PROBLEM1.) A 0.0856 𝑚3 drum contains saturated water and saturated vapour at
370°C. (a) Find the mass of each if their volumes are equal. What is the quality? Find the
volume occupied by each if their masses are equal.
Solution:
Let Vv = the volume of saturated vapour
VL = the volume of saturated liquid
mv = the mass of saturated vapour
mL = the mass of saturated liquid
𝑚3
𝑣𝑔 𝑎𝑡 370℃ = 0.004925
𝑘𝑔
𝑚3
𝑣𝑓 𝑎𝑡 370℃ = 0.002213
𝑘𝑔
(a)
Vv
VL
𝑉𝑉 =
Saturated Vapour
Saturated Liquid
0.0856 3
𝑚 = 0.0428𝑚3
2
𝑉𝑉
𝑚𝑉 =
=
𝑣𝑔
0.0428𝑚3
= 8.69𝑘𝑔
𝑚3
0.004925
𝑘𝑔
𝑉𝑉
𝑚𝐿 =
=
𝑣𝑓
0.0428𝑚3
= 19.34𝑘𝑔
𝑚3
0.002213
𝑘𝑔
Quality, 𝑥 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑢𝑟
8.69𝑘𝑔
=
= 0.31 𝑜𝑟 31%
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
19.34𝑘𝑔 + 8.69𝑘𝑔
(b)
𝑚𝑉
Vv
𝑚𝐿
VL
Saturated Vapour
Saturated Liquid
𝑚𝑉 = 𝑚𝐿
𝑉𝑉 + 𝑉𝐿 = 0.0856
𝑉𝑉 = 𝑚𝑉 𝑣𝑔 = 0.004925𝑚𝑉
𝑉𝐿 = 𝑚𝐿 𝑣𝑓 = 0.002213𝑚𝐿
Substituting in equation 2,
0.004925𝑚𝑉 + 0.002213𝑚𝐿 = 0.0856
0.004925𝑚𝑉 + 0.002213𝑚𝑉 = 0.0856
𝑚𝑉 = 11.9 𝑘𝑔
𝑚𝐿 = 11.9 𝑘𝑔
𝑚3
𝑉𝑉 = 𝑚𝑉 𝑣𝑔 = 11.9 𝑘𝑔 (0.004925 ) = 0.05905𝑚3
𝑘𝑔
𝑚3
𝑉𝐿 = 𝑚𝐿 𝑣𝑓 = 11.9 𝑘𝑔 (0.002213 ) = 0.02653𝑚3
𝑘𝑔
PROBLEM 2.) Air at 200 kPa, 30°C is contained in a cylinder/piston arrangement with
initial volume 0.1 m3. The inside pressure balances ambient pressure of 100 kPa plus an
externally imposed force that is proportional to V0.5. Now heat is transferred to the system
to a final pressure of 225 kPa. Find the final temperature and the work done in the process.
Solution:
C.V. Air. This is a control mass. Use initial state and process to find T2
𝑃1 = 𝑃0 + 𝐶𝑉 0.5
200 = 100 + 𝐶(0.1)0.5
𝐶 = 316.23
225 = 100 + (316.23)𝑉2 0.5
𝑉2 = 0.156, 𝑚3
𝑃2 𝑉2 = 𝑚𝑅𝑇2 =
𝑇2 = (
𝑃1 𝑉1 𝑇2
𝑇1
𝑃2 𝑉2
225(0.156)(303.15)
= 532𝐾 = 258.9°𝐶
) (𝑇1 ) =
𝑃1 𝑉1
200(0.1)
𝑊12 = ∫ 𝑃𝑑𝑉 = ∫(𝑃0 + 𝐶𝑉 0.5 )𝑑𝑉
𝑊12 = 𝑃0 (𝑉2 − 𝑉1 ) +
𝐶2
3
3
3 0.5
(316.23)(2)
2
𝑊12 = 100(0.156 − 0.1) +
(0.156 − 0.12 ) = 11.9 𝑘𝐽
3