Rigid Tank Venting Saturated Vapor at Constant Temperature: 04-BS-10, December 2013, Q8

Question 8: Rigid Tank Venting Saturated Vapor at Constant Temperature (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rigid tank, $V=0.85$ m³, constant $T=260\ ^\circ$C (hence constant $P=P_{sat}(260\ ^\circ\text{C})=4692.3$ kPa) throughout the process. Initial state: two-phase mixture, quality $x_1=0.7$. Final state: saturated vapor filling the entire tank. Mass leaves continuously as saturated vapor at the (constant) tank state.

Property (260°C sat.)	Liquid	Vapor
$v$ (m³/kg)	0.001276	0.042173
$u$ (kJ/kg)	1128.97	2598.72
$h_g$ (kJ/kg)	2796.60

Find. $Q$, total heat transfer (kJ).

Approach

Compute the initial and final mass in the tank from $V$ and the specific volumes at states 1 (two-phase, $x_1=0.7$) and 2 (saturated vapor, filling $V$). The mass that leaves equals the difference, and it exits at a constant enthalpy $h_g(260\ ^\circ\text{C})$ throughout (the valve always sees saturated vapor at the fixed tank state). A transient (uniform-state, uniform-flow) energy balance on the tank, with zero boundary work (rigid tank) and one exit stream, then gives $Q$.

Initial state (two-phase mixture, $x_1=0.7$). $$v_1=v_f+x_1(v_g-v_f)=0.001276+0.7(0.042173-0.001276)=0.029904\ \text{m}^3/\text{kg}.$$ $$u_1=u_f+x_1(u_g-u_f)=1128.97+0.7(2598.72-1128.97)=2157.79\ \text{kJ/kg}.$$ $$m_1=\frac{V}{v_1}=\frac{0.85}{0.029904}=28.424\ \text{kg}.$$
Final state (saturated vapor filling the tank). $$m_2=\frac{V}{v_g}=\frac{0.85}{0.042173}=20.155\ \text{kg},\qquad u_2=u_g=2598.72\ \text{kJ/kg}.$$
Mass withdrawn. $$m_{out}=m_1-m_2=28.424-20.155=8.269\ \text{kg}.$$
Energy balance (rigid tank, no work, one exit at constant $h_g$). $$Q=(m_2u_2-m_1u_1)+m_{out}h_g$$ $$Q=(20.155\times2598.72-28.424\times2157.79)+8.269\times2796.60$$ $$Q=\boxed{14{,}169\ \text{kJ}}.$$

Quantity	Result
$m_1$	28.424 kg
$m_2$	20.155 kg
$m_{out}$	8.269 kg
$Q$ (heat transfer)	14,169 kJ

Concept & Theory

Topic. Transient (uniform-state, uniform-flow) energy balance on a rigid tank venting saturated vapor at constant pressure and temperature.

Key relations. USUF mass balance $m_2-m_1=-m_{out}$; USUF energy balance (one exit, no work, negligible KE/PE): $m_2u_2-m_1u_1=Q-m_{out}h_e$, with $h_e$ constant at the exit condition throughout the process since the tank state itself never changes.

Why this works. Even though mass and energy are both leaving the tank, its intensive state stays fixed at the saturation point (260°C, 4692 kPa) for the entire process, which is precisely why the exit property $h_e=h_g(260\ ^\circ\text{C})$ can be pulled outside the integral $\int h_e\,dm_e=h_g\,m_{out}$ — a shortcut that would not be valid if the tank pressure or temperature drifted during venting.

Common pitfall. Trying to track "heat in" and "vapor out" as two separate, sequential steps (heat first, then vent, or vice versa) instead of writing one combined transient energy balance over the whole process; also, using $h_1$ (mixture enthalpy) instead of $u_1$ in the initial-state term, since it is $u$ (not $h$) that belongs in the closed-tank-contents energy accounting, with $h$ reserved for the enthalpy of the mass that actually crosses the boundary.

Source. Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 8th ed., Ch. 4 (Control Volume Analysis, transient/USUF processes).

Question 8 of 9: Rigid Tank Venting Saturated Vapor at Constant Temperature

Question 8: Rigid Tank Venting Saturated Vapor at Constant Temperature (15 marks)

Approach