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TER Assignment 6

The document provides calculations to determine the temperature of water exiting a heat exchanger. It gives the initial temperature of the water, the material properties, and calculates the heat transfer rate, exit temperature, and other values using equations of heat transfer and fluid dynamics. The calculations are shown step-by-step and include the use of an external software for material property data.

Uploaded by

Michał Owczarek
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21 views6 pages

TER Assignment 6

The document provides calculations to determine the temperature of water exiting a heat exchanger. It gives the initial temperature of the water, the material properties, and calculates the heat transfer rate, exit temperature, and other values using equations of heat transfer and fluid dynamics. The calculations are shown step-by-step and include the use of an external software for material property data.

Uploaded by

Michał Owczarek
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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6 Michal Owczarek

Assignment
-

Ts 0°C
2cm MsTEAm= 0,15k¥
Water
F- 10°C
→
v
¥-24K
✓= 4 MIS K $
5m

"
To __
10%2+24 = 17°C

EES Data : ( water at 17°C )

f- 998,7¥ Cp =

4187k¥
0=1,081.10 -6M¥ 1<=0,594%5
Pr -

-7,614

4%
Re=✓jd
•
01012m
(Turbulent)
11081.10-6%2=44403
=
4440398 7,614°
" "" '
"
Nu =
0,023 Re Pr =
0,023 . .

H
Nu =
27016

✓ = V. A =
4% .

¥10,012m/ 2=4,5210-4 k§-


in water =

f. ✓ =
998,7 k¥ •
4,5210 -4k¥ =
0,452 k¥

Q = in -

Cp (Te
. -
Ti ) 01452k¥
= •
4187 ¥3 -

124 -101°C
Q =
2615kW

6- 20
OTLN = =
11162°C
%)
Oi
Oi =
h As .
.
0 Ten h =

As 0Th
.

2615kW
h =

IT .

9012m 5m 11 , 62°C
. .
=
1211 ¥1

hfg
=

2430k¥ ( EES water at 30°C )


① total __
Ñsteam hfg
-

-0115k¥ -2430k¥
-

Q total =
364,5kW

36415kW
n=Q+
I
13,75 n=#
= =

26,5kW
-2cm
d- constantheatflux

Water

Ti -10°C
-
→

K
µ
V
)→Te=
A
80°C

Ñ=8¥n 7m

Q' = ??? Tsexit -_ ???

Tb=É= 45°C 2

EES -

water at 45°C

Cp= 4180T¥ g= 990,2k¥


1<=0,6372%1 Pr= 3,91

0=6,0210 -7m£
✓ =¥= ÉÉ =
0,424%
•

( 0 02m12
,

É
Re =
¥ =

6,02 -
to
-

7m£
=
14100 (Turbulent)

= in Cp -0T
-

?1÷ ?m
I
in -_
g. i =
990,2¥ 84min . .
.

*
in =
0,132k¥

¥E
✓
°Q= 0,132k¥ .
4180 •

( 80°C -10°C )
*
É=38,6#
"
""
0,023.1410098 3,91°
°" '
Nu =
0,023 Re Pr =
.

Nu =
82,79

82179 96372 m÷c


hid
Nu -

K .

Nu
hay
= = =

,
0102M
I

hang
=
2637,7 .mWFc
Oi hay
= •

As '

Itsexit -

Time exit
an
)
^

µ As IT d L
= =
IT 0,02m 7m =
0 44 m2
•
. -
.

38,6 kW =
2,638 k¥Fc •
944m
?
/ Tse it *
-
80°C)
I

Tse✗it=M3|26#

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