Mechanical Engineering Lecture Notes: IC Engines and Compressors

• IC ENGINES AND COMPRESSORS:

Internal Combustion Engines are the basis of human transportation and without them human civilization will come to halt. The main characteristics of an IC engines is its combustion chamber where fuel is burnt at elevated temperature and pressure to develop rotational kinetic energy.

IC engines, or internal combustion engines, are a type of heat engine that convert the heat generated by the combustion of fuel into mechanical energy. This mechanical energy is used to power a vehicle or machine. IC engines can be categorized into two types: spark ignition engines and compression ignition engines. In spark ignition engines, a spark is used to ignite the fuel-air mixture, while in compression ignition engines, the fuel-air mixture is compressed until it spontaneously ignites.

Compressors, on the other hand, are mechanical devices that are used to increase the pressure of a gas. They are commonly used in a variety of applications, such as in air conditioning and refrigeration systems, as well as in the compression of natural gas and other gases for transport or storage. There are several types of compressors, including positive displacement compressors and dynamic compressors. Positive displacement compressors work by trapping a fixed amount of gas and then compressing it, while dynamic compressors use a rotating impeller to increase the velocity of the gas and then convert that velocity into pressure.

While IC engines and compressors have some similarities in terms of their use of internal combustion, they are fundamentally different types of devices with different applications and operating principles.

Compressors are mechanical devices that are used to increase the pressure of a gas. There are several types of compressors, each with its own unique characteristics and applications. Here are some of the most common types of compressors and their functions:

• Types of Compressors and Functions:

Reciprocating compressors: This type of compressor uses a piston and cylinder to compress the gas. The piston moves back and forth within the cylinder, creating a vacuum that draws in the gas, which is then compressed as the piston moves back up.

Rotary screw compressors: These compressors use two interlocking screws to compress the gas. The screws rotate in opposite directions, drawing in the gas and compressing it as it moves through the screw threads.

Centrifugal compressors: Centrifugal compressors use a rotating impeller to increase the velocity of the gas, which is then converted into pressure by a diffuser. This type of compressor is commonly used in high-flow, low-pressure applications.

Axial compressors: Axial compressors use a series of rotating blades to compress the gas. The blades are arranged in a row, with each row of blades increasing the pressure of the gas as it moves through the compressor.

Scroll compressors: Scroll compressors use two interleaved spirals to compress the gas. One spiral remains stationary while the other moves in a circular motion, trapping and compressing the gas as it moves through the spirals.

The function of a compressor is to increase the pressure of a gas, which can then be used for a variety of purposes, such as powering tools and machinery, refrigeration, air conditioning, and gas transport and storage. Compressors are used in a wide range of industries, including manufacturing, healthcare, energy, and transportation.

• Reciprocating compressors: and its functions:

Reciprocating compressors are a type of positive displacement compressor that use a piston and cylinder to compress a gas. The piston moves back and forth within the cylinder, creating a vacuum that draws in the gas, which is then compressed as the piston moves back up. Reciprocating compressors are commonly used in applications where a high-pressure output is required, such as in natural gas processing, petroleum refining, and chemical processing.

The functions of reciprocating compressors include:

Compression of gas: The primary function of a reciprocating compressor is to compress a gas to a higher pressure, which can then be used for various industrial processes.

Gas transport: Reciprocating compressors can be used to transport gases through pipelines or other systems. The compressed gas can be moved over long distances without losing pressure, making it an efficient method of transport.

Storage: Compressed gas can be stored in tanks or other vessels for later use, and reciprocating compressors can be used to fill these storage containers.

Power generation: Reciprocating compressors can be used to generate power by compressing gas and then using that compressed gas to power a turbine or other type of engine.

Refrigeration: Reciprocating compressors can be used in refrigeration systems to compress refrigerant gases and remove heat from a space or product.

Overall, reciprocating compressors are a versatile type of compressor that can be used for a wide range of industrial applications where high-pressure gas output is required.

• Components of a reciprocating compressors:

A reciprocating compressor is a complex machine made up of several components that work together to compress a gas. The main components of a reciprocating compressor include:

Cylinder: The cylinder is the main body of the compressor where the piston moves back and forth to compress the gas.

Piston: The piston is a cylindrical component that moves back and forth within the cylinder to compress the gas. The piston is typically made of a metal alloy and is attached to a connecting rod.

Connecting rod: The connecting rod connects the piston to the crankshaft and converts the linear motion of the piston into rotational motion of the crankshaft.

Crankshaft: The crankshaft is a shaft that rotates to convert the linear motion of the connecting rod into rotational motion. The crankshaft is typically driven by an electric motor or a combustion engine.

Valves: The compressor has two sets of valves, intake and discharge, that control the flow of gas into and out of the cylinder. The intake valve opens to allow gas to enter the cylinder, and the discharge valve opens to allow compressed gas to exit the cylinder.

Crankcase: The crankcase is a compartment in the compressor that houses the crankshaft and connecting rods. It is typically filled with oil to lubricate the moving parts and reduce wear and tear.

Pressure relief valve: The pressure relief valve is a safety feature that opens when the pressure in the compressor exceeds a certain threshold. This prevents the compressor from being damaged or exploding due to excess pressure.

Overall, each component of a reciprocating compressor plays a critical role in the compressing process and must be designed and maintained to ensure reliable and efficient operation.

• Efficiency of a reciprocating compressor:

The efficiency of a reciprocating compressor is a measure of how effectively it can compress gas while consuming the least amount of energy. The efficiency of a reciprocating compressor can be affected by several factors, including the compressor design, operating conditions, and maintenance practices.

The following are some of the factors that can affect the efficiency of a reciprocating compressor:

Compression ratio: The compression ratio is the ratio of the discharge pressure to the suction pressure. The higher the compression ratio, the lower the compressor efficiency.

Clearance volume: The clearance volume is the volume of gas remaining in the cylinder after the piston has reached the end of its stroke. A larger clearance volume can decrease the efficiency of the compressor.

Gas properties: The physical properties of the gas being compressed, such as its molecular weight, specific heat ratio, and heat capacity, can affect the efficiency of the compressor.

Operating conditions: The operating conditions of the compressor, such as the suction and discharge pressures and temperatures, can also affect its efficiency.

Maintenance: Proper maintenance practices, such as keeping the compressor clean and lubricated, can help maintain its efficiency over time.

To improve the efficiency of a reciprocating compressor, it is important to properly size and design the compressor for the specific application and operating conditions. Proper maintenance practices, such as regular cleaning, lubrication, and inspection, can also help to maintain the efficiency of the compressor over time.

SECOND SESSIONAL TEST (odd SEMESTER 2009-10) B.Tech…first Semester Sub Name: Engineering Mechanics

SECOND SESSIONAL TEST (odd SEMESTER 2009-10)

B.Tech…first Semester

Sub Name: Engineering Mechanics                                         Max. Marks: 30

Sub Code: EME-201                                                          Max. Time: 2: 00 Hr

Group A

Q.1 Choose the correct answer of the following questions 1x6=6

(i) If two forces of equal magnitudes have a resultant force of the same magnitude then the angle between them is

(a) 0⁰          (b) 90⁰         (c) 120⁰          (d) 135⁰

Ans: ( c)

Explanation: P² = P² + P² +2P.P.cos θ

ð      P²= P² (1+2cos θ)

ð      cos θ = -½

ð      θ = 120°

(ii) If a ladder is kept at rest on a vertical wall making an angle θ with horizontal. If co-efficient of the friction in all the surfaces be µ, then the tangent of the angle θ will be equal to

(a) (1-µ²)/2µ                                  (b) (1-µ)² /2µ

(c) (1-µ)/2µ                                   (d) none of the above

Explanation: (a) (1-µ²)/2µ

(iv) Varignon’s theorem is related with _____________ .

Answer: moment

(v) If two forces of equal magnitudes P having an angle (90°- θ) between them, then their resultant force will be equal to ________.

Answer: √2P (1+sin θ)

(vi) A fixed joint produces

(a) 1             (b) 2             (c) 3               (d) 4         reactions

Answer: (c ) 3

(vii) The equilibrium conditions of concurrent force system is_________.

Answer: ∑Fx =0; ∑Fy=0.

Group B                                                                                                                  8x3=24

Attempt any three questions

Q.2. State and explain Varignon’s theorem of moment. Three forces of magnitudes 3 KN, 4 KN and 2KN act along the three side of an equilateral triangle ∆ABC in order. Find the position, direction and magnitude of the resultant force.         4+4

Answer: resultant force: √3 KN

Q.3  (a) Two cylindrical rollers are kept at equilibrium inside a jar or channel as shown in the figure. The channel width is 1000 mm, where as the rollers have diameters 600 mm and 800 mm respectively. The weights are 2 kN and 5 kN respectively. Find all the reactions at contacts.

                                                                                            4+4

(b) What is pure bending? If a stone is thrown with a velocity 400 m/s then find the maximum height that the stone can reach.

Q:4)  a) Classify different types of joints in beam with proper explanations.

(b) Find the reactions at the support for the beam as shown in the figure.                                           4+4

ENGINEERING. MECHANICS:

FORCE AND FORCE SYSTEM

Topic: FORCE SYSTEM

1) What is a FORCE SYSTEM? Classify them with examples and diagrams.

Ans: A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces.

Different types of force system:


(i) COPLANAR FORCES:

If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.


(ii) CONCURRENT FORCES:

A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.

(iii) LIKE FORCES:

A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.

(iv) UNLIKE FORCES: If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".


(v) NON COPLANAR FORCES:
The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

_________________________________________________________________________________
N.B. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. One can progressively resolve pairs or small groups of forces into resultants. Then another resultant of the resultants can be found and so on until all of the forces have been combined into one force. This is one way to save time with the tedious "bookkeeping" involved with a large number of individual forces. Resultants can be determined both graphically and algebraically.The Parallelogram Method and the Triangle Method. It is important to note that for any given system of forces, there is only one resultant.


It is often convenient to decompose a single force into two distinct forces. These forces, when acting together, have the same external effect on a body as the original force. They are known as components. Finding the components of a force can be viewed as the converse of finding a resultant. There are an infinite number of components to any single force. And, the correct choice of the pair to represent a force depends upon the most convenient geometry. For simplicity, the most convenient is often the coordinate axis of a structure.


A force can be represented as a pair of components that correspond with the X and Y axis. These are known as the rectangular components of a force. Rectangular components can be thought of as the two sides of a right angle which are at ninety degrees to each other. The resultant of these components ...


is the hypotenuse of the triangle. The rectangular components for any force can be found with trigonometrical relationships: F_x = Fcosθ, Fy = Fsinθ. There are a few geometric relationships that seem to common in general building practice in North America. These relationships relate to roof pitches, stair pitches, and common slopes or relationships between truss members. Some of these are triangles with sides of ratios of 3-4-5, 1-2-sqrt3, 1-1-sqrt2, 5-12-13 or 8-15-17. Committing the first three to memory will simplify the determination of vector magnitudes when resolving more difficult problems.


When forces are being represented as vectors, it is important to should show a clear distinction between a resultant and its components. The resultant could be shown with color or as a dashed line and the components as solid lines, or vice versa. NEVER represent the resultant in the same graphic way as its components.


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. A non-concurrent or a parallel force system can actually be in equilibrium with respect to all of the forces, but not be in equilibrium with respect to moments.
__________________________________________________________________________________


2) What is STATIC EQUILIBRIUM? 
    What are the conditions of static equilibrium for
            (i) concurrent force system
            (ii) coplanar non concurrent force system.

Ans: A body is said to be in 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 be zero. Therefore the general condition of any system to be in static equilibrium we have to satisfy two conditions

(i) Net force on the body must be zero ie, ΣF_i = 0;
(ii) Net moment on the body must be zero ie, ΣM_i = 0.

Now we can apply these general conditions to different types of Force System.

For concurrent force system total moment about the concurrent point is always zero as all the forces pass through the point, and we know the moment of a force passing through the point about which we shall take moment is always zero. Hence, the conditions of equilibrium for concurrent forces will be  

Net force on the body must be zero ie, ΣF_i = 0; and we can resolve it along X axis and along Y axis, ie.  (i) ΣF_x = 0; and  (ii) ΣF_y = 0.

for coplanar non concurrent force system, the equilibrium conditions are

(i) ΣF_x = 0; and  (ii) ΣF_y = 0.  (iii)  ΣM_i = 0.

 Moment on a plane:

For a force system the total resultant moment about any arbitrary point due to the individual forces are equal to the moment produced by the resultant about the same point. Now if the system is at equilibrium condition, then the resultant force would be zero. Hence, the moment produced by the resultant about any arbitrary point is zero. In case of coplanar & concurrent force system, as the forces are concurrent ie. each of the force passes through a common point. Hence, about that common point total moment of all the forces will be zero.

3) What are different types of joint? discuss them in details.

Answer: The Concepts of Joints. In Engineering terminology any force carrying linear member is called as links. Links can be attached to each other by the fasteners or joints. Hence, we can say to prevent the relative motion between two links completely or partially we use fasteners or joints.



Basically there are three types of joints which we shall discuss and they are named as,
(i) pin/ hinged joints, 
(ii) roller joints and 
(iii) fixed joints.


PIN JOINTS:

They are classified according to the degrees of freedom of the links they would allow. Like a pin or hinge joint is consisted of two links joined by the insertion of a pin at the pivot hole. A pin joint doesn't allow a vertical or horizontal relative velocities between the two links.

For better understanding of the mechanism of pin joint we would like to make a simplest type of pin joints. Suppose we would take two links and make holes at one of the ends of each link. Now if we insert a bolt through the holes of both the links, then what we get is an example of pin/hinge joints.

A pin joint although restricts any kind of horizontal or vertical displacement but they can not restrict rotation about an axis passing through the hole, in clockwise or anti clockwise direction. Hence it provides two reactions one vertical and one horizontal to restrict any kind of movement along that direction.

ROLLER JOINTS:


 

MULTIPLE CHOICE QUESTIONS: sub: engg. mechanics. Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech

Q.1) The example of Statically indeterminate structures are,
a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.2) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3, 

c. m > 2j - 3,
d. m > 2j + 3

Q.3) The property of a material to withstand a sudden impact or shock is called,
a. hardness 
b. ductility, 
c. toughness, 
d. elasticity of the material

Q.4) The stress generated by a dynamic loading is approximately _____ times of the stress developed by the gradually applying the same load.

Q.5) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus

Q.6) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above

Q.7) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces

Q.8) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body

Q.9) The variation of shear force due to a triangular load on simply supported beam is
a. uniform 
b. linear 
c. parabolic 
d. cubic

Q.10) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be

a. 1/√3

b. √3

c. 1

d. 0

11. A force F of 10 N is applied on a mass of 2 kg. What is the acceleration of the mass?

A. 2 m/s²

B. 5 m/s²

C. 10 m/s²

D. 20 m/s²

Answer: B

12. What is the moment of a force of 50 N applied at a distance of 2 meters from a fixed point?

A. 25 Nm

B. 50 Nm

C. 100 Nm

D. 200 Nm

Answer: C

13. A 2000 kg car traveling at 20 m/s collides with a 500 kg car traveling at 10 m/s in the opposite direction. What is the velocity of the cars after the collision?

A. 6.7 m/s

B. 10 m/s

C. 13.3 m/s

D. 16.7 m/s

Answer: A

14. A 500 N force is applied to a 100 kg object on a flat surface. What is the coefficient of static friction if the object is just about to move?

A. 0.5

B. 0.7

C. 0.8

D. 1.0

Answer: D

15. A beam of length 4 m and moment of inertia of 1000 kg/m² is supported at each end. What is the maximum load that the beam can support if it is uniformly loaded?

A. 500 N

B. 1000 N

C. 2000 N

D. 4000 N

Answer: C

16. A block of mass 2 kg is hanging from a string. What is the tension in the string if the block is stationary?

A. 19.6 N

B. 20 N

C. 29.4 N

D. 30 N

Answer: B

17. A roller coaster car of mass 500 kg is traveling at 20 m/s at the bottom of a  loop-the-loop. What is the minimum radius of the loop required for the car to remain in contact with the track?

A. 40 m

B. 50 m

C. 60 m

D. 70 m

Answer: D

18. A body of mass 10 kg is moving with a velocity of 5 m/s. What is the kinetic energy of the body?

A. 50 J

B. 100 J

C. 125 J

D. 250 J

Answer: B

19. A body of mass 5 kg is placed on an inclined plane which makes an angle of 30° with the horizontal. What is the force acting on the body parallel to the plane?

A. 4.9 N

B. 7.5 N

C. 8.7 N

D. 10 N

Answer: B

20. A force of 100 N is applied on a body of mass 20 kg. What is the work done by the force in moving the body through a distance of 5 meters?

A. 250 J

B. 500 J

C. 1000 J

D. 2000 J

Answer: B

21. What is the principle of moments?

A. The sum of the moments about any point of a system in equilibrium is zero.

B. The sum of the forces acting on a system in equilibrium is zero.

C. The sum of the torques acting on a system in equilibrium is zero.

D. The sum of the accelerations of a system in equilibrium is zero.

Answer: A

22. What is the difference between static and dynamic equilibrium?

A. In static equilibrium, there is no motion, while in dynamic equilibrium, there is motion.

B. In static equilibrium, the forces are balanced, while in dynamic equilibrium, the forces are unbalanced.

C. In static equilibrium, the sum of the forces and moments is zero, while in dynamic equilibrium, the sum of the forces and moments is not zero.

D. In static equilibrium, the sum of the forces and moments is not zero, while in dynamic equilibrium, the sum of the forces and moments is zero.

Answer: C

23. What is the moment of inertia?

A. The resistance of an object to angular acceleration.

B. The force required to rotate an object.

C. The distance between the center of mass and the axis of rotation.

D. The angular velocity of an object.

Answer: A

24.What is the difference between stress and strain?

A. Stress is the deformation per unit length, while strain is the force per unit area.

B. Stress is the force per unit area, while strain is the deformation per unit length.

C. Stress is the force applied to an object, while strain is the resulting deformation.

D. Stress is the resistance of an object to deformation, while strain is the resistance of an object to stress.

Answer: B

25. What is Hooke's Law?

A. The stress applied to an elastic material is proportional to the strain produced.

B. The strain produced in an elastic material is proportional to the stress applied.

C. The deformation produced in an elastic material is proportional to the force applied.

D. The force applied to an elastic material is proportional to the deformation produced.

Answer: A

26.What is the difference between a beam and a truss?

A. A beam is a one-dimensional structure, while a truss is a two-dimensional structure.

B. A beam is made up of several members connected at their ends, while a truss is made up of several members connected at their joints.

C. A beam is used to support loads that are perpendicular to its axis, while a truss is used to support loads that are parallel to its axis.

D. A beam is a rigid structure, while a truss is a flexible structure.

Answer: B

27. What is the difference between a force and a moment?

A. A force is a vector quantity, while a moment is a scalar quantity.

B. A force is a scalar quantity, while a moment is a vector quantity.

C. A force is a push or a pull, while a moment is a twist or a turn.

D. A force is a linear motion, while a moment is a rotational motion.

Answer: C

28. What is the center of mass?

A. The point where the weight of an object is concentrated.

B. The point where the forces acting on an object are balanced.

C. The point where the moments acting on an object are balanced.

D. The point where the acceleration of an object is zero.

Answer: A

29. What is the method used to determine the forces in a truss?

A. Method of joints

B. Method of sections

C. Both A and B

D. None of the above

Answer: C

30. In a truss, which members are in tension and which members are in compression?

A. All members are in tension.

B. All members are in compression.

C. Members with angled force vectors are in tension, and members with vertical force vectors are in compression.

D. Members with vertical force vectors are in tension, and members with angled force vectors are in compression.

Answer: C

31. What is the difference between a simple truss and a compound truss?

A. A simple truss is made up of one triangle, while a compound truss is made up of two or more triangles.

B. A simple truss is made up of straight members only, while a compound truss may have curved members.

C. A simple truss is statically determinate, while a compound truss may be statically indeterminate.

D. A simple truss is used for short spans, while a compound truss is used for long spans.

Answer: A

32.How many unknown forces are there in a simple truss?

A. 2

B. 3

C. 4

D. It depends on the number of joints in the truss.

Answer: B

33. What is the method used to analyze a truss with multiple loadings?

A. Superposition method

B. Substitution method

C. Iterative method

D. None of the above

Answer: A

34. What is the maximum number of reactions that can be present in a truss?

A. 1

B. 2

C. 3

D. 4

Answer: B

35. What is the difference between a statically determinate and a statically indeterminate truss?

A. A statically determinate truss has only one solution for the unknown forces, while a statically indeterminate truss may have more than one solution.

B. A statically determinate truss has more unknown forces than the number of equations available to solve them, while a statically indeterminate truss has fewer unknown forces than the number of equations available to solve them.

C. A statically determinate truss is easier to analyze, while a statically indeterminate truss requires more advanced techniques.

D. A statically determinate truss is always more efficient than a statically indeterminate truss.

Answer: C

36. What is the difference between a pinned support and a roller support?

A. A pinned support allows rotation but not translation, while a roller support allows translation but not rotation.

B. A pinned support allows both rotation and translation, while a roller support allows neither.

C. A pinned support is used for horizontal loads, while a roller support is used for vertical loads.

D. A pinned support is always more stable than a roller support.

Answer: A

37. What is the maximum number of members that can be present in a simple truss?

A. 2n-2, where n is the number of joints

B. 2n-3, where n is the number of joints

C. n-1, where n is the number of joints

D. n+1, where n is the number of joints

Answer: B

©subhankar_karmakar

more OBJECTIVE QUESTIONS on ENGINEERING MECHANICS 

ASSIGNMENT ON THERMODYNAMICS

Numericals on Thermodynamics:

1.     Mass enters an open system with one inlet and one exit at a constant rate of 50 kg/min. At the exit, the mass flow rate is 60 kg/min. If the system initially contains 1000 kg of working fluid, determine the time when the system mass becomes 500 kg.

2.     Mass leaves an open system with a mass flow rate of c*m, where c is a constant and m is the system mass. If the mass of the system at t = 0 is m₀, derive an expression for the mass of the system at time t.

3.     Water enters a vertical cylindrical tank of cross-sectional area 0.01 m² at a constant mass flow rate of 5 kg/s. It leaves the tank through an exit near the base with a mass flow rate given by the formula 0.2h kg/s, where h is the instantaneous height in m. If the tank is empty initially, develop an expression for the liquid height h as a function of time t. Assume density of water to remain constant at 1000 kg/m³.

4.     A conical tank of base diameter D and height H is suspended in an inverted position to hold water. A leak at the apex of the cone causes water to leave with a mass flow rate of c*sqrt(h), where c is a constant and h is the height of the water level from the leak at the bottom. (a) Determine the rate of change of height h. (b) Express h as a function of time t and other known constants, rho (constant density of water), D, H, and c if the tank was completely full at t=0.

5.     Steam enters a mixing chamber at 100 kPa, 20 m/s, with a specific volume of 0.4 m³/kg. Liquid water at 100 kPa and 25^oC enters the chamber through a separate duct with a flow rate of 50 kg/s and a velocity of 5 m/s. If liquid water leaves the chamber at 100 kPa and 43oC with a volumetric flow rate of 3.357 m³/min and a velocity of 5.58 m/s, determine the port areas at the inlets and exit. Assume liquid water density to be 1000 kg/m³ and steady state operation.

6.     Air is pumped into and withdrawn from a 10 m³ rigid tank as shown in the accompanying figure. The inlet and exit conditions are as follows. Inlet: v₁= 2 m³/kg, V₁= 10 m/s, A₁= 0.01 m²; Exit: v₂= 5 m³/kg, V₂= 5m/s, A₂= 0.015 m². Assuming the tank to be uniform at all time with the specific volume and pressure related through p*v=9.0 (kPa.m³), determine the rate of change of pressure in the tank.

7.     A gas flows steadily through a circular duct of varying cross-section area with a mass flow rate of 10 kg/s. The inlet and exit conditions are as follows. Inlet: V₁= 400 m/s, A₁= 179.36 cm²; Exit: V₂= 584 m/s, v₂= 1.1827 m/kg. (a) Determine the exit area. (b) Do you find the increase in velocity of the gas accompanied by an increase in flow area counter intuitive? Why?

8.     Steam enters a turbine with a mass flow rate of 10 kg/s at 10 MPa, 600^oC, 30 m/s, it exits the turbine at 45 kPa, 30 m/s with a quality of 0.9. Assuming steady-state operation, determine (a) the inlet area, and (b) the exit area. 

Answers: (a) 0.01279 m² (b) 1.075 m²