The velocity in the larger diameter
v1S1=v2S2→v2=v1S1S2=2⋅3.14⋅(0.25/2)23.14⋅(0.4/2)2=0.78m/sv_1S_1=v_2S_2\to v_2=\frac{v_1S_1}{S_2}=\frac{2\cdot3.14\cdot(0.25/2)^2}{3.14\cdot(0.4/2)^2}=0.78 m/sv1S1=v2S2→v2=S2v1S1=3.14⋅(0.4/2)22⋅3.14⋅(0.25/2)2=0.78m/s
The volume flow rate
QV=Vt=S2⋅v2⋅tt=3.14⋅(0.42)2⋅0.78=0.098m3/sQ_V=\frac{V}{t}=\frac{S_2\cdot v_2\cdot t}{t}=3.14\cdot(\frac{0.4}{2})^2\cdot0.78=0.098m^3/sQV=tV=tS2⋅v2⋅t=3.14⋅(20.4)2⋅0.78=0.098m3/s
The mass flow rate
Qm=QV⋅ρ=0.098⋅1000=98kg/sQ_m=Q_V\cdot\rho=0.098\cdot1000=98kg/sQm=QV⋅ρ=0.098⋅1000=98kg/s
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments