3. A 2-kg steam water mixture at 1.0 MPaa is contained in an inflexible tank. Heat is
added until the pressure rises to 3.5 MPaa and the temperature to 400 o C. Determine:
a) Heat transferred (kJ), b) ΔH (kJ), c)ΔS (kJ/k)
m=2kgP1=1.0 MPaP2=3.5 MPaU1=2401.41 kJ/kgU2=2926.4 kJ/kgT=400°Cm = 2kg\\ P_1 = 1.0\ MPa\\ P_2 = 3.5\ MPa\\ U_1 = 2401.41\ kJ/kg\\ U_2= 2926.4\ kJ/kg\\ T = 400°Cm=2kgP1=1.0 MPaP2=3.5 MPaU1=2401.41 kJ/kgU2=2926.4 kJ/kgT=400°C
Heat transferred = m(U2−U1)=m(U_2-U_1) =m(U2−U1)=
2(2926.4−2401.41)=2(524.99)2(2926.4-2401.41) = 2(524.99)2(2926.4−2401.41)=2(524.99)
=1049.98 kJ= 1049.98\ kJ=1049.98 kJ
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