Charging a Rigid Tank from a Supply Line (R-134a): 04-BS-10, May 2013, Q7

Question 7: Charging a Rigid Tank from a Supply Line (R-134a) (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, $V=0.2$ m³. Initial state: $T_1=8\ ^\circ\text{C}$, quality $x_1=0.6$ (60% vapor by mass). Supply line: $P_{line}=1$ MPa, $T_{line}=120\ ^\circ\text{C}$ (superheated). Filling stops when tank pressure reaches 800 kPa with the tank contents entirely vapor (i.e. the final state is saturated vapor at 800 kPa — the instant the last liquid droplet flashes away).

Find. (a) final tank temperature $T_2$; (b) mass of refrigerant $m_{in}$ that entered; (c) heat transfer $Q$ between the tank and its surroundings during filling.

Approach

This is an unsteady (transient) uniform-flow first-law problem on a closed control volume with one inlet and no exit: $m_2u_2-m_1u_1=Q+m_{in}h_{line}$ (no boundary work, rigid tank). Fix state 1 (given $T_1,x_1$) and state 2 (saturated vapor at the stated final pressure) to get $m_1,m_2,u_1,u_2$ from $V$ and the specific volumes, then $m_{in}=m_2-m_1$ and $Q$ follow directly from the energy balance.

Initial state 1. At $T_1=8\ ^\circ\text{C}$, $x_1=0.6$: $P_1=387.61$ kPa, $v_1=0.031998$ m³/kg, $u_1=313.85$ kJ/kg. $$m_1=\frac{V}{v_1}=\frac{0.2}{0.031998}=6.2504\ \text{kg}.$$
Final state 2 — saturated vapor at 800 kPa (part a). $$\boxed{T_2=T_{sat}(800\ \text{kPa})=31.33\ ^\circ\text{C}},\qquad v_2=0.025625\ \text{m}^3/\text{kg},\qquad u_2=394.96\ \text{kJ/kg}.$$ $$m_2=\frac{V}{v_2}=\frac{0.2}{0.025625}=7.8050\ \text{kg}.$$
Mass entered (part b). $$m_{in}=m_2-m_1=7.8050-6.2504=\boxed{1.5546\ \text{kg}}.$$
Supply-line enthalpy. At 1 MPa, 120°C: $h_{line}=504.21$ kJ/kg.
Heat transfer (part c). Unsteady-flow energy balance on the tank (rigid, no boundary work, one inlet, no exit): $$Q=(m_2u_2-m_1u_1)-m_{in}h_{line}=\left[7.8050(394.96)-6.2504(313.85)\right]-1.5546(504.21)$$ $$Q=\boxed{337.1\ \text{kJ}}\ (\text{positive}\Rightarrow\text{heat is transferred }\mathit{into}\text{ the tank from the surroundings}).$$

Quantity	Result
(a) $T_2$	31.33 °C
(b) $m_{in}$	1.5546 kg
(c) $Q$	337.1 kJ (into the tank)

Concept & Theory

Topic. Unsteady (transient) uniform-flow filling of a rigid tank from a pressurized supply line.

Key relations. Mass balance $m_{in}=m_2-m_1$; unsteady energy balance for a single-inlet, no-exit, no-work control volume: $m_2u_2-m_1u_1=Q+m_{in}h_{line}$, using the line's enthalpy (not internal energy) for the flowing mass since it carries flow work into the boundary.

Why this works. "Entirely vapor phase at 800 kPa" pins the final state exactly to $x_2=1$ at $P_2=800$ kPa (saturated vapor), which together with the known tank volume fixes $m_2$ independently of any assumption about the filling process itself; the energy balance then closes the problem for $Q$ once $m_{in}$ and $h_{line}$ are known, without needing to track the filling process in detail.

Common pitfall. Using the line's internal energy $u_{line}$ instead of its enthalpy $h_{line}$ in the energy balance — the flow entering the boundary carries flow work $Pv$ in addition to internal energy, and omitting it (using $u_{line}$) understates the energy added and gives the wrong sign or magnitude for $Q$.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis of Control Volumes — Unsteady-Flow Processes).

Question 7 of 9: Charging a Rigid Tank from a Supply Line (R-134a)

Question 7: Charging a Rigid Tank from a Supply Line (R-134a) (15 marks)

Approach