VSRD International Journal of Mechanical, Civil, Automobile and Production Engineering, Vol. 3 No. 7 July 2013 / 245 e-ISSN : 2249-8303, p-ISSN : 2319-2208 © VSRD International Journals : www.vsrdjournals.com

RESEARCH ARTICLE

DESIGN AND DEVELOPMENT OF MECHANICALLY OPERATED CONTROL TOILET DISCHARGE TOILET SYSTEM

1Vishwasinha Vinayak Bhosale* and 2W.S. Rathod 1Research Scholar, 2Assistant Professor, 1,2Department of Mechanical Engineering,

VJIT, Mumbai, Maharashtra, INDIA. *Corresponding Author : vishwasinhabhosale@gmail.com

ABSTRACT The project titled “Mechanically operated Control Discharge Toilet System” is part of a research work which is in process at VJTI Mumbai. In this system discharge of human waste is get controlled by using mechanically operated mechanism. This mechanism is designed for purpose of keeping Railway station platform clean & odorless.

In this project, One collecting tank is placed below toilet discharge pipe, Opening and closing of collecting tank is controlled by the mechanical linkage. Closing of collecting tank is depends on Air suction used for braking system of railway and opening of tank is depends on wind pressure, as train reaches speed up to 30Km/hr, tank will get open and collected human waste will get discharged away from of platform of railway station.

Keywords : Toilet System, Indian Railway, Platform, Air Brake System, Wind Pressure.

1. INTRODUCTION The existing toilet system in the Indian railway coaches discharges human excreta directly to the ground and the railway tracks at railway station platform. The consequences include unacceptable hygienic conditions, particularly in the railway stations, and damage to rails. IR is making efforts to introduce environment friendly toilet discharge system, for which three options are being considered; (i) Modular, (ii) Vacuum, (iii) Chemical and (iv) Bio Toilet.

The "Mechanically operated Control Discharge Toilet System” is part of a research work which is in process at Indian Railway workshop, Kurduwadi. In these system discharge of human waste is get controlled by using mechanically operated mechanism. This mechanism is designed for purpose of keeping Railway station platform clean & odorless.

In this project, waste collecting tank is placed below to toilet discharge pipe, Opening and closing of collecting tank is controlled by the mechanical mechanism. Closing of collecting tank is depends on Braking system and opening of tank is depends on rotation of aerodynamic foil which depends on train speed.

2. WORKING PRINCIPLE Mechanical operated C.D.T.S. controls discharge of human waste is get controlled by using mechanically operated mechanism. Waste collecting tank is placed below to toilet discharge pipe, Opening and closing of collecting tank is controlled by the mechanism. Closing of collecting tank is depends on Braking system and opening of tank is due to

impact of lever on obstacle which has been placed at non residence area away from platform.

Closing of Collecting Tank : When train comes to the platform brakes are get applied, Railway trains are working on the two types of brake i)Air Brake ii)Vacuum Brake. In today time railway coaches air brakes are used, in air brake system braking application is mainly depends upon pressure drop in braking pipe. ‘Brake pipe’ is a connected to compressor and brake shoe through Distribution Valve and air pressure in braking pipe is maintained up to 5bar. When driver wants to apply brake he opens release valve connected to brake pipe and pressure in the brake pipe is starts decreasing, when pressure is decreases up to 3.5bar brakes are applied and trains get stopped.

Fig. : Layout of Railway Air Brake System

Waste collecting tank is placed below to toilet discharge pipe. In colleting tank one piston cylinder arrangement is

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 246

provided. One end of cylinder is connected to brake pipe through bypass pipe, when brakes are get applied pressure in brake pipe starts decreasing which creates suction in cylinder, piston is get pulled upwards which leads to pulling of bottom plate upwards and getting lock with the help of spring loaded plunger which place in connection with wire rope, which is wounded on pulley, pulley is mounted on a shaft, which is rotating when train reaches speed of 30km/hr.

When brakes are applied pressure in brake pipe starts decreasing which creates suction in cylinder, piston is get pulled upwards which leads to pulling of bottom plate upwards and getting lock with the help of plunger which place in connection with vertical side.

After closing of waste collecting tank, human waste is get collected in tank when train reaches to platform. This helps in keeping platform clean and odorless.

Fig. : 3D Layout of Mechanically operated Control Discharge Toilet System

Opening of Collecting Tank : After stopping train at platform brakes are get release, driver closes ‘Release valve’ and starts compressor, compress air is get feed into the brake pipe and pressure in brake pipe is maintained up to 5bar.when pressure in brake pipe reaches up to 5bar pressure force is exerted on piston and tries to push piston in downward direction. So there are main force acting on bottom plate of tank for opening condition, which is Pressure force on piston cylinder

Opening of collecting tank is depends on train speed, when train reaches speed of 30km/hr collecting tank is get open. For opening of collecting tank one ‘Aerodynamic foil’ which is a rectangular foil is placed on a shaft, this shaft is attach to the locking plunger through pulley. As train reaches speed of 30km/hr shaft will rotate up to 45º which leads to pull the plunger and collecting tank is get open and human waste is get discharged. Collecting tank remains open unless brakes are applied, after applying brakes collecting tank is get closed. This way ‘Mechanically operated Control Toilet Discharge System works and keeps Railway station platform clean and odorless.

3. DESIGN OF PROTOTYPE OF MECHANICALLY OPERATED CONTROL DISCHARGE TOILET SYSTEM

One of the most important thing scientists do is to model the physical phenomenon. Virtually every phenomenon in nature can be described with the aid of laws of physics or the other fields in terms of engineering concepts relating various quantities of interest in this analysis.

The following steps are included in Design of ‘CDTS model’ : Selection of Collecting Tank Selection of Pneumatic Piston Cylinder Design of Spring Design of Shaft Design of Bearing Design of Aerodynamic foil Selection of Pulley

Selection of Collecting Tank : In this first step of designing of CDTS Model, it is require to determine the dimensions of waste collecting tank which is to placed at below of toilet discharge pipe. While determining of dimension of colleting tank, designer has to think space available at below of bogie chassis and tank capacity. By considering this factors collecting tank dimensions is fixed as 50cmX50cmX30cm, its capacity is 75lit.

As per Indian Railways rules most of the train’s takes halts of 30 minutes on major stations. By considering this capacity of collecting tank is sufficient to collect the human discharge at railway station platform.

Selection of Pneumatic Piston Cylinder : Closing mechanism of collecting tank of CDTS is working on air suction created for applying air brakes whose operating pressure is 5 bar. When train arrives at platform driver opens a release valve which is connected to the air pressure feed pipe whose pressure is maintained up to 5 bar throughout the feed pipe. When release valve is get open pressure in the brake pipe decreases, due this depression in air pressure, air suction is get created, by using this air suction brake shoes are get pushes with help of mechanical linkage and brakes are get applied. It is observed that when the pressure in brake pipe is decreases up to 3.5bar train has to stop.

For closing of collecting tank air suction due to decrease in air pressure is used. One pipe has bypassed from brake pipe whose other end connects to the inlet of pneumatic piston cylinder. Pneumatic piston cylinder has to be designed in such a way that when pressure is 5bar, piston should be at BDC position and when pressure is decreases up 4bar piston has to be at TDC. Connecting rod of piston is attached to the bottom plate of collecting tank, as the piston moves from BDC to TDC bottom plate of collecting tank is get lift upward and tank is get closed.

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 247

Selection of pneumatic cylinder is mainly depends on Operating pressure Stroke of piston

As per standard catalogue available in market pneumatic piston cylinder is selected which fulfils the requirements of design, whose dimensions is as follows : Bore Diameter = 50mm Stroke = 120mm Cylinder length = 235mm

Design a compression spring : For designing of compression spring for ‘CDTS’ the following requirements to meet:

O.D. = 1.89 in. (48 mm), free length Lf = 6.10 in(155 mm) solid height Ls = 2.16 in(55 mm) load Pf= 218.15 lb. (98.17N) Ps = 174.51lb. (78.53N), ends closed and ground, maximum. The material is oil-tempered wire.

For designing of spring following steps are to be followed:

The required spring rate (R) is determined in : R = P/L = ((Ps − Pf)/(Lf − Ls))

= (174.51-218.15)/(6.10 - 2.16) =11.07lb/in.(1.964N/mm)

In this case, assume a trial mean diameter(d) 0.2 in. (5.08mm) less than the nominal diameter and a trial design stress of 0.45 of Ultimate tensile Stress

(St)= 0.45 X (200,000) = 90,000 psi (621 MPa)

Where St = 200,000 psi (1379MPa) is in the high-diameter range for oil-tempered wire.

Stress formula for a trial wire diameter.

d= ( )/ )

Where Ps = LoadatSolidheightofspring D = Difference between Outer Diameter (O.D.) and Coil Diameter (d) D = 1.89inch - 0.2inch

= 1.69 inch = 42.31mm S = Maximum allowable Design Stress

= 0.45 of Ultimate tensile Stress (St) = 0.45 X 200000 = 90,000 psi (621 MPa)

d = (8X174.51X1.792)/πX90000)3

d = 0.206 inch = 5.25mm

Since this is only 10 percent over the assumed S, = 200,000 (1379MPa) there is no need to recalculation of ‘d’. Using the 0.206inch. (5.25 mm) temporarily as the acceptable d, Solve for D = O.D.- d = 1.89 - 0.206 = 1.684 inch. (20.32 mm)

Solve for Number of active coils(na) by following formula :

na = (GXd4)/(8XRXD3) = (11.5X106X0.2064)/(8X11.07X1.6843) = 11.39 = 12

With closed and ground ends, solid height (H) is determined in

H = d (na + 2) = 0.206(12 + 2) = 2.884 inch. (72.21 mm)

Even this solid height is larger than the 2.16 in(55 mm) specified, d = 0.206 in. (5.15 mm) can be accepted with regard to this space limitation.

As per results get by design calculations, specifications of spring is as follows :

Outer Diameter of Spring (O.D.) = 48cm, Coil Diameter of Spring (d) = 4mm, Number of Active Coils (na) = 12, Free length (Lf) = (200mm), Solid Height (H) = 48 mm

Design of Shaft : One important approach of designing a shaft is to use A.S.M.E. code. According to this code, the permissible shear stress τmax for the shaft without keyways is taken as 30% of yield strength in tension or 18% of the ultimate tensile strength of the material, whichever is minimum. Therefore,

τmax=0.30XSyt or τmax=0.18 X Sut (whichever is minimum)

If keyways are present, these values are to be reduced by 25 per cent. According to A.S.M.E. code, t he bending and torsional moments are to be multiplied by factors kb and kt respectively, to account for shock and fatigue in operating condition. Eq. for τmax is written as :

τmax= 16πXd3 (kbXMb)2 + (ktXMt)2

Where, kb = combined shock and fatigue factor applied to bending moment, and kt = combined shock and fatigue factor applied to torsional moment.

The values of kband kt for rotating shaft are as follows in table :

Sr.No. Loading Conditions 퐤퐛 퐤퐭

01. Load applied gradually 1.5 1.0

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 248

02 Load applied suddenly (minor shock) 1.5-2.0 1.0-1.5

03. Load applied suddenly (heavy shock) 2.0-3.0 1.5-3.0

For designing of shaft, which is placed at below of collecting tank following conditions are to be consider ● Weight of human exerta is 75kg (is to consider as udl),

which is stored in collecting tank of bottom plate size of 50 X 50.

● Load applied by the connecting rod of pneumatic piston cylinder due to 5bar pressure (98.17N) air on bottom plate.

For above requirements shaft can be design as per A.S.M.E. Code by following steps:

τmax=0.30 X Syt

Where, Syt= 380 N/mm2 τmax= 0.30 X 380 τmax= 114 N/mm2

Torsional moment = Load X Perpendicular distance Mt = 98.17 X 125

Mt = 12271.25 Nmm

Bending moment = (Load + Weight of Human exerta) X Perpendicular distance

Mb = (98.17 + 75) X 125 Mb = 21646.25 Nmm

Shaft diameter can be calculated by using following formula,

τmax= 16πXd3 (kbXMb)2 + (ktXMt)2

Where, kb = 1.5, as load applied suddenly but in minor range, and kt = 1.0, as load applied suddenly but in minor range.

τmax= 16πXd3 (1.5X21646.25)2 + (1.0X12271.25)2

114 = 16πXd3 X 28413.02

d = 9.23mm

As diameter of shaft calculated is 9.23mm, it can be taken as 10mm, as there is variation up to 0.75% is allowed.

Design of Bearing : While selecting the proper size of bearing, it is necessary to specify the expected life of the bearing for the given application. The information regarding the life expectancy is generally vague and values based on past experience are used. For all kinds of vehicles, the speed of rotation is not constant and the desired life is expressed in terms of millions of revolutions. In other applications, the speed of rotation is relatively constant and the desired life is expressed in terms of hours of service. The recommended bearing life for some of the application is given in Table:

Sr. No. Type of Using Life in Hours

01. Machines used intermittently 4000-8000 hr

02. Machines used 8 hours per day 12000-20000 hr

03. Machines used for continuous operation (24 hr per day) 40000-60000 hr

Selection of Bearing from Manufacturer’s Catalogue : The basic procedure for the selection of bearing from manufacturer’s catalogue consists of the following steps:

Calculate the radial and axial forces acting on the bearing and determine the diameter of the shaft where the bearing is to be fitted.

Select the type of bearing for given application.

Determine the values of X and Y, the radial and thrust factors, from the catalogue.

Calculate the equivalent dynamic load from the equation,

P = XFr + YFa

Make decision about the expected bearing life and express the life L in million revolutions.

Calculate the dynamic load capacity from equation,

L = CP

P

Check whether the selected bearing of series 60 has the required dynamic capacity. If not, select the bearing of the next series and go back to step (iii) and continue.

Ball bearings are thus selected by trial and error procedure. The above procedure is applicable to other types of bearings.

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 249

In order to explain the selection of bearing in CDTS, let us see the requirements of set up.

Diameter of shaft = 10mm,

Rotation speed 10 rpm,

The bearing subjected to radial load consist mainly two types :

Load applied by the air of 5bar on piston = 98.17N,

Weight of human exerta, collected in to the tank (udl) = 75 X 0.25 X 0.50=9.375N

Therefore total load applied on bearing is 98.17N+9.375N = 107.545N

Life of bearing is 4000 hr.

For design of ball bearing following steps are to be followed:

Selection of load Fr = 107.545N, Fa= 0

Selection of type of bearing Type: deep groove ball bearing

Selection of X and Y factors When there is no axial load, X = 1, Y = 0

Calculation of Resultant load Resultant load on bearing is to be calculated by using following formula :

P = XFr+ YFa = 1(107.545) + 0 = 107.545N

Calculation of Bearing life Life of bearing is calculated by using following formula L = 60XnXLh

106

= 60X10X4000106

= 2.4 million rev.

Calculation of Dynamic load capacity C = PXL

13

C = 107.545 X 2.413

C = 143.98 C =144 N

It is observed from SKF General Catalogue No.3200 E (Courtesy of Associated Bearing Company Limited, Pune). For selection of above bearing, following options are available with 10mm shaft diameter,

No.61800 (C = 1480N) No.6000 (C = 4620N) No.6200 (C = 5070N) No.6300 (C = 8060N)

From above options, No.61800 bearing can be used as it is more suitable (C = 1480N) for this application.

In above table following specifications are given for selected bearing : Bore diameter of bearing (d)= 10mm Dynamic load capacity (C) = 1480N. Outer diameter of bearing (D)= 19mm Axial width of bearing (B) = 5mm.

Selection of Pulley : In CDTS two M shape pulleys are used for the purpose of to pull spring loaded locking plunger through wire rope, so there in no much power transmission through. As per requirement of design two pulleys has been selected of diameter 25mm and 100mm, material of pulleys are selected as Aluminium.

Design of Aerodynamic Foil Theory : The equations for calculating lift and drag are very similar. The lift that an airfoil generates depends on the density of the air, the velocity of the airflow, the viscosity and compressibility of the air, the surface area of the airfoil, the shape of the airfoil, and the angle of the airfoil's angle of attack. However, dependence on the airfoil's shape, the angle of attack, air viscosity and compressibility are very complex. Thus, they are characterized by a single variable in the lift equation, called the lift coefficient. Due to the complexities of the lift coefficient, it is generally found via experimentation in a wind tunnel where the remaining variables can be controlled. Therefore,

The lift equation is given by :

CL =12ρV2SCL

Where L is the lifting force, ‘ρ’ is the density of air, V is the relative velocity of the air flow S is the area of the airfoil as viewed from an overhead perspective, ‘CL’ is the lift coefficient. As with lift, the drag of an airfoil depends on the density of the air, the velocity of the airflow, the viscosity and compressibility of the air, the surface area of the airfoil, the shape of the airfoil, and the angle of attack. The complexities associated with drag and the airfoil's shape, angle of attack, the air's viscosity, and air's compressibility are simplified in the drag equation by use of the drag coefficient. The drag coefficient is generally found through testing in a wind tunnel, where the drag can be measured, and the drag coefficient is calculated by rearranging the drag equation.

D =12ρV2ACD

In the drag equation,

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 250

D is the drag force, is the density of the air, V is the velocity of the air, A is a reference area, and CD is the drag coefficient.

Procedure : The donated wind tunnel used in this experiment was constructed out of cardboard, epoxy, and masking tape. The walls of the wind tunnel where the lid met the rest of the tunnel were lined with strips of soft foam, which acted to insulate the airflow in the tunnel from outside turbulence. The airflow velocity at the test section was found to be 14.4 m/s using a Dwyer Transparent pressure gauge. Shown in Figure 3 are six different airfoils used in the experiment. Each airfoil was unique in camber and thickness so as to induce various lift and drag characteristics. airfoils have similar chords, but it was difficult to produce precise sizes, the materials available were not ideal for airfoil construction.

Design of Aerodynamics foil of rectangular plate shape to meet the requirements of opening mechanism of CDTS : When there is relative motion between a real fluid and a body, the fluid exerts a equal and opposite force on the fluid. If the body is moving at a constant velocity in a stationary fluid, the fluid motion is unsteady, because at a given point in space, the velocity changes with time. However if the body is stationary and fluid flows is same in both the cases. A body wholly immersed in a real fluid may be subjected to following two kinds of forces due to relative motion between the body and the fluid.

Drag force : The component of force in the direction of flow (free stream) on a submerged body is called the drag force,(FD);

Lift force : The component of force at right angles to the direction of flow is called the lift force,(FL).

For a body moving through a fluid of mass density ‘ρ’, at a uniform velocity U, the mathematical expression for the calculation of the drag and the lift may be written as follows :

FD = CD XAXρXU2

2

FL = CLXAXρXU2

2

Where, CD = Co-efficient of drag, CL = Co-efficient of lift, ρ = Density of fluid, U = Relative velocity of fluid w.r.t. the body A = Projected area

For design of ‘Aerodynamic foil’ which is a rectangular plat, dimensions (Length of the plate (l), Width of plate (b), Thickness of the plate (t)) of plate has to be find. Among this three dimensions length (l) is taken as 150mm, Width

(b) is taken as 100 mm and thickness (t) is to be find.It is require that at the train speed of 30 Km/hr plate should be rotate up to 45º. So projected length will be same, but at 45º projected height will be measured as 70.71mm.For designing of Aerodynamic foil, experiment is to be conduct at a train speed 30km/hr on a plate size 150mm X 70.71mm at an angle 45º made by plate with horizontal. as the plate is kept at 45º to the horizontal the co-efficient of lift and drag are 0.75 and 0.15, respectively.

For calculation of dimensions of Aerodynamic foil (rectangular plate)

Step 1: Calculation of Drag force(FD): FD = CD XAXρXU2

2

FD = 0.15X0.01X1.2X8.332

2

Where, CD = 0.15, ρ = 1.2 kg/m3, U = 30 km/hr, 8.33 m/s, A = 100mm X 70.71mm = 0.01m2. FD = 0.062N

Step 2: Calculation of Lift force(FL):

FL = CL XAXρXU2

2

FL = 0.75X0.01X1.2X8.332

2

Where, CD = 0.75, ρ = 1.2 kg/m3, U = 30 km/hr, 8.33 m/s, A = 100mm X 70.71mm = 0.01m2. FL = 0.312N

Step 3:Calculation of Resultant force FR: FR = FD

2 + FL2

FR = 0.0622 + 0.3122 FR = 0.39N

Step 4: Calculation of mass of plate: Resultant force (FR) = Mass of plate (load) X Projected area of plate (A) => Mass of plate (load) = Resultantforce(FR)

Projectedareaofplate(A)

Mass of plate (load) = 0.390.01

Mass of plate (load) = 0.405 Kg.

Step 4 :Calculation of Volume of plate (t):

Mass of plate = Volume X Density of plate

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 251

Where, Density of plate = 7860 kg/m3 => Volume = 0.405

7860

Volume =5.15 X 10 5 m3= 5.15 X 104 mm3

Also

Volume = Length (l) X Width (b) X Thickness (t) 5.15 X 104 = 150 X 70.71 X Thickness (t) => Thickness (t) = 4.85mm,

So Aerodynamic foil (Rectangular steel plate) is selected with dimensions as Length (l) = 150mm, Width (b) = 100mm and Thickness (t) = 5mm

Specifications of Standard Parts: Sr. No. Particular Description Value

1 Shaft Diameter Material

10mm Mild steel

2 Bearing Ball Bearing 6201

3 Pneumatic piston cylinder Bore Stroke

50mm 120mm

4 Spring

o.d nominal diameter

inner diameter wire dia

free length no. of turns

48mm 44mm 40mm 4mm

200mm 12

5 Pulley Outer diameter Inner diameter

60mm 26mm

6 Eye bolt Standard

7 s- type hook Standard

8 Rope wire Length Wire diameter

500mm 5mm

9 M.S sheet plate Length Width

Thickness

500mm 400mm 2.5mm

10 Connecting rod Length Diameter

200mm 20mm

4. MANUFACTURING PROCESS Mechanically operated Control Discharge Toilet System is manufactured by performing following operations shown in Table:

Sr. No

Operation Sequence M/C Tool Tooling Holding

Device Gauges Time(S)

1. Cutting shaft Power Hacksaw Hacksaw cutter Work table Measuring Tape 60

2. Pulley turning Lathe machine Single point cutting tool Chuck Vernier calliper 300

3. Drilling of pulley Lathe machine Drilling tool Chuck Vernier calliper 120

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 252

4 Boring of pulley Lathe machine Boring tool Chuck Vernier caliper 300

5 Welding Welding m/c Electrode Work table - 3600

Total 4380

Costing of Manufacturing : As per following table costing of manufacturing processes is calculated Sr. No.

Operation Sequence M/C Tool Tooling Holding

Device Gauges Time(S) Cost

1. Marking - - - Measuring tape 60 -

2. Cutting Shearing m/c

Shearing wheel Work Table - 1500 50/-

3. Grinding Hand Grinder

Grinding Wheel - - 800 20/-

4. Drilling Drilling m/c HSS Drill Bit - - 1200 50/-

5. Welding Arc Welding Electrode - - 1500 70/-

Total 5060 190/-

Cost of operations required for Manufacturing of Mechanically operated Control Discharge Toilet System (CDTS) is only Rs 190/-

Cost of Standard Parts Sr. No. Part Name Quantity Cost (Rs.)

1 Shaft (axle of bike) 2 60/-

2 Bearings (ball) 3 120/-

3 Pneumatic piston cylinder 1 1900/-

4 Pulley 2 240/-

5 Spring 3 300/-

6 Eye bolt 1 50/-

7 S- type hook 1 20/-

8 Rope wire 2 feet 101/-

9 M.S sheet plate 5 1200/-

10 Connecting rod 1 40/-

11 Lock plunger 1 180/-

12 Bijagiri 2 30/-

Total Cost 4241/-

Cost of standard part require for manufacturing of mechanically operated Control Discharge Toilet System (CDTS) is only Rs 4221/-

Total cost of Manufacturing of Mechanically operated Control Discharge Toilet System (CDTS) is only Rs 190 + Rs 4241 = Rs 4431

5. ACKNOWLEDGEMENT I take this opportunity to express my deep sense of gratitude to my guide, Prof.W.S.Rathod for his continuous guidance and encouragement throughout the course of this study. Without his valuable suggestion and encouragement this would not have been possible. It is because of his experience and wonderful knowledge I can fulfill the

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 253

requirement of completing the work within the stipulated time.

6. CONCLUSION Mechanically operated Control Discharge Toilet

System is successfully pass trial and can be useful for keeping railway platform clean and hygienic by collecting human waste at railway platform.

As per research going in Indian Railway, Bio toilet has successfully implemented in some long root trains, but cost of Bio toilet system is 11.5 Lack per coach. As per costing of Mechanically operated Control Discharge Toilet System, it cost only Rs 4500/-

As the Mechanically operated Control Discharge Toilet System is working on purely mechanical linkage, hence it is more reliable.

Mechanically operated Control Discharge Toilet System (CDTS) is having less maintenance.

7. REFERENCES [1] ‘Toilet and Trains’, Prof. G. Raghuram, Indian Institute of

Management, Ahmedabad, August 2007. [2] Machine Design (V. B. Bhandari) [3] Mechanical System design (R. B. Patil.) [4] Indian railway coach manual. [5] www.drdo.org.in [6] www.indianrailways.org.in

Vishwasinha Vinayak Bhosale and W.S. Rathod VSRDIJMCAPE, Vol. III (VII), July 2013 / 254