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Report on"Design of Ventilation System for HEAT ENGINE LAB (Basement)" Submitted by
Mr. Dhiraj Satish Khinvasara (121425004) Mr. Roshan Ramchandra Makre (121425005)
M.Tech First Year (Thermal Engineering)
A report submitted in partial fulfillment of the requirements for Air Conditioning System Design
Guided byDr. S.N.SapaliFaculty of Mechanical EngineeringCollege of Engineering Pune
TABLE OF CONTENT
SR. NOCONTENTPAGE NO.
1.Layout of Heat Engine Lab Basement4
2.Specification of Heat Engine Lab Basement5
3.Outdoor Conditions & Indoor Condition6
4.Cooling Load Calculation Solar & Conduction heat gain through wall Occupants Lighting Other Equipment7
5.Design Sheet8
6.Ventilation by Air Circulation9
7.Duct Design10
8.Blower Selection18
9.References20
LIST OF FIGURES
SR. NOCONTENTPAGE NO.
1.Layout of Heat Engine lab4
2.Layout of Duct Design (For Maximum CFM)14
3.Layout of Duct Design (For Minimum CFM)16
2 . Specification of Heat Engine Lab (Basement)
The dimensions are as below Length 43.2 m
Width 18 m
Height3.44 m
a) Windows & Door
Sr. No.FaceDimension in (m)Nos.(Windows)
1North East1.99 m x 3.44 m10
2South West1.99m x 3.44 m
10
b) Appliances Sr. No.ParticularsNos.
1Peoples40
2Computers25
3Fans 10
4CFL60
c) Equipment (6)
Engine TypesModelRPMPower (kW)
4 Cylinder ,4 Stroke, water cooledBajaj380034
4 Cylinder, 4 stroke water cooledPremier500028
Research Engine Test Set up (Petrol)18004.5
Variable compression Research EngineRicardos30004.4
3.Outdoor & Indoor Conditions
Outdoor Condition and Indoor Conditions for summer (1)
Parameter DBTRHLat.
Outdoor Condition 39C30%18N
Indoor Condition 22C50-%18N
Air Velocity outside=4 m/s (5)Air Velocity inside = 0.5 m/s (5)
4. Cooling Load Calculation Cooling load can be calculated by using three methods CLTD Method ( Corrected Cooling Load Temperature Difference ) ASHRAE Method
For this project design will be done with CLTD Method Solar & Conduction heat gain through wall ( North-East, Roof)
Heat Gain Roofs, walls, and ceiling by conduction (4)Q = U x A x
U = Overall heat transfer coefficient for roof or wall A = Area of roof, wall, or glass in =Corrected cooling load temperature difference, roof, wall, or glassCorrected cooling load temperature:= CLTD +LM+ (78 - ) + ( - 85)
Calculation of Overall Heat Transfer Coefficient(U)
(2)
Thickness of stone =0.8 mThermal conductivity of stone = 1.7 W/m.Khi( Inside heat transfer coefficient ) = 72 x V (2)= 72 X 0.5=36 kJ/hr.m2.K=10 W/m2K
ho(Outside heat transfer coefficient ) = 72 x V (2) = 72 x 4 = 288 kJ/hr.m2.K = 80 W/m2K U (NE wall) =1.95 W/m2.K U (Roof) = 0.528 W/m2.K (4 Table 6.1)1. NE WALL Q = U x A x =1.95 x 43.2 x 0.65 x (39-22) =930.85 W
2. ROOF Q= U x A x = 0.5280 x 43.2 x 18 x (28-22) = 2463.43 W
WallHeat Load (kW)
North-East930.85
Roof2463.43
d) Occupants Q (sensible) = N x (Sensible heat gain) x CLF N = number of people in space, from best available source (4) Q (sensible) = 40 x 635 x 0.5 =12700 Btu/h =3722 W
Q (latent) = N x (Latent heat gain) (4) = 40 x 965 Btu/h =11310 W
e) Lighting Q (lighting) = 3.4 x W x BF x CLF (4)Q = Cooling load from lighting (BTU/hr)W = lighting capacity= 40 x 60 WBF = ballast factor =1.25CLF = Cooling load factor for lighting=1 Q=3.4 x W x BF x CLF =18000 W
f) Other (1)In other appliances like Equipments, computers, Fans and other appliances Computers =180 x 30 = 5400 WFan =70 x 20 = 1400 W
g) Equipment load (7) Power generated = Wshaft + Q exhaust + Q loss + W acc
W shaft = brake output power off of the crankshaftQ exhaust = energy lost in the exhaust flowQ loss = all other energy lost to the surroundings by heat transferW acc = power to run engine accessories
Wshaft =25 40 %Q exhaust = 20 45 %Q loss = 10 30 %W acc = 5 - 10 % Heat loss can be subdividedQloss = Qcoolant + Qoil + QambientQcoolant = 10 -30 %Qoil = 5 - 15 %Qambient = 2 - 10 %
Cooling Load Calculation ( Design Sheet)
DBTWBTRH(%)
Outside Conditions3924.430
Inside Condition221850
Room DimensionsLWHNos.
m43.2183.44
North East side windowsm1.990.810
South west side windowsm1.990.810
Volumem32674.9
External Cooling Load
WallArea U Load
m2w/m2k To(C)Ti(C)W
North East28.081.953922930.85
Roof777.60.52828222463
Total External Cooling Load3366.85
Internal Cooling Load.
PeopleCLFNos
Sensible Heat (BTU/hr)6350.5403722
Latent Heat(BTU/hr)9654011310
Equipments( Engine)Type Of EngineRated PowerPower loss to Surrounding (IAEME)
4 cylinder,4 stroke, Bajaj Engine34 kW @3800 rpm10 %13400
4 cylinder ,4 stroke, Premier28 kW @5000 rpm10%12800
Research Engine Test Set up (Petrol)4.5 kW @1800 rpm10 %1450
Ricardos E6 Type Variable compression Research Engine(Petrol)8.2 kW @3000 rpm10%1820
Ricardos E6 Type Variable compression Research Engine(Diesel)6.7 kW@3000rpm10 %1670
Mini Computer180 W305400
Fans70 W201400
LightingB.FCLFNos
CFL1.2540 W16018000
Tube lights140 W1203200
Total Internal Cooling Load61172
Total Cooling Load on systemW62102.85
kW62.102
TR19.67
5) By Air CirculationTo determine no of air changes per hour
Volume of lab(m3)No. of people workingM3/(min x person)(Ventilation standards IS3103)Air changes per hour
2674.94252011.21
No. of air changes per hourCFM
1015742
1218890.45
1422038.86
1523613.06
1625187.26
1726761.47
1828335.67
1929909.88
2031484.08
CFM = =10 x 43.2 x 18 x 4.33 x 35.31/60 =15742
6) Duct DesignFor Maximum CFM = 31484Considering Total CFM = (15742 from each branch)FPM=984 (5 Table 3)
Layout of Duct Design
First Branch
SectionCFMLength of Duct (m)Friction LossFPMDuct diameter (m)A(m)B(m)
AB1547290.029841.291.440.960
BC12972120.029201.161.290.8636
CD9472100.028901.111.280.8263
DE447212.20.027500.860.9630.6420
Second branchSectionCFMLength of Duct (m)Friction LossFPMDuct diameter (m)A(m)B(m)
PF1547280.029841.291.440.96
FG11972110.029101.161.290.863
GH9472130.028901.111.280.826
HI497211.20.027800.931.030.692
For Rectangular Cross section equivalent to Circular Cross Section
Friction Pressure drop = (8) = (0.02/100) x (43.2+43.2+18) x3.28 = 0.07 of H2O Static Pressure
There are 11 fittings: eight grill, two duct turns, one damper on the wall of the Heat Engine Lab. The total pressure drop for fittings isNumber of Bends = 2Number of Grills = 8Damper = 1
The total pressure drop for fittings is: 0.08 x (2+8+1) (9) = 0.88
Therefore, the total pressure drop is: 0.88 +0.07 = 0.95 of H2O Static PressureFor Minimum no of CFM = 15742Considering Total CFM = (7871 from each branch)FPM= 984 (5 Table 3) Layout of Duct Design
First BranchSectionCFMLength of Duct (m)Friction LossFPMDuct diameter (m)A(m)B(m)
AB787190.0289841.011.1250.75
BC6371120.0288900.931.0380.692
CD4371100.0288000.780.870.580
DE187112.20.0286700.550.61350.409
SectionCFMLength of Duct (m)Friction Loss(in of water per 100 ft)FPMDuct diameter (m)A(m)B(m)
PF787180.0289841.011.1250.75
FG11972110.0288700.861.410.9420
GH9472130.0288000.780.870.580
HI497211.20.0287700.680.7590.506
Second branch
For Rectangular Cross section equivalent to Circular Cross Section
Friction Pressure drop = (8) = (0.028/100) x (43.2+43.2+18) x3.28= 0.1 of H2O Static Pressure
There are 11 fittings: eight grill, two duct turns, one damper on the wall of the Heat Engine Lab. The total pressure drop for fittings isNumber of Bends = 2Number of Grills = 8Damper = 1
The total pressure drop for fittings is: 0.08 x (2+8+1) (9) = 0.88
Therefore, the total pressure drop is: 0.88 +0.1 =0.98 of H2O Static Pressure7) Blower SelectionBlower selection for Heat Engine Lab Basement from VENTILATIONCatalogue (9)
CFMSP (Static Pressure)Type of blower
1000-23000Up to 3.0Inline Duct Blower
1000-30000Up to 4.0Backward Incline Belt Drive Blowers
700-10000Up to 1.0Inline Duct Blower, Forward Curve
For Maximum CFM = 31484Static Pressure = 0.95For minimum CFM = 15742Static Pressure = 0.98
For maximum CFMWe are selecting Backward Incline Belt Drive Blower having CFM 1000-30000 Static Pressure Drop up to 4.0
HP2
Fan RPM1725
For minimum CFMWe are selecting Incline Duct Blower having CFM 1000-23000 Static Pressure Drop up to 3.0HP1.5
Fan RPM1500
References :1. Cooling Load Calculation by ASHRAE 19972. Refrigeration and Air Conditioning by Manohar Prasad3. Refrigeration and Psychometric Properties (Tables & Charts ) M.L. Mathur and F.S. Mehta 4. Air Conditioning Principles and system - G.Pita 5. Code of Practice For Industrial Ventilation IS :31036. I.C. Engine Specifications Catalogue, Dept.of Mechanical ,IIT Guwahati7. International Journal Of Mechanical Engineering And Technology (IJMET)8. Refrigeration and Air Conditioning by C P Arora9. Ventilation Fundamentals By Dayton
9
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