# Enthalpy-Internal Energy Conservation in an Open System

In preparation for the FE, I've come across a thermodynamic scenario in Michael R. Lindeburg, PE's review manual that I cannot seem to wrap my head around. Though I have a full solution to the following problem, it is the conceptual logic surrounding the transfer of enthalpy into internal energy that I cannot grasp.

In short: Steam with negligible velocity is allowed into a rigid, insulated tank with zero initial absolute pressure and zero elevation change. Once the tank is full, the temperature of the steam has nearly doubled. How is this possible? If the final internal energy of the gas is increased, the enthalpy must also increase, and I fail to see where the added energy comes from. The solution suggests that the initial enthalpy is converted to internal energy, but the enthalpy must increase proportionally. I am at a loss.

For those interested in the entire problem:

Statement

The absolute pressure in a rigid, insulated tank is initially zero. The tank volume is 0.04 m^3. The tank and steam inlet are at the same elevation. A valve is opened allowing 250°C and 600 kPa steam with negligible velocity to slowly fill the tank. Most nearly, what is the temperature of the steam in the tank after the tank is full.

Solution

Use the First Law of Thermodynamics for Open Systems. The control volume is the tank. Subscript $$s$$ refers to the substance in the tank. Subscript $$i$$ refers to the tank inlet, at the valve. The tank does not have an exit, so all subscript $$e$$ variables are zero. Subscript $$1$$ refers to what is initially in the tank. Since the tank is initially evacuated, it has no contents, so all subscript $$1$$ variables are zero. Subscript $$2$$ refers to the steam in the tank after filling. Since the tank is insulated, $$Q_{in}$$ is zero. There is no work done on or by the steam, so $$W_{net}$$ is zero. Although pressure does not appear explicitly in the First Law equation, the final pressure in the tank is $$p_2 = p_i$$.

$$\sum\left(\dot{m}_i\left(h_i + \dfrac{v_i^2}{2} + g z_i\right)\right) - \sum\left(\dot{m}_e\left(h_e + \dfrac{v_e^2}{2} + g z_e\right)\right) + \dot{Q}_{in} - \dot{W}_{net} = \dfrac{\partial(m_s u_s)}{\partial t}$$

\begin{align} m_i h_i &= m_2 u_2 - m_1 u_1 \\ m_i h_i &= m_2 u_2 \end{align}

Use a mass balance to relate $$m_i$$ to $$m_2$$:

\begin{align} \sum m_i - \sum m_e &= (m_2 - m_1)_{system} \\ m_i &= m_2 \end{align}

Combine the two equations and solve for the internal energy of the system:

$$u = h_i = 2957.2\text{ kJ/kg}$$

250°C, 600 kPa steam is superheated. From the superheated steam table, its enthalpy is $$h = 2957.2\text{ kJ/kg}$$. Since this is the final internal energy of the steam in the tank, interpolate to find the temperature of superheated steam with $$u = 2957.2\text{ kJ/kg}$$:

$$T = 397.0°C$$

• I found that the increase in the internal energy of the steam in the chamber exactly equals the $pv$ term of the initial enthalpy of the steam,. This suggested to me that the increase in internal energy of the steam is due to flow work done on the steam moving it into the chamber. Perhaps I'm misinterpreting the results of my calculation? Apr 27 '20 at 19:56
• So, is it then correct to say that the increase in internal energy (and temperature) of the steam is due to the $v\Delta p$ flow work moving the steam into the chamber? That's the way it works out mathematically. Apr 28 '20 at 13:18