the duct 1 2 2 stagnation enthalpy 2 2 푉2 푉1 푞 − 푤 = ℎ2 + + 𝑔푧2 − ℎ1 + + 𝑔푧1 2 2 • It is called ‘‘stagnation’’ enthalpy because at a stagnation point velocity is zero and the = 0 enthalpy of the gas is equal to ℎ0. Heat transfer is zero for adiabatic flow. Also there is no For gas flows potential energy change work done other than the flow is usually negligibly small compared to work. enthalpy and kinetic energy changes. If the fluid is sucked into the duct from a ‘‘large’’ reservoir, the reservoir can be assumed to be at the stagnation state. 2 2 푉1 푉2 • Energy equation becomes ℎ1 + = ℎ2 + 2 2 Large reservoir with negligible velocity 4-17 4-18

Stagnation State Stagnation State (cont’d)

• Stagnation state is an important reference state for compressible flow calculations. • Isentropic deceleration can be shown on a ℎ − 푠 diagram as follows

• It is the state achieved if a fluid at any other state is brought to rest isentropically. ℎ • For an isentropic flow there will a unique stagnation state. Stagnation state 푝0 푉 = 0, ℎ , 푝 , ℎ 0 0 0 0 푇 , 푠 , etc. State 1 State 2 0 0 푉1, ℎ1, 푝1, 푇1 , etc. 푉2, ℎ2, 푝2, 푇2, etc. 푉2 Isentropic ∆ℎ = 2 deceleration 푝 Any state 1 2 ℎ 푉, ℎ, 푝, 푇, 푠, etc.

Hypothetical 푠 Hypothetical isentropic deceleration isentropic • During isentropic deceleration entropy remains constant. deceleration 0 Unique stagnation state 02 푉2 푉2 • Energy conservation: ℎ + = ℎ + → ∆ℎ = ℎ − ℎ = 푉0 = 0, ℎ0, 푝0, 푇0 , etc. 0 2 2 0 2 4-19 4-20 Stagnation State (cont’d) Stagnation State (cont’d) Exercise : For the isentropic flow of an ideal gas, express the following ratios as a Exercise : An airplane is crusing at a speed of 900km/h at an altitude of 10km. function of Mach number and generate the following plot for air with 푘 = 1.4. Atmospheric air at −60℃comes to rest at the tip of its pitot tube. Determine the temperature rise of air.

−1 Read about heating of space shuttle during its reentry to the earth’s atmosphere. 푇 푘 − 1 1.0 = 1 + 푀푎2 푇 http://en.wikipedia.org/wiki/Space_Shuttle_thermal_protection_system 푇표 2 푇0 휌 휌 푝 −푘/(푘−1) 0 Exercise : An aircraft cruises at 12 km altitude. A pitot-static tube on the nose of 푝 푘 − 1 0.5 푝0 = 1 + 푀푎2 푝 2 the aircraft measures stagnation and static of 2.6 kPa and 19.4 kPa, 표 respectively. Calculate

−1/(푘−1) a) Mach number of the aircraft 휌 푘 − 1 = 1 + 푀푎2 0 1 2 3 4 5 b) speed of the aircraft 휌 2 표 푀푎 c) stagnation temperature that would be sensed by a probe on the aircraft.

4-21 4-22

Stagnation State (cont’d) Simple Area Change Flows (1D Isentropic Flows)

Exercise : Using the incompressible form of the Bernoulli’s equation, derive an Exercise : Consider the differential control volume shown below for 1D, isentropic expression for 푝0/푝 for incompressible flows. Compare it with the one given in flow of an ideal gas through a variable area duct. Using conservation of mass, linear Slide 4-21. Determine the Mach number below which two equations agree within momentum and energy, determine the engineering accuracy. a) Change of pressure with area b) Change of velocity with area 2.0 1.8 푥 1.6 Compressible 푝0 (Slide 4-21) 푝 1.4 Incompressible 푑푥 1.2 (BE) 1.0 푝 푝 + 푑푝 0 0.2 0.4 0.6 0.8 1 휌 휌 + 푑휌 푀푎 푉 푉 + 푑푉 ℎ ℎ + 푑ℎ 퐴 퐴 + 푑퐴 4-23 4-24 Simple Area Change Flows (cont’d) Simple Area Change Flows (cont’d)

• Results of the previous exercise are • Sonic flow is a very special case. It can occur

푑퐴 푑푝 푑퐴 푑푉 • when the cross sectional area goes through a minimum, i.e. 푑퐴 = 0 = 1 − 푀푎2 , = − 1 − 푀푎2 퐴 휌푉2 퐴 푉

Subsonic Flow 푀푎 < 1 Supersonic Flow 푀푎> 1

Diffuser Sonic flow may occur at the throat. 푑퐴 > 0 푑퐴 < 0 푑푝 > 0 • or at the exit of a subsonic or a supersonic diffuser 푑푉 < 0 푀푎 < 1 Nozzle Sonic flow may occur at these exits. 푑푝 < 0 푑퐴 < 0 푑퐴 > 0 푑푉 > 0 푀푎 > 1 4-25 4-26

de Laval Nozzle (C-D Nozzle) Critical State

Exercise : The nozzle shown below is called a converging diverging nozzle (C-D nozzle • Critical state is the special state where the Mach number is unity. or Con-Div nozzle or de Laval nozzle). • It is a useful reference state, similar to the stagnation state. It is useful even if there is Using the table of Slide 4-25 show that it is the only way to no actual critical state in a flow. • isentropically accelerate a fluid from subsonic to supersonic speed. • It is shown with an asterisk, like 푇 ∗, 푝∗ , 휌∗, 퐴∗, etc. • isentropically decelerate a fluid from supersonic to subsonic speed. • Ratios derived in Slide 4-21 can also be written using the critical state.

푘/(푘−1) 푘/(푘−1) 푝표 푘 − 1 푝표 푘 − 1 = 1 + 푀푎2 = 1 + 푝 2 푝∗ 2 de Laval nozzle 푀푎 = 1 푇표 푘 − 1 푇표 푘 − 1 = 1 + 푀푎2 = 1 + 푇 2 푇 ∗ 2 1/(푘−1 ) 1/(푘−1) 휌표 푘 − 1 휌표 푘 − 1 = 1 + 푀푎2 = 1 + 휌 2 휌∗ 2

4-27 4-28 Critical State (cont’d) Isentropic Flow Table

Exercise : Similar to the ratios given in the previous slide, following area ratio is also a • It provides the following ratios at different Mach numbers for a fixed 푘 value. function of 푀푎and 푘 only. Derive it. 푇 푝 휌 퐴 푚 푅푇0 For 푘 = 1.4 ∗ 3.0 푇0 푝0 휌0 퐴 퐴푝0 2.5 푘+1 푘 − 1 2 2(푘−1) 퐴 1 1 + 푀푎 퐴 2.0 = 2 퐴∗ 푀푎 푘 + 1 퐴∗ 1.5 2 1.0 0.5 0 0.5 1 1.5 2 2.5 3 푀푎

Exercise : Derive an expression in terms of 푀푎and 푘 for the following non- dimensional mass flow rate.

푚 푅푇0 퐴푝 Aksel’s Fluid Mechanics textbook 0 4-29 4-30

Exercises for Simple Area Change Flows Exercises for Simple Area Change Flows (cont’d)

Exercise : A converging duct is fed with air from a large reservoir where the Exercise : (Fox) Air flows isentropically in a channel. At an upstream section 1, Mach temperature and pressure are 350 K and 200 kPa. At the exit of the duct, cross- number is 0.3, area is 0.001 m2, pressure is 650 kPa and temperature is 62 ℃. At a sectional area is 0.002 m2 and Mach number is 0.5. Assuming isentropic flow downstream section 2, Mach number is 0.8. a) Determine the pressure, temperature and velocity at the exit. a) Sketch the channel shape. b) Find the mass flow rate. b) Evaluate properties at section 2. c) Plot the process between sections 1 and 2 on a 푇 − 푠 diagram. Exercise : Air is flowing isentropically in a diverging duct. At the inlet of the duct, pressure, temperature and velocity are 40 kPa, 220 K and 500 m/s, respectively. Inlet and exit areas are 0.002 m2 and 0.003 m2. a) Determine the Mach number, pressure and temperature at the exit. b) Find the mass flow rate.

4-31 4-32 Shock Waves Shock Waves (cont’d)

• Sound wave is a weak wave, i.e. property changes across it are infinitesimally small. • ∆푝 across a sound wave is in the order of 10−9 − 10−3 atm.

• Shock wave is a strong wave, i.e. property changes across it are finite. • Shock waves are very thin, in the order of 10−7 m. • Fluid particles decelerate with millions of 𝑔’s through a shock wave. • Shock waves can be stationary or moving. Oblique shock wave attached to the Detached curved shock wave ahead sharp nose of a bullet moving at of a blunt nosed object moving at • They can be normal (perpendicular to the flow direction) or oblique (inclined to the supersonic speed. supersonic speed. flow direction). • A shock wave can be thought as a tool used for a flow to adjust itself to downstream conditions. Normal shock wave in a supersonic nozzle. Flow is from left to right. Extra waves are due to surface roughness • In ME 306 we’ll consider stationary normal shock waves for 1D flows inside ducts.

4-33 White’s Fluid Mechanics textbook 4-34

Formation of a Strong Wave Formation of a Strong Wave (cont’d)

• A strong wave is formed by the accumulation of several weak compression waves. • First wave will cause an increase in temperature behind it. • Compression waves are the ones across which pressure increase and velocity decrease • Second wave will move faster and eventually may catch the first one. in the flow direction. 푐2 > 푐1 • Sound wave is a weak compression wave. • A third one, which is not shown, will move even faster and may catch the first two waves.

• Consider a piston pushed with a finite velocity 푉 in a cylinder filled with still gas. 푐3 > 푐2 > 푐1 • We can decompose piston’s motion into a series of infinitesimally small disturbances. • Weak compression waves have a chance to accumulate into a strong wave of finite strength. • Weak compression waves will emerge from the piston, one after the other. • Weak expansion waves that’ll be generated by pulling the piston to the left will not • The first two of such waves are sketched below. form such a strong wave. Accumulated Second wave front First wave front strong wave 푝 + 푑푝 푝 푝 + ∆푝 푝 푉 푐2 푐1 푉 푇 + 푑푇 푇 푇 + ∆푇 푇 푑푉 푉 = 0 ∆푉 푉 = 0 4-35 4-36 Normal Shock Wave Stagnation State of a Non-isentropic, Adiabatic Flow 푦 • Consider a stationary normal shock wave in a 푥 • Stagnation state concept can also be used for non-isentropic flows, but there will be multiple such states. duct of variable cross sectional area. • If the flow is adiabatic ℎ0 , 푇0 and 푐0 will be unique, but not other stagnation • Upstream and downstream states are denoted properties such as 푝0 or 휌0. by 푥 and 푦. State 1 State 2 • Due to very sudden, finite property changes, the process across the wave is 푉1, 푝1, ℎ1, 푇1, etc. 푉2, 푝2, ℎ2, 푇2, etc. considered to be non-isentropic. But it is considered to be adiabatic. 1 2 • There are two different stagnation states, state 0푥for the flow before the shock and state 0푦for the flow after the shock. Non-isentropic, Isentropic Isentropic adiabatic flow 푝0푥 ≠ 푝0푦 and 휌0푥 ≠ 휌0푦 (such as the one deceleration deceleration • However, because of the adiabatic assumption, stagnation and across a shock) Stagnation state of state 2 푉 = 0 푝 휌 ℎ 푇 , . enthalpies are the same. Stagnation state of state 1 0 , 02, 02, 0, 0 etc 푉0 = 0, 푝01, 휌01, ℎ0, 푇0 , etc. 푇0푥 = 푇0푦 = 푇0 and ℎ0푥 = ℎ0푦 = ℎ0 4-37 4-38

Shock Wave on the 푇 − 푠 Plane Property Changes Across a Shock Wave

• Governing equations for the 1D flow inside the control volume enclosing the shock 푇 푝0푥 푝0푦 wave are 푇0푥 = 푇0푦 푥 푦

푝푦 푦 푇푦 푥 푦 Shock

푇 푥 푥 • Continuity : 푚 = 휌푥푉푥퐴 = 휌푦푉푦퐴 where 퐴 = 퐴푥 = 퐴푦 푝푥

푠 • Momentum : 푝푥 − 푝푦 퐴 = 푚 푉푦 − 푉푥 푠푦 − 푠푥 푉2 푉2 • Energy : ℎ = ℎ + 푥 = ℎ + 푦 0 푥 2 푦 2

• A similar one can be drawn on the ℎ − 푠 plane. • Second Law : 푠푦 > 푠푥

4-39 4-40 Property Changes Across a Shock Wave (cont’d) Property Changes Across a Shock Wave (cont’d) 푘 1 • For the flow of an ideal gas with constant specific heats, we can work on the equations 푘+ 1 2 푘−1 푝 푀푎푥 2푘 푘 − 1 1−푘 of the previous slide to get the following (study the details from text books) • Stagnation pressure ratio : 0푦 = 2 푀푎2 − 푘 − 1 푥 푝0푥 1+ 푀푎2 푘 + 1 푘 + 1 2 푥 푘 − 1 푀푎2 + 2 푥 퐴∗ 푝 • Donwstream Mach number : 푀푎푦 = 2 푦 0푥 2푘푀푎푥− (푘− 1) • Critical area ratio : ∗ = 퐴푥 푝0푦 푘 − 1 2푘 1 + 푀푎2 푀푎2 − 1 푠 − 푠 푝 푇푦 2 푥 푘 − 1 푥 푦 푥 0푦 • Temperature ratio : = • Entropy change : = −푙푛 푇 푘 + 1 2 푅 푝0푥 푥 푀푎2 2(푘− 1) 푥 푝 2푘 푘 − 1 푦 2 • These relations are functions of 푀푎푥 and 푘 only. They are usually plotted or tabulated. • Pressure ratio : = 푀푎푥− 푝푥 푘 + 1 푘 + 1

2 • Flow before the shock and after the shock are isentropic, but not across the shock. 휌푦 (푘+ 1)푀푎푥 • Density ratio : = 2 휌푥 2 + (푘 − 1)푀푎푥 • 푇0 is the same before and after the shock, due to adiabatic assumption (푇0푥 = 푇0푦).

• 푝0 changes across the shock (푝0푥 ≠ 푝0푦). 푉푦 휌푥 • Velocity ratio : = ∗ ∗ 푉푥 휌푦 • Critical states before and after the shock are different. This is seen from 퐴푥 ≠ 퐴푦) 4-41 4-42

Property Changes Across a Shock Wave (cont’d) Property Changes Across a Shock Wave (cont’d)

• Tabulated form of these normal shock relations look like the following • Across a normal shock wave For 푘 = 1.4 6 ∗ 푝 퐴 푝 푀푎, 푉, 푝0 decreases 푦 푦 = 0푥 푝 퐴∗ 푝 5 푥 푥 0푦 푝, 푇, 휌, 퐴∗, 푠 increases

푉푥 휌푦 푇0 remains the same 4 = 푉푦 휌푥

3 푇푦 • Kinetic energy of the fluid after the shock 푇 푥 wave is smaller than the one that would 2 푝0푦 be obtained by a reversible compression 1 푝0푥 between the same pressure limits. 푀푎푦 • Lost kinetic energy is the reason of 0 1 1.5 2 2.5 3 3.5 4 temperature increase across the shock Aksel’s book 푀푎 푥 wave.

4-43 4-44 Normal Shock Wave (cont’d) Normal Shock Wave (cont’d)

Exercise : (Munson) A total pressure probe is inserted into a supersonic air flow. A Exercise : (Fox) Air with speed 668 m/s, temperature 5 oC and pressure 65 kPa goes shock wave forms just upstream of the stagnation point. The probe measures a through a normal shock wave. stagnation pressure of 410 kPa. The stagnation temperature at the probe tip is a) Determine the Mach number, pressure, temperature, speed, stagnation pressure measured with a thermocouple and found to be 555 K. The static pressure upstream and stagnation temperature after the shock wave, of the shock is measured with a wall tap to be 83 kPa. Determine the Mach number b) Calculate the entropy change across the shock wave. and the velocity of the flow. c) Show the process on a 푇 − 푠 diagram.

Exercise : Supersonic air flow inside a diverging duct is slowed down by a normal shock wave. Mach number at the inlet and exit of the duct are 2.0 and 0.3. Ratio of the exit to inlet cross sectional areas is 2. Pressure at the inlet of the duct is 40 kPa. Assuming adiabatic flow determine the pressure after the shock wave and at the exit of the duct.

4-45 4-46

Operation of a Converging Nozzle Operation of a Converging Nozzle (cont’d)

• Consider a converging nozzle. • First set 푝푏 = 푝0. There will be no flow.

• Gas is provided by a large reservoir with stagnation properties, 푇0 and 푝0. • Gradually decrease 푝푏. Following pressure distributions will be observed.

• Back pressure 푝푏 is adjusted using a vacuum pump to obtain different flow conditions inside the nozzle.

• Exit pressure 푝푒 and back pressure 푝푏 can be equal or different.

푝

푝0 1 : No flow (푝푏 = 푝0) 푝 푇0 푏 ∗ Subcritical 2 : 푝 < 푝푏 < 푝0 푝0 regime 푉 = 0 푝푒 ∗ ∗ 푝 3 : Critical (푝푏 = 푝 ) Supercritical 푀푎 = 1 ∗ 4 : 푝푏 < 푝 (choked flow) regime 4-47 푥 4-48 Operationof a Converging Nozzle (cont’d) Operation of a Converging Nozzle (cont’d)

• When air is supplied from a stagnation reservoir, flow inside a converging nozzle always Variation of 푝푒 with 푝푏 Variation of 푚 with 푝푏 remains subsonic. 푝푒 푚 • For the subcritical regime, mass flow rate increases as 푝푏 decreases. 푝0 1 4 3 푚 2 푚푎푥 • State shown with * is the critical state. When 푝 is lowered to the critical value 푝∗, exit 2 푏 4 3 Mach number reaches to 1 and flow is said to be choked.

1 • If 푝푏 is lowered further, flow remains choked. Pressure and Mach number at the exit do 푝푏 푝 ∗ ∗ 푏 푝 푝0 푝 푝0 not change. Mass flow rate through the nozzle does not change.

∗ • Case 1 is the no flow case. • For 푝푏 < 푝 , gas exits the nozzle as a supercritical jet with 푝푒 > 푝푏. Exit jet undergoes a

non-isentropic expansion to reduce its pressure to 푝푏. • From case 1 to case 3, lowering 푝푏 decreases 푝푒 and increases 푚 .

푘/(푘−1 ) • Case 3 is the critical case with minimum possible 푝푒 and maximum possible 푚 . 푝∗ 2 푝∗ • From slide 4-28 = . For air (푘 = 1.4)choking occurs when = 0.528. 푝0 푘+1 푝0 • Cases 3 and 4 are with 푀푎푒 = 1. 4-49 4-50

Operation of a Converging Nozzle (cont’d) Operation of a Conv-Div Nozzle

Exercise : (Aksel’s book) A converging nozzle is fed with air from a large reservoir • We first set 푝푏 = 푝0 and then gradually decrease 푝푏. where the pressure and temperature are 150 kPa and 300 K. The nozzle has an exit cross sectional area of 0.002 m2. Back pressure is set to 100 kPa. Determine Throat a) the pressure, Mach number and temperature at the exit. b) mass flow rate through the nozzle. 푝

Exercise : (Aksel’s book) Air flows through a converging duct and discharges into a 푝0 1 : No flow (푝푏 = 푝0) 2 : Subsonic Flow region where the pressure is 100 kPa. At the inlet of the duct, the pressure, 3 : Initial choked flow (푀푎푡ℎ푟표푎푡 = 1) temperature and speed are 200 kPa, 290 K and 200 m/s. Determine the Mach 4 : Flow with shock ∗ 5y : Shock at the exit Choked number, pressure and temperature at the nozzle exit. Also find the mass flow rate per 푝 flow. unit area. 6 : Overexpansion Same 푚 7 : Design condition 8 : Underexpansion 4-51 푥 4-52 Operation of a Conv-Div Nozzle (cont’d) Operation of a Conv-Div Nozzle (cont’d) 푝 • Flow inside the converging section is always subsonic. Throat 푝0 1 • At the throat the flow can be subsonic or sonic. 2 3 • The flow is choked if 푀푎throat = 1. This corresponds to the maximum 푚 that can pass 4 5y 푝∗ through the nozzle. 6 • Under choked conditions the flow in the diverging part can be subsonic (case 3) or 7 8 푥 supersonic (cases 6, 7 ,8).

• Depending on 푝푏 there may be a shock wave in the diverging part. Location of the Variation of 푝푒 with 푝푏 Variation of 푚 with 푝푏

shock wave is determined by 푝푏. 푝푒 푚 • Design condition corresponds to the choked flow with supersonic exit without a shock. 푝0 1 3 2 7 5y 3 • Overexpansion (푝푒 < 푝푏): Exiting jet finds itself in a higher pressure medium and 4 푚 5y 푚푎푥 8 6 4 contracts. Underexpansion (푝푒 > 푝푏): Exiting jet finds itself in a lower pressure 8 2 푝 5x 푑푒푠𝑖𝑔푛 7 6 medium and expands. For details and pictures visit http://aerorocket.com/Nozzle/Nozzle.html 1 푝푏 푝푏 푝0 and http://www.aerospaceweb.org/question/propulsion/q0224.shtml 4-53 푝0 4-54

Operation of a Conv-Div Nozzle (cont’d) Operation of a Conv-Div Nozzle (cont’d)

Exercise : Air is supplied to a C-D nozzle from a large reservoir where stagnation 2 & 3 pressure and temperature are known. Determine

푝 a) the Mach number, pressure and temperature at the exit Throat 4 b) the mass flow rate 푝0 1 2 3 4 5 5y 푝∗ 2 2 6 퐴푡 = 0.0022 m 퐴푒 = 0.0038 m 7 6 8 푇 = 318K 푥 0 푝푏 = 30 kPa 푝0 = 327 kPa 7

8

http://aerorocket.com 4-55 4-56 Operation of a Conv-Div Nozzle (cont’d)

Exercise : Air flows in a Conv-Div nozzle with an exit to throat area ratio of 2.1. Properties at a section in the converging part are measured as follows. a) Determine the ranges of back pressure for different flow regimes. b) If a shock wave is observed where the area is twice of the throat area, determine the back pressure.

푝1 = 128kPa

푇1 = 294K

푉1 = 135m/s

1

4-57