· web viewconversions. pressure. 1 pascal = 1 n/m2. 1 kg/cm2 = 98066.5 n/m2. 1 kn/m2 = 1kpa= 103...

FMM Skill Test Question bank Academic Year: 2016-17

Question Bank for Skill Test of

Fluid Mechanics & Machineries

(For Experiment No. 1,2,3,4,5,6,7,8,9)

1


Conversions

a) Pressure1 Pascal = 1 N/m2

1 kg/cm2 = 98066.5 N/m21 kN/m2 = 1kPa= 103 N/m2

1 MN/m2 = 1 Mpa = 106 N/m2

1 bar = 105 N/m2

1 bar = 1 kg/cm2 (Approx.)

b) Pressure head1 m = 100 cm = 1000 mm1 inch = 25.2 mm1 kg/cm2 = 10 m (head)

Atmospheric Pressure:1 bar = 1 kg/cm2 (Approx.)1.01325 bar = 101325 N/m2 = 101.325 KN/m2 = 10.3 m of water = 760 mm of Hg

c) Volume1m3 = 1000 litre1 litre = 10-3 m3

d) Discharge / flow rate1m3/sec = 1000 litre/sec (lps)1 litre/sec = 10-3 m3/sec1m3/sec = 1000 X 60 litre/min (lpm)

1 litre/min = 1

1000 X60 m3/sec

e) Power1 kw = 1000 w

2

FMM Skill Test Question bank Academic Year: 2016-171. To measure fluid pressure by using manometer & pressure gauges and discharge of

water by using measuring tank and stop watch.

1. Write down title of the experiment.

2. Note down the apparatus needed

3. Draw the diagram of differential manometer and constructional diagram of Bourdon tube pressure gauge.

4. Note down the stepwise procedure.

5. Observations:

Pressure Gauge readings:(i) kg/cm2 =……….N/m2

(ii) kg/cm2 =……….N/m2

(iii) kg/cm2 =……….N/m2

S1- Specific gravity of liquid flowing through pipe (water) = 1 S2- Specific gravity of manometer fluid(mercury) = 13.6 Area of tank A = 0.175 m2

6. Prepare observation table and note down readings.Sr. No.

Height of Hg in left limb (h1) cm

Height of Hg in right limb (h2) cm

DifferenceX=(h1-h2)/100 m

Differential head in meters of waterH= X(S2-S1)/ S1

Differential head in N/m2

(P)

Rise inWater level in Tank Hw

meter(m)

Time to collect the water in measuring tanktsecond

123

7. Sample Calculation.

8. Result a) Pressure intensity by using manometer for first flow rate :

1. For Bourdon tube Pressure Gauge = N/m2 2. For U-tube manometer = N/m2

3. Discharge = m3/sec

b) ) Pressure intensity by using manometer for second flow rate :1. For Bourdon tube Pressure Gauge = N/m2 2. For U-tube manometer = N/m2


c) ) Pressure intensity by using manometer for third flow rate :1. For Bourdon tube Pressure Gauge = N/m2 2. For U-tube manometer = N/m2


3


Experiment No. 1

Important Formulae

1) Conversion of pressure 1 kg/cm2 = 98066.5 N/m2

2) Difference in head in metre of mercury X=(h1-h2)/100

3) Differential head in meters of water H= X(S2-S1)/ S1where S1- Specific gravity of liquid flowing through pipe (water) = 1

4) S2- Specific gravity of manometer fluid(mercury) = 13.6

5) Pressure intensity p = w. H (N/m2)where Specific weight w = 9810 N/m3

6) Discharge Q = A .Hw

t (m3/sec)

where Area of tank = A in m2, Rise of water level = Hw in metres, Time to collect the water in measuring tank = t in seconds

4

FMM Skill Test Question bank Academic Year: 2016-172. Calibration of Bourdon Tube Pressure Gauge with the help of Dead Weight Pressure Gauge.


2. Note down the apparatus needed

3. Draw diagram of set up needed to conduct experiment


5. Observations :-

(i) Diameter of piston (d) = 0.55 cm

6. Prepare observation table and note down readings.

Sr. No

Dead Weight

Dead weight generated pressure

(True value) Pc

Test gauge readings Pgi

(Increasing)

Test gauge readings Pgd

(Decreasing)

Average Pg

=Pgi+Pgd

2

Absolute errorPc - Pg

% Error

Pc−Pg

Pg

kg kg/cm2 kg/cm2 kg/cm2 kg/cm2 kg/cm2 %1 0.237 1

2 0.237 +0.237 2

3 0.237 +0.475 3

4 0.237 +0.712 4

50.237 +0.237 +0.712

5


8. Draw the graph of Pc (Kg/cm2) Vs Pg (Kg/cm2)

9. Result1. Maximum error is _____________ for dead weight _______ kg2. Average % error is ______________

5

FMM Skill Test Question bank Academic Year: 2016-17Experiment No. 2

Calibration of Bourdon Tube Pressure Gauge with the help of Dead Weight Pressure Gauge.

Important Formulae

(i) Dead Weight (Pressure Generated) 0.119 kg – (0.5 kg/cm2) 0.237 kg – (1 kg/cm2) 0.475 kg – (2 kg/cm2) 0.712 kg – (3 kg/cm2)

ii) Area of piston a = Π4 d2 (cm2)

Where d = diameter of piston = 0.55 cm

Dead weight generated pressure = Dead weightArea of piston (kg/cm2)

6


3. To study & verify Bernoulli’s Theorem


2. Note down the apparatus needed.



5. Observations: (Note: For this experiment all measurements are in cm) Area of measuring tank A = 1500 cm2

Rise of water level in tank in sec. h = 5 cm Time required to collect the water , t = ---------- sec V2/2g – Velocity head P/W – Pressure head Z – Potential head g = 981 cm/sce2


Sr. No.

Tube No.

Cross-sectional

area(a)

DischargeQd

(A X h)/t

VelocityV

(Qd/a)

Velocity head

(V2/2g)

Pressure head(p/w)

Total head =Velocity head

+ Pressure head

cm2 cm3/sec cm/sec cm cm cm1 1 2.5 X 2.52 2 2.2 X 2.53 3 1.9 X 2.54 4 1.6 X 2.55 5 1.3 X 2.56 6 1.5 X 2.57 7 1.7 X 2.58 8 1.9 X 2.59 9 2.1 X 2.510 10 2.3 X 2.511 11 2.5 X 2.5

7. Sample Calculation.8. Result :

The total maximum head = cm of water.The total minimum head = cm of water.Loss of head is = cm of water

9. Draw the graph of : (i) H. G. L. Vs No. of tubes

(ii) T. E. L. Vs No. of tubes

Experiment No. 3

7

FMM Skill Test Question bank Academic Year: 2016-17To study & verify Bernoulli’s Theorem.

Important Formulae

1. Actual discharge Qd = A .rt (cm3/sec)

where

Area of tank A in cm2

Rise of water level r in cm

Time to collect the water in measuring tank t in seconds

2. Velocity V = Qd

aWhereA = area of corresponding tube section

3. Velocity head = V2

2.gwhere g = 981 cm/sec2

4. Pressure head = pw cm

(Actual measured from scale)5. H. G. L. = Pressure head6. Total head (T. E. L.) = Velocity head + Pressure head

8

FMM Skill Test Question bank Academic Year: 2016-174. To determine coefficient of discharge for a given Venturimeter.



3. Draw diagram of venturimeter.


4. Observations :

(i) Diameter of inlet pipe d1 = 2.6 cm = m

(ii) Area of inlet pipe a1 = π4 d1

2 = m2

(iii) Diameter of throat d2 = 1.4 cm = m

(iv) Area of inlet pipe a2 = π4 d2

2 = m2

(v) Cross section area of measuring tank A = 50 x 35 cm2= m2

(vi) S1- Specific gravity of liquid flowing through pipe (water) = 1 (vii) S2- Specific gravity of manometer fluid (mercury) = 13.6


Sr.No.

Rise of water Level of measuringTank

Deflection ofthe mercury columns of the manometer.

Converted water column height (m)H = HHg X 12.6

Time to collect the water in measuring tank

Hthroat Hpipe Hhg = (H throat – H pipe)

100cm m cm cm m m Sec

1 52 53 54 5


7. Result Average coefficient of discharge is _______

9


To determine coefficient of discharge for a given Venturimeter.

Important Formulae

1. Volume of water collected = Area of tank X Rise in height of water2. Actual Discharge Qact

Qact =Volumeof water collectedTimerequired for collection (m3/sec)

3. Calculation of theoretical discharge Qth.

Qth = a1a2

√a12−a2

2 √2gH (m3/sec)

Where

Diameter of inlet pipe of venturimeter d1 in metres

Diameter of throat of venturimeter d2 in metres

a1 = π4 d1

2

a2 = π4 d2

2

Hhg = (Hpipe – Hthroat) / 100 (metre)

Differential head in metre of water H = HHg X 12.6

4. Calculation of coefficient of discharge Cd

Cd = Actual Discharge

Theoretical Discharge

10

FMM Skill Test Question bank Academic Year: 2016-175.To determine coefficient of Discharge, Coefficient of contraction and Coefficient of

Velocity of sharp edged circular orifice.

1. Write down title of the experiment. 2. Note down the apparatus needed.3. Draw diagram of set up needed to conduct experiment 4. Note down the stepwise procedure. 5. Prepare observation table and note down readings.

1. Shape of orifice: Circular 2. Diameter of orifice (d)= 0.009 m

3. Area of cross section of Orifice (a) = m2

4. Area of cross section of measuring tank (A) = 0.175 m2

5. Initial reading on horizontal scale (X1) = 0 m6. Initial reading on vertical scale (Y1) = 0.092 m7. Initial reading on piezometric scale (h1) = 0 m

Reading for calculating Cd

Sr. No.

Reading on piezometric scale of intake tank (h2)

h = h2-h1

Rise in level of eater in measuring tank (H)

Time (T)

Actual DischargeQact = A X HT

Theoretical DischargeQth = a√2gh

Cd =Qact

Qth

Average Cd

m m m sec m3/sec m3/sec1234

Reading for calculating Cv & Cc

Sr. No.

X =X2 – X1

Y =Y2-Y1

h = h2 – h1

Cv

= X

√4.Y .h

Cc

=

Cd

CV

Average Cv Average Cc

1234

6. Sample Calculations7. Result : For Circular Orifice

1. Coefficient of Discharge (Cd) =2. Coefficient of Velocity (Cv) =3. Coefficient of Contraction (Cc) =

11


To determine coefficient of Discharge, Coefficient of contraction and Coefficient of Velocity of sharp edged circular orifice.

Important Formulae

1. Area of orifice(a)

a = π4 d2

2. Actual Discharge Qact

Qact =A .HT (m3/sec)

3. Theoretical Discharge (Qth)Qth = a √2gh

4. Coefficient of discharge Cd

Cd = Qact

Qth

5. Coefficient of velocity Cv.

Cv = X

√4.Y .h

6. Coefficient of contraction Cc

Cc = Cd

C v

12

FMM Skill Test Question bank Academic Year: 2016-176.To determine darcy’s friction factor ‘f’ in pipe for four different discharge.

1. Write down title of the experiment. 2. Note down the apparatus needed.3. Draw diagram of set up needed to conduct experiment 4. Note down the stepwise procedure. 5. Prepare observation table and note down readings.

1. Material of pipe: Copper2. Diameter of pipe : 0.017 m

2. Area of Measuring tank A= 0.50 X 0.30 m2 = m2

3. Distance between tappings = 1.035 m4. Specific gravity of fluid in pipes S1 = 1 (for water)5. Specific gravity of fluid in manometer S2 = 13.6 (for mercury)

Observation table: Readings for calculation of Darcy’s friction factor

Sr.No.

Diameter ofpipe

Manometer Readings hf =12.6 x cm ofwater

Rise in height of water in measuring tank Hm

Time for collection of water in measuring tank t

Q =A .Hm

t

Darcy’s friction factorf

x1 x2 x = x1 – x2

m m m m m m sec m3/sec1234

6. Sample Calculations

7. Result : For pipe of material Copper average darcy’s friction factor is ______

13


To determine darcy’s friction factor ‘f’ in pipe for four different discharge

Important Formulae

1. Difference in mercury in cm x = x1 – x2 where Mercury reading on right limb x1 in cmMercury reading on right limb x2 in cm

2. Actual loss of head in metre of water hf = x X ( s2−s1s1 ) (m)

3. Actual Discharge Q

Q = Area of tank (A )∈m2X rise∈liquid level(H m)∈m

time taken ( t )∈sconds

Q = A .Hm

t (m3/sec)

4. hf = f lQ2

3d5

where hf = Actual head loss in metre of waterf = Darcy’s friction factorl = length of pipeQ= Actual discharge im m3/secd = diameter of pipe in metre

So Darcy’s friction factor f = 3hf d

5

l Q2

14

FMM Skill Test Question bank Academic Year: 2016-177. To determine minor losses for flow through pipes.

1. Write down title of the experiment. 2. Note down the apparatus needed. 3. Draw diagram of set up needed to conduct experiment 4. Note down the stepwise procedure.5. Observations :

1. Measuring tank dimensions, length l = 0.6 m, breadth, b= 0.3 m, 2. Time interval for measuring riser in water level, t = 20 sec3. For sudden enlargement :

Dia. of pipe at entry (d1) = 15.99 mm = mDia. of pipe at outlet (d2) = 25.91 mm = m

4. For sudden contractionDia. of pipe at entry (d1) = 25.91 mm = mDia. of pipe at outlet (d2) = 15.99 mm = m

5. Angle of bend , Dia. of pipe at entry (d5) = 12.6 mm = m6. Angle of elbow, Dia. of pipe at entry (d6) = 12.6 mm = m


Sr. No.

Nature of pipe fittings

Inlet &Exit Diameters of pipe (m)

Manometer readings

Pressure difference hg = (H2 - H1) ‘m’ of Hg (mercury)

Actual head lost in meters of water column (m) H= 12.6 hg

Rise of water level in tank (Hw) metre

H1(m) H2(m)

1 Sudden Enlargement

d1=d2=

2 Sudden Contraction

d3=d4=

3 Bend d5=4 Elbow d6=

5. Sample calculation.

6. Result :

1. For sudden enlargementa. Actual loss of head = m of waterb. Theoretical loss of head= m of water

2. For sudden contractiona. Actual loss of head = m of water b. Theoretical loss of head= m of water

3. For Elbow k =4. For Bend k=

15


To determine minor losses for flow through pipes.

Important Formulae

1. Discharge of water Q = Quantity of water collection / time taken in second

Q = l X b X H w

t (m3/sec)

where l = length of measuring tank in metreb= breadth of measuring tank metreHw = Rise of water level in tank in metret = interval for measuring riser in water level

2. Case 1: Loss of head due to sudden enlargement1. Velocity of water at entry V1 = Q/A1

Where Q = discharge in m3/sec

A1 = area of pipe at entry = Π4 d1

2

d1= diameter at inlet in metre

2. Velocity of water at outlet V2 = Q/A2


A2 = area of pipe at exit = Π4 d2

2

d2= diameter at outlet in metre

3. Theoretical loss of head due to sudden enlargement, He

He = ¿¿¿ m of water3. Case 2: Loss of head due to sudden contraction

1. Velocity of water at entry V3 = Q/A3


A3 = area of pipe at entry = Π4 d3

2

d3= diameter at inlet in metre

2. Velocity of water at outlet V2 = Q/A4


A4 = area of pipe at exit = Π4 d4

2

d4= diameter at outlet in metre

3. Theoretical loss of head due to sudden contraction, Hc

16


Hc =0.5 V 42

2g m of water

4. Case 3: Loss of head at bend1. Actual loss of head for given bend = m of water (from observation

table)2. Velocity of water flowing through bend V5 = Q/A5


A5 = area of pipe at entry of enlargement = Π4 d5

2

d5= diameter at bend in metre

3. Theoretical loss of head at bend, Hb

Hb =K V 42

2g m of water

4. Equating Actual loss of head = Theoretical loss of head =

Constant for bend K =

5. Case 4: Loss of head at elbow1. Actual loss of head for given elbow = m of water (from observation

table)2. Velocity of water flowing through elbow V6 = Q/A5


A6 = area of pipe at entry of enlargement = Π4 d6

2

d6= diameter at elbow in metre

3. Theoretical loss of head at elbow, Hb

Heb =K V 42

2g m of water

4. Equating Actual loss of head = Theoretical loss of head =

Constant for elbow K =

17

FMM Skill Test Question bank Academic Year: 2016-178. To determine overall efficiency of a Pelton wheel turbine by conducting a trial test on it.





5. Observations :-

1. Diameter of brake drum = 0.276 m

2. Venturimeter constant = Cd a1. a2

√a12−a2

2 √2g (m3) = 0.02498 m =

6. Prepare observation table and note down readings for ½ opening

Sr. No.

Pressure Gauge P

Manometer reading Load applied on turbineSpeed of turbine

N

Left Pressure Gauge of pipe (h1)

Right Pressure Gauge of throat (h2)

Pressure headh=

(h1−h2 ) X12.61000

Dead weight

W1

Spring Load

S2

kg/cm2 mm of Hg

mm of Hg m of water kg kg rpm

1234


8. Result : The average overall efficiency of Pelton wheel is %

9. Draw the graph of Efficiency Vs Power.

18

FMM Skill Test Question bank Academic Year: 2016-17Experiment No.8

Determine Overall Efficiency of Pelton wheel:-

Important Formulae

1. Overall Efficiency = Output power/Input power X 100.

2. Output Power = 2Π N T

60 (watt)

WhereRPM N = RPM measured by Tachometer

Torque T = (W1-S2) X 9.81 X D2

Weight W1 in kg Spring Weight S2 in kg Effective diameter D = 0.276 m

3. Input Power = w X Q X H (watt)Where, Specific Weight of Water w= 9810 (N/m3)

Discharge Q= Cd a1. a2

√a12−a2

2 √2gh (m3)

Venturimeter Constant = Cd a1. a2

√a12−a2

2 √2g = 0.02498 m

h = (h1−h2 ) X12.6

1000 (m of water)

where h1 is Manometer reading connected to pipe of venturimeter in mmh2 is Manometer reading connected to throat of venturimeter in mm

Nozzle Pressure Head H = 10 . p (m)where p = Nozzle pressure in kg/cm2

19

FMM Skill Test Question bank Academic Year: 2016-179.To determine overall efficiency of a Centrifugal pump by conducting a trial test on it.





5. Observations :

1. Cross sectional area of rectangular measuring tankA = L X B = 0.5 X 0.35 m2

= m2

2. Efficiency of electric (Power factor Φ) motor used = 0.853. Specific weight of water = 9810 N/m3

4. Level difference between suction and delivery pressure gauge X = 0.21 m

6. Prepare observation table and note down readings. Sr.No.

SuctionPressure(p1)

PressureHead at inletH1 =

p1∗13.6

1000

Delivery Pressure(p2)

PressureHeadat OutletH2 = p2 X 10

Rise inWater level in Tank H

Time of waterCollection inTank T

Current Voltage

mm Hg M kg/cm2 m cm Seconds Amp Volt1234

7. Sample Calculations.

8. Result The average overall efficiency of centrifugal pump is %

9. Draw the graph of Efficiency Vs Output Power.

20


Experiment No. 9

To determine overall efficiency of a Centrifugal pump by conducting a trial test on it.

Important Formulae

1. Overall Efficiency = Output power/Input power X 100.2. Output Power = w X Q X Hm (watt)

Where, Specific Weight of Water w= 9810 N/m3

Discharge Q= Volume/ Time=( A X H ) / T m3/sec.Tank Area A = L X B = 0.5 X 0.35 = m2

Rise in water level - H is in metre in time T sec.Total Manometric Head Hm = Suction head(Hs) + Delivery head(Hd) + Pressure Level difference (X)Hs= (P1 X 13.6)/1000 (m)Suction Pressure P1 is in mm HgHd= P2 X 10 (m)X = 0.21 mDelivery Pressure P2 is in kg/cm2

3. Input Power = Current X Voltage X Power factor (Watt).Power factor = 0.85

21

· web viewconversions. pressure. 1 pascal = 1 n/m2. 1 kg/cm2 = 98066.5 n/m2. 1 kn/m2 = 1kpa= 103...

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