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## Mechanical Engineering

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Mechanical engineering combines engineering physics and mathematics principles with materials science to design, analyze, manufacture, and maintain mechanical systems. The mechanical engineering field requires an understanding of core areas including mechanics, dynamics, thermodynamics, materials science, structural analysis, and electricity.

Thumbnail: Archimedes' screw was operated by hand and could efficiently raise water, as the animated red ball demonstrates. (CC BY-SA 2.5; Silberwolf via Wikipedia )

## Mechanical Engineering Assignment Help | Mechanical Engineering Homework Help

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Popular subjects for mechanical engineering homework help.

• Automobile Engineering
• Biomechanics
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## Recently Asked Mechanical Engineering Questions

• Q 1 : 4.9 As shown in Fig. P4.9, river water used to irrigate a field is controlled by a gate. When the gate is raised, water flows steadily with a velocity of 75 ft/s through an opening 8 ft by 3 ft. If the gate is raised for 24 hours, determine the volume of water, in gallons, provided for irrigation. Assume the density of river water is 62.3 lb/ft³. See Answer
• Q 2 : 4.13 WP Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -4°C and quality of 20% at a velocity of 7 m/s. At the exit, the refrigerant is a saturated vapor at a temperature of -4°C. The evaporator flow channel has constant diameter. If the mass flow rate of the entering refrigerant is 0.1 kg/s, determine a. the diameter of the evaporator flow channel, in cm. b. the velocity at the exit, in m/s. See Answer
• Q 3 : The aim of this reading assignment is to give you the opportunity to learn about net-zero energy buildings (NZEBs). NZEBS are buildings that produce (at least) as much energy as they con- sume over a one year period. NZEBs integrate various forms of on-site renewable energy gener- ation (e.g., solar, geothermal) with energy efficient building system components (e.g., heat- ing/cooling, lighting, envelopes, electrical appliances). NZEBS are the future of the building sec- tor and are becoming more common. The papers listed below are published in the journal of the American Society of Heating, Refrigerating, and Air Conditioning Engineers (ASHRAE) - the leading international organization for research on high performance buildings. Each paper is a case study on an actual NZEB. PDF copies of each paper are available in Canvas. See Answer
• Q 4 : An inventor claims to have developed a heat engine that receives 700 k J of heat from a source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim? Why? See Answer
• Q 5 : Process 2-3: Expansion with pV = constant, U3 = U₂. See Answer
• Q 6 : 1-18 A 4-kW resistance heater in a water heater runs for3 hours to raise the water temperature to the desired level.Determine the amount of electric energy used in both kW hand kJ. See Answer
• Q 7 : 4-20 A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle. See Answer
• Q 8 : 13 Water is the working fluid in a Rankine cycle. Steam enters the turbine at 1400 lbf/in² and 1000°F. The condenser pressure is 2 lbf/in? Both the turbine and pump have isentropic efficiencies of 85%. The working fluid has negligible pressure drop in passing through the steam generator. The net power output of the cycle is 1x10° Btu/h. Cooling water experiences a temperature increase from 60°F to 76°F, with negligible pressure drop, as it passes through the condenser. Determine for the cycle o the mass flow rate of steam, in lb/h. the rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator. ) the thermal efficiency. the mass flow rate of cooling water, in lb/h. See Answer
• Q 9 : 1-23E A man goes to a traditional market to buy a steak for dinner. He finds a 12-oz steak (1 lbm = 16 oz) for \$5.50.He then goes to the adjacent international market and finds a300-g steak of identical quality for \$5.20. Which steak is the better buy? See Answer
• Q 10 : 1-22 The weight of bodies may change somewhat from one location to another as a result of the variation of the gravitational acceleration g with elevation. Accounting for this variation using the relation in Prob. 1-12, determine the weight of an 80-kg person at sea level (z = 0), in Denver (z 1610 m),and on the top of Mount Everest (z = 8848 m).!! See Answer
• Q 11 : 1-21 A pool of volume V (in m') is to be filled with water using a hose of diameter D (in m). If the average discharge velocity is V (in m/s) and the filling time is (in s), obtain a relation for the volume of the pool based on considerations of quantities involved. See Answer
• Q 12 : The rocket is in free flight along an elliptical trajectory A – A'. The planet has no atmosphere,and its mass is 0.70 times that of the earth. If the rocket has an apogee (apoapsis) and perigee(periapsis) as shown in the figure, determine the speed of the rocket when it is at these two extreme points (in km/sec). Also calculate the time required to complete one revolution (in hour). See Answer
• Q 13 : NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.Both a gauge and a manometer are attached to a gas tank to measure its pressure. The reading on the pressure gauge is 80kPa. Determine the distance between the two fluid levels of the manometer if the fluid is mercury (p = 13,600 kg/m). The distance between the two fluid levels ism. See Answer
• Q 14 : NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.Both a gage and a manometer are attached to a gas tank to measure its pressure. The reading on the pressure gage is 80kPa. Determine the distance between the two fluid levels of the manometer if the fluid is water (p = 1000 kg/m). m.The distance between the two fluid levels is See Answer
• Q 15 : The absolute pressure is Determine the absolute pressure at the same depth in a different liquid whose specific gravity is 0.85 and that is at the same localatmospheric pressure found in part (a). Determine the absolute pressure at the same depth in a different liquid whose specific gravity is 0.85 and that is at the same localatmospheric pressure found in part (a). See Answer
• Q 16 : An inventor claims that at steady state the device shown in the figure below develops power from entering and exiting streams of water at a rate of 1638 kW. The accompanying table provides data for inlet 1 and 2 and exit 4. The pressure at exit 3 is 3 bar. Stray heat transfer and kinetic and potential energy effects are negligible. Evaluate the inventor's claim.(Hint: Write the first law first to find the unknown quantities) See Answer
• Q 17 : Determine the local atmospheric pressure. (Please provide an answer before moving on to the next part.) The local atmospheric pressure is------kPa. See Answer
• Q 18 : A house is losing heat at a rate of 1500 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Express the rate of heat loss from this house per K, 'F, and R difference between the indoor and the outdoor temperature. The rate of heat loss from this house per K difference isKJ/h. The rate of heat loss from this house per 'F difference is1kJ/h. The rate of heat loss from this house per R difference is kJ/h. See Answer

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Mechanical engineering: evolution and disciplines.

Considered the most intricate branch of engineering, mechanical engineering plays a pivotal role across industries, from designing massive machines to intricate biotechnological mechanisms. Early applications of mechanics trace back to ancient Greece and China, with notable contributions from European and Arabic scientists during the medieval ages.

The industrial revolution in Europe propelled the study of mechanics, becoming a cornerstone supporting millions of livelihoods. Mechanical engineering spans various sub-disciplines, including mechanics, statics, dynamics, fluid mechanics, and more.

## Sub-disciplines Covered in Mechanical Engineering Assignment Help:

Mechanical engineering encompasses a broad spectrum of sub-disciplines, including mechanics, statics, dynamics, mechanics of materials, fluid mechanics, kinematics, continuum mechanics, and fundamental principles such as mechanical electronics.

## Principles and Applications:

Mechanical electronics, often known as Mechatronics, explores the integration of mechanics and electronics. It influences various sectors, exemplified by applications in CD-ROM drives, robotics, and space exploration.

Structural analysis delves into the performance and failure analysis of objects and machinery, examining static and fatigue failures. Thermodynamics, a fundamental aspect, studies the principles and applications of energy on machinery, influencing power plants, engines, and various systems.

## Design and Drafting:

Design and drafting, crucial components of mechanical engineering, involve precise technical illustrations guiding machinery and product design. Skilled individuals in technical drawing, known as draftsmen, play a pivotal role in this field.

## Mechanical Engineering Coursework:

Students pursuing mechanical engineering undergo coursework covering mathematics, instrumentation and design, manufacturing engineering, statics and dynamics, thermodynamics, strength of materials, and more. This coursework is consistent globally, providing a foundation for practical and analytical thinking.

## Benefits and Promising Future:

Mechanical engineering offers promising career prospects, with 15 percent of engineers employed in the USA being mechanical engineers in 2009. The field's growth was anticipated at 6 percent annually. Mechanical engineering appeals to those seeking innovation, provides diverse career options, lucrative salaries, and serves as a core engineering discipline, ensuring versatility for future pursuits.

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Less academic interest in students: One significant drawback of the current education system is its extensive coursework, leaving minimal time for in-depth learning, especially in advanced courses like Mechanical Engineering. The limited time hinders teachers from devoting sufficient attention to each topic, leading students to believe that memorization is the key to mastering the subject. However, in advanced courses like Mechanical Engineering, understanding real-life implications is crucial, making mere memorization ineffective. The lack of guidance further hampers students' interest in subjects, contributing to the challenge of diminished academic enthusiasm, emphasizing the need for specialized support such as Mechanical Engineering Homework Help.

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## What Is Mechanical Assignment Help?

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## How Long Does It Take To Get Mechanical Help From Experts?

The number of days for mechanical engineering homework completion is not fixed, though we have a moderate time chart. We try to submit homework on or before the deadline. However, the time limit to complete homework depends on the difficulty level and the proximity of the deadline.

## Get Answers In Few Hours

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## Mechanical Engineering

Homework help & tutoring.

Our name 24HourAnswers means you can submit work 24 hours a day - it doesn't mean we can help you master what you need to know in 24 hours. If you make arrangements in advance, and if you are a very fast learner, then yes, we may be able to help you achieve your goals in 24 hours. Remember, high quality, customized help that's tailored around the needs of each individual student takes time to achieve. You deserve nothing less than the best, so give us the time we need to give you the best.

If you need assistance with old exams in order to prepare for an upcoming test, we can definitely help. We can't work with you on current exams, quizzes, or tests unless you tell us in writing that you have permission to do so. This is not usually the case, however.

We do not have monthly fees or minimum payments, and there are no hidden costs. Instead, the price is unique for every work order you submit. For tutoring and homework help, the price depends on many factors that include the length of the session, level of work difficulty, level of expertise of the tutor, and amount of time available before the deadline. You will be given a price up front and there is no obligation for you to pay. Homework library items have individual set prices.

We accept credit cards, debit cards, PayPal, Venmo, ApplePay, and GooglePay.

Mechanical engineering combines principles of physics, mathematics and materials science to design and manufacture mechanical systems. Mechanical engineers use their specialized scientific knowledge and problem-solving skills to create impactful machinery for a wide range of industries.

The coursework for college students studying mechanical engineering typically involves advanced math and science principles. If you're struggling with the mechanical engineering concepts you're learning, our highly qualified tutors can help you understand them on a deeper level. You can come to us 24/7 for help with virtually any type of mechanical engineering topic or assignment. We'll help you expand your knowledge and find the solutions you need.

## Online Mechanical Engineering Tutors

Our mechanical engineering tutors provide homework help and tutoring sessions on many mechanical engineering topics. You'll get personalized support with step-by-step guidance to help you overcome your academic challenges.

## Tutoring Sessions

Scheduling a live, online tutoring session with one of our mechanical engineering tutors is the best way to get comprehensive guidance on difficult concepts. You can take advantage of our state-of-the-art whiteboard platform that offers video, audio, file upload support and desktop sharing for a productive tutoring session.

We recommend uploading relevant documents like study materials and old quizzes ahead of time — doing so will give your tutor the chance to review the materials so you can get the most out of your session. They'll tailor their instruction to ensure it readily relates to what you're learning.

## Homework Help

You can also come to us for mechanical engineering homework help. If you're tackling a challenging problem or assignment, our expert mechanical engineering tutors are here to assist you. They'll help you better understand the question and concepts involved so you can create an original solution.

If you need something more immediate, we have a  Homework Library  of solved problems you can search. Our Homework Library includes detailed examples and explanations to help you understand the process of going from question to answer.

## Mechanical Engineering Topics

• Manufacturing engineering:  Manufacturing engineering is concerned with the development and optimization of machines, tools and manufacturing techniques.
• Thermal engineering:  Thermal engineering deals with heat energy and transfer in mechanical systems.
• Acoustical engineering:  Acoustical engineering deals with the design, analysis and control of vibration and sound.
• Design engineering:  Design engineering uses mechanical engineering principles to improve machine design.

## Why Choose 24HourAnswers for Mechanical Engineering Help?

At 24HourAnswers, our goal is to help you succeed. We have a meticulous screening process to ensure we bring on only the best tutors. Many of our mechanical engineering tutors have advanced degrees and experience, including Ph.D.s and positions at top companies and universities. While other online tutoring services employ mainly college students, 24HourAnswers gives you unique access to highly qualified mechanical engineering professionals.

You can request a tutoring session or homework help from us 24/7. Our tutors respond to student queries as soon as possible, often in just minutes. You'll receive their quote based on your unique needs, and you're free to discuss the price with them. You can also specify your budget in advance when submitting your request.

We offer fair pricing with no minimum payments or monthly fees. You won't have to pay anything until you're sure our service is right for you.

## Find an Online Tutor for Mechanical Engineering

24HourAnswers is your go-to source for tutoring in mechanical engineering. Since our company began in 2005, we've helped over 1 million students around the globe. We have a student satisfaction rating of 99.5%, and we are A+ rated by the Better Business Bureau.

Get the help you need today and request a  tutoring session  or  homework help  for your mechanical engineering assignments. All you have to do is submit your request, upload any relevant documents, choose a delivery date and  create an account  if you don't already have one. The process takes just a few minutes to complete.

## Sample Problems and Solutions

Question 1:

Since a refrigeration system operates more efficiently when the condensing temperature is low, evaluate the possibility of cooling the condenser cooling water of the refrigeration system with another refrigeration system. Will the combined performance of the two systems be better, the same, or worse than one individual system? Explain why.

The answer is the combined performance will be worse than one individual system. The following is proof.

Consider the two systems as described above:

where Q L and Q H refer to the heat removed from cooled space and rejected to warm space, respectively.  Note that in the combined system, Q L2 = Q H1  because the heat rejected from REF1 (Q H1 ) is received as the cooling load for REF2 (Q L2 ).

The efficiency of refrigerator is normally defined through the coefficient of performance (COP):

The COP for the combined system is:

Note that we have the same desired output as single system Q L1.   But Q L2 = Q H1 , substitute into 2:

We see that because:

Q H2 = Q H1 + Q W2

Q H2 > Q H1

By comparing equations 1 and 3, we derrive that  COP 2 < COP 1

REFERENCES

Rajput, R.K., 2000.   Thermal Engineering .  S.Chand Publishers.

Question 2:

A two-stage ammonia system using flash-gas removal and intercooling operates on the cycle shown below.  The condensing temperature is 35 degrees C.  The saturation temperature of the intermediate-temperature evaporator is 0 degrees C, and its capacity is 150KW.  The saturation temperature of the low-temperature evaporator is -40 degrees C, and its capacity is 250KW.  What is the rate of refrigerant compressed by the high-stage compressor?

Add an adiabatic/isentropic efficiency of the compressors of 85% and THEN solve the problem. Should your answer be higher or lower than the one given in the book?

Find specific enthalpies at all points shown on the above diagram, using saturated and superheated ammonia vapor tables.

Point 5. Here the ammonia is saturated liquid. From saturated vapor table at 35 o C we get the enthalpy, and pressure, respectively:

P 5 = 1350.4 kPa

Point 6. The throttle valve is isenthalpic, so the enthalpy is

h 6 = h 5 = 346.8 kJ/kg

Point 3. Here the ammonia is saturated gas. From saturated vapor table at 0 o C we get the enthalpy, entropy, and pressure respectively:

h 3 = h g = 1442.2 kJ/kg

s 3 = s g = 5.3309 kJ/(kg*K)

P 3 = 429.6 kPa

Point 7. Here the ammonia is saturated liquid. From saturated vapor table at 0 o C we get the enthalpy

h 7 = h f = 180.36 kJ/kg

Point 8. The throttle valve is isenthalpic, so the enthalpy is

h 8 = h 7 = 180.36 kJ/kg

Point 1. Here the ammonia is saturated gas. From saturated vapor table at -40 o C we get the enthalpy and entropy, respectively:

h 1 = h g = 1388.8 kJ/kg

s 1 = s g = 5.9569 kJ/(kg*K)

Point 2. The high-stage compressor is isentropic so the entropy at point 2 is

s 2 = s 1 = 5.9569 kJ/(kg*K)

The pressure at point 2 is

P 2 = P 3 = 429.6 kPa

Now, from the superheated vapor table and using linear interpolation between pressures 400 kPa and 500 kPa we get for pressure 429.6 kPa near entropy 5.9569 kJ/(kg*K):

h = 1636.7+(1633.1-1636.7)*(429.6-400)/(500-400) = 1635.6 kJ/kg

s = 5.9907+(5.8744-5.9907)*(429.6-400)/(500-400) = 5.9563 kJ/(kg*K)

h = 1682.8+(1679.8-1682.8)*(429.6-400)/(500-400) = 1681.9 kJ/kg

s = 6.1179+(6.0031-6.1179)*(429.6-400)/(500-400) = 6.0839 kJ/(kg*K)

Now using linear interpolation, we find enthalpy at point 2 for an ideal low-stage compressor

h 2 = 1635.6+(1681.9-1635.6)*(5.9569-5.9563)/(6.0839-5.9563) = 1635.8 kJ/kg

Let h 2rfeal be the specific enthalpy at point 2 for real low-stage compressor.  Then its isentropic efficiency is given by

= (h 2real - h 1 )/(h 2 - h 1 )

From which we get

h 2real   = h 1 + *(h 2 - h 1 )

h 2real   = 1388.8+0.85*(1635.8-1388.8) = 1598.8 kJ/kg

Point 4. The low-stage compressor is isentropic so the entropy at point 4 is

s 4 = s 3 = 5.3309 kJ/(kg*K)

The pressure at point 4 is

P 4 = P 5 = 1350.4 kPa = 1.3504 MPa

Now from the superheated vapor table using a linear interpolation between pressures 1.2 MPa and 1.4 MPa we get for pressure 1.3504 MPa near entropy 5.3309 kJ/(kg*K):

h = 1606.8+(1598.8-1606.8)*(1.3504-1.2)/(1.4-1.2) = 1600.8 kJ/kg

s = 5.3916+(5.2994-5.3916)*(1.3504-1.2)/(1.4-1.2) = 5.3223 kJ/(kg*K)

h = 1658.0+(1651.4-1658.0)*(1.3504-1.2)/(1.4-1.2) = 1653.0 kJ/kg

s = 5.5325+(5.4443-5.5325)*(1.3504-1.2)/(1.4-1.2) = 5.4662 kJ/(kg*K)

Now, using linear interpolation, we find enthalpy at point 4 for an ideal high-stage compressor

h 4 = 1600.8+(1653.0-1600.8)*(5.3309-5.3223)/(5.4662-5.3223) = 1603.9 kJ/kg

Similarly to point 2, we get the enthalpy at point 4 for real high-stage compressor

h 4real   = h 3 + *(h 4 - h 3 )

h 4real   = 1442.2+0.85*(1603.9-1442.2) = 1579.64 kJ/kg

To find the mass flow rates of refrigerant through intermediate-temperature and low-temperature evaporators, respectively:

m 1 = Q 1 /(h 3 - h 6 ) = 150/(1442.2-346.8) = 0.13693 kg/s

m 2 = Q 2 /(h 1 - h 8 ) = 250/(1388.8-180.36) = 0.20688 kg/s

Letting m h be mass flow rate of refrigerant through the high-stage compressors.  Then the steady flow energy equation for the intercooler and flash tank is

m 2 *h 2real + (m h - m 1 )*h 6 = m 2 *h 7 + (m h - m 1 )*h 3

Solving for m h gives the answer

m h = m 1 + m 2 *(h 2real - h 7 )/(h 3 - h 6 )

m h = 0.13693+0.20688*(1598.8-180.36)/(1442.2-346.8) = 0.4048 kg/s

This is lower than the former value 0.4118 kg/s for ideal compressors

Question 3:

A R22 flash-gas removal system has a capacity of 180KW at an evaporating temperature of -30 degrees C when the condensing pressure is 1500KPa. Compute the power requirement for a system with a single compressor. Compute the total power required by the two compressors in the system where there is no intercooling but there is flash-gas removal at 600KPa.

Q in = 180 kW

t 1 = -30 o C

P 2 = P 4 = P 5 = 1500 kPa

P 3 = P 6 = P 7 = 600 kPa

Find the specific enthalpies at all points on the Figure 16-17.

From saturated table we have for saturated liquid near 1500 kPa

P              h f

Now using linear interpolation find the specific enthalpy at point 5

h 5 = h f = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) =  248.486 kJ/kg

The throttle valve at point 6 is isenthalpic, so the specific enthalpy at point 6 is

h 6 = h 5 = 248.486 kJ/kg

From saturated table at -30 o C we get the specific enthalpy and entropy at point 1 for saturated vapor

h 1 = h g = 393.138 kJ/kg

s 1 = s g = 1.80329 kJ/(kgK)

The saturated table gives near 600 kPa

Using linear interpolation we get for 600 kPa:

t = 5+(6-5)*(600-583.78)/(602.28-583.78) = 5.87675 o C

h g = 407.143+(407.489-407.143)*(600-583.78)/(602.28-583.78) = 407.446 kJ/kg

h f = 205.899+(207.089-205.899)*(600-583.78)/(602.28-583.78) = 206.942 kJ/kg

s g = 1.74463+(1.74324-1.74463)*(600-583.78)/(602.28-583.78) = 1.74341 J(kgK)

Now we get he enthalpy and entropy at point 3, and enthalpy at point 7:

h 3 = h g = 407.446 kJ/kg

s 3 = s g = 1.74341 J(kgK)

h 7 = h f = 206.942 kJ/kg

The throttle valve at point 8 is isenthalpic, so the specific enthalpy at point 8 is

h 8 = h 7 = 206.942 kJ/kg

The flash-gas compressor is isentropic, so the specific entropy at point 4 is

s 4 = s 3 = 1.74341 kJ/(kgK)

From saturated table at pressure 1500 kPa we find the saturation temperature 39 o C. Now from superheated table using columns with saturation temperatures 38 o C and 40 o C find the enthalpies and entropies at column with saturated temperature 39 o C near the entropy 1.74341 kJ/(kgK):

h = (427.155+425.871)/2 = 426.513 at s = (1.7365+1.7287)/2 = 1.73260

h = (431.568+430.374)/2 = 430.971 at s = (1.7501+1.7426)/2 = 1.74635

Now using liner interpolation find the specific enthalpy at point 4 for ideal flash-gas compressor

h 4 = 426.513+(430.971-426.513)*(1.74341-1.73260)/(1.74635-1.73260) = 430.018  kJ/kg

Let h 4rfeal be specific enthalpy at point 4 for real flash-gas compressor. Then its isentropic efficiency is given by

= (h 4real - h 3 )/(h 4 - h 3 )

h 4real   = h 3 + *(h 4 - h 3 ), which is

h 4real = 407.446+0.85*(430.018-407.446) = 426.63 kJ/kg

The evaporator compressor is isentropic, so the specific entropy at point 2 is

s 2 = s 1 = 1.80329 kJ/(kgK)

Similarly to the above from the superheated table we get near the entropy 1.80329 kJ/(kgK):

h = (448.703+447.771)/2 = 448.237 at s = (1.8008+1.7940)/2 = 1.7974

h = (452.901+452.019)/2 = 452.46 at s = (1.8127+1.8061)/2 = 1.8094

Now we can get the enthalpy at point 2 for an ideal evaporator compressor

h 2 = 448.237+(452.46-448.237)*(1.80329-1.7974)/(1.8094-1.7974) = 450.301 kJ/kg

The specific enthalpy at point 2 for a real evaporator compressor is given by

h 2real = h 1 + *(h 2 - h 1 )  which is

h 2real = 393.138+ 0.85*(450.301-393.138) = 441.73 kJ/kg

Let m 1 and m 2 be mass flow through evaporator and flash-gas compressors, respectively.  The mass flow m 1 is given by

m 1 = Q in /(h 1 - h 8 ) = 180/(393.138-206.942) = 0.96672 kg/s

The steady flow energy equation for points 3-6-7 gives

(m 1 + m 2 )*h 6 = m 1 *h 7 + m 2 *h 3

From which we get m 2

m 2 = m 1 *(h 6 -h 7 )/(h 3 -h 6 )  which is

m 2 = 0.96672*(248.486-206.942)/(407.446-248.486) = 0.25265 kg/s

The power required by the evaporator and flash-gas compressors are given by, respectively:

W 1 = m 1 *(h 2real - h 1 )

W 2 = m 2 *(h 4real - h 3 )

Now we can find the total power required by two compressors

W = W 1 + W 2 = m 1 *(h 2real - h 1 ) + m 2 *(h 4real - h 3 )

W = 0.96672*(441.73-393.138)+0.25265*(426.63-407.446) = 51.82 kW

This is lower than the former value 60.96 kW for ideal compressors.

Question 4:

In a R22 refrigeration system the capacity is 180KW at a temperature of -30 deg C. The vapour from the evaporator is pumped by one compressor to the condensing pressure of 1500KPa. Later the system is revised to two-stage compression operating on a cycle with intercooling but no removal of flash gas at 600KPa. Calculate the power required from a single compressor in the original system?  Calculate the power required from two compressors in the revised system?

P 2 = 1500 kPa

Find the specific enthalpies at all points in the above original one evaporator - one compressor refrigeration system, using Table A-6 for saturated and Table A-7 for superheated Refrigerant 22

From the saturated table we have for saturated liquid near 1500 kPa

P              h f

Using linear interpolation find the specific enthalpy at point 3:

h 3 = h f = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) = 248.486 kJ/kg

The throttle valve is isenthalpic, so the specific enthalpy at point 4 is

h 4 = h 3 = 248.486 kJ/kg

The compressor is isentropic, so the specific entropy at point 2 is

From the saturated table at pressure 1500 kPa we find the saturation temperature 39 o C. Now from the superheated table using columns with saturation temperatures 38 o C and 40 o C find the enthalpies and entropies at column with saturated temperature 39 o C near the entropy 1.80329 kJ/(kgK):

h = (448.703+447.771)/2 = 448.237 for s = (1.8008+1.7940)/2 = 1.7974

h = (452.901+452.019)/2 = 452.460 for s = (1.8127+1.8061)/2 = 1.8094

Now using liner interpolation find the specific enthalpy at point 2

h 2 = 448.237+(452.460-448.237)*(1.80329-1.7974)/(1.8094-1.7974) = 450.310 kJ/kg

Find the mass flow rate of refrigerant

m = Q in /(h 1 - h 4 ) = 180/(393.138-248.486) = 1.24437 kg/s

Now we can find the power of the compressor required

W = m*(h 2 - h 1 ) = 1.24437*(450.310- 393.138) = 71.1 kW

P 4 = 1500 kPa

P 2 = P 3 = P 6 = 600 kPa

Now find the specific enthalpies at all points in the above figure.

h 5 = h f = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) =  248.486 kJ/kg

The throttle valve is isenthalpic, so the specific enthalpy at points 6 and 7 is

h 6 = h 7 = h 5 = 248.486 kJ/kg

The low-stage compressor is isentropic, so the specific entropy at point 2 is

Now we get he enthalpy and entropy at point 3:

From the superheated table using columns with saturation temperatures 5 o C and 10 o C find the enthalpies and entropies at column with saturated temperature 5.87675 o C near the entropy 1.80329 kJ/(kgK):

h = 425.562+(423.974-425.562)*(5.87675-5)/(10-5) = 425.284

for s = 1.8080+(1.7894-1.8080)*(5.87675-5)/(10-5) = 1.80474

h = 429.229+(427.724-429.229)*(5.87675-5)/(10-5) = 428.965

for s = 1.8200+(1.8017-1.8200)*(5.87675-5)/(10-5) = 1.81679

h 2 = 425.284+(428.965-425.284)*(1.80329-1.80474)/(1.81679-1.80474) =  424.841 kJ/kg

The high-stage compressor is isentropic, so the specific entropy at point 4 is

s 4 = s 3 = 1.74341 J(kgK)

From the saturated table at pressure 1500 kPa we find the saturation temperature 39 o C. Now from the superheated table using columns with saturation temperatures 38 o C and 40 o C find the enthalpies and entropies at column with saturated temperature 39 o C near the entropy 1.74341 kJ/(kgK):

Now using liner interpolation find the specific enthalpy at point 4

h 4 = 426.513+(430.971-426.513)*(1.74341-1.73260)/(1.74635-1.73260) = 430.018  kJ/kg

Let m 1 and m 2 be mass flow through low-stage and high-stage compressors, respectively.

The mass flow m 1 is given by

m 1 = Q in /(h 1 - h 7 ) = 180/(393.138-248.486) = 1.24437 kg/s

The steady flow energy equation for intercooler gives

(m 2 - m 1 )*h 6 + m 1 *h 2 = m 2 *h 3

m 2 = m 1 *(h 2 -h 5 )/(h 3 -h 5 )

m 2 = 1.24437*(424.841-248.486)/(407.446-248.486) = 1.38054 kg/s

The powers required by low-stage and high-stage compressors are given by, respectively:

W 1 = m 1 *(h 2 - h 1 )

W 2 = m 2 *(h 4 - h 3 )

W = W 1 + W 2 = m 1 *(h 2 - h 1 ) + m 2 *(h 4 - h 3 )

W = 1.24437*(424.841-393.138) + 1.38054*(430.018-407.446) = 70.6 kW

Question 5:

A refrigeration system using R22 is to have a refrigerating capacity of 80KW. The cycle is standard vapour-compression cycle in which the evaporating temperature is -8 deg C and the condensing temperature is 42 deg C.

Determine the volume flow of refrigerant measured in cubic meters per second at the inlet to the compressor.

Calculate the power required by the compressor.

At the entrance to the evaporator what is the fraction of vapour in the mixture expressed both on a mass basis and a volume basis?

Q in = 40 kW

T 1 = -8 o C

T 3 = 42 o C

Refer to above diagram of the refrigeration system, and find the specific enthalpies at all points.

From the saturated vapor table at 42 o C find the specific enthalpy at point 3 for saturated liquid

h 3 = h f = 252.352 kJ/kg

The throttle valve is isenthalpic, so the same specific enthalpy is at point 4

h 4 = h 3 = 252.352 kJ/kg

From the saturated vapor table at -8 o C we get the specific enthalpy, specific entropy, specific volume, and pressure at point 1 for saturated vapor, respectively:

h 1 = h g = 402.341 kJ/kg

s 1 = s g = 1.76394 kJ/(kgK)

v 1 = v g = 61.0958 L/kg = 0.0610958 m 3 /kg

P 1 = 380.06 kPa

s 2 = s 1 = 1.76394 kJ/(kgK)

From the superheated vapor table at saturation temperature 42 o C we have the enthalpy and entropy near the entropy 1.76394 kJ/(kgK):

h                     s

Using linear interpolation we get the specific enthalpy at point 2

h 2 = 438.062+(442.449-438.062)*(1.76394-1.7618)/(1.7747-1.7618) = 438.790 kJ/kg

Now we get:

(a) The specific heat received by evaporator is given by

q in = h 1 - h 4 = 402.341-252.352 = 149.989 kJ/kg

So the mass flow of refrigerant is

M = Q in /q in = 40/149.989 = 0.266686 kg/s

Now we can get the volume flow rate at the inlet of compressor

V = m*v 1 = 0.266686*0.0610958 = 0.0162934 or 0.0163 m 3 /s

(b) Find specific power of the compressor

w = w = h 2 - h 1 = 438.790-402.341 = 36.4490 kJ/kg

Now we can find power of the compressor

W = w*M = 36.4490*0.266686 = 9.72044  9.72 kW

(c) The pressure at point 4 is

P 4 = P 1 = 380.06 kPa

From the saturated vapor table at 380.06 kPa we get the specific enthalpies and specific volumes for saturated liquid and vapor, respectively:

h f = 190.718 kJ/kg

h g = 402.341 kJ/kg

v f = 0.76253 L/kg

v g = 61.0958 L/kg

The fraction of vapor in the mixture at the entrance of the evaporator, at point 4, expressed on a mass basis is given by:

x 4m = (h 4 - h f )/(h g - h f ) = (252.352-190.718)/(402.341-190.718) = 0.29124 or  0.291

(d) Find specific volume at point 4

v 4 = (1 - x 4m )*v f + x 4m *v g = (1-0.29124)*0.76253+0.29124*61.0958 = 18.3340 L/kg

The fraction of vapor in the mixture at the entrance of the evaporator expressed on a volume basis is given by:

x 4v = (1/v 4 -1/v f )/(1/ g -1/v f ) = (1/8.3340-1/0.76253)/(1/61.0958-1/0.76253) = 0.971

Question 6:

If in a standard vapor compression cycle using R22 refrigerant the evaporating temperature is -5 deg C and the condensing temperature is 30 deg C, sketch the cycle on pressure-enthalpy coordinates and calculate the work of compression, the refrigerating effect, and the heat rejected in the condenser in KJ/Kg and the corresponding COP as well.

T 1 = -5 o C

T 3 = 30 o C

The diagram of the vapor-compression cycle is shown above, where: P 1 =P 4 ; P 2 =P 3

The corresponding cycle on pressure-enthalpy coordinates is drawn below

Find enthalpies at all points using Table A-6 for saturated Refrigerant 22 and Table A-7 for superheated Refrigerant 22

From the saturated vapor table at 30 o C we get the specific enthalpy and pressure, respectively, at point 3 for saturated liquid:

h 3 = h f = 236.664 kJ/kg

P 3 = 1191.9 kPa

Since the throttle valve is isenthalpic the specific enthalpy at point 4 is

h 4 = h 3 = 236.664 kJ/kg

From saturated vapor table at -5 o C we get the specific enthalpy and entropy at point 1 for saturated vapor:

h 1 = h g = 403.496 kJ/kg

s 1 = s g = 1.75928 kJ/(kgK)

Since the compressor is isentropic the specific entropy at point 2 is

s 2 = s 1 = 1.75928 kJ/(kgK)

P 2 = P 3 = 1191.9 kPa

From superheated vapor table at saturation temperature 30 o C we have the enthalpy and entropy near entropy 1.75928 kJ/(kgK):

h                     s

h 2 = 427.378+(431.549-427.378)*(1.75928-1.7534)/(1.7664-1.7534) = 429.265 kJ/kg

(a) The work of compression is given by

w = h 2 - h 1

w = 429.265-403.496 = 25.769  25.8 kJ/kg

(b) The refrigerating effect is given by

q in = h 1 - h 4

q in = 403.496-236.664 = 166.832  167 kJ/kg

(c) The heat rejected in the condenser is given by

q out = h 2 - h 3

q out = 429.265-236.664 = 192.601  193 kJ/kg

(d) The coefficient of performance is given by

COP = q in /w

COP = 166.832/25.769 = 6.47

## Sample Problems and Solutions in Advanced Dynamics

Derive the response of viscously damped SDOF system to the force F(t) by means of the convolution integral and plot the response for the system.

The response of the system is

Plot the response of a viscously damped force free system.

x(0) = 2 cm

ω n = 5 rad/s = frequency of undamped oscillation

ς = 0.1

We have the differential equation for oscillations

x'' + 2ςω n x' + ω n 2 x = 0                                                                    (1)

The general solution is given by

x (t) = X o exp(-ςω n t) cos(ω n √(1- ς 2 )   t + φ)                         (2)

Differentiating gives

x'(t) = -X o ω n  exp(ςω n -t) [ς cos(ω n √(1- ς 2 )   t + φ) + √(1- ς 2 ) sin(ω n  √(1- ς 2 ) t + φ)]

Substituting the initial conditions gives:

x'(0) = -X o ω n (ς cos φ + √(1- ς 2 sin φ) = 0

x (0) = X o cos φ = 2

From which we get:

The graph of x (t) for damping factor =0.1  is drawn below

(b) Now the formula (2) in part (a) gives the general solution for ς = 1

x (t) = (X o cos φ) exp(-ω n t)

Substituting the initial conditions gives the system of equations:

x'(0) = -X o ω n  cos φ = 0

x (0) = X o cos φ = 2

From the first equation we get

cos φ = 0

Substituting it into the second equation we get that the X o is infinitely large. So the system of equations has no physical solution for the given initial conditions

(c) Now the differential equation (2) in part (a) gives for ς = 2 and ω n  = 5

x'' + 20x' + 25 x = 0

Find the solution in form

x = exp(-pt)

Put it into the above equation and get the equation for p

p 2 - 20p + 25 = 0

Solving for p gives

p = 10± √(10 2 -25) = 10 ± √75

Thus the general solution is

x (t) = X 1 exp[-(10 + √75)t] + X 2 exp[-(10 - √75)t]

The initial conditions give the system of equations:

x'(0) = -X 1 (10 + √75) - X 2 (10 - √75) = 0

x (0) = X 1 + X 2 = 2

Solving for X 1 and X 2 gives:

X 1 = -0.155

Thus the equation is

x (t) = -0.155exp[-(10 + √75)t] + 2.15exp[-(10 - √75)t]

The graph is drawn below

References

Meirovitch, L., 2010, Fundamentals of Vibrations ,  Waveland Pr Inc

## Sample Problems in Mechanical Engineering

Traditionally mechanical engineering is considered to be made of three mainstream engineering aspects: manufacturing engineering, thermal engineering and design engineering. To excel in the domain a thorough knowledge of all of these is essential. A practicing mechanical engineer must solve many problems every day in all these subdomains.

Shaper  Cutting Time

Consider a shaper machine is working to smooth a surface plate of length 400mm. Assume the feed of the shaper tool is 2mm/stroke and it is working at a speed of 50 strokes per minute.  Determine the time it takes for the shaper to complete machining of the surface of the shaper machine.

The length of the travel needed on the job, L w is 400mm

Assuming a reasonable approach (A) and overrun (O) as 2 mm each, the length of cut required for shaper machine is Lw+A+O = 400 + 2+ 2 = 404 mm.

The feed of the shaper (S o ) is 2mm/stroke

The speed of the shaper is (N s ) = 50 strokes/minute, hence the time it takes to complete the cutting process = (L w +A+O)/N s *S o = 404/2*50 = 4.04 minutes.

Milling  Cutting Time Calculation

Determine the milling time required for milling a rectangular plate of length 100mm, width 50mm by a helical fluted plain HSS milling cutter of diameter 60mm and length 75mm and 6 teeth.

Assume A = O = 5mm, Vc = 40m/min and So = 0.1mm/teeth.

Cutting time Tc = Length of travel (Lc)/Table Feed (Sm)

Length of travel (Lc) = Length of work piece (Lw) + Approach (A) + Overrun (O) + (1/2 x Diameter of the rod)

Table feed(Sm) = So x Zc x N

So = Feed per teeth

Zc = Number of teeth on the cutter

N = Speed of the cutter

Lc = 100mm + 5mm + 5mm + 30mm = 140mm

Sm = 0.1 x 6 x N

N = (1000 x Vc)/(∏ x Dc) = (1000 x 40)/(∏ x 60) = 212.31rpm or ~213rpm

Sm = 0.1 x 6 x 213 = 127.8mm/min

Hence the Cutting time = Lc/Sm = 140/127.88 = 1.09minutes.

Casting and Metal Flow Problems

Consider the case of a metal mold with a sprue length of 20cm and a cross sectional area at the base of the sprue of 2.5 cm 2 . If the sprue feeds a horizontal runner leading into a mold cavity whose volume is 1560 cubic centimeter. Determine

• a) the velocity of the molten metal at the base of the sprue;
• b) volume flow rate of the molten metal; and
• c) the time it takes to fill the mold.

For determining the flow rate of the molten metal at the bottom of the sprue pin, Bernoulli’s principle is required. As per Bernoulli’s principle the pressure at the top of the sprue should be converted to the velocity at the bottom of the sprue, so it will be given by the equation:

V = sqrt (2*g*h), where h is the height of the sprue, V is the velocity of the molten metal at the bottom of the sprue,

So in the given problem

• A) Flow velocity of the molten metal, V = sqrt (2*9.81*100*20) = 198.1 cm/s
• B) The flow rate of the molten metal at the bottom of the sprue pin is given by Q = A * V

Where A is the area of cross section of the sprue pin and V is the velocity of the molten metal,

dQ/dt = 2.5 * 198.1 = 495.25 cubic centimeters/s.

• C) Time it takes to fill the mold = Q’/(dQ/dt) = 1560/495.25 = 3.15 seconds

## Thermal Engineering

Steady flow Energy Equation (first law of thermodynamics) and the principle of energy conservation

0.5 kg/s of fluid flows in a steady state process. The properties of fluid at the entrance are measured as

p 1 = 1.4bar, and the density of the fluid is 2.5kg/m3.

U i = 920kJ/kg while at exit the properties are P 2 = 5.6bar, density = 5kg/m3, U 2 = 720KJ/kg.

The velocity at entrance is 200m/s, while at exit it is 180m/s. It rejects 60KW of heat and raises through 60m during the flow process. Determine the change of enthalpy as well as the work done during the process.

The key to the solution is the fact that the total energy remains constant during the process.

Mass flow rate of fluid in the process = mf = 0.5kg/s

Pressure at the inlet to the process = 1.4bar

Density of the fluid = 2.5kg/m3

Initial internal energy of the fluid = 920KJ/kg

Final pressure of the fluid = 5.6bar

Final density at the exit = 5kg/m3

Final internal energy of the fluid =720kJ/kg

Velocity of the fluid at the entrance = 200m/s

Velocity of the fluid at the exit = 180m/s

The amount of the heat given to the fluid during the process = 60KW

The rise in height of the fluid during the process = 60m

Enthalpy change during the process (h2-h1) = ∆(U2-U1) + ∆PV

Therefore, ∆PV = p2/ρ2 – p1/ρ1.

Specific enthalpy change = (720 - 920)  x 10 3 + (5.6/5 - 1.4/2.5) x 10 5 = -144KJ/kg

Hence Enthalpy change = ∆H = mf x (h2-h1) = 0.5 x (-144) = -72 KJ/s

The steady flow energy equation (SFEE) can be applied as:

SFEE - Q - WS = mf [(h2 – h1) + ½( V 2 2 –V 1 2 ) + g(Z2 –Z1)]

60 x 10 3 - Ws = 0.5 x (-144 + 0.5 x (180 2 -200 2 ) + 9.81 x 60)

Hence Ws = 13605.7W = 136.1KW

Principle of Energy Conservation

0.25kg/s of water is heated from 30 o C to 60 0 C by hot gasses that enter at 180 0 C and leave at 80 0 C. Calculate the mass flow rate of gases when its specific heat at constant pressure is 1.08 KJ/kg - K. Find the entropy change of water and of hot gases. Take the specific heat of water as 4.186KJ/kg - K

The total energy entering the system should be equal to the energy leaving the system. The system is being heated by the hot gases entering and it is being cooled by the water leaving the system.

The temperature of the exhaust gas entering into the system = 180 degree Celsius

The temperature of the exhaust gas leaving the system = 80 degree Celsius.

The specific heat at constant pressure of the exhaust gases = 1.08KJ/kg - K

The specific heat of water = 4.1868KJ/kg - K

Water flow rate = 0.25kg/s

The heat intake from the exhaust gases = mg x Cpg x (Tg1-Tg2) (1)

.mg is the mass of the exhaust gases entering into the system

Cpg is the specific heat of the exhaust gases

Tg1 is the temperature of the exhaust gases entering into the system

Tg2 is the temperature of the exhaust gases leaving the system

The heat carried away by water = mw x Cpw x (Tw2 - Tw1) (2)

Mw is the mass of water = 0.25kg/s

CPw is the specific heat of water = 4.1868KJ/kg-K

Tw1 = 30 o C

Tw2 = 60 0 C

Equating equations 1 and 2, Mg x 1.08 x (180-80) = 0.25 x 4.1868 x (60 - 30), which yields that the mass flow rate of exhaust gases = 0.291kg/s.

Entropy change of hot water is calculable by: mw x Cpw x ln(Tw2/Tw1) = 0.25 x 4.1868 x ln(273 + 60/273 + 30)

S2 – S1 = 0.25 x 4.186 x ln(Tw2/Tw1) = 0.099KJ/kg-K

Entropy change of Exhaust gases = 0.291 x 1.08 x ln(80+273)/(180+273) = -0.0783KJ/kg-K.

Machine Design

You are required to design a pulley, B, given that the diameter of a pulley A is 300mm and the speed of rotation of pulley A is 250rpm. The belt thickness is 6mm. Determine the diameter of the pulley B, if it is rotating at a speed of 750rpm

For the belt drive to function the surface velocity of the pulley A should be same as the surface velocity of the pulley B.

The surface velocity of the pulley A = ∏ x 0.3 x 250/60 = 3.925m/s

Considering the surface velocity of the pulley as same, its diameter will be given by ∏ x D x 750/60 = 3.925m/s, where D = 100mm.

Suppose a shaft is made of a material with shear stress of 50Mpa and if the maximum torque that can act on the shaft is 200N-m, design the shaft diameter considering a factor of safety of 3.

Let T be the torque of the shaft = 200N-m,

The shear stress of the shaft material (fs) will be given by, T = ∏/16 x fs x d 3

In the given scenario diameter d is 2.73cm = 3 cm.

Considering the factor of safety of 3, diameter of the shaft = 3 x 3 = 9cm.

References:

Grote, K.H. and Antonsson, E.K., 2009.  Springer handbook of mechanical engineering  (Vol. 10). Springer Science & Business Media.

Kreith, F., 1999. Mechanical engineering handbook.

Jain, R.K., 1995.  Machine design . Khanna Publishers.

Ballaney, P.L., 1981.  Thermal Engineering . Khanna.

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## Browse Course Material

Course info, instructors.

• Prof. David Wallace
• Prof. Douglas Hart

## Departments

• Mechanical Engineering

## As Taught In

Learning resource types, mechanical engineering tools, assignments, matlab®assignments.

You must turn in your assignments at the end of each class period.

Note there is no lecture on MATLAB® Day 3, but rather you are to work on your own to solve Problem Set 3.

## Getting Help

If you need help,

- Contact your instructors and/or UA’s in person

- Or see the self-guided problem set tutorials:

Problem Set 1 Tutorial ( PDF ) Problem Set 2 Tutorial Problem Set 3 Tutorial ( PDF )

- Or ask other students in the class. Go on, help each other out!

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## 100 Best universities for Mechanical Engineering in Russia

Updated: February 29, 2024

• Art & Design
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• Engineering
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• Liberal Arts & Social Sciences
• Mathematics

Below is a list of best universities in Russia ranked based on their research performance in Mechanical Engineering. A graph of 714K citations received by 136K academic papers made by 158 universities in Russia was used to calculate publications' ratings, which then were adjusted for release dates and added to final scores.

We don't distinguish between undergraduate and graduate programs nor do we adjust for current majors offered. You can find information about granted degrees on a university page but always double-check with the university website.

## 1. Moscow State University

For Mechanical Engineering

## 42. Novgorod State University

43. belgorod state university.

## 47. Russian State University of Oil and Gas

48. siberian state aerospace university.

## 59. Tomsk State University of Control Systems and Radioelectronics

60. south-russian state university of economics and service.

## 89. Chuvash State University

90. ivanovo state power university.

## 100. Kaliningrad State Technical University

The best cities to study Mechanical Engineering in Russia based on the number of universities and their ranks are Moscow , Tomsk , Saint Petersburg , and Ufa .

## Engineering subfields in Russia

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