Apply Fluid and Thermodynamics Principles in Engineering - Assignment Part B

This assignment covers the application of fluid and thermodynamics principles in engineering. It is a part of the Advanced Diploma of Engineering (Mechanical) course with course code MIET7547C. The assessment is an individual assignment and must be submitted online through CANVAS. The summary and purpose of the assessment, assessment instructions to students, and additional instructions to students are provided in the content.

STUDENT
STUDENT - KNOWLEDGE ASSESSMENT TASK
Task Number 3 of 4 Task Name Assignment Part B
National unit(s) code MEM23006A National unit(s) title Apply fluid and
thermodynamics principles
in engineering
National qualification
code
MEM60112 National qualification
title
Advanced Diploma of
Engineering (Mechanical)
RMIT Program code C6130 RMIT Course code MIET7547C
Section A - Assessment Information
Student Name)
Student ID
Date
Duration and/or due date
Task instructions
 Summary and Purpose of Assessment
This course applies to fluid and thermodynamic devices and systems used in industry. It is suitable for
people working as technicians in engineering or related fields using hydrostatic, hydrodynamic, fluid
power or heating, ventilation and air conditioning (HVAC) equipment and those pursuing careers and
qualifications in engineering or related disciplines.
 Assessment Instructions to students
What: This is an individual assessment and must be submitted online through CANVAS (unless
otherwise instructed by your teacher).
Where:
How:
Conditions for assessment
Instructions to Students:
1. Students found in breach of assessment conditions can be charged with academic misconduct,
have their results cancelled, be excluded from the program and receive other penalties. Penalties can
also apply if a student’s assessment material is copied by others.
2. Plagiarism is the presentation of the work, idea or creation of another person as though it is one’s
own. It is a form of cheating and is a very serious academic offence that may lead to expulsion from
the University. Plagiarised material can be drawn from, and presented in, written, graphic and visual
form, including electronic data, and oral presentations. Plagiarism occurs when the origin of the
material used is not appropriately cited.
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STUDENT
Task Number 3 of 4 Task Name Assignment Part B
National unit(s) code MEM23006A National unit(s) title Apply fluid and
thermodynamics principles
in engineering
National qualification
code
MEM60112 National qualification
title
Advanced Diploma of
Engineering (Mechanical)
RMIT Program code C6130 RMIT Course code MIET7547C
Section A - Assessment Information
3. Late submission approval will be in line with Application for extension of submittable work (7
calendar days or less) form and is available at building 57, level 5, SVEHS reception.
4. RMIT special consideration is to enable you to maintain your academic progress despite adverse
circumstances. The process for special consideration can be found at
http://www1.rmit.edu.au/students/specialconsideration
5. Students with a disability or long-term medical or mental health condition can apply for adjustments
to their study and assessment conditions (Reasonable Adjustments and Equitable Assessment
Arrangements) by registering with the Equitable Learning Services (ELS) at
https://www.rmit.edu.au/students/support-and-facilities/student-support/equitable-learning-services
If you already registered with ELS and your study plan is approved, please inform your teacher if this
assessment task is not adjusted in line with approved study plan.
6. If a student’s result is NYS, feedback and an agreed deadline date for the student to demonstrate
competency will be advised.
Additional Instructions to Students:
1. This is an individual assessment and must be submitted online through CANVAS (unless otherwise
instructed by your teacher).
2. Performance requirement:
a. Satisfactory (S) performance – met minimum assessment criteria
b. Not Yet Satisfactory (NYS) performance – did not meet minimum assessment criteria
3. Your performance in this assessment will count towards the final grade only if the result is
Satisfactory (S) and will be worth 30% of the final grade.
Equipment/resources students must supply
(if applicable):
Equipment/resources to be provided by RMIT or
the workplace (if applicable):
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STUDENT
Section B - Student Answer Sheet
Questions
Provide your responses in the boxes below each question
Q1
:
BASIC FLUID DYNAMICS (Continuity of Flow & Bernoulli’s
Equation):
(a) Oil of R.D. 0.9 and dynamic viscosity 0.06 Pas flows through a pipe of
diameter 75mm at a flow rate of 5 L/s Determine:
i. The flow velocity
ii. The Reynold's No.
iii. The flow regime (2+2+1=5
marks)
Satisfactory
response
Yes
☐
No
☐
Answer: Given data:
Relative density R . D.=0.9 , Dynamic viscosity μ=0.06 Pas , Diameter D=75 mm∨0.075 m ,∧Discharge Q=5 liter
s =5
Step 1: Discharge Q= Area × Velociy= AV = π
4 × D2 ×V = π
4 × 0.0752 × V
Or 5 ×10−3=4.418 ×10−3 × V
Or V = 5 ×10−3
4.418× 10−3 =1.1317 m
s
Step 2: Reynolds number Re= ρvd
μ =0.9 ×1000 × 1.1317× 0.075
0.06 =1273.1625
Step 3: Flow regime - Since the Reynolds number, Re is less than 2000, the flow is a laminar
flow.
(b) A liquid of relative density (R.D) 0.9 is pumped through a pipe fitted
with an inclined transition piece which reduces in diameter from 200mm to
180mm, as shown in the diagram below. The pressure and velocity of the
fluid at position ➀ are 40kPa and 5m/s respectively. The elevations at ➀
and ➁ are 2.5m and 5.0m respectively.
Satisfactory
response
Yes
☐
No
☐
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STUDENT
For the transition piece shown above, assuming there are no frictional
losses in the flow, determine:
i. the Mass Flow Rate in kg/s;
ii. the Volumetric Flow Rate in L/s;
iii. the Velocity at position ➁ in m/s;
iv. the Pressure at position ➁ in kPa
v. the total system Head in m.
vi. The Fluid Power required to be delivered by a pump to maintain this
flow.
vii. the input Power required if the pumping efficiency if 65%
viii. The time and cost of filling a 250 ML tank situated at the outlet ➁ of
the pipe. (the cost of electrical power is $0.30 per kWH)
(2+2+2+4+2+2+2+4=20
marks)
Answer:
Given data:
Relative density R . D.=0.9 , Density=09.× 1000=900 kg
m3 , D1=200 mm=0.2m , D2=180 mm=0.18m , Pressure p1=4
Step 1: Mass flow rate ˙m=ρ A1 V 1=900 × ( π
4 )× 0.22 × 5
Or ˙m=141.37 kg
s
Step:2 Volumetric flow rate Q= A1 V 1= ( π
4 ) × 0.22 × 5=0.157 m3
s
Step 3: Velocity at section 2, V 2 ,Q=A1 V 1=A2 V 2
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STUDENT
Or V 2= A1
A2
× V 1= ( π
4 )× 0.22
( π
4 )× 0.182
×5=6.173 m
s
Step 4: Using Bernoulli’s equation at sections 1 and 2, we get
p1
ρg + v1
2
2 g + z1= p2
ρg + v2
2
2 g + z2
Or p2=ρg ( p1
ρg + v1
2
2 g − v2
2
2 g + z1−z2 )=900 × 9.81× ( 40× 103
900 ×9.81 + 52−6.1732
2 ×9.81 +2.5−5 )
Or p2=34.1 kPa
Step 5: Total head In the system H= p1
ρg + v1
2
2 g + z1= p2
ρg + v2
2
2 g + z2
Or H= 40 ×103
900 × 9.81 + 52
2× 9.81 +2.5=125.58 m
Step 6: Fluid Power P= presssure × discharge × Totalhead × acceleration due ¿ gravity=ρgHQ
Or P=900× 9.81 ×125.58 ×0.157=174.073 KW
Step 7: Pump input power for a given efficiency value of 65%
P¿= Pout
η = 174.073
0.65 =267.804 KW
Step 8: The time and cost of filling a 250 ML tank situated at the outlet ➁ of the pipe
Volume v=250 ML=250 ×106 L=250× 103 m3
Discharge Q= v
t
Or time t= v
Q
Or t= ( 250 ×103 )
0.157 ×3600 =442.32hrs
Cost at $0.30 per kWH¿ 267.804 KW ×442.32 hrs × $ 0.30=$355,36.52
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STUDENT
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STUDENT
(c) Air at 120 kPa and 40°C flows upward through a 6-cm-diameter
inclined duct at a rate of 55 L/s, as shown on the next page. The duct
diameter is then reduced to 5 cm through a reducer. The pressure change
across the reducer is measured by a water manometer. The elevation
difference between the two points on the pipe where the two arms of the
manometer are attached ➀ and ➁) is 200 mm. Determine the differential
height between the fluid levels of the two arms of the manometer.
(Density of the air may be neglected; Take Density of water to be 1000
kg/m3
Satisfactory
response
Yes
☐
No
☐
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STUDENT
(2+2+4+2=10 marks)
(Total Q1: 5+20+10 =
35marks)
Answer:
Given data:
Step 1:
Using the ideal gas equation
P= ρRT -----(1)
Where R is the gas constant = 0.287 kJ/kg, ρair is the density of air, T is the temperature of
air = 40 = 313.15 K, and the pressure of air is℃ P = 120 kPa.
Substituting the above values in (1), we get
120= ρair × 0.287 ×313.15
Or ρair=1.335 kg
m3
Step 2:
Apply continuity equation at section 1, we get
Discharge Q= A1 V 1 -----(2)
Given d1 = 6 cm = 0.06 m and Discharge Q = 55 litres/s = 55 ×10-3 m3/s
Substituting in (2), we get v1 = Q
A1
= Q
π
4 ×d1
2
= 55 ×10−3
π
4 × 0.062
=19.45 m
s
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STUDENT
Step 3:
The velocity at section 2
v2= Q
A2
= Q
π
4 ×d2
2 ----(3)
Substituting d2 = 5 cm = 0.05 m and Discharge Q = 55 litres/s = 55 ×10-3 m3/s in (3), we get
v2= 55 ×10−3
π
4 ×0.052
=28.01 m
s
Step 5 of 5
Apply the Bernoulli’s equation between sections 1 and 2, we get
p1
ρair g + v1
2
2 g +z1= p2
ρair g + v2
2
2 g + z2 ----(4)
Given the difference in the head Z2− Z1 = 200 mm = 0.2 m. Substituting the respective values in (4), we
get
The pressure difference P1−P2= ρair g ( V 2
2−V 1
2
2 g +0.2 )=273.8 Pa
Also, manometric pressure P1−P2= ρw gh -----(5)
Where ρw = 1000 kg/m3, solving for h, we get
h= P1−P2
ρw g = 273.8
1000 ×9.81 =0.0279 m
Therefore, the differential height of the manometer h = 0.0279 m
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STUDENT
Q2
:
BASIC FLUID DYNAMICS (Fluid Power & Forces) Developed By
Flowing Fluids:
(a) The fluid power transferred by a pump to water flowing in a horizontal
pipe at a rate of 40L/s is 8 kW. Determine:
i. The increase in fluid head and state which fluid head is involved;
ii. The pressure at the pump outlet if the pressure at the pump inlet is -
10kPa;
iii. The efficiency of the pump if the input power is 12.6 kW;
iv. The yearly cost of running this pump if it is powered by an electric
motor and is running continuously for this period (the cost of electrical
power is $0.27 per kWH)
(4+4+2+2= 12 marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
Step 1:
Fluid power P= ˙m gH ----(1)
Where P is the fluid power = 8 kW, ˙m is the mass flow rate = 40 kg/s,
( 1000 cc=1 liter , density=1 gm
cc , 1000 cc × 1 gm
cc =1000 gm=1 kg∨1liter=1 kg)
g = acceleration due to gravity = 9.81 m/s2, and H is the total head.
Substituting the above values in (1), we get
Total increase in head H= 8 ×103
40× 9.81 =20.4 m
Since in hydraulics pressure head predominates kinetics head, pressure is involved in
increase of head.
Step 2:
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STUDENT
Since, H= ( P2−P1 )
ρg
Pressure at pump outlet P2=Hρg + P1
Or P2=20.4 × 1000× 9.81+ ( −10 ) =190.124 kPa
Step 3:
The efficiency of the pump η= Pout
P¿
×100= ρgHQ
P¿
×100= ( 1000 × 9.81× 20.4 × 40× 10−3 )
12.6 × 103 × 100
Or η=63.53 %
Step 4: Yearly cost
P¿ × number of hours per day ×number of days∈the year × cost per unit=12.6 × 24 ×365 ×0.27=$ 29801.52
(b) A jet of steam 30mm in diameter is discharged from a nozzle at a
speed of 90m/s. The steam has a specific volume of 2.5 m3/kg. The jet
strikes a curved turbine blade, which deflects the steam through an angle
of 30°
(as shown below).
Satisfactory
response
Yes
☐
No
☐
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STUDENT
Determine:
i. the resultant force exerted on the blade by the stream if the blade is
stationary
ii. the resultant force exerted on the blade if it is moving horizontally at a
speed of 45m/s
iii. the power developed by the moving blade in part (b)
(Note: for all parts of this question, ignore frictional losses)
(4+4+2 = 10 marks)
(Total Q2: 12+10 = 22
marks)
Answer:
Step 1: Resolving forces in X and Y directions.
Fx=ρQ ( V 2x−V 1x )
Where ρ is the density of water = 1000 kg/m3 and Q is the discharge = Area×Velocity
Q= π
4 × d2 ×V = π
4 ×0.032 ×90=0.0636 m3
s
Fx=1000 ×0.0636 ( −90 cos 300−90 )=−106,81.13 N
F y=ρQ ( V 2 y−V 1 y )
F y=1000 ×0.0636 ( −90sin 300−0 ) =−4957.13 N
Step 2: Resultant force F= √ Fx
2 + F y
2 = √ ( −106,81.13 ) 2 + ( −4957.13 ) 2=4958.2 N
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STUDENT
Case ii: The resultant force exerted on the blade if it is moving horizontally at a speed of
45m/s
Step 1: Relative velocity V =V 1−V 2=90−45=45 m
s
Step 2: Resolving forces in X and Y directions.
Fx=ρQ ( V 2x−V 1x )
Where ρ is the density of water = 1000 kg/m3 and Q is the discharge = Area×Velocity
Q= π
4 × d2 ×V = π
4 ×0.032 ×45=0.032 m3
s
Fx=1000 ×0.032 ( −45 cos 300−45 ) =−2687.08 N
F y=1000 ×0.032 ( −45 sin 300−0 ) =−720 N
Resultant force F= √ Fx
2 +F y
2 = √ (−2687.08 )2+ (−720 )2=2781.7 N
Power produced in Case ii:
Power P=Force × velocity vector
P=F ×Vcosθ
P=2781.7 × 45 cos 300=108.413 kW
Q3
:
BASIC THERMODYNAMICS CONCEPTS:
(a) A fluid has a temperature of 25°C. What is its absolute temperature?
(b) How much heat is required to raise the temperature of 10 kg of Steel
from a temperature of 15°C to 150°C if it has a Specific Heat Capacity of
0.466 kJ/kg
(c) How much heat is required to melt 20 kg of ice (the Latent Heat of
Fusion for water is 334 kJ/kg)
(d) What is the definition of Entropy (Research)?
(Total Q3: 2+3+3+2 = 10
marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
a) Absolute temperature=Temperature∈℃+ 273.15 K =25+273.15=298.15 K
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STUDENT
b) Heat required Q=mass× Heat capacity ( c ) × ∆ T =10× 0.466 × ( 150−15 ) =629.1 kJ
c) Heat required to melt ice Q=mass× leatent heat of fusion=20× 334=6680 kJ
d) Entropy: Entropy in Greek, entropia, refers to transformation. The term entropy was
first used by Rudolph Clausius. In thermodynamics, the second law defines entropy as
the measure of irreversibility or disorderliness. Entropy increases as the disorderliness
of the process increases. For example, flow of water in water in water falls, flow of
heat, etc., are irreversible processes with, which involve entropy change. Generally,
reversible processes have constant entropy as these processes occur in a predictable
order.
Q4 CLOSED & OPEN SYSTEMS: Satisfactory
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STUDENT
: (a) In a closed industrial process, a fluid of total mass 2.5 kg and initial
temperature 30°C is continually stirred whilst being heated for a total; time
of 45 minutes. If the rate of heat input to the system is 3 kW and the
mechanical stirrer has shaft power of 0.5 kW, the heat loss rate being
0.75 kW, determine the internal energy change of the system and the final
temperature. (Take Cv for the fluid to be 2.3 kJ/kgK.)
response
Yes
☐
No
☐
Answer:
Step 1: Net work W =Shaft Power ×time=0.5× 45 × 60=1350 kJ
Step 2: Qtotal=Q¿−Qloss = ( 3−0.75 ) kW =2.25 kW =2.25 × 45 ×60=6075 kJ
Step 3: From the first law of thermodynamics, Q=du+W
Where du is the change in internal energy.
Therefore, du=6075−1350=4725 kJ
Step 4: du=m cv dt
4725 × 103=2.5 ×2.3 ×103 ×(t2−30)
Or t2=851.74 ℃
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STUDENT
(b) The heat loss per second from the steam turbine shown in the figure
below is 120 kW when the steam flow rate is 1.6 kg/s. The specific
enthalpy of the steam at inlet is 4035 kj/kg and at outlet is 2950 kj/kg.
Determine the power output from the turbine if changes in potential and
kinetic energy are neglected.
(Total Q3: 8 + 7 = 15 marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
Step 1:
Using steady flow energy equation at sections 1 and 2,
h1 + v1
2
2000 + dQ
˙m =h2+ v2
2
2000 + dW
˙m
dW
˙m =h1 −h2 + v1
2
2000 − v2
2
2000 + dQ
˙m
Since changes in kinetic and potential energies are neglected,
dW
˙m = ( 4035−2950 ) + (−120
1.6 )=1,010 kJ
kg
dW =1010 ×1.6=1,616 kW
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STUDENT
Q5
:
GASES & GAS PROCESSES:
(a) Determine the mass of air contained in a pressurised container if the
temperature is 120° C, the pressure is 70kPa (gauge) and the volume is
2 litres.
(Take R = 287J/kgK for air) (4
marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
Using the ideal gas equation Pv=mRT
Absolute pressure= Atmospheric pressure+ gauge pressure=101.325+70=171.325 kPa
m= Pv
RT = ( 171.325× 103 ×2 ×10−3 )
( 287 ×(120+273.15) ) =3.036 × 10−3 kg
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STUDENT
(b) A quantity of gas occupies a volume of 3.5 m3 at a pressure (absolute)
of 120 kPa and a temperature of 20° C. The volume is then reduced to 0.5
m3 and the temperature increased to 100°C. Determine the final pressure
if the mass of gas is 0.345 kg. What gas is it most likely to be? (HINT:
Select R values from the following table:) (6 marks)
OXYGEN NITROGE
N
HYDROGE
N
“AIR”
R(J/KgK) 260 297 4157 287
Satisfactory
response
Yes
☐
No
☐
Answer:
Using the ideal gas equation, Pv=mRT
P1
P2
× v1
v2
=T 1
T 2
120
P2
× 3.5
0.5 = 20
100
P2=4200 kPa
Since, R= Pv
mT = 120 ×103 ×3.5
0.345 ×(20+273.15) =4157 J
kg−K
Hence the gas is Hydrogen.
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STUDENT
(c) In an isobaric (constant pressure) process the initial volume of air is
1.2 litres at a temperature of 20°C and a pressure of 50kPa. If the final
temperature is 200° C determine:
i. The final volume
ii. The mass of air
iii. The work done
iv. The heat transferred (8
marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
(d) A cylinder of air is equipped with a sliding piston so that the volume
can be varied. If the pressure is 480kPa at a volume of 4 litres and
temperature 100° C and the piston is moved so that the cylinder's volume
is now 6 litres, compare the work done if the process is: •
i. Isothermal
ii. Adiabatic (1 for air is 1.4)
iii. Polytropic with an index of 1.6 (8
marks)
(Total 4+6+8+7= 25
marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
I. For isothermal process PV =C
Work done W 12=P1 V 1∫
v1
v2
dv
v
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STUDENT
W12=P1 V 1 ln ( V 2
V 1 )
W 12=480 ×103 ×4 ×10−3 ln ( 6
4 )
W12=778 J
II. For adiabatic process, P V γ =C P1 V 1
1.4=P2 V 2
1.4
P2=P1 × ( V 1
V 2 )
1.4
=480× ( 4
6 )
1.4
=272.09 kPa
Work done W 12= ( p1 v1− p2 v2 )
γ −1 = ( 480 × 4 ) −(272.09 × 6)
(1.4−1) =718.65 kJ
III. For polytropic process, P V n=C
P1 V 1
1.6=P2 V 2
1.6
P2=P1 × ( V 1
V 2 )1.6
=480 × ( 4
6 )1.6
=250.89 kPa
Work done W 12= ( p1 v1− p2 v2 )
n−1 = ( 480 × 4 ) −(250.89 × 6)
(1.6−1) =691.1 kJ
W Adiabatic >W Polytropic >W Isothermal
Q6
:
HEAT ENGINES:
(a) A heat engine has a minimum combustion temperature of 1700°C. The
exhaust gas temperature is 480°C. What is the maximum possible
("Carnot") efficiency of the engine? When the exhaust gas supercharger is
fitted, the exhaust gas temperature drops to 230°C. What effect does this
have on the maximum possible thermal efficiency?
(6 marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
Given T 1=1700 °C
T 2=480 ° C
Efficiency of Carnot cycle ηT = ( 1− T2
T1 ) ×100=( 1− 480
1700 ) × 100=71.76 %
When a super charger fitted,
Efficiency of Carnot cycle ηT = (1− T2
T1 )×100=(1− 230
1700 )× 100=86.47 %
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STUDENT
With supercharger at the exhaust, the efficiency of the heat engine increases.
(b) An internal combustion engine uses fuel of energy content 44.4 MJ/kg,
at a rate of 8 kg/h. If the thermal efficiency is 45%, determine the power
output and the rate of heat rejection.
(7 marks)
(Total: 6 + 7 = 13
marks)
Satisfactory
response
Yes
☐
No
☐
Answer:
Thermal efficiency η= ˙W
˙Q
Power ˙W = ˙Q× η= ˙m× Energy content × η=8 × 44.4 ×106 × 0.45=159.84 MW
Heat added ˙Q1= ˙m× Energy content=8 × 44.4 × 106=355.2 MJ
For rate of heat rejection , ˙W = ˙Q1− ˙Q2
159.84=355.2− ˙Q2
˙Q2=195.3 MW
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