Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that
hence , actual velocity will be equal to 8.39 m/s
Answer:
a) the amount of energy produced in kJ/K is 0.73145 kJ/K
b) the amount of energy produced in kJ/K is 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Explanation:
Draw the T-s diagram.
a)
= 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280
R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar
Δs =
Substitute all parameters in the equation
Δs =
Δs = 5 kg × 0.14629 kJ/kg.K
= 0.73145 kJ/K
b)
Δs =
Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K
T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K
R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar
Δs =
= 5 kg (0.13795 kJ/kg.K)
= 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.