Rigid Tank Charged from a Steam Supply Line: 04-BS-10, May 2018, Q5

Question 5: Rigid Tank Charged from a Steam Supply Line (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, initial: $m_1=0.5$ kg, $P_1=700$ kPa, $T_1=320\ ^\circ$C. Supply line: $P_{line}=1.5$ MPa, $T_{line}=320\ ^\circ$C (constant). Final: $P_2=1.0$ MPa, $T_2=320\ ^\circ$C. Insulated valve, uniform-flow (charging) process.

State	Description	$v$ (m³/kg)	$u$ (kJ/kg)	$h$ (kJ/kg)
1	Tank, initial (0.7 MPa, 320°C)	0.38523	2831.68	—
2	Tank, final (1.0 MPa, 320°C)	0.26786	2826.50	—
line	Supply line (1.5 MPa, 320°C)	—	—	3082.43

Find. (a) $m_2$ [kg]; (b) $Q$ [kJ].

Approach

The rigid tank fixes the volume from the known initial mass and specific volume; the same volume with the final specific volume gives the final mass, and hence the mass that entered. A uniform-flow (charging) energy balance on the tank — with the incoming stream carrying its flow enthalpy (not internal energy, since flow work crosses the boundary with it) — then isolates the heat transfer.

Fixed tank volume and final mass (part a). From steam tables at the initial state (0.7 MPa, 320°C): $v_1=0.38523$ m³/kg, so $$V=m_1v_1=0.5\times0.38523=0.19261\ \text{m}^3.$$ At the final state (1.0 MPa, 320°C): $v_2=0.26786$ m³/kg, so $$m_2=\frac{V}{v_2}=\frac{0.19261}{0.26786}=\boxed{0.7191\ \text{kg}}.$$ The mass that entered from the supply line is $m_{in}=m_2-m_1=0.7191-0.5=0.2191$ kg.
Uniform-flow energy balance (part b). With no work crossing the rigid tank's boundary and the incoming mass carrying the supply line's enthalpy (flow work already accounted for in $h$, not $u$): $$Q=(m_2u_2-m_1u_1)-m_{in}h_{line}.$$ Internal energies: $u_1=2831.68$ kJ/kg (0.7 MPa, 320°C), $u_2=2826.50$ kJ/kg (1.0 MPa, 320°C). Supply enthalpy: $h_{line}=3082.43$ kJ/kg (1.5 MPa, 320°C). $$Q=(0.7191\times2826.50-0.5\times2831.68)-0.2191\times3082.43$$ $$=(2032.46-1415.84)-675.36=616.62-675.28=\boxed{-58.7\ \text{kJ}}.$$ The negative sign shows the tank actually loses a small amount of heat to the surroundings during the fill — even though $T_2=T_1$, the tank's internal energy per unit mass barely changes while a comparatively hotter (higher-enthalpy) stream is added, so the energy balance still calls for a net heat loss to hold the final state at the stated $(P_2,T_2)$.

Quantity	Result
(a) $m_2$	0.7191 kg
(b) $Q$	−58.7 kJ (heat rejected)

Concept & Theory

Topic. Uniform-flow (unsteady-flow charging) process on a rigid tank connected to a constant-condition supply line through an insulated valve.

Key relations. Rigid-tank volume constraint: $V=m_1v_1=m_2v_2$. Uniform-flow energy balance with one inlet, no exit, no boundary work: $Q=(m_2u_2-m_1u_1)-m_{in}h_{line}$, where $m_{in}=m_2-m_1$.

Why this works. The incoming stream still carries flow work ($Pv$) across the valve even though the valve itself is insulated and the tank is rigid, so its energy contribution to the tank must be its enthalpy $h_{line}$, not its internal energy $u_{line}$ — the flow-work term is exactly what converts $u$ to $h$ for a flowing stream crossing a control-volume boundary.

Common pitfall. Assuming that because $T_1=T_2$ the process must be adiabatic ($Q=0$) — the tank's stored specific internal energy barely changes ($u_1\approx u_2$) while a higher-enthalpy stream is added at higher pressure, so the energy balance still requires a net (small) heat loss to close; equal initial/final temperatures says nothing about $Q$ by itself.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis of Control Volumes, uniform-flow processes).

Question 5 of 9: Rigid Tank Charged from a Steam Supply Line

Question 5: Rigid Tank Charged from a Steam Supply Line (15 marks)

Approach