An insulated tank is divided into two parts by a partition. One part of the tank contains 2.5 kg of compressed liquid water at 60°C and 600

Question

An insulated tank is divided into two parts by a partition. One part of the tank contains 2.5 kg of compressed liquid water at 60°C and 600 kPa while the other part is evacuated. The partition is now removed, and the water expands to fill the entire tank. Determine the final temperature of the water and the volume of the tank for a final pressure of 10 kPa. Use data from the steam tables

in progress 0
Verity 5 years 2021-08-09T09:50:13+00:00 2 Answers 1695 views 0

Answers ( )

  1. linhdan
    1
    2021-08-09T09:51:20+00:00
    Reply

    Answer:

    a) T = 45.81\,{\textdegree}C, b) V_{tank} = 0.919\,m^{3}

    Explanation:

    First, initial temperature, specific volume and specific enthalpy are obtained from steam tables:

    Water – Initial state (Compressed Liquid)

    P = 600\,kPa

    T = 60\,^{\textdegree}C

    \nu = 0.001017\,\frac{m^{3}}{kg}

    h = 251.18\,\frac{kJ}{kg}

    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:

    h_{1} = h_{2}

    Final properties of water are presented herein:

    Water – Final state (Liquid-Vapor Mixture)

    P = 10\,kPa

    T = 45.81\,{\textdegree}C

    \nu = 0.3677\,\frac{m^{3}}{kg}

    h = 251.18\,\frac{kJ}{kg}

    x = 0.025

    a) The final temperature is:

    T = 45.81\,{\textdegree}C

    b) The volume of the tank is:

    V_{tank} = m\cdot \nu

    V_{tank} = (2.5\,kg)\cdot (0.3677\,\frac{m^{3}}{kg} )

    V_{tank} = 0.919\,m^{3}

  2. ngocdiep
    -1
    2021-08-09T09:51:49+00:00
    Reply

    Answer:

    Explanation:

    The energy balance:

    E_{in} -E_{out} = ΔE_{system}

    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\frac{m^{3} }{kg}, u1 = uf → 60°c = 251.16\frac{kj}{kg}

    verified if we get quality between 0 and 1.

    P2 = 10kpa

    (u2 – u1)

    we have vf = 0.001010, v = 14.670\frac{m^{3} }{kg}

    uf = 191.79, uf_{g} = 2245.4\frac{kj}{kg}

    so therefore

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )