sample graph-exp 1

7/23/2019 Sample Graph-Exp 1

http://slidepdf.com/reader/full/sample-graph-exp-1 1/19

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Pi$e?

(1%mi')

? 1/:8

(m2%)

31

(mm)

3

(mm)

A

(m1/:8)

V

(m%)

R e

( 1/2)

f theo

(E@ > !r

E@" !r

M!!-4-ia6ram)

h f.theo

(E@" 8)

h f.exp

331:3)

(m)

f exp

(E@" 1/)

%

exp,%

LV Dgh f f

=

A

%< 3.'' F%= '63.F1

=.EE% 13.3< =.1<% =.3>' =.EE6 =.'='

%% '.<> >>= 1663.F1

=.>3> 1%.%6 =.1<E =.'6% =.<16 =.%F3

1E '.== <<= %3=3.F1

=.<11 1=.=% =.1>> =.%3E =.3%= =.'==

13 %.'' 6<6 '%=3.F1

=.3>6 >.>F= =.1EF =.1<= =.%36 =.%F=

1% %.== 6'= '6=3.F1

=.3=> <.<>= =.1F< =.1%% =.1E= =.%EF

B

%< 3.'' F16 1E63.'3

=.FFE 16.3> =.16F =.<%6 =.>'= =.1E<

%% '.<> >E= %363.'3

=.E3< 1'.11 =.1<< =.3<F =.6'6 =.1EF

1E '.== <<= '== 3.'3 =.<F1 1=.>1 =.1>3 =.'%E =.'<= =.1F1

13 %.'' 6>= '6=3.'3

=.6'> E.'%= =.1E6 =.%11 %.%%= =.1F'

1% %.== 6'6 '>=3.'3

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Pi$e?

(1%mi')

? 1/:8

(m2%)

31

(mm)

3

(mm)

A

(m1/:8)

V

(m%)

R e

(1/2)

f theo

(E@ > !r

E@" !r

M!!-4

-ia6ram)

h f.theo

(E@" 8)

h f.exp

331:3)

(m)

f exp

(E@" 1/)

%

exp,%

LV

Dgh f

f =

1A %E 3.<> 'F=.% '== 3.F1 =.F61 %E.'' =.1'> =.%<= =.=F =.=3>

%3 3.== '>' '16 3.F1 =.E16 %3.%F =.13% =.1FE =.=< =.=3'

%= '.'' '<= ''= 3.F1 =.<>E %=.%= =.13F =.133 =.=' =.='1

1< %.<> '6' '3= 3.F1 =.633 1<.%1 =.16> =.=FE =.=1' =.=%1

1% %.== '3F '36 3.F1 =.3=> 1%.1' =.1<F =.=6F =.==3 =.=11

1B %E 3.<> 'E3 '=6 3.'3 1.=>< %F.6F =.1'6 =.'3F =.=>F =.='1

%3 3.== '>= '16 3.'3 1.1<E '%.1% =.1'% =.3=% =.=66 =.=1E

%= '.'' '<6 '1> 3.'3 1.3=% 'E.6< =.1%< =.66' =.=3E =.=11

1< %.<> '<' '%= 3.'3 1.>3F 3E.1= =.1%= =.E%= =.=3' =.==<

1% %.== '<6 '%6 3.'3 %.''6 <3.%1 =.111 1.'6% =.=3 =.=='

Ta+5e *!r Data !* S#--e' E'5ar6eme't Pi$e

5

(1+min)

5x1=3

(m'+s)

h1

(mm)

h%

(mm)∆h

(m)

1

(m%x1=3)

%

(m%x1=3)

-1

(m+s)

-%

(m+s)

h4,theo

(m)?*. %

h4,exp

(m)?*. F

D

h4,exp +(-1%+%g)

1>



%E 3.<> 6'6< 6%6 =.=1 1.'F 3.%< '.'<= 1.=F< =.%<1 =.6%3 =.F11

%3 3.== 3F= 3E6 =.==6 1.'F 3.%< %.E>E =.F'F =.1F% =.'E% =.F=6

%= '.'' 336 33= =.==6 1.'F 3.%< %.'F< =.>%E =.1'' =.%<< =.F=F

1< %.<> 3=6 3== =.==6 1.'F 3.%< 1.F%1 =.<%> =.=E6 =.1>' =.F%=

1% %.== '<6 '<' =.==% 1.'F 3.%< 1.3'F =.3<F =.=3E =.=F< =.F1=

Ta+5e *!r Data !* S#--e' C!'tra&ti!' Pi$e

5

(1+min)

5x1=3

(m'+s)

h1

(mm)

h%

(mm)∆h

(m)

%

(m%x1=3)

%

(m%x1=3)-1

(m+s)

-%

(m+s)

h4,theo

(m)

?*. 6

h4,exp

(m)?*. F

D

h4,exp +(-%%+%g)

%E 3.<> E36 %E= =.6<6 3.%< 1.'F 1.=F< '.'<= =.%EE =.=61 =.=EF

%3 3.== ><6 '== =.3<6 3.%< 1.'F =.F'F %.E>E =.%11 =.=EE =.%=E

%= '.'' <'6 ''6 =.'== 3.%< 1.'F =.>E% %.'F< =.13< =.='F =.1''

1< %.<> 6<= '3= =.%%= 3.%< 1.'F =.<%> 1.F%1 =.=F3 =.=6% =.%><

1% %.== 3E6 '<= =.1%6 3.%< 1.'F =.3<F 1.3'F =.=6' =.='1 =.%F3

Ta+5e *!r Data !* =/! Be'- Pi$e

5

(1+min)

5x1=3

(m'+s)

h1

(mm)

h%

(mm)∆h

(m)

(m%x1=3)

-

(m+s)

h4,theo

(m)?*. <

h4,exp

(m)?*. F

D

h4,exp +(-%+%g)

1E



%E 3.<> E%6 %%=1 =.<=6 1.%> '.<>> =.%<% 1.%F3 1.E>E

%3 3.== >'6 %6= =.3E6 1.%> '.16= =.1F% =.FF1 1.F<=

%= '.'' <3= %F6 =.'36 1.%> %.<%% =.1'' =.<F6 1.FE'

1< %.<> 66= ''= =.%%= 1.%> %.1=% =.=E< =.336 1.F><

1% %.== 3>6 '66 =.1%= 1.%> 1.6>6 =.=3E =.%3< 1.F3<

Ta+5e *!r Data !* E5+!7 Pi$e

5

(1+min)

5x1=3

(m'+s)

h1

(mm)

h%

(mm)∆h

(m)

(m%x1=3)

-

(m+s)

h4,theo

(m)

?*. <

h4,exp

(m)

?*. F

D

h4,exp +(-%+%g)

%E 3.<> E<= E= =.>E= 1.%> '.<>> =.%<% 1.3<F %.1'%

%3 3.== >66 166 =.<== 1.%> '.16= =.1F% 1.1=< %.1E>

%= '.'' <3= %'= =.31= 1.%> %.<%% =.1'' =.><= %.1<F

1< %.<> 636 %F6 =.%6= 1.%> %.1=% =.=E< =.3>6 %.1=F

1% %.== 3>6 '3= =.1'6 1.%> 1.6>6 =.=3E =.%<1 %.=<3

1F



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A"/ La+!rat!r4 re$!rt

1. Bee handout (4aboratory ;eport &ormat)

%. dditional report re*uirement

i. 3lot a graph o f theo and f exp versus Re on the same graph andcomment on the results.

ii. 3lot a graph between e0perimental h and on a log8log paper toobtain the values o K and n in e+. () or turbulent fow in a pipe.9se log8log paper and remember that n, the slope o the straightline, is given as n = (log hf 1 - log hf 2 ) / (log V1 - log V 2 ). *he y8intercept gives the value o log K.

a. :alculate the value of n. Theoretically, the head loss due to friction is

proportional to the velocity of the flow (i.e. hf / "-%+%). :omment on theobtained value of n.

b. 0iscuss the effect of fluid velocity, pipe roughness and pipe diameter on the

value of loss coefficient (D) and hence friction loss in pipe.

iii. 3lot (h)th and (h )e0p versus : on the same graph. ;ompare thedi<erence between the e0perimental and theoretical results anddiscuss on the e<ect o fuid fowrate on energy loss.

iv. Alot on the same graph paper graphs of (h )e0p versus %#%g (sudden

enlargement), %%%g (sudden contraction), %%g (&o bend) and

%%g (&o elbow)

a. 1stimate the value o loss coe"cient (K) , i.e. slope o the

graph, or each fow conditionb. ;ompare the K values and briefy discuss the effect of pipe geometry

on the value of loss coefficient and hence energy loss in pipe. c. ;ompare the average value o K with the theoretical value or

the e0periments involving sudden pipe enlargement and &o bend.

v. riefly discuss factors contributing to errors or inaccuracy in experimental data

and propose recommendation to improve the results

%1



9;A$ ;?BI4T7

%%



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%3



%6



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1.value of n7

n = (log hf # 8 log hf % ) (log # 8 log % )

pipe %7 (log =.E3 J log =.'%)+ (log =.E3 J log =.3<)

/ 1.3<EE

pipe %7 (log =.63 J log =.1<)+ (log =.E% J log =.33)

/ 1.=6%6

pipe 17 (log =.=>6 J log =.=1)+ (log =.F J log =.6)

/ =.1<1%

pipe 17 (log =.=6 J log =.=3)+ (log =1.'6 J log %.1)

/ =.=%'3

%>



%E



%F



'=



'1



'%



''



;esult and discussion7

1.

'3

sample graph-exp 1

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