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goldfiish [28.3K]
4 years ago
7

The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso

lute pressure at a depth of 5 m in liquid whose specific gravity is 0.8 at the same location.
Engineering
1 answer:
ser-zykov [4K]4 years ago
8 0

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

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Answer:

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8 0
3 years ago
What is software certification? Discuss its importance in the changing scenario of software industry. ​
bekas [8.4K]

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<h3>What is software certification?</h3>

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6 0
2 years ago
Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work
Naily [24]

Answer:

w_{out}=319.1\frac{BTU}{lbm}

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

s_2=s_1

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}

Best regards.

3 0
3 years ago
Over 30 day period, a lake surface area is 1260 acres. The inflow is 36 cfs, thee outflow is 30 cfs. Seepage loss is 1.5 in. The
Elis [28]

Answer:

  -0.1 inches

Explanation:

The net inflow is ...

  36 cfs -30 cfs = 6 cfs

The number of seconds in 30 days is ...

  (3600 s/h)(24 h/da)(30 da) = 2,592,000 . . . . seconds/(30 days)

Then the volume of inflow is ...

  (6 ft^3/s)(2,592,000 s) = 15,552,000 ft^3

The number of square feet in 1260 acres is ...

  (1260 ac)(43560 ft^/ac) = 54,885,600 ft^2

So, the increase in depth due to the inflow is ...

  (15,552,000 ft^3)/(54,885,600 ft^2) ≈ 0.283353 ft ≈ 3.4002 in

__

The net change in water level is then ...

  inflow - seepage + precipitation - evaporation

  3.4 in -1.5 in +4.0 in -6.0 in = -0.1 in

The water level change in the period is -0.1 inch.

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3 years ago
What branch of chemistry is credited with the development of silicone?
mariarad [96]

Answer:

C). Inorganic Chemistry

Explanation:

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8 0
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