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

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