The document summarizes an experiment on a multi-pass counter-flow liquid-to-liquid heat exchanger. It includes readings from 4 trials measuring inlet and outlet temperatures on the hot and cold sides. Calculations were shown to determine the heat transfer (q) and overall heat transfer coefficient (U) for each trial. While q transferred was expected to be equal between the hot and cold sides, some trials showed a deviation, possibly due to heat loss. The heat exchanger performance improved with increasing flow rate as both q and U increased.

1. Objectives:
To Study the performance of Liquid to liquid ,counter flow heat exchanger
with multi pass.
Readings:-
Heat Exchanger: Multi-Tube Hot tube/Cold Shell
Flow Arrangement: Counter flow All air Purged
Immersion Heater Setting: 9Kw Thermocal-10%
Hot Thermostat Setting: 50 C
Cp= 4.18Kj/kg.k
Hot Sides
Test
no.
Flow m T1 (in) T2 Diff. Velocity in tubs
(L/min
)
(kg/s) o
C (out) o
C
o
C
1 5 0.083 52 40 12 0.052031093
2 10 0.167 52.5 44.5 8 0.104689067
3 15 0.25 53 46.5 6.5 0.15672016
4 20 0.333 53 48 5 0.208751254
Cold Sides
Test
no.
Flow m T1 (in) T2 Diff. Velocity
in tubs
(L/min
)
(kg/s) o
C (out) o
C
o
C
1 10 0.166 3 9 6 0.1
2 10 0.166 3 11 8 0.1
3 10 0.166 4 13 9 0.1

2. 4 10 0.166 5 15 10 0.1
Trial ∆Τ1 ∆Τ2 ∆Τµ
1 52 40 45.73793624
2 52.5 44.5 48.38983419
3 53 46.5 49.67914876
4 53 48 50.45871888
Trial q(W) U(W/m2
.o
C)
1 4163.28 439.732647
2 5584.48 557.517187
3 6792.5 660.518758
4 6959.7 666.321696
Calculations:
To Calculate q for trial 1:
q = m*Cp*∆T
q = 0.083*4180*12 = 4163.28 W = 4.1632 KW
To Calculate the velocity:
m = 5*ρ*V*A, ρ=997kg/m3
Area of 1tube= .25* π *.022
=.00032m2
V=0.052 m/s
To find U:
q=U A ∆Tm, A=20.7*10-2
m2
4.1632KW = U *A * 45.73793624
U= 439.732647 (W/m2
.o
C)
Cold Sides

3. Test
no.
Flow m T1 (in) T2 Diff. Velocity
in tubs
(L/min
)
(kg/s) o
C (out) o
C
o
C
1 10 0.166 3 9 6 0.1
2 10 0.166 3 11 8 0.1
3 10 0.166 4 13 9 0.1
4 10 0.166 5 15 10 0.1
Trial q(W) U(W/m2
.o
C)
1 4163.28 3682.63331
2 5551.04 4355.29699
3 6244.92 3950.94265
4 6938.8 3682.63331
Calculations:
To Calculate q for trial 1:
q = m*Cp*∆T
q = 0.1667 *4180* 6 = 4163.28 W
To find U:
q=U A ∆Tm
Same as the hot side., where A = 20.7*10-2
m2
U = 4163.28 / (20.7*10-2
* 45.73793624) = 3682.63331 (W/m2
.o
C)
Discussion:
The water flow rate in the hot side will transfer heat to the cold side, where
theoretically heat rejected by hot side equals heat gained by the cold side.
In counter flow heat exchanger the flow of the hot fluid is in opposite direction of
the cold fluid.

4. Increasing the mass flow rate will increase q on both sides as shown in the figures.U
will also increase with the increase of mass flow rate.
q rejected must equal q absorbed. In this experiment there was some deviation from
the law of conservation of energy. For example q for the hot & cold sides in the 2nd
trial:
This means that there is heat loss during heat transfer.
Sources of errors are due to heat loss,error in taking reading and in calculation
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f;ow rat ( L/min)
Q(W)