Question

A rigid tank initially contains 0.5 kg of steam at 800 kPa and ${300}^{o}C$ (u = 2790.6 kJ/kg, $v_1=0.3241m^3/kg$) and is connected through an insulated valve to a steam supply line that is capable of supplying steam at a constant condition of 1.4 MPa, ${300}^{o}C$ ( $h=3039.7kJ/kg$ ). The valve is opened so that the supply steam flows slowly into the tank until the pressure and temperature inside are 1.2 MPa and ${300}^{o}C(u=2788.6kJ/kg,v=0.2138m^3/kg)$. The final mass of steam in the tank and the heat transfer to (or from) the steam in the tank during the process respectively are

• A. 0.76 kg, 69.3 kJ (to the system)
• B. 0.76 kg, 69.3 kJ (from the system)
• C. 1.76 kg, 79.3 kJ (to the system)
• D. 1.76 kg, 79.3 kJ, (from the system)

Answer 1

The correct option is B 0.76 kg, 69.3 kJ (from the system)


$\text{Tank volume is constant}\Rightarrow m_2{v}_2=m_1{v}_1$

$\therefore m_2=m_1\frac{v_1}{v_2}$

$m_2=0.5\left(\frac{0.3241}{0.2138}\right)=0.76kg$

Form energy balance,

$\frac{d}{dt}(m_i{h}_i+Q)-\frac{d}{dt}(m_e{h}_e+W)={\left(\frac{dU}{dt}\right)}_{c.v}$

$h_i\frac{d{m}_i}{dt}+\frac{dQ}{dt}={\left(\frac{dU}{dt}\right)}_{c.v}$

On integrating the above expression we get,

$h_i(m_2-m_1)+Q=m_2{u}_2-m_1{u}_1$

$3039.7\times (0.76-0.5)+Q=0.76\times 2788.6-0.5\times 2796.6$

$=-69.3kJ\text{(HT from the system)}$