MOCK CLASS TEST: THERMODYNAMICS Sub: Code: EME-303; Mahamaya Technical University

Time: 2 hrs                                                                                                   Maximum Marks: 50 

Attempt all the questions: 

SECTION A: 

1) Attempt the following questions:                                                                        (5 x 2 = 10) 

a) Define system, surroundings and universe. 

b) Distinguish between Heat pump and Refrigerator. 

c) What is Exergy and Anergy? 

d) Explain the law of degradation of energy. 

e) What is triple point of water? 

SECTION B: 

2) Attempt any three questions:                                                                               (3 x 5 = 15) 

a) Distinguish between macroscopic and microscopic approaches of thermodynamics. 

b) Discuss the neccessity of 2nd law of thermodynamics. 

c) 2 kg of a gas at 10 bar expands adiabatically and reversibly till its pressure drops to 5 bar. During the process 120 kJ of non-flow work is done by the system and the temperature drops from 377°C centigrade to 257°C. Calculate the value of the index of expansion and the characteristics gas constants. 

d) Steam at a pressure of 4 bar absolute and having dryness fraction 0.8, is heated at constant volume to a pressure of 8 bar absolute. Find the final temperature of the steam. Also, find the total heat absorbed by 1 kg of steam. 

e) 2 kg of air at NTP is heated at constant volume untill the pressure becomes 6 bar. Find the change of entropy of the system. 

SECTION C: 

Attempt part (a) or part (b) of the following questions                                                 (5 x 5 = 25) 

3) (a) Explain the thermodynamic equilibrium and quasi-static process. 

(b) Prove the equivalence of Kelvin-Planck statement and Clausius statement. 

4) (a) A steam turbine developing 110 kW is supplied steam at 17.5 bar with an internal energy of 2600 kJ/min, specific volume = 15.5 m³/kg and velocity of 275 m/s. Heat loss from the steam turbine  37.6 kJ/kg. Neglecting the changes in potential energy, determine the steam flow rate in kg/hr. 

(b) A reversible engine takes 2400 kJ/min from a reservoir at 750 K develops 400 kJ/min of work during cycle. The engine rejects heat at two reservoir at 650 K and 550 K. Find the heat rejected to each sink. 

5) (a) Explain the causes of internal and external irreversibility. 

(b) Explain the importance of Gibb's function and Gibb's free energy. 

6) (a) 5 kg steam at pressure 8 bar and temperature 300°C is adiabatically mixed with 4 kg steam at 6 bar and 250°C. Find the final condition of the mixture. Also find the change in entropy. 

(b) Hot steam is flowing through a perfectly adiabatic pipe. At point A the temperature of the steam is 250°C and pressure is 4 bar, while at the point B, its temperature is 275°C and pressure is 3.5 bar. Find the direction of the flow. 

7) (a) 5 kg of Oxygen is enclosed within a vessel of 0.05 m³ at a temperature 200°C, is being supplied 120 kJ of energy through heating. Find the final pressure and temperature. 

(b) One kg of an ideal gas is heated from 18.3°C to 93.4°C. Assuming R = 287 J/kg-K and  γ  = 1.18 for the gas. Find out (i) specific heats, (ii) change in internal energy, and (iii) change in enthalpy and entropy.

QUESTIONS BANK 5 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 4 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 3 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 2: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

1)      Explain the principle of Super-position.

Ans: The principle of superposition states that “The effect of a force on a body does not change and remains same if we add or subtract any system which is in equilibrium.”

In the fig 4 a, a force P is applied at point A in a beam, where as in the fig 4 b, force P is applied at point A and a force system in equilibrium which is added at point B. Principle of super position says that both will produce the same effect.

2)      What is “Force-Couple system?”

Ans: When a force is required to transfer from a point A to point B, we can transfer the force directly without changing its magnitude and direction but along with the moment of force about point B.

As a result of parallel transfer a system is obtained which is always a combination of a force and a moment or couple. This system consists of a force and a couple at a point is known as Force-Couple system.

      In fig 5 a, a force P acts on a bar at point A, now at point B we introduce a system of forces  in equilibrium (fig 5 b), hence according to principle of superposition there is no change in effect of the original system. Now we can reduce the downward force P at point A and upward force P at point B as a couple of magnitude Pxd at point B (fig 5 c).

3) What do you understand by Equivalent force systems?

Ans: Two different force systems will be equivalent if they can be reduced to the same force-couple system at a given point. So, we can say that two force systems acting on the same rigid body will be equivalent if the sums of forces or resultant and sums of the moments about a point are equal.

4)      What is orthogonal or perpendicular resolution of a force?

Ans: The resolution of a force into two components which are mutually perpendicular to each other along X-axis and Y-axis is called orthogonal resolution of a force.

If a force F acts on an object at an angle θ with the positive X-axis, then its component along X-axis is F_x = Fcosθ, and that along Y-axis is F_y = Fsinθ

5) What is oblique or non-perpendicular resolution of a force?

Ans: When a force is required to be resolved in to two directions which are not perpendiculars to each other the resolution is called oblique or Non-perpendicular resolution of a force.

   
       F_OA = (P sin β)/ sin (α +β)

 F_OB = (P sin α)/ sin (α +β)

QUESTION BANK 1: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTION BANK: ENGINEERING MECHANICS

by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force and force system?

Ans: A force is a physical quantity having magnitude as well as direction. Therefore, it is a   vector quantity. It is defined as an "external agency" which produces or tends to produce or destroys or tends to destroy the motion when applied on a body.

Its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.

When two or more forces act on a body or particle, it is called force system. Therefore, a force system is a collection of two or more forces.

2)      What is static equilibrium? What are the different types of static equilibrium?

Ans: A body is said to be in static equilibrium when there is no change in position as well as no rotation exist on the body. So to be in equilibrium process, there must not be any kind of motions ie there must not be any kind of translational motion as well as rotational motion.

We also know that to have a linear translational motion we need a net force acting on the object towards the direction of motion, again to induce an any kind of rotational motion, a net moment must exists acting on the body. Further it can be said that any kind of complex motion can be resolved into a translational motion coupled with a rotating motion.

“Therefore a body subjected to a force system would be at rest if and only if the net force as well as the net moment on the body is zero.”

There are three types of Static Equilibrium

1.      Stable Equilibrium

2.      Unstable Equilibrium

3.      Neutral Equilibrium

3)      What are the characteristics of a force?

Ans: A force has four (4) basic characteristics.

·         Magnitude: It is the value of the force. It is represented by the length of the arrow that we use to represent a force.

·         Direction: A force always acts along a line, which is called as the “line of action”. The arrow head we used to represent a force is the direction of that force.

·         Nature or Sense: The arrow head also represent the nature of a force. A force may be a pull or a push. If a force acts towards a particle it will be a push and if the force acts away from a point it is pull.

·         Point of Application: It is the original location of a point on a body where the force is acting. 

4)      What are the effects of a force acting on a body?

Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.

·         A force may change the state or position of a body by inducing motion of the body. (External effect)

·         A force may change the size or shape of an object when applied on it. It may deform the body thus inducing internal effects on the body.

·         A force may induce rotational motion into a body when applied at a point other than its center of gravity.

·         A force can make a moving body into an equilibrium state at rest.

5)      What is composition and resolution of forces?

Ans: Composition of forces: Composition or compounding is the procedure to find out single resultant force of a force system

Resolution of forces: Resolution is the procedure of splitting up a single force into number of components without changing the effect of the same.

6)      What is Resultant and Equilibrant?

Ans: Resultant: The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

Equilibrant: Any concurrent set of forces, not in equilibrium, can be put into a state of equilibrium by a single force. This force is called the Equilibrant. It is equal in magnitude, opposite in sense and co-linear with the resultant. When this force is added to the force system, the sum of all of the forces is equal to zero.

7)      Explain the principle of Transmissibility?

Ans: The principle of transmissibility states “the point of application of a force can be transmitted anywhere along the line of action, but within the body.”

The fig 3 a shows a force F acting at a point of application A and fig 3 b, the same force F acts along the same line of action but at a different point of action at B and both are equivalent to each other.

Steam Turbine and Power Plants

Steam turbine

A steam turbine is a device that extracts thermal energy from pressurized steam and uses it to do mechanical work on a rotating output shaft. Its modern manifestation was invented by Sir Charles Parsons in 1884.

Because the turbine generates rotary motion, it is particularly suited to be used to drive an electrical generator – about 90% of all electricity generation in the United States (1996) is by use of steam turbines. The steam turbine is a form of heat engine that derives much of its improvement in thermodynamic efficiency through the use of multiple stages in the expansion of the steam, which results in a closer approach to the ideal reversible process.

Back Pressure Steam Turbine

Steam turbines are the prime movers in generating electricity. Back pressure steam turbines are a type of steam turbine that is used in connection with industrial processes where there is a need for low or medium pressure steam.

The high pressure steam enters the back pressure steam turbine and while the steam expands – part of its thermal energy is converted into mechanical energy. The mechanical energy is used to run an electric generator or mechanical equipment, such as pumps, fans, compressors etc.

The outlet steam leaves the back pressure steam turbine at “overpressure” and then the steam returns to the plant for process steam application such as heating or drying purposes.

Steam Turbine Power Plants:

Steam turbine power plants operate on a Rankine cycle. The steam is created by a boiler, where pure water passes through a series of tubes to capture heat from the firebox and then boils under high pressure to become superheated steam. The heat in the firebox is normally provided by burning fossil fuel (e.g. coal, fuel oil or natural gas). However, the heat can also be provided by biomass, solar energy or nuclear fuel. The superheated steam leaving the boiler then enters the steam turbine throttle, where it powers the turbine and connected generator to make electricity. After the steam expands through the turbine, it exits the back end of the turbine, where it is cooled and condensed back to water in the surface condenser. This condensate is then returned to the boiler through high-pressure feedpumps for reuse. Heat from the condensing steam is normally rejected from the condenser to a body of water, such as a river or cooling tower.

Steam turbine plants generally have a history of achieving up to 95% availability and can operate for more than a year between shutdowns for maintenance and inspections. Their unplanned or forced outage rates are typically less than 2% or less than one week per year.

Modern large steam turbine plants (over 500 MW) have efficiencies approaching 40-45%. These plants have installed costs between $800 and$2000/kW, depending on environmental permitting requirements.

Combustion (Gas) Turbines:

Combustion turbine plants operate on the Brayton cycle. They use a compressor to compress the inlet air upstream of a combustion chamber. Then the fuel is introduced and ignited to produce a high temperature, high-pressure gas that enters and expands through the turbine section. The turbine section powers both the generator and compressor. Combustion turbines are also able to burn a wide range of liquid and gaseous fuels from crude oil to natural gas.

The combustion turbine’s energy conversion typically ranges between 25% to 35% efficiency as a simple cycle. The simple cycle efficiency can be increased by installing a recuperator or waste heat boiler onto the turbine’s exhaust. A recuperator captures waste heat in the turbine exhaust stream to preheat the compressor discharge air before it enters the combustion chamber. A waste heat boiler generates steam by capturing heat form the turbine exhaust. These boilers are known as heat recovery steam generators (HRSG). They can provide steam for heating or industrial processes, which is called cogeneration. High-pressure steam from these boilers can also generate power with steam turbines, which is called a combined cycle (steam and combustion turbine operation). Recuperators and HRSGs can increase the combustion turbine’s overall energy cycle efficiency up to 80%.



Combustion (natural gas) turbine development increased in the 1930’s as a means of jet aircraft propulsion. In the early 1980’s, the efficiency and reliability of gas turbines had progressed sufficiently to be widely adopted for stationary power applications. Gas turbines range in size from 30 kW (micro-turbines) to 250 MW (industrial frames). Industrial gas turbines have efficiencies approaching 40% and 60% for simple and combined cycles respectively.

The gas turbine share of the world power generation market has climbed from 20 % to 40 % of capacity additions over the past 20 years with this technology seeing increased use for base load power generation. Much of this growth can be accredited to large (>500 MW) combined cycle power plants that exhibit low capital cost (less than $550/kW) and high thermal efficiency.

STRESS, STRAIN AND YOUNG'S MODULUS

STRESS:

When a material is subjected to an external force, it will either totally comply with that force and be pushed away, like a liquid or powder, or it will set up internal forces to oppose those applied from outside. Solid materials generally act rather like a spring – when stretched or compressed, the internal forces come into play, as is easily seen when the spring is released.

A material subjected to external forces that tend to stretch it is said to be in tension, whereas forces which squeeze the material put it in compression.

An important aspect is not so much the size of the force, as how much force is applied per unit of cross-sectional area. The term ‘stress’, symbol σ (Greek letter sigma), is used for the force per unit area, and has the units of pascals (Pa) with 1Pa being one newton per square metre.

Because the reference area is so large, it is normally necessary to use high multiples such as the megapascal (MPa = 10⁶ Pa) and gigapascal (GPa = 10⁹ Pa). However, when we bear in mind that, in electronics, the area over which forces are applied is generally very much smaller, it is useful to keep in mind that one MPa is equivalent to a force of 1 newton applied on a square millimetre of area.



STRAIN:

A material in tension or compression changes in length, and the change in length compared to the original length is referred to as the ‘strain’, symbol ε (Greek letter epsilon). Since strain is a ratio of two lengths it has no units and is frequently expressed as a percentage: a strain of 0.005 corresponds to a ½% change of the original length.



HOOKE'S LAW:

As you know from a spring, if you gradually stretch it, the force needed increases, but the material springs back to its original shape when the force is released. Materials which react in the same way as a spring are said to be ‘elastic’. Typically if we measure the extension of different forces and plot the graph of this, we will find that the extension is proportional to the force applied. Materials that obey Hooke’s Law exhibit a linear relationship between the strain and the applied stress (Figure 1).

Figure 1: Stress-strain graph for an elastic solid

Many metals follow Hooke’s Law until a certain level of stress has been applied, after which the material will distort more severely. The point at which straight line behaviour ceases is called the limit of proportionality: beyond this the material will not spring back to its original shape, and is said to exhibit some plastic behaviour (Figure 2). The stress at which the material starts to exhibit permanent deformation is called the elastic limit or yield point.

Figure 2: Stress-strain graph for a typical metal

 
As Figure 2 shows, if the stress is increased beyond the yield point the sample will eventually break. The term (ultimate) tensile strength is used for the maximum value of tensile stress that a material can withstand without breaking, and is calculated at the maximum tensile force divided by the original cross-sectional area.

Note that there may be substantial differences between the stress at the yield point and on breaking – for example, one source quotes the ‘ultimate tensile strength’ for AISI304 stainless steel as 505 MPa, and the ‘yield tensile strength’ as 215 MPa. For most engineering purposes, metals are regarded as having failed once they have yielded, and are normally loaded at well below the yield point.

With some materials, including mild steel, the stress/strain graph shows a noticeable dip beyond the elastic limit, where the strain (the effect of the load) increases without any need to increase the load. The material is said to have ‘yielded’, and the point at which this occurs is the yield point. Materials such as aluminium alloys on the other hand don’t show a noticeable yield point, and it is usual to specify a ‘proof’ test. As shown in Figure 3, the 0.2% proof strength is obtained by drawing a line parallel to the straight line part of the graph, but starting at a strain of 0.2%.

Figure 3: Stress-strain graph for an aluminium alloy

YOUNG'S MODULUS:

As you will appreciate from the shapes of Figure 2 and Figure 3, the slope of the stress/strain graph varies with stress, so we generally take only the slope of the initial straight-line portion. The stress/strain ratio is referred to as the modulus of elasticity or Young’s Modulus. The units are those of stress, since strain has no units. Engineering materials frequently have a modulus of the order of 10⁹Pa, which is usually expressed as GPa. 

Introduction To the Combustion of Fuels

Combustion:

Principle of Combustion:

Combustion is the conversion of a substance called a fuel into chemical compounds
known as products of combustion by combination with an oxidizer. The combustion
process is an exothermic chemical reaction, i.e., a reaction that releases energy as it
occurs.

Thus combustion may be represented symbolically by:
Fuel + Oxidizer = Products of combustion + Energy

Here the fuel and the oxidizer are reactants, i.e., the substances present before the
reaction takes place. This relation indicates that the reactants produce combustion
products and energy. Either the chemical energy released is transferred to the
surroundings as it is produced, or it remains in the combustion products in the form of
elevated internal energy (temperature), or some combination thereof.

Fuels are evaluated, in part, based on the amount of energy or heat that they
release per unit mass or per mole during combustion of the fuel. Such a quantity is
known as the fuel’s heat of reaction or heating value.

Heats of reaction may be measured in a calorimeter, a device in which chemical
energy release is determined by transferring the released heat to a surrounding fluid.
The amount of heat transferred to the fluid in returning the products of combustion to
their initial temperature yields the heat of reaction.

In combustion processes the oxidizer is usually air but could be pure oxygen, an
oxygen mixture, or a substance involving some other oxidizing element such as
fluorine. Here we will limit our attention to combustion of a fuel with air or pure
oxygen.

Chemical fuels exist in gaseous, liquid, or solid form. Natural gas, gasoline, and
coal, perhaps the most widely used examples of these three forms, are each a complex
mixture of reacting and inert compounds. We will consider each more closely later in
the chapter. First let’s review some important fundamentals of mixtures of gases, such
as those involved in combustion reactions.

Therefore, combustion refers to the rapid oxidation of fuel accompanied by the production of heat, or heat and light. Complete combustion of a fuel is possible only in the presence of an adequate supply of oxygen.

Oxygen (O₂) is one of the most common elements on earth making up 20.9% of our air. Rapid fuel oxidation results in large amounts of heat. Solid or liquid fuels must be changed to a gas before they will burn. Usually heat is required to change liquids or solids into gases. Fuel gases will burn in their normal state if enough air is present.

Most of the 79% of air (that is not oxygen) is nitrogen, with traces of other elements. Nitrogen is considered to be a temperature reducing dilutant that must be present to obtain the oxygen required for combustion.

Nitrogen reduces combustion efficiency by absorbing heat from the combustion of fuels and diluting the flue gases. This reduces the heat available for transfer through the heat exchange surfaces. It also increases the volume of combustion by-products, which then have to travel through the heat exchanger and up the stack faster to allow the introduction of additional fuel air mixture.

This nitrogen also can combine with oxygen (particularly at high flame temperatures) to produce oxides of nitrogen (NO_x), which are toxic pollutants.

Carbon, hydrogen and sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and sulphur dioxide, releasing 8084 kcals, 28922 kcals & 2224 kcals of heat respectively.

Under certain conditions, Carbon may also combine with Oxygen to form Carbon Monoxide, which results in the release of a smaller quantity of heat (2430 kcals/kg of carbon) Carbon burned to CO2 will produce more heat per pound of fuel than when CO or smoke are produced.

C + O₂ → CO₂ + 8084 kCals/kg of Carbon

2C + O₂ → 2 CO + 2430 kCals/kg of Carbon

2H₂ + O₂ → 2H₂O + 28,922 kCals/kg of Hydrogen

S + O₂ → SO₂ + 2,224 kCals/kg of Sulphur

3 T’s of Combustion:

The objective of good combustion is to release all of the heat in the fuel. This is accomplished by controlling the "three T's" of combustion which are

1. Temperature high enough to ignite and maintain ignition of the fuel,
2. Turbulence or intimate mixing of the fuel and oxygen, and
3. Time sufficient for complete combustion.

Commonly used fuels like natural gas and propane generally consist of carbon and hydrogen. Water vapor is a by-product of burning hydrogen. This robs heat from the flue gases, which would otherwise be available for more heat transfer.

Natural gas contains more hydrogen and less carbon per kg than fuel oils and as such produces more water vapor. Consequently, more heat will be carried away by exhaust while firing natural gas.

Too much, or too little fuel with the available combustion air may potentially result in unburned fuel and carbon monoxide generation. A very specific amount of O2 is needed for perfect combustion and some additional (excess) air is required for ensuring complete combustion. However, too much excess air will result in heat and efficiency losses.

Not all of the heat in the fuel are converted to heat and absorbed by the steam generation equipment. Usually all of the hydrogen in the fuel is burned and most boiler fuels, allowable with today's air pollution standards, contain little or no sulfur. So the main challenge in combustion efficiency is directed toward unburned carbon (in the ash or incompletely burned gas), which forms CO instead of CO₂.

                                                                                                                             Subhankar Karmakar

MOCK QUESTION PAPER: APPLIED THERMODYNAMICS (2 units only)

                                                                   Paper Code: EME-402

B.  Tech - ME

(SEM.IV) Sessional Examination, 2011 – 12

Applied Thermodynamics

Time:   3hrs                        Total Marks:  100

    Note:   (1)           Attempt all questions.

         (2)  Be precise in your answer.

SECTION-A:

Q.1: Answer the following questions as per the instructions.           

2X10=30

 (i) What is the importance of feed pump in steam engine?

(ii) What is reversible adiabatic process?

(iii) Explain the term isothermal compressibility?

(iv) What is missing quantity?

(v) What is Work Ratio in Carnot vapour cycle?

(vi) Explain the term “Specific steam consumption.”

(vii) What is thermal efficiency of a steam engine?

(viii) What is indicated power?

(ix) What is mean effective pressure of a steam engine?

(x) What is inversion temperature?

SECTION-B:

Q.2: Answer any three parts of the followings:     

                                                                                                               3X10=30

a) Derive the Tds equations.

b) Derive the expressions of mass discharge of steam through a Nozzle.

c) A single cylinder double acting steam engine is supplied with dry and saturated steam at 11.5 bar and exhaust occur at 1.1 bar. The cut-off occurs at 40% of the stroke. If the stroke equals 1.25 times the cylinder bore and engine develops 60 kW at 90 rev/min. Determine the bore and the stroke of the engine. (Assume hyperbolic expansion and diagram factor of 0.79.)

Also calculate the theoretical steam consumption

d) Dry saturated steam enters a steam nozzle at a pressure of 12 bar and is discharged at a pressure of 1.5 bar. If the dryness factor of the discharged steam is 0.95, what would be the final velocity of the steam? Neglect initial velocity of steam.

If 12% heat drop is lost in friction, find the % reduction in the final velocity.

SECTION C:
Q.3: Answer any two parts of the following: 

                                                                                         5X2=10

a) Explain the term “Joule-Thomson coefficient.”

b) With proper diagrams explain the term nozzle efficiency.

c) Explain the Clausius Clapeyron equation. Also write their field of application.

Q.4: Answer any one part of the following:   

                                                                                           1X10=10

a) Explain the effect of velocity and pressure in the flow of a nozzle. What is a choked flow? Also explain the concept of critical pressure in isentropic flow through nozzle.

b) Steam at a pressure of 20 bar, 250°C expands in a convergent-divergent nozzle up to the exit pressure of 2 bar. Assuming a nozzle efficiency of 0.94 for supersaturated flow up to the throat and nozzle efficiency as 90%, find (i) velocity at throat, (ii) mass flow rate if the throat diameter is 1 cm and (iii) velocity and diameter of the nozzle.

Q.5: Answer any three questions: 

                                                                         3X10=30

a) Derive the Maxwell’s Equations

b) Prove that C_p - C_V = -T(∂V/∂T)_p²(∂p/∂V)_T.

c) Steam at a pressure of 10 bar, dry saturated enters the nozzle when exit pressure is 0.3 bar. The nozzle efficiency for the convergent position is 96% and that of the divergent portion is 92%. The throat diameter for each nozzle is 6 mm. Find the mass flow rate of steam and the exit diameter required.

d) Air enters a nozzle at 5 bar, 350°C and comes out at 0.95 bar. The efficiency of expansion through the nozzle is 92%. If the mass flow rate of air is 1.5 kg/s, determine the exit diameter of the nozzle and velocity of air at exit.