Fluid Flow Friction and Fitting Loss

Experiment 3Fluid Flow Friction and Fitting Loss

ObjectiveTo determine the pressure head loss in different diameters pipes, joints and valves

TheoryWhen a fluid flows through pipes, energy is lost inevitably due to frictions which occurs as a result of viscous drag. Fluid friction produces eddies and turbulence, and these form of kinetic energy are eventually converted into thermal energy. Losses in energy can be expressed in term of pressure or head loss.The total head, H , for a fluid flowing across a pipe is being derived based on Bernoulli’s principle and is expressed as follow:

H= pρg

+V2

2 g+z

where pρg is the pressure head, V

2

2g is the dynamic or velocity head, z is the elevation head

For laminar flow (Re < 2000), energy loss is given by Hagen-Poiseuilles Equation:

∆ P f=128μlQπ d4 or ∆ hf=

128 μlQρπ d4

Where,∆ P f = Pressure loss due to frictionμ = fluid viscosityl = pipe lengthd = pipe diameterQ = volumetric flow-rate

For turbulent flow (Re > 4000), the pressure loss can be calculated based on Darcy’s Equation:

∆ P f=8 ( RρV 2 )

ld( ρV

2

2)

Where,R = Shear stress acting on the wallρ = fluid densityV = fluid velocity

The dimensionless (RρV 2 ) can be group as ∅ , known as the friction factor

Thus the equation would be simplified to:

∆ P f=8∅ ld( ρV

2

2) or ∆ hf=8∅ l

d(V

2

2)

Effect of pipe diameter on energy lossesDifferent pipe diameters would result in different amount of energy losses depending on the regime of flow. The head loss is inversely proportional to the diameter of the pipe.

Energy losses due to sudden change in pipe diameterConsider a sudden enlargement in pipe flow area from A1 to A2, the head loss is

∆ hf=V 1

2

2 g(1−

A1

A2)

2

Head loss at sudden contractionEddies that formed between the ‘vena contracta’ and the pipe wall caused the most energy dissipation. Between the vena contracta and the downstream section (2) a flow pattern similar to that occurring after an abrupt enlargement is formed and thus loss occurs once again.

∆ hf=V 2

2

2 g(A2

Ac−1)

2

=K cV 2

2

2 g

Where,K c = coefficient of friction for contractionEnergy losses in fittingsEnergy is lost whenever the direction of flow in a pipe is altered. The magnitude of these losses is mainly dependent on the radius of curvature of the bend. A pipe bend, elbow or junction therefore causes an additional head loss. The extra loss is expressed as

∆ hf=K❑ (V❑2

2g)

Where,

K = coefficient of frictions for fittings

Losses in valvesValves that are installed in a piping system are causing additional losses of head. For turbulent flow, the head loss can be represented by

∆ hf=K❑(V❑2

2g)

ApparatusWater tank ,water pump , flowmeter , pressure meter , 1 m straight pipe of 8 mm diameter (copper tube) , 1 m straight pipe of 12 mm diameter (PVC) full , 1 m straight pipe of 15.5 mm diameter (PVC), sudden enlargement pipe , sudden contraction pipe , 90° bend , 90° elbow , 90° T-joint , 45° Y-joint , gate valve , ball valve , globe valve & in-line strainer.

Procedure

1. The water tank was ensured to be 34 full.

2. All the valves of the trainer was shut off.3. The trainer main power supply was switched on. The water pump was ensured to run.4. The by-pass (BV) and flow regulating valve (FRV) was adjusted to obtain the desired

liquid flow rate.5. All valves except V 1

❑ was turned off. The pressure meter was connected across the 8 mm copper pipe to measure the head loss.

6. V 1❑ was turned off while V 2

❑ was turned on, with the rest of the valves remained closed, the head loss was measured with the pressure meter including across the contraction , 12 mm PVC pipe and the enlargement portion.


❑ was turned on, the pressure meter was used to measure head loss across 15.5 mm pipe.


❑ was turned on, the pressure meter was used to measured head loss across 18 mm pipe, ball valve, 45° Y-joint and 90° bend.

9. V 4❑ was turned off while the globe valve was fully turn on, the pressure meter was used to

measure the head loss in 90° elbow, 90° T-joint, in-line strainer, gate valve and globe valve.

10. The experiment was repeated using different flow rate 5 time respectively.11. All the data was obtained and recorded in the table.

Results

FittingPressure Drop (mH20)

3(GPM) 4(GPM) 5(GPM) 6(GPM) 7(GPM)straight pipes8 mm (copper tube) 0.64 1.00 1.51 2.02 2.80

Fitting 5(GPM) 7(GPM) 9(GPM) 11(GPM) 13(GPM)12 mm (PVC) full 0.16 0.29 0.45 0.60 0.80sudden enlargement 0.00 0.01 0.01 0.02 0.02sudden contraction 0.15 0.28 0.45 0.62 0.80

Fitting 6.5(GPM) 8.5(GPM) 10.5(GPM)

12.5(GPM)

14.5(GPM)

15.5 mm (PVC) 0.10 0.16 0.22 0.29 0.34Fitting 7(GPM) 9(GPM) 11(GPM) 13(GPM) 15(GPM)

18 mm (PVC) 0.17 0.25 0.38 0.46 0.55Bends

Fitting 6(GPM) 8(GPM) 10(GPM) 12(GPM) 14(GPM)90° Bend 0.04 0.10 0.17 0.21 0.2945° Y-joint -0.09 -0.07 -0.05 -0.03 0.01

Fitting 5(GPM) 6(GPM) 7(GPM) 8(GPM) 9(GPM)90° Elbow 0.01 0.01 0.06 0.07 0.0990° T-joint 0.01 0.05 0.07 0.08 0.11Valve

Fitting 5(GPM) 6(GPM) 7(GPM) 8(GPM) 9(GPM)Gate 0.06 0.08 0.12 0.16 0.20Globe 1.05 1.48 2.04 2.88 3.03In-line strainer 2.68 3.56 4.65 5.14 5.71

Fitting 6.5(GPM) 8.5(GPM) 10.5(GPM)

12.5(GPM)

14.5(GPM)

Ball 0.22 0.36 0.50 0.68 0.86Table 1

Fitting Flowrate : 3(GPM)straight pipes ∆ h(m) l /d V 2/2g(m) ∅ k

8 mm (copper tube) 0.64 125 0.72 9.06x10-5 -Fitting Flowrate : 5(GPM)

12 mm (PVC) full 0.16 83.3 0.40 6.11x10-5 -sudden enlargement 0.00 - 0.40 - 0sudden contraction 0.15 - 0.40 - 0.4

Fitting Flowrate : 6.5(GPM)15.5 mm (PVC) 0.10 64.5 0.24 8.23 x10-5 -

Fitting Flowrate : 7(GPM)18 mm (PVC) 0.17 55.6 0.15 2.59 x10-4 -Bends ∆ h(m) l /d V 2/2g(m) ∅ k

Fitting Flowrate : 6(GPM)90° Bend 0.04 - 0.11 - 0.445° Y-joint -0.09 - 0.57 - -0.2

Fitting Flowrate : 5(GPM)90° Elbow 0.01 - 0.14 - 0.190° T-joint 0.01 - 0.14 - 0.1Valve ∆ h(m) l /d V 2/2g(m) ∅ k

Fitting Flowrate : 5(GPM)Gate 0.06 - 0.14 - 0.4Globe 1.05 - 0.14 - 7.5In-line strainer 2.68 - 0.14 - 19.1

Fitting Flowrate : 6.5(GPM)Ball 0.22 - 0.67 - 0.3

Table 2


8 mm (copper tube) 1.00 125 0.25 4.08 x10-4 -Fitting Flowrate : 7(GPM)

12 mm (PVC) full 0.29 83.3 0.78 5.69 x10-5 -sudden enlargement 0.01 - 0.78 - 0.01sudden contraction 0.28 - 0.78 - 0.4





Fitting Flowrate : 6(GPM)Gate 0.08 - 0.11 - 0.7Globe 1.48 - 0.11 - 13.5In-line strainer 3.56 - 0.11 - 32.4


Table 3



12 mm (PVC) full 0.45 83.3 1.28 5.38 x10-5 -sudden enlargement 0.01 - 1.28 - 8 x10-3

sudden contraction 0.45 - 1.28 - 0.4Fitting Flowrate : 10.5(GPM)

15.5 mm (PVC) 0.22 64.5 0.63 6.90 x10-5 -Fitting Flowrate : 11(GPM)

18 mm (PVC) 0.38 55.6 0.38 2.29 x10-4 -Bends ∆ h(m) l /d V 2/2g(m) ∅ k



Fitting Flowrate : 7(GPM)Gate 0.12 - 0.28 - 0.4Globe 2.04 - 0.28 - 7In-line strainer 4.65 - 0.28 - 17


Table 4



12 mm (PVC) full 0.60 83.3 1.92 4.78 x10-5 -sudden enlargement 0.02 - 1.92 - 0.01sudden contraction 0.62 - 1.92 - 0.3






Fitting Flowrate : 12.5(GPM)Ball 0.68 - 0.49 - 1

Table 5



12 mm (PVC) full 0.80 83.3 2.68 4.57 x10-5 -sudden enlargement 0.02 - 2.68 - 7 x10-3

sudden contraction 0.80 - 2.68 - 0.3Fitting Flowrate : 14.5(GPM)

15.5 mm (PVC) 0.34 64.5 1.20 5.60 x10-5 -Fitting Flowrate : 15(GPM)

18 mm (PVC) 0.55 55.6 0.70 1.80 x10-4 -Bends ∆ h(m) l /d V 2/2g(m) ∅ k

Fitting Flowrate : 14(GPM)90° Bend 0.29 - 0.61 - 0.545° Y-joint 0.01 - 0.61 - 0.02



Fitting Flowrate : 14.5(GPM)Ball 0.86 - 0.66 - 1

Table 6

Graph of head loss against flow rate

0.0002 0.0004 0.0006 0.0008 0.0010

1

2

3

4

5

6

7

8

9

8mm (copper tube)Linear (8mm (copper tube))12mm(PVC) fullLinear (12mm(PVC) full)15.5 mm (PVC)Linear (15.5 mm (PVC))18mm(PVC)Linear (18mm(PVC))

flow rate(m^3/s)

head

loss

(m)

8mm (copper tube) Gradient of graph= 10936.84 y-intercept= -2.4768412mm (PVC)full Gradient of graph = 845.614 y-intercept= -0.1322815.5mm (PVC)Gradient of graph = 363.1579 y-intercept= -0.0931618mm (PVC)Gradient of graph = 586.8421 y-intercept= -0.14184


0.0002 0.0004 0.0006 0.0008 0.0010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

sudden enlargement Linear (sudden enlargement )sudden contractionLinear (sudden contraction)

flow rate(m^3/s)

head

loss

(m)

Sudden enlargementGradient of graph= 46.49123 y-intercept= -2.47684Sudden contractionGradient of graph = -0.00482 y-intercept= -0.25088


0.0002 0.0004 0.0006 0.0008 0.0010

0.05

0.1

0.15

0.2

0.25

0.3

0.35

90°BendLinear (90°Bend)90°ElbowLinear (90°Elbow)90°T-jointLinear (90°T-joint)45°Y-jointLinear (45°Y-joint)

flow rate(m^3/s)

head

loss

(m)

90° Bend Gradient of graph= 338.5965 y-intercept= -0.0919390° ElbowGradient of graph = 398.2456 y-intercept= -0.0449190°T-jointGradient of graph = 403.5088 y-intercept= -0.1101845° Y-jointGradient of graph = 122.807 y-intercept= -0.04614


0.0002 0.0004 0.0006 0.0008 0.0010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Gate valveLinear (Gate valve)Ball valveLinear (Ball valve)Globe valveLinear (Globe valve)In-line strainerLinear (In-line strainer)

flow rate(m^3/s)

head

loss

(m)

Gate valveGradient of graph= 485.9649 y-intercept= -0.1493Ball valveGradient of graph = 949.1228 y-intercept= -0.25246Globe valveGradient of graph = 611.4035 y-intercept= -0.13307In- line strainerGradient of graph = 1063.158 y-intercept= -0.30316

Sample of calculation

1 G (Gallon) = 1

220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For 8mm (copper tube) straight pipe, when the pressure drop is 1.16mH 2O ,

The head loss, Δh = Pƿg = 2.80×9800kgm−1 s−2

999 kgm−3×9.81ms−2

= 2.80 ml/d (pipe length per pipe diameter) = 1m/ 8mm = 1m/8×10−3m = 125V 2/2g =? , diameter of copper tube = 8mm= 8×10−3m therefore, radius of copper tube = diameter/2= 8×10−3m/2= 4×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0001893m3/s = V (π × (4×10−3m) ²)V = (0.0003m3/s) / (π × (4×10−3m) ²)V 2/2g = (0.0003m3/s) / (π × (4×10−3m) ²) ² /2(9.81) = 0.72 m Ø (friction factor) =?As mentioned above,

Δh f=8 ø ld (V

2

2)

0.72 = 8 ø (125) ((0.0001893m3/s) /(π×(4×10−3m)²) ²2

)

ø = 1.16/ (8 (125) ((0.0003m3/s) /(π×(4×10−3m)²) ²2

))

= 9.06x10-5

We know that, 1 G (Gallon) = 1

220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For sudden enlargement straight pipe, when the pressure drop is 0.01mH 2O,

The head loss, Δh = Pƿg = 0×9800kgm−1 s−2

999 kgm−3×9.81ms−2 = 0 m

V 2/2g =? , diameter of sudden enlargement = 12mm= 12×10−3m therefore, radius of copper tube = diameter/2= 12×10−3m/2= 6×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (6×10−3m) ²)V = (0.0003155m3/s) / (π × (6×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (6×10−3m) ²) ² /2(9.81) = 0.40 m

k (coefficient of frictions for fittings) =?

Enlargement: Δh = V 12

2g (1-

A1

A c) ² = V 12

2gK L

Δh = V 12

2gK L

0m= 0.40 K L

K L = K = 0

1 imp GPM (imperial Gallon per minute) =? m3/s (Cubic meter per second)

As we know, 1 G (Gallon) = 1

220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For sudden contraction straight pipe, when the pressure drop is 0.15mH 2O,


999 kgm−3×9.81ms−2 = 0.15m

V 2/2g =? , diameter of sudden enlargement = 12mm= 12×10−3m therefore, radius of copper tube = diameter/2= 12×10−3m/2= 6×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (6×10−3m) ²)V = (0.0003155m3/s) / (π × (6×10−3m) ²)V 2/2g = (0.0003155/s) / (π × (6×10−3m) ²) ² /2(9.81) = 0.40 m k (coefficient of frictions for fittings) =?

Contraction: Δh = V 12

2g ( A2

A c-1 ) ² = V 22

2gK L

Δh = V 22

2gK L

0.15m= 0.40 K L

K L = K = 0.4



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For 90° Bend, when the pressure drop is 0.03mH 2O,


999 kgm−3×9.81ms−2 = 0.03 m

V 2/2g =? , diameter of 90° Bend = 18mm= 18×10−3m therefore, radius of copper tube = diameter/2= 18×10−3m/2= 9×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)

0.0003m3/s = V (π × (9×10−3m) ²)V = (0.0003m3/s) / (π × (9×10−3m) ²)V 2/2g = (0.0003m3/s) / (π × (9×10−3m) ²) ² /2(9.81) = 0.07 m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

0.03m= 0.07 K K = 0.43



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For 90° Elbow, when the pressure drop is 0.10mH 2O,


999 kgm−3×9.81ms−2 = 0.04 m

V 2/2g =? , diameter of 90° Elbow = 15.5mm= 15.5×10−3m therefore, radius of copper tube = diameter/2= 15.5×10−3m/2= 7.75×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (7.75×10−3m) ²)V = (0.0003155m3/s) / (π × (7.75×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (7.75×10−3m) ²) ² /2(9.81) = 0.14m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

0.01 m= 0.14 KK = 0.77



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For 90° T-joint, when the pressure drop is 0.04 mH 2O,


999 kgm−3×9.81ms−2 = 0.01 m

V 2/2g =? , diameter of 90° Elbow = 15.5mm= 15.5×10−3m therefore, radius of copper tube = diameter/2= 15.5×10−3m/2= 7.75×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (7.75×10−3m) ²)V = (0.0003155m3/s) / (π × (7.75×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (7.75×10−3m) ²) ² /2(9.81) = 0.14m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

0.01m= 0.14 KK = 0.1



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For 45° Y-joint, when the pressure drop is -0.09 mH 2O,

The head loss, Δh = Pƿg = −0.09×9800 kgm−1 s−2

999 kgm−3×9.81ms−2 = -0.09 m

V 2/2g =? , diameter of 90° Y-joint = 18mm= 18×10−3m therefore, radius of copper tube = diameter/2= 18×10−3m/2= 9×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003785m3/s = V (π × (9×10−3m) ²)V = (0.0003785m3/s) / (π × (9×10−3m) ²)V 2/2g = (0.0003785m3/s) / (π × (9×10−3m) ²) ² /2(9.81)

= 0.57m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

-0.09m= 0.57KK = -0.2



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For Gate valve, when the pressure drop is 0.06mH 2O,


999 kgm−3×9.81ms−2 = 0.06 m

V 2/2g =? , diameter of Gate valve = 15.5mm= 15.5×10−3m therefore, radius of copper tube = diameter/2= 15.5×10−3m/2= 7.75×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (7.75×10−3m) ²)V = (0.0003155m3/s) / (π × (7.75×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (7.75×10−3m) ²) ² /2(9.81) = 0.14m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

0.06m= 0.14 K

K = 0.4



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For Ball valve, when the pressure drop is 0.22mH 2O,


999 kgm−3×9.81ms−2 = 0.22 m

V 2/2g =? , diameter of Ball valve = 18mm= 18×10−3m therefore, radius of copper tube = diameter/2= 18×10−3m/2= 9×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0004101m3/s = V (π × (9×10−3m) ²)V = (0.0004101m3/s) / (π × (9×10−3m) ²)V 2/2g = (0.0004101m3/s) / (π × (9×10−3m) ²) ² /2(9.81) = 0.67m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

0.22m= 0.67 KK = 0.3



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For Globe valve, when the pressure drop is 1.05mH 2O,


999 kgm−3×9.81ms−2 = 1.05 m

V 2/2g =? , diameter of Globe valve = 15.5mm= 15.5×10−3m therefore, radius of copper tube = diameter/2= 15.5×10−3m/2= 7.75×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)0.0003155m3/s = V (π × (7.75×10−3m) ²)V = (0.0003155m3/s) / (π × (7.75×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (7.75×10−3m) ²) ² /2(9.81) = 0.14m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

1.05m= 0.14 KK = 7.5



220m3 (Cubic meter)

= 4.545×10−3 m3

So,1Gmin = 4.545×10−3m3

1×60 s = 0.000076 m3/s

1mH 2O = 9800 Pa = 9800kg m−1 s−2

For In-line strainer valve, when the pressure drop is 2.68mH 2O,


999 kgm−3×9.81ms−2 = 2.68 m

V 2/2g =? , diameter of Globe valve = 15.5mm= 15.5×10−3m therefore, radius of copper tube = diameter/2= 15.5×10−3m/2= 7.75×10−3mQ (flow rate) = V (fluid velocity) A (area)Q = V (πr2)

0.0003155m3/s = V (π × (7.75×10−3m) ²)V = (0.0003155m3/s) / (π × (7.75×10−3m) ²)V 2/2g = (0.0003155m3/s) / (π × (7.75×10−3m) ²) ² /2(9.81) = 0.14m k (coefficient of frictions for fittings) =?

Δh f = K ( V2

2g)

2.68m= 0.14 KK = 19.1

Discussion

In this experiment, fluid friction trainer with the help of a pressure meter was used to determined the head loss in different diameters of pipes, also different types of joints and valves. From the data collected and recorded in the tables, in a straight pipe ,we could summarized that the bigger the diameter , the lower the head loss or in other word the diameter increases inversely proportional to head loss as in principles that there will be fewer molecules bumping against the sides of the wall for the same amount of volume and fluid flow rate while going through the pipe excluding the 18 mm (PVC) pipe as the anomaly as its head loss increases more than the other ,which can suggest that the pipe may have higher surface roughness compare to the others. All solid materials have a certain degree of surface roughness, in the naked eyes the given materials may look smooth on the surface however under a high-powered microscope, hills and

valleys that could interfere with sliding motion may exist although it only has a small effect on friction for most materials but appropriate for this case since the margin of difference are very small (this could also be apply to copper tube).

As for sudden enlargement , there are no head loss presence which suggest that there are little to resistance occur while the fluid flows through it while for sudden contraction there are positive value for head loss which mean there are resistive force that act toward the fluid flow, as there are may be a sharp corner in the end of the contraction that increases the bumping factors between the molecules and the pipe’s wall.

Moving on to bends, The 90° bend has the highest head loss value which suggest that the fluid may face a sudden and sharp corner thus again increasing the rate of bumping factors. While for 45° Y-joint, it has the lowest head loss value and furthermore by obtaining a negative value this is indicating that there is no head loss occur but the fluid are able to flow much faster. On another note, for 45° Y-joint, it is designed as a guide vanes which help to direct the flow of fluid with less unwanted swirl and disturbances.

Furthermore, for the valve, the higher the fluid flow rate, the higher is the pressure drop. As in-line strainer has higher pressure drop than the other valves, it shows that there is higher resistance of water molecules in this valve than the others.

While doing the experiment, there are certain set of precaution steps so to obtain a better result while executing it smoothly. Before installing the pressure meter, ensure the fluid friction trainer is turn off to avoid the water to burst through the porous of the installing inlet. Next, while reading the pressure meter, ensure the value is settled for a while since fluid may flow inconsistently in reality.

ConclusionWithout exterior disturbances, the head loss of fluid increases inversely proportional to the pipe’s diameter.Head loss also influenced by the type of bends and valve including the material of pipe.Reference

Bruce Munson, T. O. (2013). Fundamentals of Fluid Mechanics. John Wiley & Sons, Inc.

Kurtus, R. (2014, May 25). Causes of Friction. Retrieved July 1, 2015, from School for champion: http://www.school-for-champions.com/science/friction_causes.htm#.VZs90_nR-M8

Documents

Fluid Flow Friction and Fitting Loss