Answer:
The total mass in the tank = 0.45524 kg
The amount of heat transferred = 3426.33 kJ
Explanation:
Given that:
The volume of the tank V = 0.06 m³
The pressure of the liquid and the vapor of H2O (p) = 15 bar
The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]
By applying the energy rate balance equation;
[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]
where;
[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]
Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]
If we integrate both sides; we have:
[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
We obtain the following data from the saturated water pressure tables, at p = 15 bar.
Since:
[tex]h_e =h_g[/tex]
Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]
[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]
[tex]v_g = 0.1318 \ m^3/kg[/tex]
Hence;
[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]
[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]
[tex]v_1 = 0.02728 \ m^3/kg[/tex]
Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar
[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]
Hence;
[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]
[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]
[tex]u_1 = 1193.428[/tex]
However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:
[tex]m_1 = \dfrac{V}{v_1}[/tex]
[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]
[tex]m_1 = 2.1994 \ kg[/tex]
From the question, given that the final quality; [tex]x_2 = 1[/tex]
[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]
[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]
[tex]v_2 = 0.1318 \ m^3/kg[/tex]
Also;
[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]
[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]
[tex]u_2 = 2594.5 \ kJ/kg[/tex]
Then the final mass can be calculated by using the formula:
[tex]m_2 = \dfrac{V}{v_2}[/tex]
[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]
[tex]m_2 = 0.45524 \ kg[/tex]
Thus; the total mass in the tank = 0.45524 kg
FInally; from the previous equation (1) above:
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]
Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]
Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]
Q = 3426.33 kJ
Thus, the amount of heat transferred = 3426.33 kJ
