¡Descarga ejercicios calor e intercambiadores y más Monografías, Ensayos en PDF de Calor y Transferencia de Masa solo en Docsity!
Module 7: Solved Problems
1. Deionized water flows through the inner tube of 30-mm diameter in a
thin-walled concentric tube heat exchanger of 0.19-m length. Hot
process water at 95C flows in the annulus formed with the outer tube
of 60-mm diameter. The deionized water is to be heated from 40 to
60 C at a flow rate of 5 kg/s. The thermo physical properties of the
fluids are:
DEIONIZED
WATER
PROCESS
WATER
kg/m3)
cp(J/kg.K
k(W/m.K
N.s/m
2
pr
(a) Considering a parallel-flow configuration of the heat exchanger,
determine the minimum flow rate required for the hot process
water.
(b) Determine the overall heat transfer coefficient required for the
conditions of part a.
(c) Considering a counter flow configuration, determine the minimum
flow rate required for the hot process water. What is the
effectiveness of the exchanger for this situation?
Schematic:
Process
water,h
Deionized
water,c
T c,i =40 C
T h,i =95 C
ΔT 1 T^2
mc=5kg/s
T c,0i =60 C
T h,o
T
x
Concentric tube
PF,L=0.19m
D=30mm
Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible
kinetic and potential energy changes.
Analysis: (a) from overall energy balances,
.
, ,
. q mc h Thi Tho mc hTco Tc i
For a fixed term , ) h will be a minimum when T h,o is a minimum.
With the parallel flow configuration, this requires that Th,o=Tc,o =60C.
Hence,
Th (^) , i
. ( m
kg s J kgK C
kg s J kgK C
c T T
mc T T mh h hi ho
c co ci
- 85 / 4197 /. ( 95 60 )
,min , ,
.
. , ,
(b)From the rate equation and the log mean temperature relation,
T
T
ln
T - T
2
1
1 2 , ,
q UA Tlm PF TlmPF
2. Water with a flow rate of 0.05kg/s enters an automobile radiator at
400K and leaves at 330 K. The water is cooled by air in cross flow
which enters at 0.75kg/s and leaves at 300K. If the overall heat
transfer coefficient is 200W/m
2
.K, what is the required heat transfer
surface area?
Schematic:
Water T (^) h,i =400K
m (^) h=0.05kg/s
Tc,i =300K
m (^) c=0.75kg/s
Air
Th,o=330K
Tc,o
Assumptions: (1) Negligible heat loss to surroundings and kinetic and
potential energy changes, (2) Constant properties.
Analysis: The required heat transfer rate is
q ( mc ) h ( Th , i Th , o ) 0. 05 kg / s ( 4209 J / kg. K ) 70 K 14 , 732 W
.
Using the -NTU method,
max
min max , ,
max
min
q q W W
and
henceC C T T W K K W
C C W K
C C W K
hi ci
c
h
From figure, NTU1.5, hence
2 2 A NTU ( C min / U ) 1. 5 210. 45 W / K ( 200 W / m. K ) 1. 58 m
Comments: (1) the air outlet temperature is
T (^) c , o Tc , i q / Cc 300 K ( 14 , 732 W / 755. 25 W / K ) 319. 5 K
(2) Using the LMTD approach, ΔTlm=51.2 K, R=0.279 and P=0.7.
Hence from fig F0.95 and
/ ( 14 , 732 )/[ 0. 95 ( 200 /. ) 51. 2 ] 1. 51.
2 2 A q FU Tlm W W m K K m
NTU -ln(1- ) ln( 1 0. 62 ) 0. 97
since C /C 0,
, /( [ ]) 3. 46 10 / 86 , 400 / ( 65 ) 0. 62
min max
6 min , ,
6 min , ,
hence q C T T W W K K
C C q T T W K W K
hi ci
c co ci
And
m C c W K J kgK kg s
A NTUC U W K W m K m
c (^) c / pc 86 , 400 / / 4178 /. 20. 7 /
,
.
min
(b) using the final overall heat transfer coefficient, find
Since C min /Cmax0,
m q h W J kg kg s
T T W K K W
NTU
h fg
hi ci
hence, q C ( ) 0. 384 ( 886 , 400 / ) 65 2. 16106
1 exp( ) 1 exp( 0. 485 ) 0. 384
6 6
.
min , ,
Comments: The significant reduction (38%) in represents a
significant loss in turbine power. Periodic cleaning of condenser
surfaces should be employed to minimize the adverse effects of
fouling.
m h
.
4. Water at 225 kg/h is to be heated from 35 to 95C by means of a
concentric tube heat exchanger. Oil at 225kg/h and 210C, with a
specific heat of 2095 J/kg.K, is to be used as the hot fluid. If the
overall heat transfer coefficient based on the outer diameter of the
inner tube if 550W/m
2
.K, determine the length of the exchanger if the
outer diameter is 100mm.
Schematic:
Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible
kinetic and potential energy changes, (3) Constant properties.
Properties: Table for Water:
( T (^) c ( 35 95 ) C / 2 338 K ): cp , c 4188 J / kg. K
_
Analysis: From rate equation with Ao=DoL, L=q/U oDoΔT m
The heat rate, q, can be evaluated from an energy balance on the cold
fluid,
J kgK K W s h
kg h q mc cc Tc Tci 4188 /. ( 95 35 ) 15 , 705 3600 /
.
In order to evaluate ΔT m, we need to know whether the exchanger is
operating in CF or PF. From an energy balance on the hot fluid, find
C
kgK
J
s h
kg h T (^) h o Thi q mhch C W
- 1 .
.
, ,
5. Consider a very long, concentric tube heat exchanger having hot and
cold water inlet temperatures of 85 and 15C. The flow rate of the hot
water is twice that of the cold water. Assuming equivalent hot and
cold water specifies heats; determine the hot water outlet temperature
for the following modes of operation (a) Counter flow, (b) Parallel
flow.
Schematic:
Th,i=85 C
Tc,i =15 C
Cc
Ch=2Cc
Assumptions: (1) equivalent hot and cold water specific heats, (2)
Negligible Kinetic and potential energy changes, (3) No eat loss to
surroundings.
Analysis: the heat rate for a concentric tube
Heat exchanger with very large surface area
Operating in the counter flow mode is
q q max C min( Th , i Tc , i )
Combining the above relation and rearranging, find
hi ci h i h
c hi ci hi h
h o T T T C
C
T T T
C
C
T , , , , , ,
min , (^ ) ( )
Substituting numerical values
T (^) h o ( 85 15 ) C 85 C 50 C 2
,
For parallel flow operation, the hot and cold outlet temperatures will
be equal; that is T c,o=Th,o. Hence
C (^) c ( Tc , o Tc , i ) Ch ( Th , i Th , o )
Setting Tc,o=Th,o and rearranging
T C C
C
C
T
C
C
T T
ho
h
c ci h
c ho hi
,
, , ,
Comments: Note that while =1 for CF operation, for PF operation
find = q/q max=0.67.
q mhcphThi Tci kg s J kgK C W
5
. ,
. , ( ) 0. 5 / ( 2650 /. )( 100 60 ) 0. 53 10
5 5
q q max
(b)
C
kg s J kgK
C
m c
q T T
c pc
c o ci
5
.
,
, ,
Since T c,o<Th,o, a parallel flow mode of operation is possible.
However, with (Cmin/Cmax ) = ( / , )=0.63,
.
,
. m (^) hcp (^) h mccp c
From fig (NTU) PF0.95, (NTU) CF0.
Hence
(ACF/APF)= (NTU) CF/ (NTU) PF (0.75/0.95)=0.
Because of the reduced size requirement, hence capital investment,
the counter flow mode of operation is preferred.
