Electric Heating of a Gas at Constant Pressure

Example 5-5 Electric Heating of a Gas at Constant Pressure

A piston-cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 3.7 kJ occurs.

a)Shown that for a closed system, the boundary work $W_b$and the change in internal energy $∆U$ in the first-law relation can be combined into one term, $∆H$, for a constant pressure process. 

b) Determine the final temperature of the steam.

 

Solution: Saturated water vapor in a piston-cylinder device expands at constant pressure as a result of heating. It is to be shown that $∆U+W_b=∆H$, and the final temperature is to be determined.

Assumption: 1. The tank is stationary and thus, the kinetic and potential energy changes are zero, $∆KE=∆PE=0.$

Therefore, $∆U=∆E$ internal energy is the only form of energy that may change during this process. 

2. Electrical wires constitute a very small part of the system, and thus, the energy changes of the wire can be neglected.

 

Analysis: We take the content of the cylinder, including the resistance wires, as the the (Fig. 5-13). This is a closed system since no mass crossed the system boundary during the process. We observed that a piston-cylinder device typically involves a moving boundary and thus boundary work $W_b$. The pressure remains constant during the process and thus $P_2=P_1$. Also, heat is lost from the system and electrical work $W_e$

is done on the system.

 

a) This part of the solution involves a general analysis for a closed system undergoing a quasi-equilibrium constant-pressure process, and thus, we consider a general closed system. We take the direction of heat transfer Q to be to the system and the work W to be done by the system. We also express the work as the sum of boundary and other forms of work (such as electrical and shaft). Then, the energy balance can be expressed as:

$$\begin{gathered} E_{in}-E_{out}=∆E_{system} \\ Q-W=∆U+∆KE+∆PE (∆KE=∆PE=0) \\ Q-W_{other}-W_b=U_2-U_1 \end{gathered}$$

For a constant pressure process, the boundary work is given as $W_b=P_0(V_2-V_1)$

Substituting this into the preceding relation gives:

$$\begin{gathered} Q-W_{other}-P_0(V_2-V_1)=U_2-U_1 \\ However, \\ {P}_0=P_2=P_1\to Q-W_{other}=(U_2+P_2{V}_2)-(U_1+P_1{V}_1) \end{gathered}$$

Also, H=U+PV, and thus:

$Q-W_{other}=H_2-H_1 \left(kJ\right) (5-18)$

which is the desired relation. This equation is very convenient to use in the analysis of closed systems undergoing a constant-pressure quasi-equilibrium process since the boundary work is automatically taken care of by the enthalpy terms, and one on longer needs to determine it separately.

 

b) The only other form of work in this case is the electrical work, which can be determined from:

$$\begin{gathered} W_e=VI∆t=\left(120V\right)*(0.2A)*\left(300s\right)=7.2 kJ \\ State 1: P_1=300 kPa, sat.vapor \Rightarrow h_1=h_{g@300kPa}=2724.9 kJ/kg \\ (Table A-5) \end{gathered}$$

The enthalpy at the final state can be determined directly from Eq. 5-18 by expressing heat transfer from the system and work done on the system as negative quantities (since their directions are opposite to the assumed directions). 

Alternately, we can use the general energy balance relation with the simplification that the boundary work is considered automatically by replacing $∆U by∆H$for a constant-pressure expansion or compression process:

$$\begin{gathered} E_{in}-E_{out}=∆E_{system} \\ {E}_{in}-E_{out} (Net energy transfer by heat, work, and mass) \\ ∆E_{system} (Change in internal, kinetic, potential, etc. energies) \\ {W}_{e,in}-Q_{out}-W_b=∆U \\ {W}_{e,in}-Q_{out}=∆H=m(h_2-h_1) (\sin ce P=cons\tan t) \\ 7.2 kJ-3.7 kJ=(0.025 kg)*(h_2-2724.9) kJ/kg \\ {h}_2=2864.9 kJ/kg \end{gathered}$$

Now, the final state is completely specified since we know both the pressure and the enthalpy. The temperature at this state is

$State 2: P_2=300 kPa, h_2=2864.9 kJ/kg \Rightarrow T_2=200 C$(Table A-6)

Therefore, the steam will be at 200 C at the end of this process.

Discussion: Strictly speaking, the potential energy changes of the steam is not zero for this process since the center of gravity of the steam rose somewhat. Assuming an elevation change of 1 m (which is rather unlikely), the change in the potential energy of the steam would be 0.0002 kJ, which is very small compared to other terms in the first-law relation. Therefore, in problems of this kind, the potential energy is always neglected.