Mechanical Engineering: Detailed Solutions to Fluid Mechanics Problems

This document presents a comprehensive set of solutions to a mechanical engineering homework assignment covering fluid mechanics and thermodynamics. The solutions address five distinct tasks, including calculations of water volume and weight in a hinged water storage tank, determination of upthrust on a sea water float using Archimedes' principle, calculations of water velocity and pipe diameter changes, energy calculations for melting aluminum ingots in a furnace, and analysis of pressure, temperature, and mass changes in pressurized air tanks and pistons. Each task is broken down step-by-step, including formulas, calculations, and diagrams where appropriate, to provide a clear understanding of the concepts and problem-solving techniques involved. The solutions cover a range of topics such as pressure transmission, buoyancy, fluid dynamics, heat transfer, and the application of the ideal gas law.

SOLUTIONS TO THE QUESTIONS
Task1
A hinged water storage is shown below. Its top side is denoted as C, the narrow side as B and the
long side as A. Assume water density is1000 kg/m3 and gravity is 9.81 m/s2
Side A is hinged at the bottom edge and secured using a clasp at its top edge.
Dimensions are: L = 1.25m, H = 0.6m & W = 0.5m
Answers:
(a) Volume of water = lxwxh= 1.25x0.6x 0.5 = 0.375m3
The weight of water in the storage = ρgV= 9.81 x1000x 0.375 = 3.68kN
Based on principle of transmission of pressure in Fluids: In an enclosed system of fluid, pressure
transmission occurs uniformly across all points within the system
Hence the force would act at the centroid of the side A given by intersection of diagonals as
shown:
Point of force action (3.68kN)

(b)
R
W 0.6
0.3
(c) Now, taking moments about the hinge (base)
The system is at static equilibrium hence:
Wx0.3-Rx0.6= 0
R= 3.68x0.3/0.6 = 1.84kN acting at 0.6m from the base
Task2
A sea water float consists of a cylinder and cone combined as
shown here below. The density of sea water is 1026 kg/m3 and g =
9.81 m/s2
Calculate the total upthrust in kN which is exerted on the float
when it is totally immersed in the ocean.
Draw a labelled diagram showing the float, the water level and all forces acting on the float.
Dimensions are: D = 0.5m, H = 1.6m, h = 0.8m

Answers:
In part a, we invoke Archimedes principle which states that: An object immersed in a fluid either
wholly or partially, will experience an upthrust which is equal to the weight of fluid displaced
by the same object
Therefore, it can be deduced that: Volume of fluid displaced = Volume of object immersed
Volume of object immersed= 1/3Ah + 1/4πd2H
Substituting: 1/3 x3.142 x(0.5/2)2x0.8 + ¼ x 3.142 x0.52x1.6 = 0.052367 + 0.3142 = 0.3665m3
Weight of fluid displaced = ρVg = 1026 x 9.81x 0.3665 = 3.69kN (this is the upthrust )
Due to Fluid weight above
Due to Side pressure (Balanced on both sides)
Upthrust

Task 3: A boiler is supplied with fresh water at a rate of 3 kg/s initially through a pipe of
internal diameter D =40 mm.
(a) Calculate the velocity of the water in the pipe at diameter D.
(b) The velocity of the water just before it enters the boiler through a valve at A is increased
through a tapered section of the pipe. Calculate the required smaller internal diameter if
the velocity of the water should be 2.5 times greater at A.
Answers:
Assuming ρ=1000kg/m3
M’= 3kg/s
D= 40mm
(a) Velocity at the broader section can be determined as follows:
Normally Q= Av and M’= ρQ
M’ = ρAV
A= 1/4x3.142x 0.042
V= m’/ρ(1/4x3.142xD2) = 3/1000x1/4x3.142x0.042
= 3/1.2568 = 2.387m/s
b) Now, V2 (velocity at the narrow section) = 2.5V1= 2.5x 2.387= 5.96m/s
while m’ = ρA2V2

A2 = m’2/ ρV2= 3/1000x 5.96= 0.00050272
But A= 1/4x3.142d2
Hence substituting d in above, d= (4x0.00050272/3.142)0.5 hence d= 8mm
Task4
A smelting furnace is fuelled with propane gas to melt Aluminium ingots (shown below) for
wheel manufacturing. The furnace is loaded up with 4 aluminium ingots each of mass 22.5 kg.
(a) Calculate the energy that must be supplied by the furnace to melt the ingots completely.
Before being placed in the furnace the ingots are stored at 20°C.
(b) If 2.1 kg of propane is used to complete the melting process calculate the efficiency of the
furnace’s heat transfer process.
(c) A standard aluminium ingot is 746 mm in length at 20°C. When the ingots reach a
temperature of 500°C. Calculate:

(i) the increase in length due to the thermal expansion of an ingot assuming the ingot
is free to expand
(ii) the induced force in an ingot assuming it has become jammed in the furnace and is
NOT free to expand
Melting point of aluminium 660°C
Specific heat capacity for aluminium (cp): 0.91 kJ/kgK
Latent heat of melting for aluminium (L): 321 kJ/kg
Modulus of elasticity for aluminium: 69 GPa
Coefficient of linear expansion for aluminium: 22.2 x 10-6 /°C
Calorific value for propane 49.93 MJ/kg
Ingot cross sectional area 0.019 m2
Answers:
(a) This is given by the following formula:
H= MCT’
Now, given M= 22.5 per piece, 4 ingots, therefore total M= 22.5 x 4= 90
C= 0.91kJ/KgK
T’ = T2-T1= 660-20= 640K
Therefore H= 90 x 0.91 x 640= 52.416kJ
The latent heat of melting
= 321kJ/kg x 22.5 x 4= 28.89kJ
Therefore the total heat of melting the ingots= Heat of melting + latent heat of melting
= 52.416 + 28.89 = 81.306kJ

(b) The Heat of combustion of propane
H= Cv xM = 49.93 x106 x2.1 = 104.853MJ
(c ) Given L= 746mm, To = 20oC, T1= 500oC
Now, l’ = linear expansion coefficient x temperature change x initial length lo
L’= lo x eT’=746 x 22.2 x10-6x(500-20)
L’=7.949mm
The new length L1= lo +l’= 746+7.949 = 753.95mm
(ii) The induced stress = Strain x Modulus of elasticity
Strain = l’/lo= 7.949/746 = 0.0106555
Stress= l’/lo x Modulus of elasticity = 69x109 x 0.0106555= 735. 22MPa
The induced Force F= 0.019 x 735.22 = 13.969MN
Task5
Part 1
A pressurised air tank supplies compressed air to anair engine. When the air engine is not
running the absolute pressure in the tank is 280 bar and the temperature of the air is 70°C.

Calculate:
(a) The temperature of the air in the high pressure air tank when the air engine starts
and the absolute pressure in the tank drops to 250 bar. Volume remains constant.
(b) The mass of air in the pressurised air tank.
Note: 1 m3 = 1000 litres
1 bar = 1 x 105 Pa
The characteristic gas constant for air is 287 J/kgK
Tank volume = 450 litres
Part 2
Answers:
Part 1: Given P1= 220kPa, v1= 0.2m3, T1= 25oC and V2= 1/3x0.2 = 0.0667m3T2= T1
(temperature is constant)
Invoking the Pressure law, therefore P2= P1V1/V2= 220x103x0.2/1/3x0.2= 660kPa
The air in a pistonhas an absolute pressure of 220 kPa, volume of
0.2 m3 and a temperature of 25 °C.
A force on the piston pushes it down so the air occupies a third
of its original volume. If the temperature remains constant and
the characteristic gas constant for air is 287 J/kgK, calculate:

(b) Perfect gas equation is used:
M= P2V2/R2T2 = 660x1000x1/3x0.2x105/287 x 298
= 0.5145kg
Part 2:
5) (a) When engine is off, Po= 280bars, To= 70+273= 343K
The volume is constant hence P1/T1= Po/To
T1 = P1To/Po = 250 x(343)/ 280 = 306.25K, hence T1= 33.25oC
(b) Assume perfect gas hence PV= mRT
M= PV/RT= 0.45 x 280x105/287x343= 128kg