Known A two-phase liquid–vapor mixture of water in a closed, rigid container is heated on a hot plate. The initial pressure and quality and the final pressure are known.
Find Indicate the initial and final states on a T–υ diagram and determine at each state the temperature and the mass of water vapor present. Also, if heating continues, determine the pressure when the container holds only saturated vapor.
Schematic and Given Data:
Engineering Model
1. The water in the container is a closed system.
2. States 1, 2, and 3 are equilibrium states.
3. The volume of the container remains constant.
Analysis Two independent properties are required to fix states 1 and 2. At the initial state, the pressure and quality are known. As these are independent, the state is fixed. State 1 is shown on the T–υ diagram in the two-phase region. The specific volume at state 1 is found using the given quality and Eq. 3.2. That is,
v=(1-x) v_{ f }+x v_{ g }=v_{ f }+x\left(v_{ g }-v_{ f }\right) (3.2)
v_{1}=v_{ fl }+ x _{1}\left(v_{ g 1}-v_{ f 1}\right)
From Table A-3 at p_{1}=1 \text { bar, } v_{\text {fl }}=1.0432 \times 10^{-3} m ^{3} / kg \text { and } v_{ gl }=1.694 m ^{3} / kg. Thus,
v_{1}=1.0432 \times 10^{-3}+0.5\left(1.694-1.0432 \times 10^{-3}\right)=0.8475 m ^{3} / kg
At state 2, the pressure is known. The other property required to fix the state is the specific volume v_{2}. Volume and mass are each constant, so v_{2}=v_{1}=0.8475 m ^{3} / kg . \text { For } p_{2}=1.5 bar, Table A-3 gives v_{ f 2}=1.0582 \times 10^{-3} m ^{3} / kg \text { and } v_{ g 2}=1.59 m ^{3} / kg. Since
1 v_{ f }<v_{2}<v_{ g 2}
2 state 2 must be in the two-phase region as well. State 2 is also shown on the T–υ diagram above.
a. Since states 1 and 2 are in the two-phase liquid–vapor region, the temperatures correspond to the saturation temperatures for the given pressures. Table A-3 gives
T_{1}=99.63^{\circ} C \quad \text { and } \quad T_{2}=111.4^{\circ} C
b. To find the mass of water vapor present, we first use the volume and the specific volume to find the total mass, m. That is,
m=\frac{V}{v}=\frac{0.5 m ^{3}}{0.8475 m ^{3} / kg }=0.59 kg
Then, with Eq. 3.1 and the given value of quality, the mass of vapor at state 1 is
x=\frac{m_{\text {vapor }}}{m_{\text {liquid }}+m_{\text {vapor }}} (3.1)
m_{ gl }=x_{1} m=0.5(0.59 kg )=0.295 kg
The mass of vapor at state 2 is found similarly using the quality x_{2}. To determine x_{2}, solve Eq. 3.2 for quality and insert specific volume data from Table A-3 at a pressure of 1.5 bar, along with the known value of υ, as follows
x_{2}=\frac{v-v_{ f 2}}{v_{ g 2}-v_{ f 2}}
=\frac{0.8475-1.0528 \times 10^{-3}}{1.159-1.0528 \times 10^{-3}}=0.731
Then, with Eq. 3.1
m_{g 2}=0.731(0.59 kg )=0.431 kg
c. If heating continued, state 3 would be on the saturated vapor line, as shown on the T–υ diagram of Fig. E3.2. Thus, the pressure would be the corresponding saturation pressure. Interpolating in Table A-3 at v_{ g }=0.8475 m ^{3} / kg , \text { we get } p_{3}=2.11 bar.
1 The procedure for fixing state 2 is the same as illustrated in the discussion of Fig. 3.8.
2 Since the process occurs at constant specific volume, the states lie along a vertical line.
Skills Developed
Ability to…
• define a closed system and identify interactions on its boundary.
• sketch a T–υ diagram and locate states on it.
• retrieve property data for water at liquid–vapor states, using quality.
Quick Quiz
If heating continues at constant specific volume from state 3 to a state where pressure is 3 bar, determine the temperature at that state, in °C. Ans. 282°C
