25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 1 of 8: Tank Charging & Blower Compression

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations, May 2017 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 1: Tank Charging & Blower Compression (20 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.

(a) Tank charging — reversible expansion into an evacuated tank

Given. Tank A ($V_A=1.0\text{ m}^3$) at $P_1=700\text{ kPa}$, $T_1=20\,{}^{\circ}\text{C}=293.15\text{ K}$, connected via a valve to evacuated tank B ($V_B=0.3\text{ m}^3$). Valve opens, then closes at pressure equilibrium; overall system (A+B) is rigid and adiabatic. Air remaining in A undergoes a reversible (isentropic) process as it pushes gas through the valve.

Two rigid tanks joined by a valve — closed, adiabatic overall system.

Find. Final pressure, the mass of air in each tank, and the temperature of the air in each tank.

Approach. Treat A+B together as one closed, rigid, adiabatic system (total internal energy conserved, no boundary work) to fix $P_2$; use the isentropic relation for the gas remaining in A to fix $T_{A2}$; then split the mass by the ideal-gas law in each tank.

Total mass and overall energy balance. Since $m_A T\propto PV$ for an ideal gas at fixed $R$, and the combined system is closed/rigid/adiabatic ($U_1=U_2$), the initial-mass identity $m_{A1}T_1c_v = (P_2V_A/R)c_v+(P_2V_B/R)c_v$ collapses to $$P_1V_A = P_2(V_A+V_B)$$ $$\boxed{P_2 = P_1\dfrac{V_A}{V_A+V_B} = 700\times\dfrac{1.0}{1.3} = 538.5\text{ kPa}}$$
Temperature remaining in tank A (isentropic). The air that stays in A expands reversibly and adiabatically from $(P_1,T_1)$ to $P_2$: $$T_{A2}=T_1\left(\dfrac{P_2}{P_1}\right)^{(k-1)/k}=293.15\times\left(\dfrac{538.5}{700}\right)^{0.2857}=272.0\text{ K}=-1.2\,{}^{\circ}\text{C}$$
Mass split. Total initial mass, $m_{A1}=\dfrac{P_1V_A}{RT_1}=\dfrac{700\times1.0}{0.287\times293.15}=8.320\text{ kg}$. Mass remaining in A, $m_{A2}=\dfrac{P_2V_A}{RT_{A2}}=\dfrac{538.5\times1.0}{0.287\times272.0}=6.898\text{ kg}$. Mass that flowed into B, $m_{B2}=m_{A1}-m_{A2}=8.320-6.898=1.422\text{ kg}$.
Temperature in tank B. From the ideal-gas law applied to B's final state: $$T_{B2}=\dfrac{P_2V_B}{Rm_{B2}}=\dfrac{538.5\times0.3}{0.287\times1.422}=\boxed{395.9\text{ K}=122.7\,{}^{\circ}\text{C}}$$ Check: $m_{A2}T_{A2}+m_{B2}T_{B2}=6.898(272.0)+1.422(395.9)=2439\text{ kg}\cdot\text{K}=m_{A1}T_1$. ✓

Question 1(a) — final results
Quantity	Value
Final equilibrium pressure	$P_2 = 538.5\text{ kPa}$
Mass remaining in tank A	$m_{A2} = 6.898\text{ kg}$
Mass in tank B	$m_{B2} = 1.422\text{ kg}$
Temperature in tank A	$T_{A2} = 272.0\text{ K} = -1.2\,{}^{\circ}\text{C}$
Temperature in tank B	$T_{B2} = 395.9\text{ K} = 122.7\,{}^{\circ}\text{C}$

(b) Adiabatic blower — steady-flow energy equation

Given.

Given data
Quantity	Symbol	Value
Inlet pressure, temp., int. energy	$P_1,T_1,u_1$	90 kPa, 20°C, 209.3 kJ/kg
Inlet volumetric flow	$\dot V_1$	12.3 m³/min
Exit pressure, temp., int. energy	$P_2,T_2,u_2$	106 kPa, 30°C, 303.4 kJ/kg
Exit velocity	$V_2$	34 m/s

Find. Blower power input and the discharge (exit) flow area.

Approach. Steady-flow energy equation for an adiabatic, single-inlet/single-exit device: $\dot W_{in}=\dot m(\Delta h+\Delta ke)$, with $h=u+RT$ from the supplied $u$ values; mass flow from the inlet volumetric flow and specific volume; exit area from $\dot m=\rho_2A_2V_2$.

Specific enthalpies from the given $u$. $h_1=u_1+RT_1=209.3+0.287(293.15)=293.4\text{ kJ/kg}$; $h_2=u_2+RT_2=303.4+0.287(303.15)=390.4\text{ kJ/kg}$.
Mass flow rate. $v_1=\dfrac{RT_1}{P_1}=\dfrac{0.287\times293.15}{90}=0.9348\text{ m}^3/\text{kg}$; with $\dot V_1=12.3/60=0.205\text{ m}^3/\text{s}$, $$\dot m=\dfrac{\dot V_1}{v_1}=\dfrac{0.205}{0.9348}=0.2193\text{ kg/s}$$
Blower power (SFEE, adiabatic, $V_1\approx0$). $$\dot W_{in}=\dot m\left[(h_2-h_1)+\dfrac{V_2^2-V_1^2}{2}\right]=0.2193\left[96.97+\dfrac{34^2}{2000}\right]$$ $$\boxed{\dot W_{in}=0.2193(96.97+0.578)=21.4\text{ kW}}$$
Discharge area. $v_2=\dfrac{RT_2}{P_2}=\dfrac{0.287\times303.15}{106}=0.8208\text{ m}^3/\text{kg}$, so $$A_2=\dfrac{\dot m\,v_2}{V_2}=\dfrac{0.2193\times0.8208}{34}=\boxed{0.00529\text{ m}^2=52.9\text{ cm}^2}$$

Question 1(b) — final results
Quantity	Value
Mass flow rate	$\dot m = 0.219\text{ kg/s}$
Blower power input	$\dot W_{in} = 21.4\text{ kW}$
Discharge area	$A_2 = 0.00529\text{ m}^2\ (52.9\text{ cm}^2)$

Topic: Ideal-gas closed-system energy balances and the steady-flow energy equation (SFEE).

Key relations: $m_1T_1\propto P_1V_1/R$; isentropic relation $T_2/T_1=(P_2/P_1)^{(k-1)/k}$; SFEE $\dot Q-\dot W_s=\dot m[\Delta h+\Delta ke+\Delta pe]$; ideal-gas $h=u+Pv=u+RT$.

Why this works: Part (a) is a classic uniform-flow filling problem collapsed to a single algebraic step because the product $mT$ is proportional to $PV/R$ regardless of the (unknown) temperature split — this lets the overall energy balance fix $P_2$ without ever needing $T_{B2}$ explicitly, while the reversibility assumption for the gas remaining in A supplies the missing isentropic relation. Part (b) is routine SFEE bookkeeping once $u\to h$ is completed via $h=u+RT$, since the tabulated data gave internal energy, not enthalpy.

Common pitfall: Using the SFEE with the given $u$ values directly as if they were enthalpies (omitting the $Pv$ flow-work term) understates $\Delta h$ and the blower power; forgetting the kinetic-energy term at 34 m/s exit velocity is a smaller but still non-negligible ($\approx0.6\%$) error here.

Source: Cengel & Boles, Thermodynamics: An Engineering Approach, Ch. 4 (control-volume energy analysis) and Ch. 5 (steady-flow devices).

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