Working on the radiator of Suzuki Baleno 1999.
How to calculate the overall heat transfer coefficient (U)?
How to calculate the heat transfer area and compare it with the experimental data being collected.
Episode 59 : Introduction of Process Integration
Pinch Diagram and Heat Integration
Reference: Notes from course on “Modelling, design and control for process integration”, CAPEC, August 2000 (R. Dunn)
SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Energy balance of Diesel Production plant in refinery. Calculation of make up hydrogen requirement in the reactor. Calculation of Steam requirement in fractionator for distillation.
Episode 59 : Introduction of Process Integration
Pinch Diagram and Heat Integration
Reference: Notes from course on “Modelling, design and control for process integration”, CAPEC, August 2000 (R. Dunn)
SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Energy balance of Diesel Production plant in refinery. Calculation of make up hydrogen requirement in the reactor. Calculation of Steam requirement in fractionator for distillation.
Episode 60 : Pinch Diagram and Heat Integration
The optimal allocation of mass and energy within a unit operation, process and/or site.
Optimal allocation can be based on economic, environmental or other important objectives.
SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
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Often engineers are tasked with communicating equipment specifications with suppliers, where process data needs to be exchanged for engineering quotations & orders. Any dearth of data would need to be computed for which process related queries are sometimes sent back to the process engineer’s desk for the requested data.
The following tutorial is a refresher for non-process engineers such as project engineers, Piping, Instrumentation, Static & Rotating Equipment engineers to conduct basic process calculations related to estimation of mass %, volume %, mass flow, actual & standard volumetric flow, gas density, parts per million (ppm) by weight & by volume.
Cooling Towers in Process Industries are part of Utilities design. As the name suggests their primary purpose is to provide cooling requirements to industrial hot water from unit operations & unit processes. Examples include chillers and air conditioners. The principle of operation is to circulate hot water through a tower and allow heat dissipation to the ambient. Cooling towers can operate by natural draft or forced draft methods wherein fans are used to increase heat transfer.
A QUICK ESTIMATION METHOD TO DETERMINE HOT RECYCLE REQUIREMENTS FOR CENTRIFUG...Vijay Sarathy
Turbomachinery Engineers often conduct studies to determine if a hot gas bypass is required for a given centrifugal compressor system. This would mean building a process model and simulating it for Emergency Shutdown conditions (ESD) & Normal Shutdown conditions (NSD) to check if the compressor operating point crosses the surge limit line (SLL). A quick estimation method that uses dimensionless number called the inertia number can be used to check prior to the study, if a Hot gas bypass (a.k.a. Hot Recycle) is required in addition to an Anti-surge line (ASV or a.k.a Cold Recycle).
Exploring LPG Cylinders for Medical Oxygen - A Preliminary StudyVijay Sarathy
The following article is a study to explore the usage of LPG cylinders for medical oxygen in times of medical emergencies. The study aims at understanding how long medical oxygen can be supplied to cater to patients requiring supply between 0.5 lit/min to 2 lit/min.
Abstract The requirement of energy in any processing industry is not only a need but it is indeed a most wanted utility. In a typical processing or manufacturing industry the most common utility are steam and cooling water. However the cost of these utility are no longer cheap, in fact they are expensive. Therefore saving these utility or minimizing the usage of these utilities is one of the most needed practice in a processing industry. Pinch technology is the most common method, which is aimed at minimizing the requirement of utilities by maximizing the process to process heat transfer. In the present study temperature interval diagram or TID is used to identify the targets for minimum utility requirement and maximum process to process heat transfer in a processing facility. The targets for heat exchanger network are presented and minimization of number of heat exchangers are provided using stream splitting technique. Keywords: Pinch design, stream splitting, HEN synthesis, Utilities, TID
Episode 60 : Pinch Diagram and Heat Integration
The optimal allocation of mass and energy within a unit operation, process and/or site.
Optimal allocation can be based on economic, environmental or other important objectives.
SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Chemical Process Calculations – Short TutorialVijay Sarathy
Often engineers are tasked with communicating equipment specifications with suppliers, where process data needs to be exchanged for engineering quotations & orders. Any dearth of data would need to be computed for which process related queries are sometimes sent back to the process engineer’s desk for the requested data.
The following tutorial is a refresher for non-process engineers such as project engineers, Piping, Instrumentation, Static & Rotating Equipment engineers to conduct basic process calculations related to estimation of mass %, volume %, mass flow, actual & standard volumetric flow, gas density, parts per million (ppm) by weight & by volume.
Cooling Towers in Process Industries are part of Utilities design. As the name suggests their primary purpose is to provide cooling requirements to industrial hot water from unit operations & unit processes. Examples include chillers and air conditioners. The principle of operation is to circulate hot water through a tower and allow heat dissipation to the ambient. Cooling towers can operate by natural draft or forced draft methods wherein fans are used to increase heat transfer.
A QUICK ESTIMATION METHOD TO DETERMINE HOT RECYCLE REQUIREMENTS FOR CENTRIFUG...Vijay Sarathy
Turbomachinery Engineers often conduct studies to determine if a hot gas bypass is required for a given centrifugal compressor system. This would mean building a process model and simulating it for Emergency Shutdown conditions (ESD) & Normal Shutdown conditions (NSD) to check if the compressor operating point crosses the surge limit line (SLL). A quick estimation method that uses dimensionless number called the inertia number can be used to check prior to the study, if a Hot gas bypass (a.k.a. Hot Recycle) is required in addition to an Anti-surge line (ASV or a.k.a Cold Recycle).
Exploring LPG Cylinders for Medical Oxygen - A Preliminary StudyVijay Sarathy
The following article is a study to explore the usage of LPG cylinders for medical oxygen in times of medical emergencies. The study aims at understanding how long medical oxygen can be supplied to cater to patients requiring supply between 0.5 lit/min to 2 lit/min.
Abstract The requirement of energy in any processing industry is not only a need but it is indeed a most wanted utility. In a typical processing or manufacturing industry the most common utility are steam and cooling water. However the cost of these utility are no longer cheap, in fact they are expensive. Therefore saving these utility or minimizing the usage of these utilities is one of the most needed practice in a processing industry. Pinch technology is the most common method, which is aimed at minimizing the requirement of utilities by maximizing the process to process heat transfer. In the present study temperature interval diagram or TID is used to identify the targets for minimum utility requirement and maximum process to process heat transfer in a processing facility. The targets for heat exchanger network are presented and minimization of number of heat exchangers are provided using stream splitting technique. Keywords: Pinch design, stream splitting, HEN synthesis, Utilities, TID
Free heat convection |HEAT TRANSFER LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
Experiment Name: Free Heat Convection from a Horizontal Heated
Cylinder
2. Experiment Aim:
1. Calculating the free heat convection coefficient (ℎ푁퐶) for a
horizontal heated solid cylinder.
2. Find the relationship between RaNo. And NuNo. for fluid flow
around a cylinder
Heat/light/electrical energy is out today’s necessity and has scarcity also. Energy conservation is key requirement of any industry at all times.
In general, industries use heat energy for conservation of raw material to finished product. The source of heat energy is generally saturated or super heated steam. The steam generation is common use one boiler with carity of fuels. Whatever may be the fuel the generation should be as economy as possible which adds to the product cost. Further the usage of steam and recycling steam condensate back to boiler is an art depending on plant layouts.
In this project the steam generator is water tube boiler fired with rice husk. The steam is transferred to the tyre/tube moulds where tyres/tubes are cured while the heat is rejected to the tyres the condensate forms and this condensate is put back to the boiler. While doing so the steam is also stopped back to boiler without rejecting complete heat to the product. This gets flashed into atmosphere at feed water tank. The science of separation of condensate from steam saves energy. Better the separation more the fuel conservation.
In the steam generator the fuel is burnt to heat the water and form steam. This fuel burnt flue gas carries lot of energy, out through chimney. Prior to exhausting through the heat left in flue need to be recovered, through heat recovery mechanisms’. In this project an air-preheater condensate heat recovery unit is the major energy consuming station.
To demonstrate the effect of cross sectional area on the heat rate.
To measure the temperature distribution for unsteady state conduction of heat through the uniform plane wall and the wall of the thick cylinder.
The experiment demonstrates heat conduction in radial conduction models It
allows us to obtain experimentally the coefficient of thermal conductivity of some unknown materials and in this way, to understand the factors and parameters that affect the rates of heat transfer.
To understand the use of the Fourier Rate Equation in determining the rate of heat flow for of energy through the wall of a cylinder (radial energy flow).
To use the equation to determine the constant of proportionality (the thermal conductivity, k) of the disk material.
To observe unsteady conduction of heat
forced heat convection | HEAT TRANSFER LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
Experiment Name: Forced Heat Convection
2. Experiment Aim:
1. Calculating the forced heat convection coefficient (ℎ퐹퐶) for a heated cylinder
2. Find the relations between Re N, and NuNo. for fluid flow around a cylinder
3. Introduction:
Convection; is the mode of energy transfer between a solid surface and
the adjacent liquid or gas that is in motion, and it involves the
combined effects of conduction and fluid motion, convection is divided
into two types:
DESIGN AND ANALYSIS OF AN AIR COOLED RADIATOR FOR DIESEL ENGINE WITH HYDROSTA...IAEME Publication
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ABSTRACT
Heat/light/electrical energy is out today’s necessity and has scarcity also. Energy conservation is key requirement of any industry at all times.
In general, industries use heat energy for conservation of raw material to finished product. The source of heat energy is generally saturated or super heated steam. The steam generation is common use one boiler with carity of fuels. Whatever may be the fuel the generation should be as economy as possible which adds to the product cost. Further the usage of steam and recycling steam condensate back to boiler is an art depending on plant layouts.
In this project the steam generator is water tube boiler fired with rice husk. The steam is transferred to the tyre/tube moulds where tyres/tubes are cured while the heat is rejected to the tyres the condensate forms and this condensate is put back to the boiler. While doing so the steam is also stopped back to boiler without rejecting complete heat to the product. This gets flashed into atmosphere at feed water tank. The science of separation of condensate from steam saves energy. Better the separation more the fuel conservation.
In the steam generator the fuel is burnt to heat the water and form steam. This fuel burnt flue gas carries lot of energy, out through chimney. Prior to exhausting through the heat left in flue need to be recovered, through heat recovery mechanisms’. In this project an air-preheater condensate heat recovery unit is the major energy consuming station.
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Heat transfer area and Heat transfer cofficient (U)
1. Heat & Mass Transfer
Semester Project
Fall-2017 (5th
Semester, Batch-2015)
Working on the radiator of Suzuki Baleno 1999
Submitted To:
Dr. S Kamran Afaq
Submission Date:
January 15, 2018.
Sr. No Registration No. Student Name Contribution
1 15-ME-071 Muhammad Usama Experimental data
2 15-ME-075 Muhammad Abdullah Khalid Complete Project report
3 15-ME-072 Muhammad Haseeb Adil Calculations
4 15-ME-074 Khubaib Rizwan Print outs and calculation
Department of Mechanical Engineering
HITEC University Taxila Cantt.
2. Problem Statement / Project Statement:
To calculate the overall heat transfer coefficient (U).
To calculate the heat transfer area and compare it with the experimental data being collected.
Project Introduction:
For the save use of IC engine we must safely transfer the extra heat from engine through some source.
Radiator is use for transferring heat from engine. In this project we find out that how heat of engine is
transfer from radiator in atmosphere. We found that different areas of radiator are included in heat
transfer.
Objectives:
Our Group has analyzed a Radiator of a 1300 CC Suzuki Baleno.
Our Objective was to calculate the Heat Transfer Area and compare it with Experimental Data.
The Method used to calculate the Heat Transfer is “LMTD”.
Instruments Used:
We use following instruments.
Anemometer:
To Measure the Speed of the Air coming out of Radiator Fan
Digital Pyrometer Laser Temperature Sensor:
To accurately determine the Temperature at the inlet and outlet of Radiator.
Vernier Caliper and Measuring Tape:
To correctly measure the dimensions of the Radiator understudy.
3. Diagram:
Assumptions:
Radiator Type is “Cross Flow with both Fluids Unmixed”
Mass Flow Rate had to be assumed due to the unavailability of the Specifications of Water Pump.
Heat Flow is considered to be Uniform.
Frictional Effects inside the Radiator Tubes are neglected.
The Radiator is taken as Steady Flow Device.
Fluid Properties are assumed to be Constant.
No Axial Heat Transfer.
No Change in Operating Temperatures.
Fins present on the radiator is Assumed to be of Rectangular Cross Section.
Technical Data / Available Data:
Radiator Length = 36 cm = 0.36 m
Radiator Width = 62cm = 0.62 m
Radiator Depth = 2.8cm = 2.8e-2 m
𝑚̇ Of water = 0.34 kg/s
Velocity of air = 8 m/s
Tube Thickness = 0.028m
Tube Depth = 0.0135m
Fin Thickness = 1 mm = 0.001 m
Fin depth = 0.0135 m
Fin width = 0.008 m
Spacing b/w Fins = 0.002m
Total tubes= 62
4. Inlet Temperature of Water= T (hot in) = 85o
C
Outlet Temperature of Water= T (hot out) = 61o
C
Inlet Temperature of Air = T (cold in) =18o
C (atmospheric)
Outlet Temperature of Air = T (cold out) =?
Water Properties: (from table A9)
T(avg) =(85+61)/2=73o
C
ρ = 975.82 Kg/m3
Cp = 4191.8 J/kg.K
k = 0.6654 W/m.K
Pr = 2.45
ѵ = 4.43e-7 m^2/s
Air Properties
T(avg)= (33.68+18)/2=25.84 o
C
ρ = 1.2085 Kg/m3
Cp = 1005 J/Kg.K
ρ = 1.2085 Kg/m3
K = 0.0261 W/m.K
μ = 1.872 × 10-5
Kg/m.s
Pr = 0.7124
Calculations /Results:
Q̇ = ṁ Cp ΔT
As, Q̇ c = Q̇ h
Q̇ h = ṁw Cp-h (ΔT) h
Q̇ h = (0.34) (4191.8) (85-61)
Q̇ h = 34.205 kW
Now;
Q̇ h = ṁaCp-c (ΔT) c
34.205e3 = ṁaCpc (ΔT) c
ṁa = ρAV
5. ṁa= 1.225*(0.36*0.616)*8
ṁa=2.17 kg/s
34205=2.17*1005* ΔT
ΔT=15.68o
C
T(cold out)-T(cold in)= 15.68
T(cold out)= 15.68+18
T(cold out)=33.68o
C
Number of fins = length/center to center distance
=0.36/0.003 = 120
Total fins = 61*120 = 7320
Experimental area:
ATotal = AFin + AUnfin + A1 + A2 + A3
AFin = (Pl + ATip) NFins
AFin = [(0.028+0.001)*2*0.008 + (0.001*0.028)] *7320
AFin = 3.60 m2
AUnfin = (ANofin) NTubes - NFins (ATip)
AUnfin = (0.36*0.028) (62) – (7320) (0.001*0.028)
AUnfin = 0.624-0.2049 = 0.4190*2
6. AUnfin = 0.838 m2
A1 = (0.002*0.36)
A1 = 7.2e-4 m2
A1 = A2
A2 = 7.2e-4 m2
A3 = 0.0135*0.002
A3 = 2.7e-5 m2
A3 = A4
A4= 2.7e-5 m2
Now,
ATotal = AFin + AUnfin + A1 + A2 + A3 + A4
ATotal = 3.60 + 0.838 + 2*7.2e-4 + 2*2.7e-5
ATotal =4.439 m2
Theoretical area:
A = (Q̇ ) / (U × (ΔTlm))
ΔTlm cross flow both fluid un mixed
(ΔTlm) counter flow = (ΔT1 - ΔT2) / ln (ΔT1/ ΔT2)
ΔT1 = Th-in – Tc-out = 85-33.68 = 51.32 o
C
ΔT2 = Th-out – Tc-in = 61-18 = 43o
C
ΔTlm = (51.32-43)/ ln (51.32/43)
So, (ΔTlm) Counter Flow = 47.03o
C
For (F),
T1 = 85 o
C t1 = 18o
C
T2 = 61 o
C t2 =33.68 o
C
P = (t2 - t1) / (T1 - t1)
P = (33.68-18) / (85-18) = 0.234
R = (T1 - T2) / (t2 - t1)
R = (85-61) / (33.68-18) = 1.53
7. So,
F = 0.97 (From Graph Using P & R)
Now,
(ΔTlm) Cross Flow = (F) (ΔTlm) Counter Flow
(ΔTlm) Cross Flow = (0.97) (47.03)
(ΔTlm) Cross Flow= 45.619o
C
For hi:
Geometry = Rectangular
Hydraulic Diameter (Dh)
Dh = 4(Ac) / P
Dh = 4(0.002*0.0135) / (0.002+0.0135)×2
Dh = 3.48e-3m
As, ṁ = ρAv
v = ṁ / ρA
v = 0.34/(975.82*(0.002*0.013)) = 12.90m/s
Reynolds Number (Re)
Re = V Dh / v
Re = (12.90*3.48e-3)/4.43e-7
Re = 101449.06
As, Re > 10,000 So, Turbulent Flow
Nu =0.023*Re^0.8*Pr^0.3 0.3 for cooling
Nu=0.023*(101449.06^0.8)*(2.45^0.3)
Nu=304.42
Now, Nu = hi Dh / K
So, hi = Nu k / Dh
= (304.42*0.665)/3.48e-3
hi = 58172.60 W/m2
.o
C
For ho:
Hydraulic Diameter (Dh)
Dh = 4(Ac) / P
Dh = 4(0.002*0.36) / (0.002+0.36)×2
8. Dh = 3.97e-3m
Reynolds Number (Re)
Re = V Dh / v
Re = (8*3.97e-3)/15.63e-6
Re = 2036.03
As, Re <2300 So, Laminar Flow + external forced convection
From table
Nu =0.228*Re^0.731*Pr^(1/3)
Nu=0.228*(2036.03^0.731)*(0.712^0.333)
Nu=53.39
Now, Nu = h0 Dh / K
So, h0 = Nu k / Dh
= (53.39*0.0261)/3.97e-3
hi = 351 W/m2
.o
C
U =
𝟏
𝟏
𝒉 𝒊
+
𝟏
𝒉 𝟎
U =
𝟏
𝟏
𝟑𝟓𝟏
+
𝟏
𝟓𝟖𝟏𝟕𝟐
U = 348.89 W/m2
.o
C
A = 34205/(348.89*45.61)
A = 2.149 m2
Percentage Error of Theoretical & Experimental Values (if any)
Error = (4.439-2.149)/4.439
Error = 51.58%
9. Practical / Industrial Applications of the Project: (if any)
Radiators are used for cooling internal combustion engines, mainly in automobiles but also in piston-engine
aircraft, railway locomotives, motorcycles, stationary generating plants and other places where such engines
are used. Tiny radiators known as heat sinks are used to convey heat from the electronic components into a
cooling air stream. Radiators are also found as components of some spacecraft.
Project Pictures: