Answer:
Heat transferred = 175.6kJ
Explanation:
Given
m = 1.2kg
T1 = 200°C = 473K
P2 = 800kPa
From the steam table, we need u1 and u2.
u1 = 2594.2Kj/Kg, s1 = 6.4302Kj/KgK
When P2 = 800kPa and T2 = T1
u2 = 2631.1Kj/Kg, s2 = 6.8177Kj/KgK
The heat transfer is calculated as follows;
Q = T∆S
Q = T(s2 - s1)
Q = 473k (6.8177Kj/KgK - 6.4302Kj/KgK)
Q = 219.19 kJ
Calculating the Energy Balance
∆E = E, out - E,in
∆E = m(u2 - u1)
So, we have
m(u2 - u1) = Q - E
1.2(2631.1Kj/Kg -2594.2Kj/Kg) = Q - E
44.28 = Q - E ---- Make E the subject of formula
E = Q - 44.28
E = 219.19 - 44.28
E = 174.91kJ
Hence, the work done during the process is 174.91kJ
Answer:
Heat transferred = 219.9 kJ
work done = 175.6 KJ
Explanation:
We will have to use data obtained from engineering steam tables to solve this problem
At state 1, [tex]T_{1}[/tex] = 200 c
[tex]u_{1} = 2594.3KJ/kg[/tex]
[tex]s_{1} = 6.8177 KJ/kg.k[/tex]
At state 2,
{[tex]P_{2}[/tex] = 800 Kpa, [tex]T_{1}=T_{2}[/tex] (isotherm process)}
[tex]u_{2}=2631.1KJ/kg[/tex], [tex]s_{2}=6.8177KJ/kg.K[/tex]
The heat transfer from this reversible isotherm process can be determined using the formula:
[tex]Q=T\Delta S= mass\times T (s_{2}-s_{1})[/tex]
[tex]Q=(1.2Kg\times473K)(6.8177-6.4302)KJ/kg.K = 219.9KJ[/tex]
∴ The heat transferred = 219.9 KJ/kg
The piston cylinder problem in question can be considered as a closed system. This means that no mass can enter or leave the system ( most piston cylinders are considered as closed systems)
The energy balance for closed systems are given as :
Change in internal potential kinetic energies = Net energy transfer by heat, work and mass.
[tex]\Delta U = Q_{in}- W_{out}[/tex]
[tex]=m(u_{2}-u_{1})[/tex]
[tex]W_{out}= Q_{in}-m(u_{2}-u_{1})[/tex]
By substituting in values, for the other parameters the work done during this process can be obtained to be
[tex]W_{out} = 219.9KJ - (1.2kg)\times(2631.1 - 2594.2)KJ/Kg= 175.6KJ[/tex]
The work done during the process = 175.6 KJ
