Question

A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and a quality of 0.2. The tank has a pressure-regulating venting valve that allows pressure to be constant. The tank is subsequently being heated until its content becomes a saturated vapor (of quality 1.0). During heating, the pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. You can neglecting the kinetic and potential energy effects.
Required:
a. Determine the total mass in the tank at the initial and final states, in kg.
b. Calculate the amount of heat (in kJ) transferred from the initial state to the final state.

Answer 1

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 $\mathbf{x_{initial} - 0.20}$

By applying the energy rate balance equation;

$\dfrac{dU}{dt} = Q_{CV} - m_eh_e$

where;

$m_e =- \dfrac{dm_{CV}}{dt}$

Thus, $\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e$

If we integrate both sides; we have:

$\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}$

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

$h_e =h_g$

Then: $h_g = h_e = 2792.2 \ kJ/kg$

$v_f = 1.1539 \times 10^{-3} \ m^3 /kg$

$v_g = 0.1318 \ m^3/kg$

Hence;

$v_1 = v_f + x_{initial} ( v_g-v_f)$

$v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )$

$v_1 = 0.02728 \ m^3/kg$

Similarly; we obtained the data for $u_f \ \& \ u_g$ from water pressure tables at p = 15 bar

$u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg$

Hence;

$u_1 = u_f + x_{initial } (u_g -u_f)$

$u_1 =843.16 + 0.2 (2594.5 -843.16)$

$u_1 = 1193.428$

However; the initial mass $m_1$ can be calculated by using the formula:

$m_1 = \dfrac{V}{v_1}$

$m_1 = \dfrac{0.06}{0.02728}$

$m_1 = 2.1994 \ kg$

From the question, given that the final quality; $x_2 = 1$

$v_2 = v_f + x_{final } (v_g - v_f)$

$v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})$

$v_2 = 0.1318 \ m^3/kg$

Also;

$u_2 = u_f + x_{final} (u_g - u_f)$

$u_2 = 843.16 + 1 (2594.5 - 843.16)$

$u_2 = 2594.5 \ kJ/kg$

Then the final mass can be calculated by using the formula:

$m_2 = \dfrac{V}{v_2}$

$m_2 = \dfrac{0.06}{0.1318}$

$m_2 = 0.45524 \ kg$

Thus; the total mass in the tank = 0.45524  kg

FInally; from the previous equation (1) above:

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

$Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)$

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