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الكلية كلية هندسة المواد     القسم قسم هندسة المعادن     المرحلة 2
أستاذ المادة حيدر عبد حسن علوان الجبوري       6/8/2011 8:05:57 AM


Fluid flow
Derived units:
Force                              n                                kg m/s
Energy or work             joule                           kg m / s               nm
Power                            watt                            kg m/s                 j/s
Pressure                         Pascal                        kg  m/s                 n/m
Viscosity                                                          kg m/s                 ns/sm

Conversion of units     
Example            1
                     Convert   t   poise  to British engineering units and si units
Sol
                             1 poise = 1 9/cm s  =   19 /1cm * 1s
    =   19   *     ib

 

 

 


Ex 2    
               Convert 1 kw to     h.p
Sol  
             1kw = 1000 w  = 1000   j/s
             1 h.p    =  o.746    kw

   Pressure
1 ibf / in                                          6.895   kn/ m
1 atm                                               101.3 kn/ m
1 bar                                                100  kn /m
1 ft water                                         2.99kn /m
1 in water                                        249   n/m 
1 in hg                                              3.39 kn /m
1 mm hg                                           133 n/m

Pressure     
           P= f/a =mg/a         Newton s law

 

 

 

 

 

 

 

 

 

Where      m= the mass of piston , pan , and weight
                 g = local acceleration of gravity
                 a = cross – sectional area of piston
 
the mass            m= ahƒ
                 a= the cross – sectional area
                 h=  height
                 p = density
          p=f/A = mg/A = Ah  g/ A  =  h  g
late = 101325 pa = 101.325 kpa = 0.10132 mpa
lbar  = 10000 pa ……………..
ex
dead – weight gauge with a 1- cm diameter piston is used to measure pressure very accurately .in a particular instant a mass of 6.14 kg (include piston   pan) . if the local acceleration of gravity 9.82 m/s .what the gauge pressure being measured .if the barometric pressures 748 to      what the absolute pressure
f = mg = 6.14 * 9.82 = 60.295 n
gauge pressure = f / a = 60.295 /           =   76.77  n/ cm
the absolute pressure  = p = 76.77 + 748 (0.01332)
                                          =  86.74      n/cm 
Ex   at 27 c  a manometer filled with mercury reads 60.5 cm . the  local acceleration of gravity is 9.784 m/s . what pressure dose this height of mercury correspond
                 P = hg           at 27 c the density of mercury is  13.53 g/cm                 then
                        p = g 0.5 cm k  13.53 g/cm * 9.784 m/sec
                          = 8009       / ms  cm  
                         P = 8.009 kg m/s cm =  8.009  n / cm
                             P = 80.09 kpa = 0.8009 bar

 

 

 

 

 


For   incompressible fluid
                                 U1 A1  = u2 a2
   Momentum changes in a fluid

 

 

 

 

 

Pda force up stream pressure        fluid in a stream tube
P +dp /dl     
The net force along tube         to the pressure gradient  is
pdA  - [ p + dp / dl  dl ] [ dA +                       
the force acting is the weight of the fluid

the component of this force a long the stream tube is

neglected second order terms and noting that sin? = dz/ dl
 total force on fluid =   
the rate change of momentum of the fluid along the stream tube

 

 

 

 

 

 

 

 

 

 

Energy of  a fluid in motion

Internal energy
When a fluid flows from location to another , energy will converted from one for m to another , this energy is known as internal energy
dv =
dv internal energy
sw net work
sq heat added
potential energy

the potential energy of the fluid , due to its position in the earth s gravitational field .
the potential  energy is   gz
where
      g acceleration
      z  position
kinetic energy
if the fluid is moving with a velocity   u ,the kinetic energy is 1/2 u
pressure energy
this represents the work which must be done in order to introduce the fluid , without change in volume

therefore          pv
   where
        p  the pressure
       ? the volume per unit mass of fluid
  ?   The total energy of unit mass of fluid is , therefore
U + p?   +pz   +   1/2 u  ?
If the fluid flows from section   1    to section   2    and        is the net heat absorbed from the surroundings and    ws  is the net work done by the fluid on the surrounding

?  u2

?   

Where           w  = ws + ?(p?)
                      H  =  U + p?
               Where    H the enthalpy 

 

 


جريان موائع  بوليمر                                     المرحلة الثالثة       
Fluid flow
Derived units:
Force                              n                                kg m/s
Energy or work             joule                           kg m / s               nm
Power                            watt                            kg m/s                 j/s
Pressure                         Pascal                        kg  m/s                 n/m
Viscosity                                                          kg m/s                 ns/sm

Conversion of units     
Example            1
                     Convert   t   poise  to British engineering units and si units
Sol
                             1 poise = 1 9/cm s  =   19 /1cm * 1s
    =   19   *     ib

 

 

 


Ex 2    
               Convert 1 kw to     h.p
Sol  
             1kw = 1000 w  = 1000   j/s
             1 h.p    =  o.746    kw

   Pressure
1 ibf / in                                          6.895   kn/ m
1 atm                                               101.3 kn/ m
1 bar                                                100  kn /m
1 ft water                                         2.99kn /m
1 in water                                        249   n/m 
1 in hg                                              3.39 kn /m
1 mm hg                                           133 n/m

Pressure     
           P= f/a =mg/a         Newton s law

 

 

 

 

 

 

 

 

 

Where      m= the mass of piston , pan , and weight
                 g = local acceleration of gravity
                 a = cross – sectional area of piston
 
the mass            m= ahƒ
                 a= the cross – sectional area
                 h=  height
                 p = density
          p=f/A = mg/A = Ah  g/ A  =  h  g
late = 101325 pa = 101.325 kpa = 0.10132 mpa
lbar  = 10000 pa ……………..
ex
dead – weight gauge with a 1- cm diameter piston is used to measure pressure very accurately .in a particular instant a mass of 6.14 kg (include piston   pan) . if the local acceleration of gravity 9.82 m/s .what the gauge pressure being measured .if the barometric pressures 748 to      what the absolute pressure
f = mg = 6.14 * 9.82 = 60.295 n
gauge pressure = f / a = 60.295 /           =   76.77  n/ cm
the absolute pressure  = p = 76.77 + 748 (0.01332)
                                          =  86.74      n/cm 
Ex   at 27 c  a manometer filled with mercury reads 60.5 cm . the  local acceleration of gravity is 9.784 m/s . what pressure dose this height of mercury correspond
                 P = hg           at 27 c the density of mercury is  13.53 g/cm                 then
                        p = g 0.5 cm k  13.53 g/cm * 9.784 m/sec
                          = 8009       / ms  cm  
                         P = 8.009 kg m/s cm =  8.009  n / cm
                             P = 80.09 kpa = 0.8009 bar

 

 

 

 

 


For   incompressible fluid
                                 U1 A1  = u2 a2
   Momentum changes in a fluid

 

 

 

 

 

Pda force up stream pressure        fluid in a stream tube
P +dp /dl     
The net force along tube         to the pressure gradient  is
pdA  - [ p + dp / dl  dl ] [ dA +                       
the force acting is the weight of the fluid

the component of this force a long the stream tube is

neglected second order terms and noting that sin? = dz/ dl
 total force on fluid =   
the rate change of momentum of the fluid along the stream tube

 

 

 

 

 

 

 

 

 

 

Energy of  a fluid in motion

Internal energy
When a fluid flows from location to another , energy will converted from one for m to another , this energy is known as internal energy
dv =
dv internal energy
sw net work
sq heat added
potential energy

the potential energy of the fluid , due to its position in the earth s gravitational field .
the potential  energy is   gz
where
      g acceleration
      z  position
kinetic energy
if the fluid is moving with a velocity   u ,the kinetic energy is 1/2 u
pressure energy
this represents the work which must be done in order to introduce the fluid , without change in volume

therefore          pv
   where
        p  the pressure
       ? the volume per unit mass of fluid
  ?   The total energy of unit mass of fluid is , therefore
U + p?   +pz   +   1/2 u  ?
If the fluid flows from section   1    to section   2    and        is the net heat absorbed from the surroundings and    ws  is the net work done by the fluid on the surrounding

?  u2

?   

Where           w  = ws + ?(p?)
                      H  =  U + p?
               Where    H the enthalpy 

 

 

 

 

 

 

                    

 

 

 

                    


المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .
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