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:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
The reactor, generator, and the cooling towers