Mechanical Power Engineering               4th year MPE
Mansoura
                                                Department                          1st semester
    University
                                           Faculty of Engineering               Mid-Term 2020-2021
Course Title: Design of Mechanical
                                       Total Marks 25
Power Engines                                              No. of Pages: 3+Tables
                                   Allowed Time 1.25 hours
Course Code: MPE4414
Name:…………………………………..……………………………………..Sec.( )
  Assume reasonably any missing data – Give units of ALL deduced quantities
  Question No. (1) : Correct the following expressions if there are mistakes:            (8 Marks)
     1. The over square engine has lower piston speed and higher heat transfer losses compared with
         the under square engine.
                                                                                                         1
          (True)
      2. Maximum brake torque occurs at engines speeds lower than the ones at which maximum
         brake mean effective pressure occurs.                                                           1
         Maximum brake torque occurs at engines speeds equal to the ones at which maximum
         brake mean effective pressure occurs.
      3. Gas turbines are considered common examples of external combustion engines at which the
         combustion occurs in a combustion chamber outside the turbine itself.                           1
          Gas turbines are considered common examples of internal combustion engines at which
          the combustion occurs in a combustion chamber before the turbine itself.
      4. Fiber reinforcement is used to decrease thermal conductivity of the aluminum alloys used for
         piston fabrication and hence reduces heat losses from combustion chamber.                       1
          Fiber reinforcement is used to increase thermal and mechanical loads limits of the
          aluminum alloys used for piston fabrication.
      5. Advantages of sintered conrods include lower weight, higher velocity limit, and lower cost
         compared to forged conrods which make sintered rods suitable for modern engines.                1
          Advantages of sintered conrods include lower weight and higher velocity limit but higher
          cost compared to forged conrods which make sintered rods suitable for modern engines.
      6. Steel alloys are commonly used for the fabrication of crankshafts because of the lower weight
         compared with cast iron crankshafts.
                                                                                                         1
          Steel alloys are commonly used for the fabrication of crankshafts because of the higher
          strength compared with cast iron crankshafts.
      7. The connecting rod connects the piston and the flywheel and is responsible for converting the
         linear motion to a rotating motion.                                                             1
          The connecting rod connects the piston and the crankshaft and is responsible for
          converting the linear motion to a rotating motion.
      8. At least 3 journal bearings are used to hold an overhung crankshaft.
                                                                                                         1
          At least 3 journal bearings are used to hold a center crankshaft.
                                               Page 1 of 4
Question No. (2) :                                                                   (7 Marks)
The following figure represents a
performance map for a 4-stroke naturally
aspirated spark ignition engine. If the engine
provides a maximum power of 143 kW at a
speed of 5200 rpm, select proper values of
different operating parameters including
engine displacement, bore, stroke, and
number of cylinders that satisfy the engine
requirements. Use OPD parameter to verify
your result.
                     𝑷𝒃 = 𝑻. 𝝎
                               𝟐𝝅 × 𝟓𝟐𝟎𝟎
         𝟏𝟒𝟑 × 𝟏𝟎𝟎𝟎 = 𝑻 ×                                                                        1
                                  𝟔𝟎
            ∴ 𝑻 = 𝟐𝟔𝟐. 𝟔 𝑵. 𝒎
From chart:
            ∴ 𝒃𝒎𝒆𝒑 = 𝟏𝟏 𝒃𝒂𝒓                                                                      1
Typical values from table:
let
                                           𝑵𝒄 = 𝟔 𝒄𝒚𝒍𝒊𝒏𝒅𝒆𝒓𝒔                                      1
                                                   𝑺
                                                     =𝟏
                                                   𝑩
                                                                  𝑵
                                                 𝑷𝒃 = 𝒃𝒎𝒆𝒑𝑽𝒅 𝑵𝒄                                  1
                                                                  𝒏
                                                          𝝅 𝟐      𝟓𝟐𝟎𝟎
                                    𝟏𝟒𝟑 = 𝟏𝟏 × 𝟏𝟎𝟎 ×        𝑩 𝑺×𝟔×
                                                          𝟒        𝟔𝟎 × 𝟐
                                       ∴ 𝑩𝟐 𝑺 = 𝑩𝟑 = 𝟎. 𝟎𝟎𝟎𝟔𝟑𝟕                                   1
                                           ∴ 𝑩 = 𝑺 = 𝟖. 𝟔 𝒄𝒎
Engine displacement
                                                     𝝅 𝟐
                                            𝑽𝒅,𝒕𝒐𝒕 =   𝑩 𝑺𝑵𝒄
                                                     𝟒
                                                     𝝅
                                                    = × 𝟖. 𝟔𝟑 × 𝟔 = 𝟐𝟗𝟗𝟕. 𝟑𝟒𝟑 𝒄𝒎𝟑 ≈ 𝟑 𝑳          1
                                                     𝟒
Check:
                                                  𝑷𝒃      𝟏𝟒𝟑
                                      𝑶𝑷𝑫 =             =     = 𝟒𝟕. 𝟕 𝒌𝑾/𝑳                       1
                                                 𝑽𝒅,𝒕𝒐𝒕    𝟑
Which verify the OPD span in tables (40:80 kW/L)
                                                 Page 2 of 4
Question No. (3) :                                                                              (10 Marks)
It is required to design the piston and connecting rod of a 3L 4-stroke 6-cylinder SI engine. The engine
has a bore of 8.8 cm and connecting rod to crank offset ratio of 4. If the maximum allowable pressure
inside the combustion chamber is 5 MPa, find suitable dimensions for piston including its crown, barrel,
skirt, and its rings and pin. Give also proper dimensions of the conrod I-section and perform force
analysis to calculate the maximum stress upon the conrod at 5000 rpm when a right angle is created
                                                                                              1
between the conrod the crank arm. (Take the constant 𝛼 in Rankine-Gordon equation to be          for steel
                                                                                            7500
conrod)
Piston crown thickness:                  1               𝝈𝒃𝒓 = 𝟎. 𝟑 𝑴𝑷𝒂                for low speed engines
Let the piston be                                        ∴ 𝒍𝟏 ≈ 𝟓𝟖 𝒎𝒎
made of aluminum
                                                         Let total piston length = 1.5 D= 132 mm
with a design stress
(𝑺𝒅 = 𝟒𝟎 𝑴𝑷𝒂)                                            Clearance:
Grashoff’s formula                                       For crown= 0.01 D= 0.88 mm
                                                         For skirt= 0.002 D= 0.18 mm
            𝑷𝒎𝒂𝒙
𝒕 = 𝟎. 𝟒𝟑𝑫√
             𝑺𝒅                                          Piston pin (made from steel)                              1
                                                         𝑳𝒑𝒊𝒏 = 𝑫 − 𝟓 = 𝟖𝟑 𝒎𝒎
                            𝟓                            Let 𝒅𝒐 = 𝟏. 𝟐𝒅𝒊
        = 𝟎. 𝟒𝟑 × 𝟖𝟖√          ≈ 𝟏𝟒 𝒎𝒎
                            𝟒𝟎
                                                              𝑰   𝝅
                                                         𝒛=     =   (𝒅𝟒 − 𝒅𝟒𝒊 )
Piston rings (assuming CI rings)         1                    𝒄 𝟑𝟐𝒅𝒐 𝒐
                                                              𝝅           𝒅𝟒𝟎
                                                         =        (𝒅𝟒𝒐 −       )
                                                             𝟑𝟐𝒅𝒐        𝟏. 𝟐𝟒
                   𝒃 ≈ 𝟎. 𝟎𝟑𝑫 ≈ 𝟑 𝒎𝒎
                                                         = 𝟎. 𝟎𝟓𝟎𝟖𝟑𝒅𝟑𝒐
        Let:    𝒉 = 𝒃 = 𝟑 𝒎𝒎
                                                               𝑭𝒑 𝑳𝒑𝒊𝒏           𝟎. 𝟎𝟖𝟑
Check for 4 rings                                        𝑴=            = 𝟑𝟎𝟒𝟏𝟎 ×        = 𝟔𝟑𝟏 𝑵. 𝒎
                                                               𝟐 𝟐                  𝟒
                     𝑫   𝟖𝟖                                                        𝑴
          𝒉𝒎𝒊𝒏 =       =       = 𝟐. 𝟐 𝒎𝒎                 Bending stress: 𝝈𝒃 =          = 𝟏𝟎𝟎 𝑴𝑷𝒂
                    𝟏𝟎𝒏 𝟏𝟎 × 𝟒                                                     𝒛
Top land:                                                ∴ 𝒅𝒐 ≈ 𝟓𝟎 𝒎𝒎, 𝒅𝒊 ≈ 𝟒𝟐 𝒎𝒎
               𝒉𝟏 = 𝒕 = 𝟏𝟒 𝒎𝒎                                     𝑭𝒑 ⁄𝟐
                                                          𝝈𝒔 = 𝝅              = 𝟐𝟔. 𝟑 𝑴𝑷𝒂 < 𝟒𝟎𝑴𝑷𝒂
Width of ring grooves:                                           (𝒅𝟐𝒐 − 𝒅𝟐𝒊 )
                                                               𝟒
               𝒉𝟐 = 𝒉 = 𝟑 𝒎𝒎                                       𝑭𝒑
                                                         𝝈𝒃𝒓 =             = 𝟏𝟎 𝑴𝑷𝒂, ∴ 𝒍𝒆𝒏𝒅 ≈ 𝟔𝟏 𝒎𝒎
                                                                 𝒅𝒐 𝑳𝒆𝒏𝒅
Barrel and skirt             2
                                                         For conrod
𝒕𝟏 = 𝒃 + 𝟎. 𝟎𝟑𝑫 + 𝟎. 𝟒𝟔
                                                                 𝝅
= 𝟑 + 𝟎. 𝟎𝟑 × 𝟖𝟖 + 𝟎. 𝟒𝟔                                 𝑽𝒅,𝒕𝒐𝒕 = 𝑩𝟐 𝑺𝑵𝒄
                                                                 𝟒
                                                                   𝝅
≈ 𝟕 𝒎𝒎                                                   𝟑𝟎𝟎𝟎 = 𝟖. 𝟖𝟐 𝑺 × 𝟔, ∴ 𝑺 = 𝟖. 𝟑 𝒄𝒎
                                                                   𝟒
     𝑹𝒍,𝒎𝒂𝒙                                                    𝑺
𝒍𝟏 =                                                     ∴ 𝒂 = = 𝟒. 𝟏𝟓 𝒄𝒎
     𝑫𝝈𝒃𝒓                                                      𝟐
                                                                                                               1
𝑹𝒍,𝒎𝒂𝒙 = 𝟎. 𝟎𝟓𝑭𝒑 , 𝑭𝒑 = 𝑷𝒎𝒂𝒙 𝑨𝒑                          ∴ 𝒓 = 𝟒𝒂 = 𝟏𝟔. 𝟔 𝒄𝒎
               𝝅    𝟖.𝟖 𝟐
𝑭𝒑 = 𝟓𝟎𝟎𝟎 × (           ) = 𝟑𝟎. 𝟒𝟏 𝒌𝑵
                𝟒 𝟏𝟎𝟎
∴ 𝑹𝒍,𝒎𝒂𝒙 = 𝟏𝟓𝟐𝟎. 𝟓𝟑𝟏 𝑵                                                                 Best Wishes
                                                                                  Dr. Waleed M. Shaaban
                                                                                 Dr. Mahmoud A. Shouman
                                               Page 3 of 4
For I-section                         1
Let 𝑯 = 𝟓𝒕, 𝑩 = 𝟒𝒕                                     𝑰𝒙𝒙 = 𝟖. 𝟑𝟖 × 𝟏𝟎−𝟖 𝒎𝟒
𝑨 = 𝟏𝟏𝒕𝟐 ,                                                       𝑴
      𝟒𝟏𝟗 𝟒                                            𝑴′ =           = 𝟑𝟑𝟗. 𝟏𝟓 𝑵. 𝒎
                                                              𝟏 − 𝝃𝑭𝒂
𝑰𝒙𝒙 =      𝒕
       𝟏𝟐                                                 𝑯
𝒌𝒙𝒙 = 𝟏. 𝟕𝟖𝟐𝒕                                          𝒄=   = 𝟏𝟕. 𝟓 𝒎𝒎
                                                          𝟐
𝝈𝒄 = 𝟑𝟑𝟎 𝑴𝑷𝒂 𝒇𝒐𝒓 𝒔𝒕𝒆𝒆𝒍                                 Combined stress                 1
          𝝈𝒄 𝑨                                                      ′
                                                            𝑭𝒂 𝒄𝑴
𝑭𝒄 =                = 𝟓𝑭𝒑                              𝝈=     +     = 𝟏𝟐𝟕. 𝟐𝟐 𝑴𝑷𝒂
             𝒓                                              𝑨   𝑰𝒙𝒙
       𝟏+ 𝜶( )
            𝒌𝒙𝒙                                                     < 𝟑𝟑𝟎 𝑴𝑷𝒂 𝒇𝒐𝒓 𝒔𝒕𝒆𝒆𝒍
𝟑𝟑𝟎 × 𝟏𝟎𝟔 × 𝟏𝟏𝒕𝟐
                   = 𝟓 × 𝟑𝟎𝟒𝟏𝟎
     𝟏    𝟎. 𝟏𝟔𝟔
𝟏+      (        )
   𝟕𝟓𝟎𝟎 𝟏. 𝟕𝟖𝟐𝒕
∴ 𝒕 = 𝟕 𝒎𝒎, 𝑯 = 𝟑𝟓 𝒎𝒎, 𝑩 = 𝟐𝟎 𝒎𝒎
Big end depth= 1.1H= 39 mm
Small end depth= 0.8H= 28 mm
Force analysis                            2
           𝒂
∅ = 𝒕𝒂𝒏−𝟏 = 𝟏𝟒. 𝟎𝟑°
           𝒓
𝜽 = 𝟗𝟎 − ∅ = 𝟕𝟓. 𝟗𝟔°
                  𝒂
𝒂𝒑 = 𝝎𝟐 𝒂(𝒄𝒐𝒔𝜽 + 𝒄𝒐𝒔𝟐𝜽)
                  𝒓
  𝟐𝝅 × 𝟓𝟎𝟎𝟎 𝟐
=(         )
     𝟔𝟎
           × 𝟎. 𝟎𝟒𝟏𝟓(𝒄𝒐𝒔𝟕𝟓. 𝟗𝟔
           + 𝟎. 𝟐𝟓𝒄𝒐𝒔𝟐 × 𝟕𝟓. 𝟗𝟔)
           = 𝟐𝟒𝟗. 𝟕 𝒎⁄𝒔𝟐
Let weight of reciprocating parts= 2 kg
        𝑾𝑹
𝑭𝑰𝑹 =      𝒂 = 𝟐 × 𝟐𝟒𝟗. 𝟕 = 𝟒𝟗𝟗. 𝟒 𝑵
        𝒈 𝒑
       𝑭𝒑 − 𝑭𝑰𝑹 𝟑𝟎𝟒𝟏𝟎 − 𝟒𝟗𝟗. 𝟒
𝑭𝒂 =           =
        𝒄𝒐𝒔∅       𝒄𝒐𝒔𝟏𝟒. 𝟎𝟑
               = 𝟑𝟎𝟖𝟑𝟏. 𝟕𝟕 𝑵
𝑴 = 𝟎. 𝟎𝟔𝟒𝒓𝟑 𝝆𝑨𝝎𝟐 𝐬𝐢 𝐧(𝜽 + ∅)
   = 𝟎. 𝟎𝟔𝟒 × 𝟎. 𝟏𝟔𝟔𝟑 × 𝟕𝟖𝟎𝟎 × 𝟏𝟏
               𝟐(
                  𝟐𝝅 × 𝟓𝟎𝟎𝟎 𝟐
   × (𝟕/𝟏𝟎𝟎𝟎)              ) 𝐬𝐢 𝐧(𝟗𝟎)
                     𝟔𝟎
   = 𝟑𝟑𝟕. 𝟒𝟑 𝑵. 𝒎
Let E=200 GPa
     𝟎. 𝟏𝒓𝟐          𝟎. 𝟏 × 𝟎. 𝟏𝟔𝟔𝟐
𝝃=          =
      𝑬𝑰𝒙𝒙                    𝟒𝟏𝟗   𝟕 𝟒
                𝟐𝟎𝟎 × 𝟏𝟎𝟗 ×       (    )
                               𝟏𝟐 𝟏𝟎𝟎𝟎
             = 𝟏. 𝟔𝟒𝟑𝟒𝟕 × 𝟏𝟎−𝟕
                                              Page 4 of 4
                                        Mechanical Power Engineering                   4th year MPE
    Mansoura
                                                Department                              1st semester
    University
                                           Faculty of Engineering                   Mid-Term 2020-2021
Course Title: Design of Mechanical
                                       Total Marks 25
Power Engines                                              No. of Pages: 3+Tables
                                   Allowed Time 1.25 hours
Course Code: MPE4414
Name:…………………………………..……………………………………..Sec.( )
  Assume reasonably any missing data – Give units of ALL deduced quantities
  Question No. (1) : Choose the correct answer:                                    (8 Marks)
     1. The process at which the exhaust is pushed out the combustion chamber during the intake
         process in a 2-stroke ICE is called
          a) Exhaust stroke                                  b) Exhaust recycling
          c) Exhaust blowdown                                d) Scavenging
      2. The under square engine results in .... friction losses compared with an over square engine with
         similar bore
          a) Lower                                           b) Higher
          c) Equal                                           d) Unknown
      3. A common example of external combustion engines is
          a) Steam turbine                                   b) Gas Turbine
          c) SI engine                                       d) Electric motor
      4. Fiber reinforcement of aluminum alloys used for piston fabrication gives.... thermal and
         mechanical load limits
          a) Lower                                           b) Higher
          c) Equal                                           d) Unknown
      5. Advantages of fabricating conrods using casting manufacturing include
          a) Low weight                                      b) Low cost
          c) High strength                                   d) Good quality
      6. Piston functions include
          a) Sealing off the working chamber                 b) Actuating the intake and exhaust valves
          c) Exchanging power with the Flywheel              d) All choices
      7. The crankshaft is required to have high modulus of rigidity in order to withstand
          a) Normal stresses                                 b) Torsion
          c) Vibrations                                      d) Variable load
      8. Sintered conrods results in....compared with forged conrods which make sintered rods suitable
         for modern engines.
          a) Higher weight                                   b) Higher velocity limit
          c) Lower cost                                      d) All choices
                                               Page 1 of 3
Question No. (2) :                                                                         (7 Marks)
               3
A V4 2500 cm , 4-stroke diesel engine equipped with a two-stage turbocharging system (see figure
below) and operates at 4000 rpm and a volumetric efficiency of 140% with an air to fuel ratio of 21:1.
Assuming that the air density equals 1.2 kg/m3 and the specific heat at constant pressure is constant and
equals 1.2 kJ/kg.K, determine;
1. The power that both high-pressure and low-pressure turbochargers must produce
2. The temperature at the intercooler exit.
3. If the engine operates with a brake thermal efficiency of 34%, determine the engine power output
   (Assuming that the combustion efficiency to be equal 96% and the lower heating value of used fuel
   is 42.5 MJ/kg).
                                              Page 2 of 3
Question No. (3) :                                                                           (10 Marks)
It is required to design the piston and connecting rod of a 2.4L 4-stroke 4-cylinder SI engine. The engine
has a bore of 9 cm and connecting rod to crank offset ratio of 4. If the maximum allowable pressure
inside the combustion chamber is 6 MPa, find suitable dimensions for piston including its crown, barrel,
skirt, and its rings and pin. Give also proper dimensions of the conrod I-section and perform force
analysis to calculate the maximum stress upon the conrod at 3200 rpm when a right angle is created
                                                                                                1
between the conrod the crank arm. (Take the constant 𝛼 in Rankine-Gordon equation to be           for steel
                                                                                             7500
conrod)
                                                                                       Best Wishes
                                               Page 3 of 3                        Dr. Waleed M. Shaaban
                                                                                 Dr. Mahmoud A. Shouman
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              If there is a change crank angel from 30, it
              is always rounded up in the subsequent
              calculations to the larger number
inverse cos