# MTH7123 Assignment 1 Solution and Discussion

• Assignment # 1 MTH 7123
Fall 2019
Maximum Marks: 15
Due Date: November 19, 2019

INSTRUCTIONS

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Question: 1 Marks: 5

Following are some applications of either from fluid statics or fluid dynamics .Complete the table writing the appropriate field against each.

Application Fluid dynamics / Fluid statics
Basketball floating on bath tub of water
Air over the surface of an automobile
Circulation of blood
Measuring of blood pressure
Mass transfer in oil pipelines

Question2 Marks: 10

A 3kg plastic tank that has a volume of 0.2m3 is filled with liquid water. Assuming the density of water is 1000kgm-3determine the weight of combine system.

• Question2 Marks: 10
A 3kg plastic tank that has a volume of 0.2m3 is filled with liquid water. Assuming the density of water is 1000kgm-3determine the weight of combine system.

Solution:
Assumption The density of water is constant throughout.
mwater= ρV=(1000 kgm-3)( 0.2 m3)=200kg 3 Marks
mtotal= mwater + mtank=200kg + 3kg = 203kg 3 Marks
W=mg=(203 kg)(9.81ms-2)=1991.40N ≈ 1990N (up to 3 significant digits) 4 Marks

• Question: 1 Marks: 5
Following are some applications of either from fluid statics or fluid dynamics .Complete the table writing the appropriate field against each.

Application
Fluid dynamics / Fluid statics

Basketball floating on bath tub of water

Air over the surface of an automobile

Circulation of blood

Measuring of blood pressure

Mass transfer in oil pipelines

Solution:
The fluid mechanics is defined as the science that deals with the behavior of fluids at rest (fluid statics) or in motion (fluid dynamics), and the interaction of fluids with solids or other fluids at the boundaries.
Fluid statics is all about pressure. Here are the rules;

1. Pressure at any point in a fluid is the same in all directions and is transmitted through static fluids without loss (Pascal’s principle)
2. From 1, the pressure at the wall of any vessel is perpendicular to the wall
3. Pressure due to depth is P = ρgh, and is the same at any horizontal level of connected fluid.
4. The weight of a buoyant object is equal to the weight of the displaced liquid (Archemedes)
The foundational axioms of fluid dynamics are the conservation laws, specifically, conservation of mass, conservation of linear momentum, and conservation of energy (also known as First Law of Thermodynamics). They are expressed using the Reynolds transport theorem. In addition to the above, fluids are assumed to obey the continuum assumption. Fluids are composed of molecules that collide with one another and solid objects. However, the continuum assumption assumes that fluids are continuous, rather than discrete. Consequently, it is assumed that properties such as density, pressure, temperature, and flow velocity are well-defined at infinitesimally small points in space and vary continuously from one point to another. The fact that the fluid is made up of discrete molecules is ignored.
Application Fluid dynamics / Fluid statics
Basketball floating on bath tub of water Fluid statics
Air over the surface of an automobile Fluid dynamic
Circulation of blood Fluid dynamic
Measuring of blood pressure Fluid static
Mass transfer in oil pipelines Fluid dynamic

• Application Fluid dynamics / Fluid statics
Basketball floating on bath tub of water dynamic
Air over the surface of an automobile dynamic
Circulation of blood dynamic
Measuring of blood pressure dynamic
Mass transfer in oil pipelines static

• • A 3kg plastic tank that has a volume of 0.2m3 is filled with liquid water. Assuming the density of water is 1000kgm-3determine the weight of combine system.

We know that the density of water is 1000 kg/m2\text{m}^2m2. We can use the following equation to figure out the weight of the combined system.
m=ρVm=\rho Vm=ρV

(Where m is mass, ρ\rhoρ is density, and V is volume)

Substitute the values we know:
m=ρVm=\rho Vm=ρV

m=(1000)(0.2)m=(1000)(0.2)m=(1000)(0.2)

m=200m=200m=200 kg

The combined mass is the mass of the tank plus the water.
mtotal=3+200m_{total}=3+200mtotal​=3+200

mtotal=203m_{total}=203mtotal​=203 kg

The weight can then be calculated using the following equation:
W=mgW=mgW=mg

(Where W is weight, m is mass and g is force of gravity)

Substitute the values we know:
W=mgW=mgW=mg

W=(203)(9.81)W=(203)(9.81)W=(203)(9.81)

W=1991.43W=1991.43W=1991.43 N 