Answer to Question #128453 in Mechanical Engineering for Afra Syab

Question

Steam with a quality of 0.85 is contained in a rigid vessel at a pressure of 200 kPa. Heat is then added until the temperature reaches (a)400 "C and (b)140 "C. Determine the final pressures. Ans. (a) 415 kPa (h)269 kPa
2

Answer 1

Initial conditions of steam:

Pressure (p1)=200kPa=0.2MPa

Quality (X)=0.85

From steam table

vf=0.00106 m3/kg

vg=0.88568 m3/kg

initial state specific volume (v1)= vf+X(vg-vf)

=0.00106+0.85x(0.88568-0.00106)=0.7529m3/kg.

Now heat is supplied at constant volume such that temperature of system becomes

Case-(a) :

Temperature T=400⁰C,specific volume (v2)=(v1)=0.7529 m3/kg

From steam table the critical temperature Tc=373.94

Since T2=400C >Tc, Steam is super heated ,

From superheated steam table at T=400c and v=0.7530m3/kg,

we get following value of P and v at T=400C

P(MPa) v(m3/kg)

0.40 0.77264

0.45 0.68634

therefore pressure (p) at v=0.7539 m3/kg is given by linear interpolation

P₂-0.40=[(0.45-0.4)/(0.68634-0.77264)]x(0.7529-0.7739)

P₂=0.4121MPa=412.1 kPa

Case-(a) :

Temperature T=140C, The specific volume of saturated steam(vg)=0.50845 m3/kg

The specific volume of system(v2)=v1=0.7529 m3/kg is more than that of saturatred steam at 140 C.

therefore steam is superheated .

From superheated steam table at T=140C

P(MPa) v(m3/kg)

0.24 0.77616

0.26 0.71495

herefore pressure (P₂) at v=0.7539 m3/kg is given by linear interpolation

P₂-0.24=[(0.26-0.24)/(0.71495-0.77616)]x(0.7529-0.77616)

P₂=0.2461MPa=246.1KPa