Answer:
Explanation:
= Area of section 1 =
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 =
= Density of fluid =
= Area of section 2 =
Mass flow rate is given by
The mass flow rate through the pipe is
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
The speed at section 2 is .
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
