6.4 Application Of General Equation For A Closed System Analysis
Example 6.4: Application of General Equation for a Closed System
Question
A gas is contained in a cylinder fitted with a movable piston as presented in Figure E6.4.1.
Figure E6.4.1: A cylinder fitted to a moving piston
If the cylinder is placed in boiling water with the piston held in a fixed position, 5 kcal of heat is absorbed by the gas, which equilibrates at 150oC and a higher pressure. The piston is then released, and the gas does 250 J of work in moving to its new equilibrium position. The final gas temperature is 150oC. Write the energy balance equation for each of the two stages of this process, and in each case solve for the unknown energy term in the equation. Express all energies in joules.
Solution
The problem is a closed …show more content…
Therefore, equation (6.6) becomes; but W= -250 J
Therefore, Q= 250 J
Thus, the gas absorbs an additional 250 J of heat as it expands and re-equilibrates at 150oC.
6.4.2 Energy balances on open systems
As defined in 6.2.2, an open system is that in which mass crosses its boundaries as the process occur. Work must be done on such system to force mass in, and work is done on the surroundings by mass that emerges, therefore, work terms must be included in the energy balance equation.
6.4.2a Flow work and shaft work
In an open system, the net work W done on the system by its surroundings is the summation of work done by the shaft and that done by flow. This can be written mathematically as:
W=W_s+W_f (6.12)
where; Ws=shaft work, or work done on the process fluid by a moving part within the system (e.g., a pump rotor); Wf=flow work, or work done on the fluid at the system inlet minus work done by the fluid at the system outlet.
6.4.2b Flow work
Mathematical expression for the flow work had been derived by several literature and may be expressed as …show more content…
Therefore, the various terms for the open system energy balance are as follow:
Energy transfer in by mass transfer: (U_i ) ̂+(E_Ki ) ̂+(E_Pi ) ̂ (6.14a)
Energy transfer out by mass transfer: (U_o ) ̂+(E_Ko ) ̂+(E_Po ) ̂ (6.14b)
Net transfer of energy by heat flow: Q (6.14c)
Net transfer of energy by mechanical or electrical work in (shaft work):W (6.14d)
Net transfer of energy work introduced to or removed by mass transfer(flow work):
P_i V ̂_i m_i-P_o V ̂_o m_o (6.14e)
Introducing Equation (6.14a-e) into the energy balance equation, we have:
0=((U_i ) ̂+(E_Ki ) ̂+(E_Pi ) ̂ ) m_i-((U_o ) ̂+(E_Ko ) ̂+(E_Po ) ̂ ) m_o+〖Q+W〗_s+P_i V ̂_i m_i-P_o V ̂_o m_o (6.15a)
0=((U_i ) ̂m_i+(E_Ki ) ̂m_i+(E_Pi ) ̂m_i+P_i V ̂_i m_i )-((U_o ) ̂m_o+(E_Ko ) ̂m_o+(E_Po ) ̂m_o+P_o V ̂_o m_o )+〖Q+W〗_s (6.15b) But; H=U+PV
Question
A gas is contained in a cylinder fitted with a movable piston as presented in Figure E6.4.1.
Figure E6.4.1: A cylinder fitted to a moving piston
If the cylinder is placed in boiling water with the piston held in a fixed position, 5 kcal of heat is absorbed by the gas, which equilibrates at 150oC and a higher pressure. The piston is then released, and the gas does 250 J of work in moving to its new equilibrium position. The final gas temperature is 150oC. Write the energy balance equation for each of the two stages of this process, and in each case solve for the unknown energy term in the equation. Express all energies in joules.
Solution
The problem is a closed …show more content…
Therefore, equation (6.6) becomes; but W= -250 J
Therefore, Q= 250 J
Thus, the gas absorbs an additional 250 J of heat as it expands and re-equilibrates at 150oC.
6.4.2 Energy balances on open systems
As defined in 6.2.2, an open system is that in which mass crosses its boundaries as the process occur. Work must be done on such system to force mass in, and work is done on the surroundings by mass that emerges, therefore, work terms must be included in the energy balance equation.
6.4.2a Flow work and shaft work
In an open system, the net work W done on the system by its surroundings is the summation of work done by the shaft and that done by flow. This can be written mathematically as:
W=W_s+W_f (6.12)
where; Ws=shaft work, or work done on the process fluid by a moving part within the system (e.g., a pump rotor); Wf=flow work, or work done on the fluid at the system inlet minus work done by the fluid at the system outlet.
6.4.2b Flow work
Mathematical expression for the flow work had been derived by several literature and may be expressed as …show more content…
Therefore, the various terms for the open system energy balance are as follow:
Energy transfer in by mass transfer: (U_i ) ̂+(E_Ki ) ̂+(E_Pi ) ̂ (6.14a)
Energy transfer out by mass transfer: (U_o ) ̂+(E_Ko ) ̂+(E_Po ) ̂ (6.14b)
Net transfer of energy by heat flow: Q (6.14c)
Net transfer of energy by mechanical or electrical work in (shaft work):W (6.14d)
Net transfer of energy work introduced to or removed by mass transfer(flow work):
P_i V ̂_i m_i-P_o V ̂_o m_o (6.14e)
Introducing Equation (6.14a-e) into the energy balance equation, we have:
0=((U_i ) ̂+(E_Ki ) ̂+(E_Pi ) ̂ ) m_i-((U_o ) ̂+(E_Ko ) ̂+(E_Po ) ̂ ) m_o+〖Q+W〗_s+P_i V ̂_i m_i-P_o V ̂_o m_o (6.15a)
0=((U_i ) ̂m_i+(E_Ki ) ̂m_i+(E_Pi ) ̂m_i+P_i V ̂_i m_i )-((U_o ) ̂m_o+(E_Ko ) ̂m_o+(E_Po ) ̂m_o+P_o V ̂_o m_o )+〖Q+W〗_s (6.15b) But; H=U+PV