Answer:
v₂ = 1.25 m/s
Explanation:
given,
inlet velocity, v₁ = 5 ft/s
inlet Area = A = 10 ft²
outlet velocity, v₂ = ?
outlet Area = 4 A
= 4 x 10 = 40 ft²
Using continuity equation
A₁ v₁ = A₂ v₂
10 v₁ = 40 v₂
v₂ = 1.25 m/s
Hence, the exit velocity of the water through the duct is equal to 1.25 m/s.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
See explanations for step by step procedures into getting answer.
Explanation:
Given that:
A fluid particle flowing along a stagnation streamline, as shown in Video V4.9 and Fig. P4.35, slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance s = 0.8 ft upstream of the stagnation point at t = 0 is given approximately by s = 0.8e-0.61t where t is in seconds and s is in feet.(a) Factor the relation for the speed as a function of position out of the relation for speed as a function of time. What is the expression for speed V in terms of position s only
(b) For the position as a function of time given in the problem, what is the relation between the speed of the particle V and time t
(c) Symbolically, what does the derivative of position with respect to time equal.
See comolete solving at attachment
