Two rigid tanks are connected by a valve. Tank A contains $0.2\ \mathrm{m^3}$ of water at $400\ \mathrm{kPa}$ and $80\ \%$ quality. Tank B contains $0.5\ \mathrm{m^3}$ of water at $200\ \mathrm{kPa}$ and $250\ \mathrm{^\circ C}$. The valve is now opened, and the two tanks eventually come to the same state. Determine the pressure and the amount of heat transfer when the system reaches thermal equilibrium with surroundings at $25\ \mathrm{^\circ C}$. Assumptions 1: Tanks are stationary and thus the KE and PE changes are zero and there are no work interactions so $W = 0$.

To start this question I know we have to find the specific internal energy for Tank A and Tank B then use this to eventually find the total heat transfer so, I began by finding the properties of water in Tank A at $p = 400\ \mathrm{kPa}$, which are taken from Tables A-5 in Thermodynamics an Engineering Approach At $400\ \mathrm{kPa}$ the specific internal energies are retrieved to be $$u_\mathrm f = 604.22\ \mathrm{kJ/kg}$$ and then $$u_\mathrm{fg} = 1948.9\ \mathrm{kJ/kg}$$ My question is why are we using the $u_\mathrm{fg}$ or the specific internal energy of evaporation for this question and not the specific internal energy of saturated vapour, $u_\mathrm g$?

Please ask me any further questions you need to help me!

Table A-5

  • 1
    $\begingroup$ You could use the internal energy for liquid and vapour, or you could use the internal energy for liquid and the energy of evaporation since both pairs essentially give you the same information. What's better depends on how you want to calculate the next properties. $\endgroup$
    – Loong
    Commented May 29, 2021 at 19:13
  • $\begingroup$ Some data are missing. For example : 1) What is "water 80% quality" ? Is it 80% purity ? What is the residual stuff ? 2) What is the temperature of the water in container A ? 3) What is "saturated water" ? $\endgroup$
    – Maurice
    Commented May 29, 2021 at 21:11
  • $\begingroup$ @Maurice 80% quality means that 80 % of the mass is vapor and 20% is liquid. $\endgroup$ Commented May 30, 2021 at 2:58
  • $\begingroup$ I would have used $u_L$ and $u_G$. I would also have used $v_L$ and $v_G$ . $\endgroup$ Commented May 30, 2021 at 3:23
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    $\begingroup$ Ahhh I see, ultimately using either will result in the same values, I too would have used u_L and u_G, Thanks a lot for the responses! How do I proceed from here? How can I upvote your responses or mark as answered ? $\endgroup$
    – jjrrrmkaas
    Commented May 30, 2021 at 15:29

1 Answer 1


It seems that the OP left, but I liked the question so I will answer it. The notation $_{fg}$ has been addressed in the comments, so I will answer to your "how do I proceed from here?".

Disclaimer I will use Smith, Van Ness, and Abbott book for the water tables, but it doesn't matter for what we are asked. I think they have the same reference as your book, so answers will be almost the same.

Main Idea If $\ce{H2O}$ is a pure gas or liquid, it has two degrees of freedom, and its thermodynamic state is specified two intensive variables. If saturated, then we need only one.

I will talk about states $1A$ (Tank A initially), $1B$ (Tank B initially), and $2$ (final state).

Initial Calculations

$\ce{Tank 1 A}$: We have that $P_{1A} = 400 \; \pu{kPa}$ and the quality $x_{1A} = 0.8$, so it is saturated water, and the thermodynamic state is specified. They give us the quality $x$ because a general thermodynamic property for a pure fluid in LV equilibrium $M$ is obtained by doing $M = x M_{g} + (1 - x)M_l$.

\begin{align} v_{1A} &= 0.3699928 \; \pu{m^3/kg} \\ u_{1A} &= 2163.0074 \; \pu{kJ/kg} \end{align} Of interest will be the initial mass of Tank A $$ m_{A} = \dfrac{V_A}{v_{1A}} = 0.54055 \; \pu{kg} $$

$\ce{Tank 1 B}$: We have that $P_{2A} = 200 \; \pu{kPa}$ and $T_{2A} = 250 \; \ce{°C}$, thus the thermodynamic state is specified

\begin{align} v_{2A} &= 1.1989 \; \pu{m^3/kg} \\ u_{2A} &= 2731.4 \; \pu{kJ/kg} \end{align} Of interest will be the initial mass of Tank B $$ m_{B} = \dfrac{V_B}{v_{1B}} = 0.41705 \; \pu{kg} $$

The specific internal energy of state $1$ is $$ u_1 = \dfrac{m_A u_{1A} + m_Bu_{2A}}{m_A + m_B} = 2410.551 \; \dfrac{\pu{kJ}}{\pu{kg}} $$


For a closed system, where no work is done, the 1st law states $$ m\Delta u = Q \rightarrow Q = m(u_2 - u_1) \tag{1} $$ where $m = m_A + m_B$.

Also, since matter does not enter or leave the system, our whole mass resides in the tank volumes. Therefore, the final specific volume is $$ v_2 = \dfrac{V_A + V_B}{m_A + m_B} \rightarrow v_2 = 0.73099 \; \dfrac{\pu{m^3}}{\pu{kg}} \tag{2} $$

Final State

With Eq. (2) we are ready to go, since we have our two intensive variables $T_2 = 25 \; \ce{°C}$ and $v_2 = \; 0.73099 \; \pu{m^3/kg}$. However, upon looking at the water tables, this specific volume is between the saturated liquid and saturated gas at $T_2$. Thus, the pressure is informed immediately $$ \boxed{P_2 = 3.166 \; \pu{kPa}}$$ And the quality and specific internal energy of this saturated water is \begin{align} x_2 &= \dfrac{v_2 - v_{2l}(T_2)}{v_{2l}(T_2) - v_{2g}(T_2)} \rightarrow x_2 = 0.01682 \\ u_2 &= x_2 u_{2g}(T_2) + (1 - x_2)u_{2l}(T_2) = 143.5728 \; \dfrac{\pu{kJ}}{\pu{kg}} \end{align} And Eq. (1) yields $$ \boxed{Q = -2170.858\; \pu{kJ}} $$ The system has given this energy to the surroundings. Makes sense, because we started with two tanks at high $P$ and $T$, and ended at ambient temeperature.


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