Answer:
(a) 104 m/s
Explanation:
enters with specific enthalpy 'h1' = 3445.3 kJ/kg
exits with specific enthalpy 'h2' 3051.1 kJ/kg
velocity 'V1'= 100 m/s
In order to find velocity at the exit 'V2', we use this relation
0 = Qcv - Wcv + m[(h1-h2) + [tex]\frac{V1^{2}-V2^{2} }{2}[/tex] + g(z1-z2)]
0= [Qcv + m[(h1-h2) + [tex]\frac{V1^{2}-V2^{2} }{2}[/tex] ]
0=( h1-h2) + ( [tex]\frac{V1^{2}-V2^{2} }{2}[/tex] )
0= (3445.3-3051.1)+ [tex]\frac{100^{2} - V2^{2}}{2}[/tex]
0= 394.2 + [tex]\frac{100^{2} - V2^{2}}{2}[/tex]
-394.2= [tex]\frac{100^{2} - V2^{2}}{2}[/tex]
-788.4 = 10000-V2²
V2²= 10000+ 788.4
V2²= 10784.44
Taking square root on both sides
V2=103.84
V2≈ 104m/s
Therefore, The velocity at the exit is most closely 104m/s
