Answer:
Explanation:
The energy balance:
[tex]E_{in} -E_{out} =[/tex] Δ[tex]E_{system}[/tex]
so therefore
0 = ΔU = mm(u2-u1)
u1 =u2
The properties of water are table A = 4 and A-6
p1 = 600kpa
T1 = 60°c
} ⇒ v1 = vf → 60°c = 0.001017[tex]\frac{m^{3} }{kg}[/tex], u1 = uf → 60°c = 251.16[tex]\frac{kj}{kg}[/tex]
verified if we get quality between 0 and 1.
P2 = 10kpa
(u2 - u1)
we have vf = 0.001010, v = 14.670[tex]\frac{m^{3} }{kg}[/tex]
uf = 191.79, uf[tex]_{g}[/tex] = 2245.4[tex]\frac{kj}{kg}[/tex]
so therefore
Answer:
a) [tex]T = 45.81\,{\textdegree}C[/tex], b) [tex]V_{tank} = 0.919\,m^{3}[/tex]
Explanation:
First, initial temperature, specific volume and specific enthalpy are obtained from steam tables:
Water - Initial state (Compressed Liquid)
[tex]P = 600\,kPa[/tex]
[tex]T = 60\,^{\textdegree}C[/tex]
[tex]\nu = 0.001017\,\frac{m^{3}}{kg}[/tex]
[tex]h = 251.18\,\frac{kJ}{kg}[/tex]
The insulation means that there is no heat transfer from or to the tank. There is also no presence of work or mass interactions from or to the system. There are changes in volume and pressure of system due to the removal of the partition. Hence, the First Law of Thermodynamics is simplified into the following expression:
[tex]h_{1} = h_{2}[/tex]
Final properties of water are presented herein:
Water - Final state (Liquid-Vapor Mixture)
[tex]P = 10\,kPa[/tex]
[tex]T = 45.81\,{\textdegree}C[/tex]
[tex]\nu = 0.3677\,\frac{m^{3}}{kg}[/tex]
[tex]h = 251.18\,\frac{kJ}{kg}[/tex]
[tex]x = 0.025[/tex]
a) The final temperature is:
[tex]T = 45.81\,{\textdegree}C[/tex]
b) The volume of the tank is:
[tex]V_{tank} = m\cdot \nu[/tex]
[tex]V_{tank} = (2.5\,kg)\cdot (0.3677\,\frac{m^{3}}{kg} )[/tex]
[tex]V_{tank} = 0.919\,m^{3}[/tex]
