University of Highlands & Islands Thermodynamics & Fluids Assignment

This document presents a comprehensive solution to a Thermodynamics and Fluids assignment, likely for a Mechanical Engineering program. The assignment covers a range of topics including thermodynamic systems, open systems, and their interactions with surroundings. It includes diagrams and explanations of temperature-enthalpy diagrams showing phase changes of water, calculations involving steam properties, and sketching of processes on pressure-volume diagrams. The solution also delves into calculations of specific gas constants, mass and volume of air in a cylinder, and the mechanism of heat transfer by convection. It further explores static and dynamic pressure, hydrostatic pressure, and fluid viscosity at the molecular level. Additional problems address non-Newtonian fluids, gas density, and energy balance in a steam turbine condenser, including the use of steam tables and non-flow energy equations. The assignment concludes with explanations of perfect gases and adiabatic compression processes.

Thermodynamics and Fluid mechanics
Term Paper
Date
Number of Words
Name
Institution Affiliate

Question 1
With the aid of diagram, describe what a thermodynamic system is and how an open
thermodynamic system Interacts with surroundings.
Solution
SURROUNDINGS
A thermodynamic system is a certain area or universe space where one or more thermodynamic
processes occur.
Anything outside a thermodynamic system is known as the surrounding. A definite barrier called
a boundary is divide the surrounding and system.
.
SYSTEM BOUNDARY
OPEN
SYSTEM BOUNDARY
MATTER
HEAT
SURROUNDINGS

An open system is a thermodynamic system that permits the flow of both energy and mass into
and out of the system.

Question 2
a) Sketch and label a temperature-enthalpy diagram showing the phase change-transition of
H2O at 2 bar from liquid at 0oC to super-heated vapor at 250 degrees.
Solution
1
Saturated vapor line 3
5
4
Constant Pressure Line
2
V (
m3 K g−1 ¿
T
(
℃ ¿
0
100
250
Liquid vapor mix
Compressed
Liquid
Vapor region
Temperature enthalpy diagram

Key
1 Compressed liquid at 0 ℃
2 Saturated Liquid
3 Saturated vapor
4 Super-heated vapor
5 Critical point
Note: The temperature is held constant at 2 bars.
b) With the aid of the steam property table, calculate:
i. Specific internal energy of steam at 20 bar which has a dryness fraction of 0.75.
Solution
¿ the saturated steam table at P=20 ¯, ∧x=0.75
We find
Uf =907 KJ
kg , Ug=2600 KJ
Kg
Using the formula Ux=Uf +x ( Ug−hf )
Subsitutingthe values ofUf ∧hf ∈the above formula
Ux=907+0.75 ¿
¿ 907+1269.75
¿ 2176.75
Therefore , specific enrgy of the staem=2176.75 KJ
kg

ii. Specific enthalpy of steam at 25 bar, 400° C.
Solution
Given that 25¯¿ 25 atm=2525 Pa , T =400
Now , here we will use both theTemperature value∧prsusure value
because we the steam is acting beyond vapour−equilibrium
¿ the steam table
hf =535 KJ
Kg ,hg=2717 KJ
kg
using the formulah=hf + x ( hg−hf )
Substituting the values of hf ∧hg∈the formulaabove
hx=35+ 0.75 ( 2717−535 ) kJ / kg
¿ ( 535+1636.5 ) KJ /kg
¿ 2,171.5 KJ
kg
Therefore , the specific enthalpy hx=2,171.5 KJ
kg
Question 3

Sketch the following processes on pressure- volume (P-v) diagram indicating the application law
and process direction:
Isothermal expansion followed by constant pressure heating.
Solution
Application of law
In the above isothermal process, the temperature T of the gas remains constant. Meaning
∆ U =0 , Q=W
P
(
P
a
)
T
1
V (m3 ¿
S
u
p
e
r
-
h
e
a
t
e
d
r
e
g
i
o
n
T
3
Compressed
Liquid
C
r
i
t
i
c
a
l
p
o
i
n
t
Liquid +vapor
T
e
T
2
P
C
V
C
P
r

Also, according to general gas law, PV=nRT.
But n=constant and T=constant, then PV=constant

Question 4
The cylinder of an internal combustion engine with a capacity of 1 liters contains air at a
pressure and temperature of 14 bar and 80°C respectively. Taking the molecular mass, (Mr) of
air as 29 and the universal gas constant, R0 as 8.3144 kJ kmol-1 K-1, calculate:
I. The specific gas constant, R (2 Marks)
Solution
R = Ro/mr
The specific gas constant, R = the universal gas constant/ the molecular mass
The specific gas constant, R = (8.3144 kJ kmol-1 K-1)/29
The specific gas constant, R = 0.2867 kJ kmol-1 K-1
ii. The mass of air in the cylinder
Solution
PV = mRT
Where P is pressure = 14 bar
1 bar air pressure = 101 kPa
P = 14 bar/1 bar x 101 kPa = 1414 kpa
V volume = 1 liters
1 liter = 0.001 m3
V volume = (1 liter)/ (1 liters) x 0.001 m3 = 0.001 m3
R is The specific gas constant, R = 0.2867 kJ kmol-1 K-1

T is temperature = 80oC + 273 K = 353 k
m = PV/RT = (1414 kpa)( 0.001 m3)/(0.2867 kJ kmol-1 K-1)(353 k)
m = 0.01397 kg
iii. The volume of the air if it is compressed to a pressure of 23 bar and temperature
120°C.
Solution
P1V1/T1 =P2V2/T2
Pressure 1 P1 = 14 bar
Volume 1 V1 = 0.001 m3
Temperature 1 T1 = 353 k
Pressure 2 P2 = 23 bar
Volume 2 V2 = ?
Temperature T2 = 120oC = 120oC + 273 K = 393 K
(14 bar) (0.001 m3)/ (353 K) = (23 bar) (V2)/ (393 K)
V2 = [(14 bar) (0.001 m3)/ (353 K)] x (393 K)/(23bar)
V2 = 0.0006777 m3

Question 5
Briefly describe the mechanism of heat transfer by convection and list one factor that influence
the rate of heat transfer by this method.1
Solution
Convection is the transfer of heat from one place to another by movement of fluids (bulk
movement of molecules within fluids) such as air or water (Gases and liquids).
Viscosity is one factor that affects convection. Currents move quickly within liquids of low
viscosity.
Question 6
1 Michael, M, & T Tsatsaronis, Engineering thermodynamics. in, 2nd ed., CRC Press, 2017, pp.
1-122.

a) Briefly describe the difference between static and dynamic pressure (in terms of their
origin)
Solution
Static pressure originates from fluid that is no moving (stationary fluid). If we have fluid
which is not moving, the pressure in the fluid is called static therefore static pressure is
quantity of pressure in a non-moving liquid. For static pressure in all directions, pressure
is the same. Dynamic pressure originated from the fluid that is in motion, dynamic
pressure depends on movement direction.
b) Describe with the aid of a diagram, how pressure sensors can be arranged to measure
static and dynamic pressure in a flow.
Solution
Pressure sensors can be arranged as shown above to measure dynamic and static pressure in a
flow. Static pressure will be measured by Piezometer, and is given by the height h1 shown in the
diagram. Dynamic pressure is measured using the pitot tube, and Dynamic pressure is equal to
(h2-h1).
Piezometer Pitot tube
h2
h1
c) Calculate the hydrostatic pressure at the bottom a column of water 12.5m deep.

Solution
Hydrostatic pressure in a liquid can be calculated as
Hydrostatic pressure = ρgh
Where ρ is density of water = 1000kg/m3
g is gravity = 9.8 m/s2
h is height =12.5 m
Hydrostatic pressure = (1000kg/m3) (9.8 m/s2) (12.5 m)
Hydrostatic pressure = 122500 pascal
d) Explain briefly the origin of fluid viscosity at the molecular level and how it manifests
itself in the bulk fluid.
Solution
Viscosity is defined as the fluid’s resistance to flow. The origin of viscosity happens because of
interaction of different molecules in the bulk fluid. Molecules of fluid resist flow of other
molecules which are adjacent to them and cause friction in their movement. This friction in
movement of bulk solid results in viscosity
e) Explain the term non Newtonian fluid and give an example of one.
Solution
A non-Newtonian fluid do not comply with the viscosity law of Newton. viscosity law of
Newton states that shear stress is proportional to the velocity gradients between the two layers of

neighboring fluid. For a given pressure and temperature, the proportion of shear stresses to shear
rate is constant. Example starch suspensions
f) Given that the pressure of a perfect gas is 105kpa, calculate its density at 35oC.
Solution
For a perfect gas density ρ = P/RT
Where P is pressure = 105kpa
R is gas constant = 8.3144 kJ kg-1 K-1
T is temperature = 35oC + 273 K = 308 K
For a perfect gas density ρ = (105kpa)/ (8.3144 kJ kg-1 K-1) (308 K)
For a perfect gas density ρ = 0.041 kg/m3

Question 7
a) Steam from a turbine enters a condenser at a pressure of 0.035 bar, a velocity of 60
m s-1 and dryness fraction 0.99 as shown if Figure Q5a. The saturated liquid condensate
exits the system with a velocity of 10 m s-1 at the same pressure of 0.035 bar. The mass
flowrate through the condenser is 0.5 kg s-1.
I) If the exit point is 3.25m below entry, using an energy balance approach, determine
the rate of heat transfer to the cooling water.
Solution
Dryness fraction denoted by x= 0.99
Mass flow rate around condenser=0.5kgs-1
Difference in height = -3.25
At entry, the state of working fluid is steam with;
Velocity u1=60ms-1
Pressure P=0.035bar
H1=?
At exit, the state of working fluid is saturated condensate liquid
Velocity u2=10ms-1
Pressure =0.035 bar
H2= hf =112kJ/Kg

From steam tables;
At a P=0.035bar; hf=112kJ/Kg, hg=2550kJ/Kg, hfg=2438kJ/Kg , T=26.70c
H1= (1-x) hf +x hg
= (1-0.99) 112kJ/Kg +(0.99*2550) kJ/K
= 2525.62 kJ/Kg
Applying energy balance approach;
˙q+ws=∆ H +∆ EK +∆ EP
Where;
˙q = Rate of heat transfer to the cooling tower
ws =Shaft work
∆ H = Change in enthalpy across the system
∆ EK =Change in kinetic energy across the system
∆ EP =Change in potential energy across the system
(Lim et al. 2019).
There is no shaft work, therefore, ws=0
General equation reduces to;
˙q=∆ H +∆ EK +∆ EP
˙q= ˙m(H2-H1) + 1
2 ˙m(u2
2−u1
2) + ˙mg(z2-z1)

˙q=0.5* (112-2525.62) + 1
2∗0.5∗¿(10❑
2 −60❑
2 ) +0.5*9.81(-3.25)
= -1207.7 kJ
s
Negative sign implies that heat loss to the cooling
II) Sketch the temperature enthalpy diagram for the system showing the state point of the
working fluid at entry and exit
Solution
X Y
112 26.7
2525.6 26.7

112 2525.6
0
5
10
15
20
25
30
Temperature- Enthalpy Graph
y
Enthalpy kJ/Kg
Temperature (0c
b) A quantity of wet steam at 2 bar, internal energy 1315 kJ kg-1 is contained in a pistoncylinder
device as shown in Figure Q1 b). The cylinder is heated changing it in to drysaturated
steam at a pressure of 10 bar.
i. With the aid of steam property tables, calculate the dryness fraction of the steam
before
the heating process and the specific internal energy of the dry-saturated steam.
Solution
Taking a basis of one kilogram of wet steam
From steam property tables;
At 2bar; uf= 505kJ/Kg, ug= 2530kJ/Kg
At 10 bars; uf= 762kJ/kg, ug=2584kJ/Kg

U is given as 1315kJ/Kg
Let dryness fraction be represented by x
U=(1-x) uf +x ug
1315= (1-x) *505 +2530x
1315=505-505x+2530x
1315-505=2530x-505x
810=2025x
X=0.4
Let specific internal energy be represented by ^u
^u =(1-x) uf + x ug
=2762*(1-0.4) +0.4*2584
= 1490.8 kJ/Kg
II) Using non-flow energy equation and energy balance approach, determine the
specific heat input to the steam if the work transfer from the steam at the
piston is 400kJ/Kg
Solution
Q+ W = ∆ u
2 Lim, Wen Cong, Shazed Mohammad Tashrif, Yang Miang Goh, and Soo Jin Adrian Koh.
2019. “Validation of the Energy Balance Approach for Design of Vertical Lifeline
Systems.” International Journal of Occupational Safety and Ergonomics.

∆ u= (1490.8-1315) kJ/Kg
= 175.8
Therefore, Q=175.8+400
=575.8kJ/Kg.
Question 8

a) Explain what is meant by the term perfect gas?
Solution
Perfect gas also known as an ideal gas which obeys all available gas laws. In fact, no gas in this
world is ideal. These gases behave ideally under certain conditions. Perfect gas is gas that is
physically compatible with a common perfect relationship between volume, temperature and
pressure, which is known as the general gas law. The laws include both Charles’s law and
Boyle’s law.
b) A quantity of CO2 at an initial pressure, volume and temperature of 1 bar, 0.05m3 & 20°C
respectively is compressed adiabatically until the pressure is 5 bar. The CO2 is then cooled at
constant pressure until the volume drops to 0.011m3.
i. Sketch both processes on a single PV diagram
Solution
An adiabatic process is the one with no heat flow into the system (Q=0).
Recall that
U + W = Q
Where U = internal energy, Q = heat energy, W = work done.
For an adiabatic process Q = 0, hence U = - W.
This simply implies a negative work which denotes that work is done on the system (C02) by the
environment.
This negative work is mathematically by a decrease in volume (V2 to V1) and represented on the
PV diagram (find attachment) as the gas moves from point A to B.
For the second phase, the gas was cooled at a constant pressure.

When a gas is being cooled down, it volumes decreases hence bringing about a negative work on
the system by environment.
This is denoted on the PV diagram as the graph moving from point B to C.
PV diagram for a gas that is adiabatically compressed and at a constant pressure
A
P2……
P1…………………………………B……………………… C
V2 V1 V3
Compressed adiabatic Process > A B
Constant pressure > B - C
ii. Calculate the mass of the CO2 present
Solution
According to the conservation law of mass, mass cannot be destroyed nor created.
Which implies that the number of moles of CO2 at the start of the process is the same as the end
of the process.
Recall that

n = mass / molar mass and n = volume of gas/22.4 dm³
Where n = number of moles.
Since the number of moles is constant all through, we can say that
Mass/molar mass = volume of gas/ 22.4
The final volume of gas is 0.011 m³ = 11 dm³
The molar mass of CO2 = 44.01 g/mol.
By substituting the parameters, we have that
m/ 44.01 = 11/22.4
m/ 44.01 = 0.491
m = 0.491× 44.01
m = 21.61g.
iii. Calculate the volume after the adiabatic process
Solution
For an adiabatic process, we have that
P1×(V1)^γ = P2 × (V2)^γ
Where γ = ratio of heat capacities for C02 =1.29
P1 = 1 bar, V1 = 0.05 m³
P2 = 5 bar, V2 =?
By substituting the parameters, we have
1 × (0.05) ^1.29 = 5 × (V2) ^1.29
0.02097 = 5 × (V2) ^1.29
(V2) ^1.29 = 0.02097/5

(V2) ^1.29 = 0.00419.
V2 = (0.00419) ^1/1.29
V2 = 0.014 m³.
iv. Calculate the work transferred during the adiabatic process
Solution
In adiabatic process, Quantity of the work done will be given by the formulae below
W = {1/γ - 1}× (P1V1 - P2V2)
Let us break down the expression.
{1/γ - 1} = 1/ 1.29 - 1 = 1/0.29 = 3.448
(P1V1 - P2V2) = (1×0.05 - 5×0.014) = (0.05 - 0.07) = - 0.02 J
You can see that we had a negative value for work done.
v. Calculate the gas temperature after constant pressure cooling
Solution
We are assuming the CO2 gas to be an ideal gas, hence
PV = nRT is a good model of the gas equation.
The pressure value P we will be using here is the value of P2 from the the adiabatic process
which is 5 bar.
P2 = P = 5 bar.
We have to convert the pressure to pascal hence we have P2 as 5×10^5 pa
n is the number of mole of gas = 11/22.4 = 0.491 mole.
R = constant = 8.314 J/mol.K

V = 0.011 m³
5×10^5 × 0.011 = 0.491× 8.314 ×T
T = 5×10^5 × 0.011/ 0.491×8.314
T = 5,500 / 4.0822
T = 1347.3 K
vi. Calculate the heat transfer during the constant pressure process
Solution
In a constant pressure process, heat energy = work done
Q = W.
For the given constant pressure process, the volume changed from V2 of the adiabatic process
(0.014m³) to the final volume of the constant pressure process (0.011 m³)
Work done = pressure × change in volume
Where pressure (P) = 5×10^5 pa and change in volume = 0.011 - 0.014 = - 0.003 m³
W = 5×10^5 × - 0.003 = - 1500 J
We are having a negative value of work because the gas is compressed, that's there is a decrease
in volume which implies that the environment is doing work on the system (CO2).

Question 9
a) List and describe the heat transfer mechanisms throw each layer of a double glazed
window unit as heat is transferred from a room to the outside ambient air.
Solution
Heat transfer in a room involve three major mechanisms that is; convection, radiation and
conduction. In a room heat transfer occurs through the roof, walls and windows 3. Heat loss
through the windows is controlled using the use of the double glazed windows. Double glazing
is can be through installing an already sealed window or adding an extra layer of glass that is put
in front of a window pane with an air gap or a vacuum between the two glass 5.
Heat from the room is transferred to the window through convection, through a free
connective cell formed by air in the room. The heat then moves from the first glass pane to the
next through conduction. If the gap between the double glazed glass has air, then then there will
be reduced conduction of the heat due to minimized room for air movement, but if the gap is
3 Awbi, H. B. (1998). Calculation of convective heat transfer coefficients of room surfaces
for natural convection. Energy and buildings, 28(2), 219-227.
5 De Giorgi, L., Bertola, V., & Cafaro, E. (2011). Thermal convection in double glazed windows
with structured gap. Energy and Buildings, 43(8), 2034-2038.

made of a vacuum then there will be no conduction 4 (Collins etal.1995). Heat is then lost
through the glass to the ambient air or the outside air through convection.
b) A 150mm thick block wall of height 2.4m and width 4m has an inside surface temperature, t1
of 47°C and an outside ambient air temperature, tf of 15°C as shown in Figure Q3. The thermal
conductivity of the wall is 0.78 W/ m K, and the heat transfer coefficients for the inside and
outside surfaces are 37 and 8 W/ m2 k, respectively. Calculate:
i. The rate of heat transfers through the wall surface area
Solution
˙Q=kA ∆ T
d =hA ∆ T
To get The rate of heat transfers through the wall surface area. We first calculate ts
( 0.78 ) ( 9.6 ) (T s −47
0.150 )=8 ( 9.6 ) ( 15−T s )
49.92 ( T s−47 ) =76.8 ( 15−T s )
49.92 T s−2345.24=1152−76.8 T s
127.72T s =3498.24
T s=27.6 ° C
Q=8 ( 9.6 ) ( 15−27.6 ) =−967.7 W
4 Collins, R. E., Turner, G. M., Fischer-Cripps, A. C., Tang, J. Z., Simko, T. M., Dey, C.
J., ... & Garrison, J. D. (1995). Vacuum glazing—A new component for insulating
windows. Building and Environment, 30(4), 459-492.

ii. The temperature of the outside surface of the wall, ts.
Solution
( 0.78 ) ( 9.6 ) (T s −47
0.150 )=8 ( 9.6 ) ( 15−T s )
49.92 ( T s−47 ) =76.8 ( 15−T s )
49.92 T s−2345.24=1152−76.8 T s
127.72T s =3498.24
T s=27.6 ° C
iii. The temperature of the ambient air on the inside of the wall, tA
Solution
−967.7= ( 37 ) ( 9.6 ) ( 47−T A )
−967.7=355.2 ( 47−T A )
−2.7=47−T A
−49.2=−T A
T A=49.2℃