Mass Flow Rate And Volumetric Flow Rate Engineering Essay

Aims: To find the denseness of three types of solids and three types of liquids. To happen out the mass flow rate and volumetric flow rate of the cold mercantile establishments of the H2O dispensers and H2O lights-outs ( Inti academic block E thermofluid lab ) .And calculate the speed of the fluid. Measure the volume of the containers that used in the preceding experiments.

Theories and background:

A stuff ‘s denseness is defined as its mass per unit volume. It is, basically, a measuremement of how tightly affair is crammed together. The rule of denseness was discovered by the Grecian scientist Archimedes. To cipher the denseness ( normally represented by the Grecian missive “ I? ” ) of an object, take the mass ( m ) and divide by the volume ( V ) :

I? =

The SI unit of denseness is kilogram per three-dimensional metre ( kg/m3 ) . It is besides often represented in the cgs unit of gms per three-dimensional centimetre ( g/cm3 ) . One of the most common utilizations of denseness is in how different stuffs interact when assorted together. Wood floats in H2O because it has a lower denseness, while an ground tackle sinks because the metal has a higher denseness. Helium balloons float because the denseness of the He is lower than the denseness of the air.

Another of import effect of denseness is that it allows you to work out for mass and volume, if given the other measure. Since the denseness of common substances is known, this computation is reasonably straightforward, in the signifier:

V * I? = m or = V

Density is a cardinal construct in analysing how stuffs interact in fluid mechanics, conditions, geology, stuff scientific disciplines, technology, and other Fieldss of natural philosophies.

The preservation of mass is a cardinal construct of natural philosophies. Within some job sphere, the sum of mass remains changeless — mass is neither created nor destroyed. The mass of any object is merely the volume that the object occupies times the denseness of the object. For a fluid ( a liquid or a gas ) the denseness, volume, and form of the object can all alter within the sphere with clip. And mass can travel through the sphere. On the figure, we show a flow of gas through a constricted tubing. There is no accretion or devastation of mass through the tubing ; the same sum of mass leaves the tubing as enters the tubing. At any plane perpendicular to the halfway line of the tubing, the same sum of mass base on ballss through. We call the sum of mass passing through a plane the mass flow rate. The preservation of mass ( continuity ) tells us that the mass flow rate through a tubing is a changeless. We can find the value of the mass flow rate from the flow conditions.

Derivation for Flow in a Pipe:

I? = Density of the fluid ( SI: kg/m3 )

A = Cross-sectional Area of Pipe ( SI: M2 )

In this experiment, we are straight use the mass ( kilogram ) of the substances to split the clip used ( s ) .

Mass Flow Rate =

The volumetric flow rate ( V ) of a system is a step of the volume of fluid go throughing a point in the system per unit clip. The volumetric flow rate can be calculated as the merchandise of the crosssectional country ( A ) for flow and the mean flow speed ( V ) .

If country is measured in square pess and speed in pess per second, volumetric flow rate measured in three-dimensional pess per second. Other common units for volumetric flow rate include gallons per minute, three-dimensional centimetres per second, litres per minute, and gallons per hr.

Derivation for Flow in a Pipe:

A = Cross-sectional Area of Pipe ( SI: M2 )

V = Velocity of the fluid in the pipe ( SI: m/s )

We besides can straight utilize the volume ( V ) divide the clip ( T ) . Q =

Procedure:

Part 1

Choose any three types of solid ( wood, gum elastic, steel ) and three types of liquid ( H2O, soft drinks, java ) . For solid portion, step the length, breadth and tallness to happen out the volume of the solid. After that step the mass of the solid with the weighing machine. Density can happen out by utilizing mass and volume of the solid. For the liquid, step the mass of cylinder before continue the experiment. After that, mensurate the mass of the cylinder with contains liquid. And enter down the volume that usage in the experiment.

Part 2

Use the two given containers ( teapot and bottle ) to happen out the mass flow rate and volumetric flow rate. Measure the weight of the containers without H2O. After that pour the cold mercantile establishments of the H2O dispensers and get down to enter the clip with the stop watch. Measure the entire weight of the containers with the cold mercantile establishments of the H2O dispensers and record down the information. Besides record down the volume of the H2O that usage in this step.Repeat the stairss by utilizing H2O lights-outs ( Inti block E thermofluid lab ) . Make the both containers ( teapot and bottle ) with follow the process above.

Part 3

Pour H2O in the both containers and record down the clip with utilizing the stop watch when the containers are full. After that usage weighing machine to weigh the mass of the containers with the H2O.

Consequence and Analysis:

Part 1

Solid

Material

Mass ( kilogram )

Volume ( M3 )

Density ( kg/m3 )

Steel

0.5615

0.000069

8137.68

Wood

0.414

0.00060146

688.325

Rubber

0.017

0.000012936

1314.162

Volume: Steel = ( 7.5x4x2.3 ) x10-6 = 0.000069m3

Wood = ( 122×5.8×0.85 ) x10-6 = 0.00060146m3

Rubber = ( 5.6×1.1×2.1 ) x10-6 = 0.000012936m3

Density: Steel = 0.5615/0.000069 = 8137.68 kg/m3

Wood = 0.414/0.00060146 = 688.325 kg/m3

Rubber = 0.017/0.000012936 = 1314.162 kg/m3

Liquid

Liquid

Cylinder Mass ( kilogram )

Cylinder mass with liquid ( kilogram )

Volume ( milliliter )

Density ( kg/m3 )

Water

0.1045

0.237

130

1019.23

Coffee

0.015

0.2595

240

1018.75

Soft Drinks ( 100 plus )

0.018

0.3545

325

1035.38

Density:

Water: ( 0.237-0.1045 ) / 0.00013 = 1019.23 kg/m3

Coffee: ( 0.2595-0.015 ) / 0.00024 = 1018.75 kg/m3

Soft Drinks: ( 0.3545-0.018 ) / 0.000325 = 1035.38 kg/m3

Part 2

Weight of Teapot ( kilogram )

Entire Weight ( kilogram )

Time ( s )

Volume ( milliliter )

Mass flow rate ( kg/s )

Volumetric flow rate ( m3/s )

Velocity ( m/s )

Cold mercantile establishments of the H2O dispensers

0.217

0.4365

5.02

220

0.04725

=4.38×10-5

0.55796

Water Taps

0.217

0.4055

3.95

190

0.04772

4.81×10-5

0.1694

Mass flow rate:

Cold mercantile establishments of the H2O dispensers: ( 0.4365-0.217 ) /5.02 = 0.043725 kg/s

Water Tapss: ( 0.4055-0.217 ) /3.95 = 0.04772 kg/s

Volumetric flow rate:

Cold mercantile establishments of the H2O dispensers: 0.00022/5.02 = 4.38×10-5 m3/s

Water Taps: 0.00019/3.95 = 4.81×10-5 m3/s

Speed:

Cold mercantile establishments of the H2O dispensers: Q = Av

4.38×10-5 m3/s = ( 7.85×10-5 M2 ) ten ( V )

V = 0.55796 m/s

Water Taps: Q = Av

4.81×10-5 m3/s = ( 2.84×10-4m2 ) ten ( V )

V = 0.1694m/s

Weight of Bottle ( kilogram )

Entire Weight ( kilogram )

Time ( s )

Volume ( milliliter )

Mass flow rate ( kg/s )

Volumetric flow rate ( m3/s )

Velocity ( m/s )

Cold mercantile establishments of the H2O dispensers

0.08

0.2995

5.02

220

0.04725

4.38×10-5

0.55796

Water Taps

0.08

0.2685

3.95

190

0.04772

4.81×10-5

0.1694

Part 2

Mass flow rate:

Cold mercantile establishments of the H2O dispensers: ( 0.2995-0.08 ) /5.02 = 0.043725 kg/s

Water Tapss: ( 0.2685-0.08 ) /3.95 = 0.04772 kg/s

Volumetric flow rate:

Cold mercantile establishments of the H2O dispensers: 0.00022/5.02 = 4.38×10-5 m3/s

Water Taps: 0.00019/3.95 = 4.81×10-5 m3/s

Speed:

Cold mercantile establishments of the H2O dispensers: Q = Av

4.38×10-5 m3/s = ( 7.85×10-5 M2 ) ten ( V )

V = 0.55796 m/s

Water Taps: Q = Av

4.81×10-5 m3/s = ( 2.84×10-4m2 ) ten ( V )

V = 0.1694m/s

Part 3

Mass of teapot ( kilogram )

Entire Mass ( kilogram )

Time ( s )

0.217

1.0445

20.72

Volume: Volumetric flow rate ten Time

Volume of Cold mercantile establishments of the H2O dispensers= 4.38×10-5 m3/s x 20.72s = 9.07536×10-4 M3

Volume of H2O lights-outs = 4.81×10-5 m3/s x 20.72s = 9.96632×10-4m3

Mass of Bottle ( kilogram )

Entire Mass ( kilogram )

Time ( s )

0.08

0.5265

10.74

Volume: Volumetric flow rate ten Time

Volume of Cold mercantile establishments of the H2O dispensers=4.38×10-5m3/s x 10.74s=4.70412×10-4 M3

Volume of H2O lights-outs = 4.81×10-5 m3/s x 10.74s = 4.48932×10-4m3

Discussions on the consequences:

The denseness of the stuffs or substances is depending on its mass and volume. Mass is rearward relative to the volume. So that, every individual stuffs or substances has the different and specific denseness for it. When the value of mass is big and the value of volume is little, the denseness will be high. Mass flow rate is the mass of the liquid flow through the container or the pipe over the clip taken. The mass of the liquid will impact the value of the mass flow rate. The less clip taken will hold the big value of the mass flow rate. Volumetric flow rate is the volume of the H2O that flow through the container in the clip taken. The volumetric flow rate is the speed of the H2O in the surface country. Volume of the containers is different to each other. To mensurate the volume of the containers, we can utilize any sort of liquids and we can acquire the same reply. The heavier liquid the longer clip taken, in contrast the lighter liquid will take a short clip to flux. From the ciphering portion can turn out that the mass is rearward relative to the clip taken to flux. We can happen the volume of the containers by multiplying the volumetric flow rate with the clip taken to pour in the liquid to the full.

Decisions:

Overall, the denseness of the solids is much different to each others. For the liquids, the denseness is non much different. The different stuffs or substances have their specific denseness. From the consequences, we can see that the solids denseness is ever higher than liquids denseness. The mass flow rate and volumetric flow rate for both teapot and bottle are the same. This can turn out that different containers but same liquids used can acquire the same rate of mass flow and rate of volumetric flow. The speed of cold mercantile establishments of the H2O dispensers for teapot is same as the speed of the bottle. The speed of H2O lights-outs for teapot is besides the same as the speed of the bottle. For the portion 3, the volume of the cold mercantile establishments of the H2O dispensers and the H2O lights-out for teapot is near to each other. It is less than 10 % different of the volume. The volume of the H2O lights-outs and the cold mercantile establishments of the H2O dispensers is besides about the same. The per centum different is less than 10 % .

Recommendations:

The experiment should be repeated for few times to acquire more accurate reading.

The stuffs should be had the complete form so that we can cipher the volume accurately and right.

The liquid flow rate should be consistency so that the volumetric flow rate will be more accurate.