HT 305HEAT TRANSFER THROUGH

SUBMERGED HELICAL COIL IN ANAGITATED VESSEL

Date of Experiment: 21st September, 2011.

Date of Presentation: 26th September, 2011.

Name of the group: B1(b)Student Information:

Sri Vivek 09002059 Report Puneet Singla 09002060 Report Amruta Rokade 09d02019 PPT Tarun Gangwar 09d02022 Presentation

Motivation and Objectives

• Determination of coil side heat transfer coefficient through submerged helical coil in the vessel under steady state conditions.

• Tube coils offer a substantial amount of heat transfer area at a considerably low cost.

• Coils have lower wall resistance & higher coil side HT coefficient.

• Helical coil heat exchanger is compact in size and less expensive.

• The tubes are coiled into helices in which inlet and outlet are conveniently located side by side.

The SiederTate equation is

NuD= 0.023(ReD)0.8(Pr)1/3(μ/μw)0.12…(1)

• The correlation describing the Nusseltno. for heat transfer to fluids in vessel with mechanical agitation heated or cooled by submerged coils isknown as Chilton, Drew and Jebens’ correlation:

hcd/k = 0.87(NρL2/μ)2/3(Cpμ/k)1/3(μ/μw)0.12 …(2)

• The general heat transfer equation is given by

1/U = 1/ho + 1/hi+x/k …(3)

• The thermal conductivity of pipe material is very high so the above expression is reduced to:

1/U = 1/ho + 1/hi …(4)

Theory

To calculate experimentally, we use the relationship between hi and v0.8. Substituting this and the Seider Tate equation in the Overall heat transfer equation, we get:

  The plot of 1/U vs 1/ v0.8 gives 1/ ho (k2) as intercept. Then, we get hi by subtracting the intercept from 1/U and reciprocating.The correlation describing the Nusselt no. for heat transfer to fluids in vessel with mechanical agitation heated or cooled by submerged coils is as follows: 

Where, Dj = inside diameter of the agitated vessel [m] hc = coil side heat transfer coefficient [kcal/hr-m2-0C] L = agitator diameter [m] N = agitator speed [rev/sec] or rev/hr K = thermal conductivity of fluid in the vessel [Kcal/hr-m-OC] μ = viscosity of fluid in the vessel [kg/m-hr] μw = viscosity of fluid in the vessel at coil wall temp. [kg/m-hr]

Experimental Setup:

Experimental Procedure: Fill the given agitated vessel with the given

test liquid to about 85-90 % of its capacity and start the agitator

motor

Start the water motor pump. Adjust the flow rate of the water at the

desired level by its speed regulator

Start the heaters in the agitated vessel and set the desired temperature (about 600C) on the thermostat. Throughout the given set

of readings keep this temperature at this level.

After steady state is attained note down

inlet and outlet temperatures of the cooling water. Also

measure the flow rate of the cooling water

Take three readings of temperatures for each flow rate and keep the

agitation speed constant.

Perform the same experiment by keeping the flowrate of water

constant (at about 400 lph) and varying the rpm of the agitator

Calculation Procedure Calculate the amount of heat transferred to

the cold fluid Q=m*C*∆T

Calculate LMTD and area of heat transfer

Calculate overall heat transfer coefficient U using U=Q/A*LMTD

Calculate velocity v and Reynolds no Re of fluid in the pipe

Plot 1/U v/s 1/v^0.8 and obtain ho from

intercept

Calculate inner heat transfer coefficient

(experimental)

Calculate theoretical heat transfer

coefficient hth from Seider Tate Equation

Plot Re vs hth and Re vs hexp

Plot 1/v^.8 vs hexp and 1/v^.8 vs hth

For Part 2 calculate Q and U

Plot log U ns log N and calculate the

error

Observation table

Sno Flow rate (cold)

t1 t2 LMTD Q=mCpdT U v 1/v0.8 1/U

1 240 27.8 32.4 34.84 1288.36 378.82 1.04 0.96 0.0026

2 300 27.8 31.6 35.26 1330.38 386.55 1.30 0.80 0.0026

3 360 27.7 31.2 35.52 1470.42 424.17 1.57 0.69 0.0024

4 420 27.7 30.5 35.88 1372.39 391.92 1.83 0.61 0.0026

5 480 27.7 30.0 36.13 1288.36 365.31 2.09 0.55 0.0027

6 520 27.7 29.6 36.34 1197.34 337.60 2.35 0.50 0.0030

7 600 27.7 29.3 36.49 1120.32 314.56 2.61 0.46 0.0032

Observation table

1/v0.8 1/U 1/ho hi exper Re Pr Nu hi theor

0.96 0.0026 0.0017 1418.83 11817 5.77 87.86 5616.97

0.80 0.0026 0.0017 1503.31 14772 5.77 104.19 6714.76

0.69 0.0024 0.0017 2027.87 17726 5.77 120.55 7769.18

0.61 0.0026 0.0017 1565.80 20681 5.77 136.37 8788.86

0.55 0.0027 0.0017 1285.35 23635 5.77 151.75 9779.72

0.50 0.0030 0.0017 1056.48 26589 5.77 166.74 10747.04

0.46 0.0032 0.0017 901.52 29544 5.77 181.41 11691.07

Graphs

Graphs

Graphs

5.2 5.4 5.6 5.8 6 6.2 6.4 6.65

5.2

5.4

5.6

5.8

6

6.2

6.4

f(x) = 0.556164238133613 x + 2.61252833045675

f(x) = 0.501509144784976 x + 2.83198245998345

Ln U and Ln ho vs. Ln N

lnU

Linear (lnU)

ln ho

Linear (ln ho)

Ln N

Ln

U a

nd

Ln

ho

Results and Discussion

• The value of h0 is obtained from the graph of 1/U vs 1/v^0.8 to be 588.23 J/(m2s oC).

• The value of inner side heat transfer coefficient hi increases with increase in the flow rate of the cold fluid, as can be observed from graph of hi vs 1/v^0.8.

• The value of inner side heat transfer coefficient hi increases with increase in the number of rotations of the agitator in the hot fluid. The slope of the graph of log U vs log N is seen to be 0.55 , whereas the expected value from the Chilton’s, Drew’s and Jeben’s corrrealtion is 0.66 with an error of 16 %

Conclusion

• With increase in flow rate, the velocity of each differential packet gets less time for heat exchange, thus less temperature gain by the cold fluid.

• With the increase in agitator speed, the turbulence increases and thus the heat exchange.

• The helical coil arrangement’s heat transfer coefficient is better than the corresponding straight coil, due to more turbulence generated inside the tube because of turns of helix geometry.

Precautions and Sources of Error

• A constant watch must be kept on the rotameter as the flow rate of the cold fluid (water) fluctuates.

• The readings are to be taken only in the relay “off” state as there are temperature fluctuations in the relay “on” state.

• The hot fluid chamber is open to air. Loss of liquid to the atmosphere might pose a problem. A closed agitator vessel would be better.

• Cooling water inlet was from tap , so inlet temperature is not possible to maintain constant