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Result and Analysis Table 1 FT1 (LPM) FT2 (LPM) TT1 (̊ C) TT2 (̊ C) TT3 (̊ C) TT4 (̊ C)

The document presents the results and analysis of an experiment involving heat transfer. It includes two tables of experimental data showing inlet and outlet temperatures and flow rates. It then shows the calculation steps to determine efficiency, heat transfer rate, log mean temperature difference, and overall heat transfer coefficient. These include converting flow rates to mass flow rates, calculating minimum heat capacity, actual heat transfer, maximum possible heat transfer, and using this to find efficiency. The heat transfer rate and log mean temperature difference are then used to calculate the overall heat transfer coefficient.

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Aeyrul Khairul
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94 views3 pages

Result and Analysis Table 1 FT1 (LPM) FT2 (LPM) TT1 (̊ C) TT2 (̊ C) TT3 (̊ C) TT4 (̊ C)

The document presents the results and analysis of an experiment involving heat transfer. It includes two tables of experimental data showing inlet and outlet temperatures and flow rates. It then shows the calculation steps to determine efficiency, heat transfer rate, log mean temperature difference, and overall heat transfer coefficient. These include converting flow rates to mass flow rates, calculating minimum heat capacity, actual heat transfer, maximum possible heat transfer, and using this to find efficiency. The heat transfer rate and log mean temperature difference are then used to calculate the overall heat transfer coefficient.

Uploaded by

Aeyrul Khairul
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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RESULT AND ANALYSIS

Table 1

FT1 FT2 TT 1 TT 2 TT 3 TT 4
(LPM) (LPM) (̊C) (̊C) (̊C) (̊C)

10.0 2.0 35.2 30.4 47.3 47.8

10.0 4.0 31.7 29.4 46.7 47.3

10.0 6.0 29.8 28.6 46.4 47.2

10.0 8.0 30.1 29.2 47.5 48.3

10.0 10.0 30.8 29.9 47.9 47.1

Table 2

FT1 FT2 TT 1 TT 2 TT 3 TT 4
(LPM) (LPM) (̊C) (̊C) (̊C) (̊C)

2.0 10.0 30.6 30.0 46.2 48.6

4.0 10.0 30.8 30.3 46.2 48.6

6.0 10.0 31.0 30.3 47.1 48.7

8.0 10.0 31.1 30.3 47.4 48.4

10.0 10.0 31.2 30.3 47.2 47.9


CALCULATION STEPS: Cp hot@48̊ C = 4.181 kJ/kg. K

VH and VC (convert LPM to m3/s) Cp cold@28̊ C = 4.179 kJ/kg. K


A = 0.067 m2
ṁH/ ṁC = V̇H/ V̇C x ρ
ρ = 988.1 kg/m3
Minimum heat capacity, Cmin = ṁC x Cp cold @ 28 °C

Heat capacity, qH = ṁH. Cp hot (THin – TH out)

Max heat transfer, q max = Cmin (TH in – TC in)

𝑞𝐻
Efficiency = x 100 %
𝑞 𝑚𝑎𝑥

𝚫𝐓𝐢𝐧− 𝚫𝐓 𝐨𝐮𝐭
Log Mean Temperature Difference, ΔTm = 𝚫𝐓 𝐢𝐧
𝒊𝒏
𝚫𝐓 𝐨𝐮𝐭

𝒒
U = 𝑨 𝒙 𝚫𝐓𝐦

For the first calculation, ( FT1 = 10 LPM, FT2 = 2 LPM )

𝑳 𝟏 𝒎𝒊𝒏 𝟏 𝒎𝟑 −𝟒 𝒎𝟑
VH = 10 𝑴𝑰𝑵 x x 𝟏𝟎𝟎𝟎 𝑳 = 𝟏. 𝟔𝟔𝟕𝟏𝟎
𝟔𝟎 𝒔 𝒔

𝑳 𝟏 𝒎𝒊𝒏 𝟏 𝒎𝟑 −𝟒 𝒎𝟑
VC = 2 𝑴𝑰𝑵 x x 𝟏𝟎𝟎𝟎 𝑳 = 𝟎. 𝟑𝟑𝟑𝟏𝟎
𝟔𝟎 𝒔 𝒔

−4 𝑀3 𝑘𝑔 𝑘𝑔
ṁH = VH x ρ = 1.66710 𝑥 988.1 = 0.1647
𝑆 𝑚3 𝑠

−4 𝑘𝑔 𝑘𝑔
ṁC = VC x ρ = 0.33310 𝑥 988.1 = 0.0329
𝑚3 𝑠

Cmin = ṁC x Cp cold @ 28 °C

𝑘𝑔 𝐾𝐽
= 0.0329 x 4.181 𝑘𝑔 x ( 320.8 – 320.3K )
𝑠

𝑘𝐽
= 0.1375 𝐾 . 𝑠

qH = ṁH. Cp hot (THin – TH out)

𝑘𝑔 𝑘𝐽
= 0.1647 𝑥 4.181 𝑥 (320.8 − 320.3𝐾)
𝑠 𝑘𝑔

= 0.3443 kW
q max = Cmin (TH in – TC in)

𝑘𝐽
= 0.1375 𝐾 . 𝑠 x (320.8 – 303.4K)

= 2.3925 kW

𝑞𝐻
Efficiency = 𝑞 𝑚𝑎𝑥 x 100 %

0.3443
= x 100
2.3925

= 14.4%

𝚫𝐓𝐢𝐧− 𝚫𝐓 𝐨𝐮𝐭 (𝟑𝟐𝟎.𝟖−𝟑𝟎𝟑.𝟒 𝑲)−(𝟑𝟐𝟎.𝟑−𝟑𝟎𝟖.𝟐𝑲)


ΔTm = 𝚫𝐓 𝐢𝐧 = 𝟑𝟐𝟎.𝟖−𝟑𝟎𝟑.𝟒 = 𝟏𝟒. 𝟓𝟗 𝑲
𝒊𝒏 𝑰𝒏 ( )
𝚫𝐓 𝐨𝐮𝐭 𝟑𝟐𝟎.𝟑−𝟑𝟎𝟖.𝟐

𝒒
U = 𝑨 𝒙 𝚫𝐓𝐦

𝟎.𝟑𝟒𝟒𝟑 𝒌𝑾
= 𝟎.𝟎𝟔𝟕 𝒙 𝟏𝟒.𝟓𝟗 = 𝟎. 𝟑𝟓𝟐𝟐 𝒎𝟑 . 𝑲

The calculation steps above were repeated for all data in table 1 and 2

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