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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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A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an
vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0
2 years ago
5) /
Tpy6a [65]

Answer:

I don't understand...

4 0
3 years ago
What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?
Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0
3 years ago
1. What is the maximum value of the linear density in a crystalline solid (linear density defined as the fraction of the line le
insens350 [35]
Number three number three number three I’m not 100% sure though
4 0
3 years ago
A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem
lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is \Delta h=9.975\,m

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature.  As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2}  \Delta v^2 =0

(\rho stands here for density, h for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

\frac{\Delta P}{\rho}+g\, \Delta h  =0

\Delta P= -g\, \rho\, \Delta h

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\

We insert that into our last equation and get:

\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0
3 years ago
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