Can some please help me with my questions. The Questions are in these two pictures below. I don't know how to do this question.
1 answer:
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Answer:
28,8 m/s
Explanation:
In a steady flow system we can say that m1=m2 which means that the mass flow in the entrance in the same in the outlet. m is flow (kg/s)
we know that where V (m/s) is velocity, A (m^2) ia area and v is specific volume (m^3/kg)
Since m1=m2 we can say
clearing the equation
we can specific volume (m^3/kg) from thermodynamic tables
for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg
In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated (vg) gas from the thermodynamic tables. we use the next equation to get (v) for the condition of interest, in this case 80% quality.
v= vf +x*(vg - vf)
where:
x: quality
vf = liquid-saturated-specific-volume
vg =steam-saturated-specific-volume.
for this problem
x = 0,8
vf = 0,00102991
vg = 3,24015
so
we get = 2,593 m^3/kg
The area is the one for a circle
r1 = 0,1 m^2 for area 1
r2=0,5 m^2 for area 2
A1 = 0,0314 m^2
A2 = 0,7853 m^2
we know that V1 is 20 m/s
replacing these values in the equation
we get V2 = 28,2 m/s.
Answer:
q = (ρg/μ)(sin θ)(h³/3)
Explanation:
I've attached an image of a figure showing the coordinate system.
In this system: the velocity components v and w are equal to zero.
From continuity equation, we know that δu/δx = 0
Now,from the x-component of the navier stokes equation, we have;
-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)
Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0
Then our eq 1 is now;
ρg(sin θ) + μ(δ²u/δy²) = 0
μ(δ²u/δy²) = -ρg(sin θ)
Divide both sides by μ to get;
(δ²u/δy²) = -(ρg/μ)(sin θ)
Integrating both sides gives;
δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)
Now, the shear stress is given by the formula;
τ_yx = μ[δu/δy + δv/δx]
From the diagram, at the free surface,τ_yx = 0 and y = h
This means that δu/δy = 0
Thus, putting 0 for δu/δy in eq 2, we have;
0 = -(ρg/μ)(sin θ)h + b1
b1 = h(ρg/μ)(sin θ)
So, eq 2 is now;
δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)
Integrating both sides gives;
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3
Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.
Thus, we now have:
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y
Factorizing like terms, we have;
u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)
The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;
∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]
q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0
q = (ρg/μ)(sin θ)[h³/2 - h³/6]
q = (ρg/μ)(sin θ)(h³/3)
