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lukranit [14]
3 years ago
10

Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on ske

tches of the p–υ and T–υ diagrams.a. At p = 3 bar, υ = 0.5 m3 /kg, find T in °C and u in kJ/kg.b. At T = 320°C, υ = 0.03 m3 /kg, find p in MPa and u in kJ/kg.c. At p = 28 MPa, T = 520°C, find υ in m3 /kg and h in kJ/kg.d. At T = 10°C, υ = 100 m3 /kg, find p in kPa and h in kJ/kg.e. At p = 4 MPa, T = 160°C, find υ in m3 /kg and u in kJ/kg.
Engineering
1 answer:
Anarel [89]3 years ago
5 0

heres a link! ;)

http://www.econsulat.ro/

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I want a problems and there solutions of The inception of cavitation?​
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The overview of the given scenario is explained in explanation segment below.

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Which of the following is not a reason to give yourself extra "cushion" when driving?
damaskus [11]

Answer:

The question is incomplete, the complete question is:

Which of the following is not a reason to give yourself extra "cushion" when driving?

A. Poor visibility B. Poor road conditions C. Inclement weather D. None of these.

The correct answer is D. None of these.

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All the options are not reasons to give yourself an extra cushion when driving, rather they are reasons that are not favorable to driving at all.

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7 0
3 years ago
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome
oksano4ka [1.4K]

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = \sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}

in this expression the densities are constant

        v = A  √h

       A =\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }

 

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = \sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = \frac{1}{454.55^2} \ v^2

       Δh = \frac{dh}{dv}   Δv

       \frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}

Suppose we have a height reading of h = 20 cm = 0.20 m

             

a) uncertainty 2.5 m / s ( 0.05)

        \frac{\delta v}{v} = 0.05

       \frac{\Delta h}{h} = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        \frac{\Delta h}{h} =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

5 0
3 years ago
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