Answer to Question #228380 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A quantity of steam at 10 bar and 0.85 dryness occupies 0.15 m3. Determine the heat supplied to raise the temperature of the steam to 300°C at constant pressure and

percentage of this heat which appears as external work.

Take specific heat of superheated steam as 2.2 kJ/kg K.

Answer 1

Solution;

Given ;

Pressure of steam ,$P_1=P_2=10$ bar

Dryness fraction,$x_1=0.85$

Volume of steam; $v_1=0.15m^3$

Final temperature of steam,$t_{p_2}=300°c$

Specific heat of superheated steam,$C_{ps}=2.2kJ/kgK$

From steam tables;

At 10 bar;

$\nu_g=0.194m^3/kg$

$t_s=179.9°c$

$h_{fg}=2013.6kJ/kg$

Mass of steam will be given as;

$m_s=\frac{V_1}{x_1\nu_g}=$ $\frac{0.15}{0.84×0.194}$ =0.909kg

Heat supplied per kg of steam is;

=$(1-x_1)h_{fg}+c_p(t_{p_2}-t_s)$

=(1-0.85)×2013.6+2.2×(300-179.9)

=566.26kJ/kg

Total heat supplied;

=566.26×0.909=514.7kJ

External work done during the process;

W=$p(\nu_{sp_2}-x\nu_g)$

W=$p[(\nu_g×\frac{t_{p_2}}{t_{p_1}})-x_1\nu_g]$

$W=10×10^5[(0.194×\frac{300+273}{179.9+273})-(0.85×0.194)]×10^{-3}$

$W=80kJ/kg$

Percentage of total heat supplied per kg is;

=$\frac{80}{566.26}=0.141$ =$14.1$ %