Solution to Thermodynamics Problems and Calculations

This document provides a comprehensive solution to a thermodynamics assignment, addressing several key problems. The solution begins with an analysis of internal energy, heat transfer, and work done within a closed system, along with a detailed explanation of constant pressure expansion. It then explores specific heat at constant pressure and volume, and proceeds to solve a problem involving saturated water vapor and power input calculations. Furthermore, the solution analyzes a steady flow system, deriving the energy balance equation and discussing the implications of steady and unsteady systems. Finally, the assignment addresses problems related to power required for a compressor, heat added during combustion, and work output calculations, including the use of heat values, heat capacity, and fuel consumption to determine the overall efficiency.

Question 1
Part a
∆ Esys=Qnet ,∈¿+W net ,out ¿
Where,
∆ Esys is change in internal energy of the closed system.
Qnet ,∈¿¿ is the net heat transferred into the system.
W net ,out is the net work done on or by the closed system.
Part b
Constant pressure expansion analysis case.
Part c
Specific heat at constant pressure c p refers to the energy needed to increase the temperature of
the unit mass of a material by one degree as the pressure is kept constant.
Specific heat at constant volume cv refers to the energy needed to increase the temperature of the
unit mass of a material by one degree as the volume is kept constant.
∆ U =cv ∆ T
∆ H =c p ∆ T
Part d
At state (1) the condition of fluid is in saturated water vapor. At 300 kPa,
u1=ug=2543.4 kJ /kg
v1 =v g=0.6063 m3 /kg
Amount of power input W =VIT
W =120V ∗0.2 A∗5∗60
W =8.0 kJ

Applying first law of thermodynamics
Q=w+du
Q=wboundary + wr +m ( u2 −u1 )
Q=mρ (v2 −v1 ) +wr +m ( u2−u1 )
Q=mρ r2−mρ r1+ m v2−m v1+ wr
−3. 5=0.0028 h2− ( 0.0028∗300∗0.6063 )− ( 0.0028 2543.4 ) −8.0
h2 =28 86.01kJ /kg
Now at P=300 kPa and h2 =2865.2kJ / kg
From the superheated water tables,
T 2 ≅ 2000 C
The final temperature of steam in the cylinder is 2000 C.
Question 2
Part a
We express the energy balance for steady flow system as;
˙Q¿− ˙W ¿+∑¿
˙mθ= ˙Qout− ˙W out +∑
out
˙mθ
Where θ=h+ V 2
2 +gz
We simplify it to
( ˙Q¿− ˙Qout )+∑¿
˙mθ= ( ˙W¿− ˙W out )+∑
out
˙mθ

˙Qnet + ˙m1 ( h1+ V 1
2
2 +g z1 )+ ˙m2 ( h2 + V 2
2
2 +g z2 )= ˙W net + ˙m3 ( h3 + V 3
2
2 + g z3 )
Rearranging,
˙Qnet − ˙W net = ˙m3 ( h3+ V 3
2
2 + g z3 )− ˙m1 ( h1 + V 1
2
2 + g z1 )− ˙m2 ( h2+ V 2
2
2 +g z2 )
Since the system is steady, the rates of flow of energy and mass are constant across the system
boundary. This means that ˙m1+ ˙m2= ˙m3
Therefore,
˙Qnet − ˙Wnet = [( ˙m1+ ˙m2) (h3 + V 3
2
2 + g z3 )− ˙m1 (h1 + V 1
2
2 + g z1 )− ˙m2 (h2+ V 2
2
2 + g z2 ) ]
Part b
For unsteady system, the rates of flow of energy and mass vary across the system boundary. This
means that ˙m1 ≠ ˙m2 ≠ ˙m3
Hence,
˙Qnet − ˙W net = ˙m3 (h3+ V 3
2
2 + g z3 )− ˙m1 (h1 + V 1
2
2 + g z1 )− ˙m2 (h2+ V 2
2
2 +g z2 )+∆ E
Question 3
Part a
Needed: Power required to drive the compressor.
T 1=50 C=278 K
m=0.3 kg/s
T 2=1500 C=423 K
losses F=25 kW

Power required P=m∆ T + F
P=0.3 ( 150−5 ) +25
¿ 68.5 kW
Part b
Needed: Amount of heat added during combustion.
Exit temperature T =12500 C
Heat value ¿ 45 MJ / kg
Heat capacity c p=1.0 kJ /kg
Amount of heat added Q¿=m cp ∆ T
Q¿=0.3∗1∗( 1250−5 )
¿ 373.5 kJ
Fuel consumption ¿ 373.5
45000 =0.00 83 kg
Part c
T ex=3000 C
F=20 kW
Work output W =0.3 ( 300−5 )+20
W =108.5 kJ /s
Net work produced ¿ 108.5−68.5=40 kJ / s
Efficiency η= 40
68.5 =0.584=58.40 %