Paper Format NCB

7/29/2019 Paper Format NCB

1/6

Available online at www.sciencedirect.com

Procedia Engineering 00 (2014) 000000

www.elsevier.com/locate/procedia

Selected papers of Mechanical, Civil and Chemical engineering Tracks of 4th Nirma UniversityInternational Conference on Engineering, 2013

Click here, type the title of your paper, Capitalize first letter

Abstract

Click here and insert your abstract text. 2014 The Authors. Published by Elsevier Ltd.

Selection and peer-review under responsibility of Institute of Technology, Nirma University, Ahmedabad, India.

Keywords: Type your keywords here, separated by semicolons ;

1. Main text

Here introduce the paper, and put a nomenclature if necessary, in a box with the same font size as the rest of the

paper. The paragraphs continue from here and are only separated by headings, subheadings, images and formulae.

The section headings are arranged by numbers, bold and 10 pt. Here follows further instructions for authors.

Nomenclature

A radius of

B position of

C further nomenclature continues down the page inside the text box

1.1. Thermal Analysis

To find out the amount of steam which is generated during the heating process, it is necessary to find out the

amount of heat utilized by the water. Experiments are conducted on different quantities of water. Readings of

initial and final temperature of water are taken after an interval of five minutes and time required for complete

evaporation is measured. The time required for complete evaporation of water is also measured. The heat utilized,

Qu is calculated as,

pm C dTQ

Time

=

(1)

1877-7058 2014 The Authors. Published by Elsevier Ltd.

Selection and peer-review under responsibility of Institute of Technology, Nirma University, Ahmedabad, India.
http://www.sciencedirect.com/science/journal/22120173http://www.sciencedirect.com/science/journal/22120173


2/6

2

Where, Qabs is the heat absorbed in watts, m is mass of water in kg/sec. dT is temperature difference between

initial and final water tank temperature which is 25oC

790 m 4187 25=

m= 0.007547 kg/sec

m= 27.14 kg/hr

System takes almost 6 hours and 30 minutes to evaporate 8kg of water. Hence 175 liters of water is required for

condensation of 8kg of water.The amount of heat transferred to through copper tube is calculated as,

mQ U A T= (2)

Where, U is overall heat transfer coefficient, whose value ranges from 1000 W/m2K to 5000W/m2K . Lowest

value of U is considered for calculating the area. Tm is the log mean temperature difference which is given by the

formula,

h, i c, o h, o c, i

h, i c, o h, o c, i

(T T ) (T T )T m

log((T T ) / (T T ))

=

V (3)

Where, Thi is the inlet temperature of steam, Tho is the outlet temperature of steam, Tci is the initial temperature

of water, Tco is the final temperature of water. As the water temperature increases in every hour, so Tm is also

calculated for each hour.

Reason

1.2. Heat losses from the condenser to the atmosphere

a) b)


3/6

3

Fig. 1. (a) 120 liter of tank; (b) Heat losses from the tank.

Fig. shows the heat losses,

Where,

Q1= Qc+QR

Qc = Heat losses by convection.QR= heat losses by Radiation

Heat losses due to convection from 0.32x0.5 wall i.e. Q1

C = Q h A T

A= Surface Area

0.5 0.32= 0.16 2 0.32.= =

C 6 .32 (44 26)= Q

=34.56

Heat losses due to radiation from 0.32 x 0.5 wall

4 4R 1 2( )Q A T T =

8 4 4R 5.67 10 0.28 0.32 (317 299 )

= Q

R 10.69 Watt=Q

1 = +C RQ Q Q


4/6

4

1 34.56 10.69 45.25 Watt= + =Q

Similarly,

Heat losses due to convection from 0.75 x 0.32 Wall (Q2)

C = Q h A T

A= Surface Area 0.75 0.32= 0.24= 0.24 2 0.48= =

C 6 .48 (44 26)Q =

=51.84 Watt

Heat losses due to radiation from 0.75 x 0.32 wall

4 4R 1 2( )Q A T T =

8 4 4R 5.67 10 0.28 0.48 (317 299 )= Q

R 16.04 WattQ =

2 C RQ Q Q= +

2 51.84 16.04 67.88 WattQ = + =

Heat losses from upper surface of water due to convection (Q3)

=

Qc h A T

A= Surface Area 0.75 0.5= 0.375=

C 6 0.375 (55 26)Q =

C 65.25 WattQ =

Heat losses due to radiation from water surface4 4

R 1 2( )Q A T T =


5/6

5

8 4 4R 5.67 10 0.9 0.375 (328 299 )= Q

R 68.54 WattQ =

3 = +C RQ Q Q

3 65.25 68.54= +Q

3 133.79 Watt=Q

Heat absorbed by water in the tank

p

liquid

m C dTQ Time

=

liquid 120 4187 24Q23400

=

liquidQ 515 Watt=

Total 1+ 2+ 3+ liquid=Q Q Q Q Q

Total + + +45.25 67.88 133.79 515.32=Q

Total 762.24 Watt=Q

So actual heat absorbed by the water is 480 watts their four we can optimized the size of the tank . To find out the

amount of steam which is generated during the heating process, it is necessary to find out the amount of heat

utilized by the water by eq. (1)

pm C dTQ

Time

=


6/6

6

1.3. CFD Analysis

Cfd analysis of 120 liter of tank is carried out with the help of Ansys Fluent. For CFD analysis a tank isgenerated of required size. Meshing of tank and tube is done the maximum number of nodes are up to 500000 with

skewness isless than 0.9.

Fig.4. Contours of 120 lit of tank

Paper Format NCB

Documents

Transcript of Paper Format NCB