Design a system to supply water flow from the reservoir to the elevated tank shown in below at a discharge of Q= 1.0 m3/s.

REPORT

Flow Chart of Report

 

 

 

 

 

I.               Problem Statement

Design a system to supply water flow from the reservoir to the elevated tank shown in below at a discharge of Q= 1.0 m³/s.

II.           Approach and Assumptions

Identifying the piping connections of the system and arranging all its auxiliaries from initial to the final discharge. Locating and calculating all the valves, gate valve and all head losses. Assuming, iterating and calculating for the achieving required flow rate at discharge of Q= 1.0 m³/s.

 

 

 

 

 

 

 

 

 

 

III.             Result and Discussion

Pipe line Connection:               

 

 

 

 

 

 

 

 

 

 

 

 

 

 

As per calculation and iterations based on all head losses considering, to attain the flow rate at the discharge of 1 m³/s

Taking pipe diameter as 15 inches = 0.381 m, the cross section area of pipe =

For required flow rate of 1 m³/s as per chosen pipe diameter, the flow velocity = (1 m³/s)/ (= 8.771m/s

Reynolds number =  =

Referring Chart

For pipe roughness (e/d) =0.001, e= 0.00038m or 0.38mm f = 0.02, and head loss for 50 m pipe length is computed using Darcy-Weishbach equations

 

 ,  = 131.6, head loss = 0.02*131.6*3.92 = 10.3 m for pipe roughness 0.38 mm

For cast iron pipe (e/d) =0.0005, e= 0.00019m or 0.19mm f= 0.017,

Head loss = 8.8 m for pipe roughness 0.19 mm or 190 micron

For cost iron pipe Surface roughness is 0.26 mm (From table 14.2)

For longer pipe of 300m length, let us chose pipe with roughness under 260 micron, then e/d = 0.00068, and f is 0.017

If we choose cost iron 50 m pipe also, Head loss = 0.017*131.6*3.92 m = 8.77 m

With same cost iron pipe for 300m long pipe, head loss = 0.017*(300/0.38)*3.92 m = 52.6 m.

The head loss in 300 m long pipe is quite large with cast iron pipe.

Loss can be minimized if we choose Smooth pipe. From the chart the f will get reduced from 0.017 to 0.0095

With smooth pipe for 300m long pipe, head loss = 0.0095*(300/0.38)*3.92 m = 29.4 m.

The head losses considered are the entrance loss, 90 deg bend elbow losses, losses at half opened gate valve, pipe friction losses, globe valve losses and exit loss.

 

ID	INTERFACES	HEAD LOSS [M]
1	RESERVOIR TO PIPE: ENTRANCE LOSS	1.92
2	GLOBE VALVE HEAD LOSS	23.52
3	50 M LONG PIPE	8.77
4	90 DEGREE ELBOW BEND	1.02
5a	300 M LONG PIPE( Cost iron)	52.6 m
5b	3000 m long pipe ( smooth pipe)	29.4
6	GATE VALVE	8.23
7	90 DEGREE ELBOW BEND	1.02
8	50 M LONG PIPE	8.77
9	PIPE TO TANK: EXIT LOSS	4.1

 

Primary head losses (Includes all the pipes, glove valve, and gate valve) = 23.52+ 8.77+ 29.4+8.23+8.77= 78.69

Secondary head loss (Includes entrance, elbow bend, exit) = 1.92+1.02+1.02+4.1 = 8.06

Total head loss = 86.75

% of secondary head loss is close to 10%

Head loss due to component are taken from class notes

After solving energy equation we can conclude that, we require a pump of 1339 KW/ 1796 HP when 15 inch (0.381 m) dia pipe and elbows are used.

To lower the Pump power requirements, we need to increase the pipe diameter. Doubling the pipe diameter reduces flow velocity by factor of 4 and thus Head loss will reduce by factor of 16.

With 30 Inches pipe diameter

Taking pipe diameter as 30 inches = 0.762 m, the cross section area of pipe =

For required flow rate of 1 m³/s as per chosen pipe diameter, the flow velocity = 2.19/s

Reynolds number =  =

For cost iron pipe Surface roughness is 0.26 mm (From table 14.2)

For longer pipe of 300m length, let us chose pipe with roughness under 260 micron, then e/d = 0.00034, and f is 0.017

With same cost iron pipe for 300m long pipe, head loss = 1.63 m

Total head loss will reduce by a factor of 32

 

Final design             

Case 1:- By taking 300m Cast Iron Pipe

ID	INTERFACES	Material	HEAD LOSS [M]
			Pipe Size 15”	Pipe Size 30”
1	RESERVOIR TO PIPE: ENTRANCE LOSS		1.92	0.06
2	GLOBE VALVE HEAD LOSS		23.52	0.735
3	50 M LONG PIPE	Cast Iron	8.77	0.274063
4	90 DEGREE ELBOW BEND		1.02	0.031875
5a	300 M LONG PIPE( Cast iron)	Cast Iron	52.6	1.64375
6	GATE VALVE		8.23	0.257188
7	90 DEGREE ELBOW BEND		1.02	0.031875
8	50 M LONG PIPE	Cast Iron	8.77	0.274063
9	PIPE TO TANK: EXIT LOSS		4.1	0.128125
		Total Head Loss	109.95	3.435938
		Power Required	1961387 Watt	655258.2 Watt
		Power Required	2630.22 HP	878.7012 HP

 

Case 2:- Taking 300m Smooth pipe

 

ID	INTERFACES	Material	HEAD LOSS [M]
			Pipe Size 15”	Pipe Size 30”
1	RESERVOIR TO PIPE: ENTRANCE LOSS		1.92	0.06
2	GLOBE VALVE HEAD LOSS		23.52	0.735
3	50 M LONG PIPE	Cast Iron	8.77	0.274063
4	90 DEGREE ELBOW BEND		1.02	0.031875
5b	3000 M LONG PIPE( Smooth)	Smooth	29.4	0.91875
6	GATE VALVE		8.23	0.257188
7	90 DEGREE ELBOW BEND		1.02	0.031875
8	50 M LONG PIPE	Cast Iron	8.77	0.274063
9	PIPE TO TANK: EXIT LOSS		4.1	0.128125
		Total Head Loss	86.75	2.710938
		Power Required	1676897	646367.9Watts
		Power Required	2248.719	866.7793 HP

 

IV.        Summary

 

a.     Component List

ID	COMPONENTS INVOLVED	DESCRIPTION/RESULT	LENGTH [M]	MATERIAL
1	RESEVOIR	ELEVATION = 450 M	***	Tank
2	GLOBE VALVE	FULLY OPEN	***	Cast Iron
3	PUMP	786 KW/ 1054 HP	***	Cast Iron
4	PIPE	DIA 0.381 M (15 INCH)	50	Cast Iron
5	90 DEGREE ELBOW BEND	DIA 0.381 M (15 INCH)	***	Cast Iron
6	PIPE	DIA 0.381 M (15 INCH)	300	SMOOTH PIPE
7	GATE VALVE	HALF OPENED	***	Cast Iron
8	90 DEGREE ELBOW BEND	DIA 0.381 M (15 INCH)	***	Cast Iron
9	PIPE	DIA 0.381 M (15 INCH)	50	Cast Iron
10	TANK	ELEVATION = 500 M	***	Tank

 

b. Head Losses

ID	INTERFACES	HEAD LOSS [M]
1	RESERVOIR TO PIPE: ENTRANCE LOSS	1.92
2	GLOBE VALVE HEAD LOSS	23.52
3	50 M LONG PIPE	8.77
4	90 DEGREE ELBOW BEND	1.02
5a	300 M LONG PIPE( Cost iron)	52.6
5b	3000 m long pipe ( smooth pipe)	29.4
6	GATE VALVE	8.23
7	90 DEGREE ELBOW BEND	1.02
8	50 M LONG PIPE	8.77
9	PIPE TO TANK: EXIT LOSS	4.1

 

V.                                Appendix

Formulae:

For all head losses:        

                        h_L = K_L*((1/2g)*v²)

                        Where, h_L= Head loss, m

                                     K_L= Loss factor,

                                     g= Acceleration due to gravity= 9.81 m/s²,

                                     v= Flow velocity, m/s

 

i.                 For Entrance Loss,

K_L= 0.5.

ii.               For Globe Valve,

K_L= 6; Ref: Fluid Mechanics by Frank M. White. (Table 6.5)

iii.             Pipe Friction Loss:

h_f = f (L/d) (v²/2g).

Where, h_f = Head loss in m due to Pipe Friction Loss               

 f= Friction factor; Friction factor depends on a) Pipe roughness b) Pipe diameter c) Flow velocity

      Value taken from Ref: Fluid Mechanics by FM White (Equation 6.48)

      L= Length of Pipe;

      d= Diameter of Pipe;

                        g= Acceleration due to gravity= 9.81 m/s²,

                         v= Flow velocity, in m/s

iv.             Elbow 90^o Bend:

K_L= 0.26.

v.               Gate Valve:

K_L= 2.1; from lecture notes.

vi.              Energy Equation:

*The Mechanical Energy Equation in Terms of Energy per Unit Mass

The mechanical energy equation for a pump or a fan can be written in terms of energy per unit mass where the energy into the system equals the energy out of the system.

Epressure,in + Evelocity,in + Eelevation,in + Eshaft = Epressure,out + Evelocity,out + Eelevation,out + Eloss                                     

              (1)

Where,

p = static pressure (Pa, (N/m²))

ρ = density (kg/m³)

v = flow velocity (m/s)

g = acceleration of gravity (9.81 m/s²

hin= elevation height at inlet (m)

hout= elevation height at outlet (m)

E_shaft= net shaft energy per unit mass for a pump, fan or similar (J/kg)

E_loss = hydraulic loss through the pump or fan (J/kg)

The energy equation is often used for incompressible flow problems and is called the Mechanical Energy Equation or the Extended Bernoulli Equation.

Efficiency

According to (1) more loss requires more shaft work to be done for the same rise of output energy. The efficiency of a pump or fan process can be expressed as:

*The Mechanical Energy Equation in Terms of Energy per Unit Volume

The mechanical energy equation for a pump or fan (1) can also be written in terms of energy per unit volume by multiplying (1) with the fluid density - ρ:

  (2)

Where,

γ = ρ g =specific weight   (N/m3)

The dimensions of equation (2) are

Energy per unit volume

The Mechanical Energy Equation in Terms of Energy per Unit Weight involving Heads

The mechanical energy equation for a pump or a fan (1) can also be written in terms of energy per unit weight by dividing with gravity - g:

                          (3)

 = net shaft energy head per unit mass for a pump, fan in (m)

= loss head due to friction (m)

The dimensions of equation (3) are energy per unit weight (Nm/N = m)

E shaft = shaft power (W)

m = mass flow rate (kg/s)

Q = volume flow rate (m3/s)

 

VI.                             Actual Calculations:

 

To attain discharge Flow Rate                       Q= 1 m3/s discharge,

Assume Diameter of Pipe/ Valve                   d= 0.381 m (15 inch);

Velocity of Fluid                                            v= Q/ (pi*d²/4)

                                                                        v= 1/ (3.14*0.381²/4)

                                                                        v= 8.77 m/s

Reservoir to Pipe: Entrance loss              h_e=

                                                                           = =1.96 m

 

Globe valve head loss: Globe valve is fully open

                                                                        h_L= ===23.52 m

1.                                    Pipe Friction Loss:

Length L= 50 m

Reynolds No. = (rho*v*d/ Dynamic Viscosity (µ)) = (998*8.77*0.381/0.001 Pa-s) =3335152

è Implies the flow is turbulent.

 

                              àPipe Material= Cast Iron à Roughness = 0.26

As per FM White Moody Chart,

                  f= 0.017

                                    h_f = f(L/d)(v²/2g)

                                        = 0.017* (50/0.381)(8.77²/2*9.81)=8.8 m

 

2.                                    90^o Elbow Bend 1 Loss:

 

h_L=0.2==1.0 m

 

3.                                    Pipe Friction Loss:

Length L= 300 m

Reynolds No. = (rho*v*d/ Dynamic Viscosity (µ)) = (998*8.77*0.381/0.001 Pa-s) =3335152

è Implies the flow is turbulent.

                              àPipe Material= Cast Iron à Roughness = 0.26

As per FM White Moody Chart,

                  f= 0.017

                                    h_f = *  =  *

                                        = 52.6 m ( With cast iron pipe)

With doubling of pipe diameter to 30”, Velocity gets reduced to 2.19m/s, Re Number 0.9 Million and head loss gets reduced to

                                    h_f = *  =  *

                                        = 1.63 m

( With cast iron pipe), quite low so we do not need to go for smooth pipe.

 

4.                                    Gate valve head loss: Gate valve is half closed

h_L= 2.1*((1/2g)*v²) = = 8.23 m

 

5.                                    90^o Elbow Bend 2 Loss:

 

h_L= 0.26*((1/2g)*v²) = 0.26* ((1/2*9.81)*(8.77) =1.019 m

 

6.                                    Pipe Friction Loss:

Length L= 50 m

Reynolds No.= (rho*v*d/ Dynamic Viscosity(µ))= (998*8.77*0.381/0.001 Pa-s)  =3335152

è Implies the flow is turbulent.

 

                  àPipe Material= Cast Iron à Roughness = 0.26

As per FM White Moody Chart,

                  f= 0.017

                                    h_f = f(L/d)(v²/2g)

                                        = 0.017* (50/0.381)(8.77²/2*9.81) = 8.8 m

 

7.                                    Pipe to tank :Exit loss

h_L= = = 4.1 m

i.                                     Sum of head loss= 114.20 m

ii.                                   Gain of Elevation (Given) = 50 m

iii.                                 Pump Head, h _shaft/pump = Sum of Head Loss + Gain of Elevation

                  = 86.75 + 50

                  =136.75 m

iv.                                  Pump Efficiency, ɳ = 0.8

v.                                    Pump Power =  =

                                                =1,338,834 watt = 1339 KWatt

In HP                                       = 1339 * 1.341 = 1796 HP

 

 

VII.                          Reference

 

1.                Fluid mechanics FM White Book.

2.               https://www.engineeringtoolbox.com/mechanical-energy-equation-d_614.html                 

 

 

A large and very high-speed turbine is to operate at an angular velocity of 18 000 rev/min and will have a rotor with a principal mass moment of inertia of 225 N.m.s2 . It has been suggested that because this turbine will be installed at the North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic loads on its bearings. Estimate the size of these additional loads due to gyroscopic effect.

 

 

  

 

  

Question 1:                                                                                                                            (9 marks)

It is required to carryout dynamic force analysis of the four bar mechanism as shown in the figure.

 

Assume the following data:

rad

𝜔2 = 20

 

𝛼2 = 160

s

rad s2

𝑂𝐴 = 250 mm

𝑂𝐺2 = 110 mm

𝐴𝐵 = 300 mm

𝐺3 = 150 mm

𝐵𝐶 = 300 mm

𝐶𝐺4 = 140 mm

𝐶  = 550 mm

∠𝐴𝑂𝐶  = 60⁰

 

Link	Mass (kg)	Mass Moment if Inetia (kgm2)
2	20.7	0.01872
3	9.66	0.01105
4	23.47	0.0277

 

a)       Draw the acceleration diagram                                                                                                          (2 marks)

b)       Calculate the angular accelerations of all links                                                                                (4 marks)

c)        Calculate the inertia forces                                                                                                                   (3 marks)

 

Solution:-

 

 

 

 

 

From velocity and acceleration analysis,

V_A=250mm x 20 = 0.25mx20 =5m/s

V_B=4m/s, V_BA=4.75m/s

 

𝛼^rA = 0.25x20² =100m/s2

𝛼^tA = 0.25x160=40m/s2

 

A^r_B=VB²/CB=4²/0.3=53.33m/s2

A^r_BA=V_BA²/BA=4.75²/0.3= 75.2m/s2

Og2=AG2=48m/s2

Og3=AG3=120m/s2

 

𝛼3=ABA/AB=19/0.3=63.13 rad/s2

 

𝛼4=Ag/CB=129/0.3=430 rad/s2

 

Inertia Forces (Accelerating Forces)

Fg2=m2xAg2=20.7x48=993.6N (in direction of Og2)

Fg3=m3xAg3=9.66x120=1159.2N (in the direction of Og3)

Fg4=m4xAg4=23.47x65.4=1534.94N (in the direction of Og4)

 

h₂=(IG₂x 𝛼₂)/F₂=(0.01872x160)/993.6=3.01x10^-3m

h₃=(IG₃x 𝛼₃)/F₃=(0.01105x63.3)/1159.2=6.03x10^-4m

h₄=(IG₄x 𝛼₄)/F₄=(0.0277x430)/1534.94=7.76x10^-3m

 

The inertia forces F_i2, F_i3, F_i4 have magnitudes equal and direction opposite to the respective accelerating forces and will be tangents to the circles of radius h2, h3 and h4 from G2, G3 and G4 so as to oppose 𝛼2,  𝛼3 and  𝛼4.

 

F_i2 =993.6N

F_i3 =1159.2N

F_i4=1534.94N

 

Question 2:                                                                                                                                                (11 marks)

a)       For the following cam follower mechanism, define: prime circle, pitch circle and trace point.

(3 marks)

 

 

Prime circle: It is the smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a roller follower, the prime circle is larger than the base circle by the radius of the roller.

 

Pitch circle: It is a circle drawn from the centre of the cam through the pitch points.

 

Trace point: It is a reference point on the follower and is used to generate the pitch curve. In a roller follower, the centre of the roller represents the trace point.

 

 

 

 

 

 

 

b)       The reciprocating radial roller follower of a plate cam is to rise 40 mm with simple harmonic motion in 180o of cam rotation and return with simple harmonic motion in the remaining 180o. If the roller radius is 7.5 mm and the prime-circle radius is 40 mm, construct the:

i.        displacement diagram                                                                                      (2 marks)

ii.        the pitch curve                                                                                                  (1 marks)

iii.        the cam profile for clockwise cam rotation.                                                   (3 marks)

If you are using the analytical method, you will need to use Excel or MATLAB and provide the screenshot of the drawing. You can access Excel and MATLAB through the EIT remote lab.

If you are using the graphical method, take a picture of your working and insert it as a screenshot.

 

 

 

 

c)        Identify the type of cam follower mechanism for the following arrangements:   (2 marks)

 

 

 

Answer:- Type of Cam Follower:-

 

a.      Knife Edge follower

b.      Flat Faced Follower

c.      Roller Follower

d.      Spherical Follower

 

Question 3:                                                                                                                                                (12 marks)

a)       Explain how planetary gear train is used for getting different gear ratios in an automobile.

 

Planetary gear sets are used and combined in a complex manner so that transmissions with seven or eight speeds forward plus reverse are possible. Shifts are made by engaging or releasing one or more internal clutches that drive a gear set member, or by engaging or releasing other clutches or bands that hold a gear set member stationary. An automatic transmission might have as many as seven of these power control units (clutches or bands). One-way clutches are also used that self-release and overrun when the next gear is engaged. The control units can operate without the interruption of the power flow.

 

PLANETARY GEAR SET OPERATION

 

Planetary gear sets are so arranged that power enters through one of the members and leaves through one of the other members while the third member is held stationary in reaction. Power flow through a planetary gear set is controlled by clutches, bands, and one-way clutches. One or more clutches will control the power coming to a planetary member and one or more reaction members can hold a gear set member stationary. The third planetary member will be the output.

(a) If the planet carrier is held with the sun gear rotating, the planet gears simply rotate in the carrier and act as idler gears between the sun and ring gears.

(b) If the sun or ring is held, the planet gears will walk around that stationary gear; they rotate on their shafts as the carrier rotates.

(c) If two parts are driven and no parts are held, the planet gears are stationary on their shafts, and the whole assembly rotates as a unit.

b)      Determine gear ratios for all the speeds for the following truck transmission gear box.

(6 marks)

 

 

Solution:-

 

T2=17T

T3=43T

T4=36T

T5=27T

T6=17T

T7=24T

T8=33T

T9=43T

T10=18T

T11=22T

 

Speed 1 = 2-3-6-9

Gear Ratio 1 = (T3/T2)x(T6/T9) = (43/17)x(17/43) = 1

 

Speed 2 = 2-3-5-8

Gear Ratio 2 = (T3/T2)x(T5/T8) = (43/17)x(27/33) = 2.06

Speed 3 = 2-3-4-7

 

Gear Ratio 1 = (T3/T2)x(T4/T7) = (43/17)x(36/24) = 3.79

 

Reverse = 2-3-6-10-11-9

 

Gear Ratio R = (T3/T2)x(T6/T10)x(T11/T9) = (43/17)x(17/18)x(22/43) = 1.22

 

Question 4:                                                                                                                            (8 marks)

Determine the bearing reactions at A and B for the system given below. If it rotates at 350 rev/min. Determine the magnitude and the angular orientation of the balancing mass if it is located at a radius of 50 mm.

 

 

 

Solution:-

 

w=360rev/min, R=50mm

w= (350x2π)/60 = 36.65 rad/ sec

 

F₁=m₁xR₁xw² = 2x0.025x36.65²= 67.16N

F₂=m₂xR₂xw² = 1.5x0.035x36.65²= 70.51N

F₁=m₃xR₃xw² = 3x0.040x36.65²= 161.18N

 

F1 =67.16 sin90 = 67.16j N

F2 = 70.51 (ang -165) = -70.51 sin75i – 70.51cosj = - 68.10i-18.24j

F3 = 161.18N (ang.-75) = 161.18cos75i-161.18sin75j = 41.71i-155.68j

 

∑F = F1+F2+F3

∑F = -26.39i – 106.76j N

 

∑F = 109.95N (ang.-103.9ᵒ)

 

Since all rotating masses are in a single plane the correction mass must be in that plane.

Fc = -∑F = 109.65N (ang.76.1)

 

Fc = Mc x Rc x W²

109.65 = Mc x 0.050 x 36.65²

Mc=1.637 1kg

Θc = 76.1ᵒ

 

Question 5:                                                                                                                                              (10 marks)

A large and very high-speed turbine is to operate at an angular velocity of 18 000 rev/min and will have a rotor with a principal mass moment of inertia of 225 N.m.s2 . It has been suggested that because this turbine will be installed at the North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic loads on its bearings. Estimate the size of these additional loads due to gyroscopic effect.        (5 marks)

Given :-

I = 225 Nms2 or kgm2

w = 18000rev/min

now, angular speed of precession:

The angular velocity of precession is normal to the earths orbital plane and its magnitude is the angular speed 50.2”/year,nearly.

Therefore, angular speed of precession = 360 ᵒ/ period of revolution in years

=360/25800 = 0.01395ᵒ/year

Importantly, the inclination of the polar axis to the normal to the ecliptic remains the same 23.4ᵒ.

i.e Wp = 0.01395ᵒ/year

=0.01395/(365x24x60x60) x (π/180)

Wp=7.720 x 10^-12 rad / sec

w= 18000/60 x (π/180) = 5.2359 rad/sec.

Gyroscopic Couple acting on the turbine

C=I x w x wp

=225x5.235x7.720x10^-12

=9.993x10^-9 Nm

Therefore, the value of couple is too small.

It might not have any effect. Hence, the additional load does not matter on the system.

 

a)     The diameter of the driver pulley is 25 cm and that for the driven pulley is 55 cm for an open belt drive. The driver pulley rotates at 345 rpm. The angle of contact for the driver pulley is

    rad. Calculate:

i.        the centre distance between two pulleys                                                       (1 mark)

ii.        the angle of contact for driven pulley,                                                            (1 mark)

iii.        length of the belt,                                                                                            (1 mark)

iv.        angular speed of the follower pulley                                                              (1 mark)

v.        The velocity of the belt.                                                                                   (1 mark)

 

Solution: - For Open belt drive

 

Driver Pulley d2= 25 cm N2 = 345 rpm

Driven Pulley d1 = 55 cm

Angle of contact = 3.054 rad

 

Centre distance between two pulleys(x)

 

Θ = 3.054 rad = 3.054 x (180/π) = 174.98ᵒ

 

Θ = 180 - 2𝛼

 174.98ᵒ = 180 - 2𝛼

𝛼 = 2.509ᵒ

 

Sin 𝛼 = d2-d1 / 2x

Sin2.509 = 55-25 / 2x

X = 342.65 cm

 

Length of belt = L = 2x + [π(D₂ + D₁)]/2 + [(D₂ + D₁)²]/4x

 

L = 733.08 cm

 

Velocity of belt, V = (π x d2 x N2) / 60

=( π x 0.25 x 345 ) / 60

V = 4.516 m / sec.

 

As we know, N2/N1 = d1/ d2

345/N1 = 55/25

N1=156.81 rpm

 

Angular speed of follower pulley,

w= (2 x π x N1) / 60

   = (2 x π x 156.81) / 60

W = 16.42 rad/sec

 

REFERENCES

1.      A textbook of mechanics by R S Khurmi S Chand and Company LTD.

2.      A textbook of theory of machine by R S khurmi .

3.      THEORY OF MACHINE by Joseph Shigley.

4.      Some engineering notes (like made easy and ace etc).

Open ended design process

Open ended design process:
The most important aspect of the design process is to ensure that the air is heated to the
appropriate temperature. Outside of this, it is up to you to come up with design criteria.
Some ideas to be considered are outlined below:
• Consider the energy required to heat the tubes to their steady state operating
temperature
• Consider the cost of materials to build the heater
• Consideration of parasitic heat losses (heat losses to the ambient environment)
• Set restrictions on the size of the heater cross-section
• Consider the temperatures of the rods. Are they realistic?
• Consider whether the rod can be lumped or if you must consider spatial temperature
variation
Again, the design process is open ended. You are not required to consider all of the above
and you should impose additional limitations and design considerations as you see fit.
I recommend developing an iterative technique to determine the temperature of the rods and
air. I recommend grouping columns of rods together (in other words, treat all rods in a
particular column as having the same temperature and heat transfer coefficient). Also,
double check that the rod can be treated as lumped in your analysis. If it cannot, additional

ENME 4430 Design Projects

Air-conditioning costs of commercial office buildings are typically high, especially in the building with large
number of occupants. In order to maintain the Indoor Air Quality (IAQ) in meeting the building code, it is
required to bring in fresh air of 15 cfm (cubic foot /minute) of fresh air per occupant. On the peak hour of the
cooling season, the out air (O.A) is at 98.10 F and humidity ratio of 0.0172 in a building with 100 occupants.
This measure raises the cooling costs of the building. In the interest of reducing the cooling cost, one of the
most widely employed energy efficiency technology is the air-to-air energy recovery as shown in Figure 2. As
shown in Figures 1 and 2 that 1, 500 cfm of fresh mixes with recirculated of mass flow rate, mr and the mixture
is sent to the air-conditioner (A/C). However, due to energy recovery, the enthalpy of the mixed air entering the
(A/C) unit in Figure 2, h2 is less than the corresponding h2 of the conventional unit of the conventional system
with no energy recovery system unit as shown in Figure 1, thereby reducing the capacity of the A/C. It may be
noted that the mass flow rate and enthalpy of the supply air (h3) to the space is same for both the cases. The
detailed schematic of the energy recovery system is shown in Figure 3. The outside air (OA) at state 1 shown in the
Figure 3, with humidity ratio of w1 and temperature of T1 enters the desiccant wheel (DW) to which regenerated air at
state “12” with humidity ratio of w12 and temperature of T12 is supplied from the other side of the desiccant wheel to
dehumidify the OA. Heat energy supplied in the heater (H) to raise the temperature T12 helps in dehumidifying the
incoming OA. Greater the value of T12 greater is the dehumidification of OA. Due to rise in temperature of the OA after
dehumidification, it passes through the Heat Wheel (HW), where it is cooled to a lower temperature T3. In order to
accomplish this, a mixture of OA at state 9 is mixed with exhaust air from the building space at state 8 in the Mixing Box
(MB1), and the mixture of these airstreams at state 10 passes through the HW wheel. Indirect evaporative cooler (Mcycle)
is employed to reduce the temperature from T3 to T4. Airstream at state 4 is now mixed with the return air from the
building space at state 7 in the Mixing Box (MB2) so that, the temperature and humidity ratio at state 5 are much lower
compared to that of the conventional system, resulting in lower cooling capacity required in the air-conditioner (AC). The
objective of the design project is to determine the following results:
(a) What is the capacity of the air-conditioner for the conventional unit as shown in the Figure 1?
(b) What is the capacity of the air-conditioner with the energy recovery unit shown in the Figure 2 ?
(c) Heat energy required in the heater (H) shown in Figure 3, between the states (11) and (12)?
(d) Assuming the heat rejected in the condenser of the air-conditioner employed in the Figure 2 is 1.2 times the
capacity of the (qA/C ), will it provide the required heat in the heater of part (c ) ?
(e) Write the executive summary of the project as per the instructions shown on the last page.
A/C
exhaust fan
supply air@ T3 = 61F, w3 = 0.0086
Building Space
@ 76F, 55 %
To exhaust O.A
1
3
4
6 h0
Peak Cooling Load = 21.29
tons, SHR =0.63
Wc, conv
h3 =24 Btu/lb
h4 =30 Btu/lb
1,500 cfm of outside air@ T1 = 98.1F, w1 = 0.0172
exhaust air@ T5 = 76F, w5 = 0.1040
5
2
mr
Figure 1 Conventional A/C System
2
A/C
exhaust fan
supply air@ T3 = 61F, w3 = 0.0086
Building Space
@ 76F, 55 %
To exhaust O.A
Energy Recovery Unit
1
1' 2 3
4
5
h1
h1'
Peak Cooling Load = 21.29
tons, SHR =0.63
h1' < h0 Wc < Wc, conv
h3 =24 Btu/lb
h4 =30 Btu/lb
1,500 cfm of outside air@ T0 = 98.1F, w0 = 0.1040
exhaust air@ T4 = 76F, w4 = 0.104
6
exhaust air@ T5 = 76F, w5 = 0.104
mr
Figure 2 A/C System with Energy Recovery Unit
6400 sqft
Building Space @
76 F, 55% r.h
100 occupants
C. Load = 21.29 tons
1500 cfm fresh air
O.A
R.A
MB2
W9 = 0.0172
T9 = 98.1 F
S.A
A/C
Indirect Evaporative Cooler
M - Cycle
HW DW
W7 = 0.0104
T7 = 76 F
W5 = 0.01014
T5 = 75.5 F
8
7
6 5
4
13
10 12
9
3
2 1
11
H
W1 = 0.0172
T1 = 98.1 F
W2 = 0.0102
T2 = 148.5. F
W12 = 0.0138
W11 = 0.0138 T12 = 178. F
T11 = 125. F
W3 = 0.0102
T3 = 117. 1 F
W6 = 0.0086
T6 = 61.0 F
W4 = 0.0102
T4 = 74.2 F
m1 =210 #/min
m8 =105 #/min
m13 =210 #/min
m7 =629 #/min
MB1
m9 =105 #/min
m6 =734 #/min
O.A
SHR = 0.63
W10 = 0.0138
T10 = 87 F
W8 = 0.0104
T8 = 76 F
Figure 3 Conventional A/C System with Energy Recovery Unit for Summer Operation
3
Typical Contents of Executive Summary
The executive summary would start with a statement relevant to the objective of the experiment or project to underscore the significance of the project or experiment. For instance, if the project is about the air conditioning system, the appropriate relevant statement would be “the energy consumed for heating and cooling of buildings is about 37 percent of total energy consumed in the U.S. The main objective of the project is specified in as detailed a manner as possible without exceeding 2 or 3 sentences. A very brief description of the apparatus or system involving 2 or 3 sentences should be presented here. Providing the Figure or picture will not be an effective replacement for the description. It can be done only in the case of an extremely complicated Figure. The underlying principle based upon which, the main work involved in the project is then specified not exceeding 1 or 2 sentences. Avoid writing equations or specifying references in the summary.
The approach to the work, which may involve numerical simulations or experimental work or a survey etc,. is clearly discussed in the second paragraph. A brief summary of the procedure not involving specific details should be presented here without exceeding 4 or 5 sentences followed by the specification of key results. Presentation of a graph or chart is not an effective replacement for the specification of key results. For instance, if the objective is to determine the coefficient of thermal conductivity at various temperatures for a fluid, you may tend to present a graph or Table to reflect this variation. This is to be avoided. Instead you may indicate that the coefficient of thermal conductivity varied from 0.2 to 1.5 W/m.K as the temperature varied from -100C to 2000C with an average value of 1.75 W/m,K.
The key concluding remarks involving the main trends exhibited by the results of the project or experiments should be covered in the last paragraph. This should be followed by a brief recommendations that may be useful to produce better quality results or that may help in avoiding delays, poor quality results or cut down on total cost, etc,. For a moderate size of the project, the executive summary should be presented within one or two pages.