Answer to Question #234651 in Molecular Physics | Thermodynamics for Peter

Question #234651
A rigid vessel contains 5.0 kg of wet steam at 0.4 MPa. After the addition of 9585 KJ, the steam has a pressure of 2.0 MPa and a temperature of 700 degrees Celsius. Determine the initial energy and the specific volume of the steam?
1
Expert's answer
2021-09-08T16:14:35-0400

Solution;

Given;

m=5.0kgm=5.0kg

QA=9585kJQ_A=9585kJ

P1=0.4MPaP_1=0.4MPa

Vessel is rigid, meaning;

V1=V2V_1=V_2

P2=2.0MPaP_2=2.0MPa

T2=700°c=973KT_2=700°c=973K

Heat added;

QA=m(h2h1)Q_A=m(h_2-h_1)

From steam tables,at 700°c and 2.0MPa;

h2=3923.81kJ/kgh_2=3923.81kJ/kg

Therefore;

9585=5×(3923.81h1)9585=5×(3923.81-h_1)

h1=19619.0595855h_1=\frac{19619.05-9585}{5}

h1=2006.81kJ/kgh_1=2006.81kJ/kg

From steam tables ,at 0.4MPa and h1=2006.81kJ/kg;

hf=604.37kJ/kgh_f=604.37kJ/kg

hg=2737.68kJ/kgh_g=2737.68kJ/kg

vf=0.001084m3/kgv_f=0.001084m^3/kg

vg=0.4624m3/kgv_g=0.4624m^3/kg

ug=2552.74kJ/kgu_g=2552.74kJ/kg

uf=603.93kJ/kgu_f=603.93kJ/kg

h1=hf1+xhg1h_1=h_{f_1}+xh_{g_1} =hf+x(hghf)h_f+x(h_g-h_f)

2006.81=604.37+x(2737.68604.37)2006.81=604.37+x(2737.68-604.37)

x=0.657x=0.657

Take;

v1=vf+x(vgvg)v_1=v_f+x(v_g-v_g)

v1=0.001084+0.657(0.46240.001084)v_1=0.001084+0.657(0.4624-0.001084)

Hence,the specific volume is;

v1=0.3043m3/kgv_1=0.3043m^3/kg

To calculate internal energy;

u=uf+x(uguf)u=u_f+x(u_g-u_f)

u=603.93+0.657(2552.74603.93)u=603.93+0.657(2552.74-603.93)

Hence the internal energy is;

u=1884.29kJ/kgu=1884.29kJ/kg






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