7 12521039 savannah yonita tugas2

8/21/2019 7 12521039 Savannah Yonita Tugas2

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Nama : Savannah Yonita C

NIM : 12521039

Kelas : A

Kelompok : 7

1. Nat!al Conve"tion

7. A 40-cm-diameter, 110-cm-high cylindrical hot water tank is located in the bathroom

of a house maintained at 20°. !he surface tem"erature of the tank is measured to be 44°.

!aking the surrounding surface tem"erature to be also 20°, determine the rate of heat loss

from all surfaces of the tank by natural con#ection.

$awab %

&iketahui %! ' 20°

!w ' 44°

k ' 0,02(0) *+m. °

# ' 1,(2710- m2+s

r ' 0,727(

&itanya % /total ' ...

$awab %

ntuk bagian selimut silinder

• Gr=gβ (Tw−T ) L3

v2

Tf =( Tw+T )

2=

44+202

=32° C

β= 1

Tf =

1

(32+273 ) K =3,2787 x10

−3/ K

110

40 cm=0,4



Gr=

9,8m

s2

x3,2787 x 10−3/ K x (44−20) K x (1,1m)3

(1,627 x 10

−5m

2

s )

2

')77411373' ),77103

ilinder tegak da"at disamakan dengan "lat tegak 5ika D(0,4m)≥

35 L

Gr1 /4

3,883 x 109 ¿

1

4

¿¿

D=35 x1,1m

¿

0,4m ≥0.1542

• 6a ' r r

' ),77103 0,727( ' 22030200

' 2,2103

• ntuk kisaran 6a 103-101) Nu=0,1 xRa L1/3

Nu=0,1 x2,82 x109

L1 /3

' 141,2

• h=

k

L Nu

h=0,02603W /m ℃

1,1m x141,28=3,34W /m

2℃

• A s=πDL

'),140,4m1,1m

'1,)2 m2

• Q=h As (Tw−T )

'

3,34W

m2

℃ x1,)2 m2844-209°'110,7*

ntuk bagian tutu" silinder

• Lc=

A s

p



¿ π D

2/4πD

= D

4

¿0,4 m

4=0,1m

• Ra=

gβ (Tw−T ) Lc3

v2

Pr

Ra=

9,8m

s2

x 3,2787 x 10−3/ K x (44−20 ) K x(0,1m)3

(1,627 x10

−5m

2

s )

2 x0,7276

'2,12)10(

• Nu=0,54 Ra

1 /4=0,54 x (2,123 x106)1/4=20,61

•h=

k

Lc

Nu

h=0,02603W /m ℃

0,1m x20,61=5,36W /m

2℃

• As=π D

2/4

¿3,14 x (0,4 m)2

4=0,1256m

2


¿5,36W

m2

℃ x 0,1256m2 x (44−20 ) ℃=16,16W

ntuk bagian bawah silinder

• Nu=0,27 Ra

1/4=0,27 x(2,123 x 106)1 /4=10,31

•h=

k

Lc

Nu

h=0,02603W /m ℃

0,1m x10,31=2,68W /m

2℃


•¿2,68W

m2

℃ x 0,1256 m2 x (44−20) ℃=8,08W

$adi, /total di seluruh "ermukaan silinder ' /selimut:/tutu" atas:/tutu" bawah



'110,7*: 16,16W : 8,08W

'1),02 *



2. #$te!nal %o!"e& Conve"tion

7. onsider a "erson who is trying to kee" cool on a hot summer day by turning a fan on

and e"osing his entire body to air flow. !he air tem"erature is °; and the fan is blowing

air at a #elocity of ( ft+s. <f the "erson is doing light work and generating sensible heat at a

rate of )00 =tu+h, determine the a#erage tem"erature of the outer surface 8skin or clothing9

of the "erson. !he a#erage human body can be treated as a 1-ftdiameter cylinder with an

e"osed surface area of 1 ft2. &isregard any heat transfer by radiation. *hat would your

answer be if the air #elocity were doubled

$awab %

! ' °;

> ' ( ft+s

/ ' )00 =tu+h

k ' 0,0123 =tu+h.ft. °;

# ' 0,10310-) m2+s

r ' 0,72(0

• ℜ=

D

v

¿

6 f!

s x 1 f!

0,1809 x10−3f!

2/s =3,317 x 10

4

• Nu=

hD

k

1+(0,4

Pr )

2/3

¿¿1/ 4¿¿

¿0,3+ 0,62ℜ0,5

Pr

1 /3

¿

D=1 ft



1+( 0,4

0,7260)2/3

¿¿1/ 4¿¿

¿0,3+0,62(3,317 x10

4)0,50,72601 /3

¿

' 107,4

• h=

k

D Nu

¿0,01529

"!u

h # f ! # ℃

1 f! x107,84

¿1,65 "!uh

# f! 2# ℃

• Q=h A s (Ts−T )

Ts=T + Q

h A s

¿85 ℃+ 300"!u /h

1,65 "!u

h # f!

2# ℃ x18 f!

2

=95,10 ℃

$ika kece"atan angin men5adi dua kali li"at, maka 2( ft+s ' 12 ft+s

• ℜ=

D

v

¿

12 f!

s x1 f!

0,1809 x10−3

f! 2/s

=6,633 x 104

• Nu=

hD

k

1+(0,4

Pr )

2/3

¿¿1/ 4¿¿

¿0,3+0,62ℜ0,5

Pr1 /3

¿



1+( 0,4

0,7260)2 /3

¿¿1/4¿¿

¿0,3+0,62(6,633 x10

4)0,50,72601 /3

¿

' 1(,3

• h=

k

D Nu

¿0,01529

"!u

h # f ! # ℃

1 f! x165,95=2,537

"!u

h # f!

2# ℃

• Q=h A s (Ts−T )

Ts=T + Q

h A s

¿85 ℃+ 300"!u /h

2,537 "!u

h # f!

2# ℃ x 18 f!

2

=91,6 ℃

$adi, 5ika kece"atan dinaikkan men5adi dua kali li"at, suhu luar badan manusia

men5adi 31,(



3. Inte!nal %o!"e& Conve"tion

7. *ater is to be heated from 20° to ° as it flows through a )-cm-internal-

diameter, 3 m-long tube. !he tube is e?ui""ed with an electric resistance heater,

which "ro#ides uniform heating throughout the surface of the tube. !he outer surface of

the heater is well insulated, so that in steady o"eration all the heat generated in the heater is transferred to the water in the tube. <f the system is to "ro#ide hot water at a rate of .

@+min, determine the "ower rating of the resistance heater. Also, estimate the inner surface

tem"erature of the "i"e at the eit.

Answer

'ik :

T$=(T%+T&)

2=20+85

2=52,5

0C

' 332,1 kg+m)

k ' 0,()1 *+m.0

# ' B + ' 0,( 10-( m2+s

" ' 4173 $+kg.0

r ' 4,)2

'it :

a. Q=' (

b. !s 'CC..

(a)a*an :

a.+ Ac ' D E &2 ' D E 80,0) m92 ' 7,0( 10-4m2

As ' pL' E & @ ' E 80,0) m9 83 m9 ' 047 m2

$ikadiketahui

' , @+min ' 0.00 m

)

+minm= ) x ' 8332,1 kg+m)9 80.00 m)+min9 ' ,4)23 kg+min ' 0,140 kg+s

Q ' m " 8 !e F !i 9

' 80,140 kg+s9 84,173 k$+kg.09 8 F 209 0

' ),1( k$+s

' ),1( k*

*.+ *s ' h 8!s F !m9 !s ' !m :

*s

h

*s '´Q + As ' ),1( k*+ 0,47 m2 ' 4,01 k*+m2



m 'v

Ac '0,0085m

3

7,065 x10−4

m2=¿

12,0) m+min ' 0,200 m+s

6e ' m x Dv ' (0,2005

m

s

)(0,03m)

0,658 x 10−6

m2/ s

=¿ 3141,)4

Gu ' 8 h &9 + k ' 0,02) 6e0, r 0,4 ' 0,02) 8 3141,)490, 84,)490,4 ' (1,0)

Hemudian,

h 'k

D Gu ' 80,()1 *+m.o9 8(1,0)9 + 80,0) m9 ' 12),(( *+m2.o

sehingga

!s ' !m : 8 *s + h 9 ' o : I 84010 *+m29 + 812),(( *+m2.o9 J ' 120,0( o

+

120 o

7 12521039 savannah yonita tugas2

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