Posted: November 15th, 2022

Thermodynamic systems

Thermodynamic systems.

In this assignment you will undertake a range of activities designed to help your understanding of the effects of energy transfer in thermodynamic systems.
Introduction
You are presented with three tasks which require you to solve thermodynamic system problems that might occur in mechanical engineering design situations.
The first task concerns calculation of the expansion of a tie bar when undergoing a rise in temperature and the compressive force that would be exerted on the bar if it is rigidly held so as to prevent it from expanding.
The second task concerns a heat exchanger that takes in feed water at 30°C and produces dry steam at 100°C. You will need to calculate the heat energy received per kilogram of steam produced and the thermal efficiency of the heat exchanger. The evidence that you produce and submit for criterion P8 will be generated by completing The first task involves calculating the expansion of a tie bar as its temperature rises and the compressive force that would be exerted on the bar if it were rigidly held to prevent it from expanding.
The second task involves a heat exchanger that takes in 30°C feed water and produces 100°C dry steam. You must compute the heat energy received per kilogram of steam produced as well as the heat exchanger’s thermal efficiency. The evidence for criterion P8 that you create and submit will be generated by completing Tasks 1 and 2.
The final task is in two parts. In the first you are asked to calculate the temperature of the air–fuel mixture after it has been compressed in an engine cylinder. In the second part you are asked to calculate the mass of air that is released from a compressed air storage cylinder from its initial and final temperature and pressure readings.

Assignment objectives:

To calculate the dimensional change that accompanies temperature change in a solid material.
To calculate the heat transfer that accompanies temperature change in a substance.
Determine the thermal efficiency of a heat transfer process,
and apply thermodynamic process equations.

Task 1
A brass tie bar of length 500 mm and diameter 20 mm undergoes a change in temperature from 25°C to 135°C. The coefficient of linear expansion for copper is 19 × 10−6 0C−1 and its modulus of elasticity is 84 GPa.
Determine the change of length that takes place.
Solution
To determine the change of length that took place, we use the equation:
dl=lo x a(t_1-t_0)
Where dl is the change in length,
lo is the initial length,
a is the linear expansion coefficient,
t_0 is the initial temperatures in (0_C )and
t_1 is the final temperatures in (0_C )

lo (meters)=500/1000=0.5 m and a=19×〖10〗^(-6 )
dl=0.5 x 19×〖10〗^(-6 ) (135℃-125℃)=1.045×〖10〗^(-3) m=1.045mm
Determine the compressive force induced in the bar if it is rigidly held and expansion is prevented.
Solution
e=s/t where e is the young’s modulus
s=tensile stress
t=tensile strain
We first determine the original area which can be expressed as:
A=πr^2=3.142 x (10^2)/1000000=3.14159×〖10〗^(-4) m^2
To determine the strain, we use the relationship
Strain=(Length of the stretch)/(Original length)=d/h
Strain=(1.045 x 〖10〗^(-3))/0.5=0.00209
To determine the stress, we use the formula
Stress=Force/area
But e=s/t
Therefore,
s=e x area=84000000000pa x 2.09 ×〖10〗^(-3)=175260000000N/m^2
Force=area x stress=3.14159×〖10〗^(-4) m^2 x 175260000000N/m^2 =550,590,000N
Force=5.5 x 〖10〗^8 N
task 2
A heat exchanger produces dry steam at 100°C from feed water at 30°C at a rate of 1.5 kgs−1. The heat exchanger receives heat energy at a rate of 600 kW from the fuel used. The specific heat capacity of water is 4187 Jkg−1K−1 and its specific latent heat of vaporisation is 2257 Jkg−1.
Determine the heat energy received per kilogram of steam produced.
Solution
Q=CM∆T
Where Q= heat added
c= specific heat
m= mass
∆T=change in temperature
The heat energy received is
Q=4187j〖kg〗^(-1)×1.5kgs^(-1) ×70℃=439635 J/S
ii) Determine the output power of the heat exchanger and its thermal efficiency.
Solution
The efficiency of the engine can be given by=w/Q_h
The thermal energy expelled can therefore be expressed as:
e=W/Q_h =Q_h-Q_c/Q_h
But e=1-Q_C/Q_h
To determine the power required,
P=W/∆t
But the efficiency of the heat exchanger can be expressed by
Efficiency=439635/600000 x 100=0.73%
Power (p)=w/∆t
Power=(25,716,933.23)/((600000-439635) )=160.365 Kwh
Task 3
A single cylinder petrol engine has a volume compression ratio of 8:1 takes in a mixture of fuel and air at a temperature of 25°C and its pressure is 101 kPa. If the pressure at the end of the compression stroke is 1.5 MPa what will be its final temperature?
Solution
We start by determining the cylinder pressure in pre ignition which can be expressed by
p=p_(o ×〖CR〗^r )
Where P=Cylinder pre ignition pressure
p_o=Cylinder pressure at the bottom
CR=Compression ratio and
r=Specific heat ratio
Therefore
1.51.5×〖10〗^6 N/m^2=10.1×〖10〗^4 N/M^2×8^r
8^r=14.85148515

〖 r log〗⁡〖8=log⁡14.851485 〗
〖r=log〗⁡14.851485/log⁡〖8 〗 =1.2975117≅1.3
1.3=C_(p/C_v )=
Where,C_p is the constant pressure and C_v is the constant volume
1.3=101,000/x
x=77692.31 m^3
Given that the compression ratio is 8:1
The initial volume=77692.31×8=621538.48 m^3

Using the thermodynamic formulae below
□(du/dt)=□(dQ/dT)=C_r
dT^2=534934
dT=731.39
25℃=731.39-x
x-25=731.39
x=756.39℃
Hence, c_(p-c_v )=T((dp )/(dT )(dv/dt)
C_p=1500000
C_v=77692.31
dp×dv=1399000×543846.17
√(dT^2 )=√(534934.035×25)=3656.95
∆T+25℃=3656.95+25℃=3681.95℃
Final temperatures x=3681.95℃
A compressed air storage cylinder has a volume of 0.5 m3 and contains air at an absolute pressure of 1.8 MPa and temperature 20°C. A quantity of the air is released during which the temperature of the remaining air falls to 15°C and the pressure to 1.5 MPa. Calculate the mass of the air released. The characteristic gas constant for air is 287 Jkg−1K−1.
Solution

pv=mRT
Where p is the absolute pressure,V is the volume,m is the mass
,R is the gas constant and T is the absolute temperature
1800000 N/M^2 x 0.5m^3=m x 287j/(〖kg〗^(-1) k^(-1) )×20℃
m=156.79 kg
Density=Mass/Volume=156.79/0.5=31,359 kg/m^3
But
p1v1/t1=p2v2/t2
(1800000 N/M^2 x 0.5m^3)/(20℃)=(1500000 N/M^2 x v2)/(15℃)
v2=0.45m^3
The air released=0.05m^3
mass=Density x volume=31,359 kg/m^3 ×0.05m^3=1567.95kg

References
Cengel, Y. A., & Boles, M. A. (2002). Thermodynamics: an engineering approach (Vol. 5). M. Kanoğlu (Ed.). New York: McGraw-Hill.
Hill, P. G., & Peterson, C. R. (1992). Mechanics and thermodynamics of propulsion. Reading, MA, Addison-Wesley Publishing Co., 1992, 764 p., 1.

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Tags: Thermodynamic systems