Types of technology include:

A pressure gage at the inlet to a gas compressor indicates that the gage pressure is 40.0 kPa. Atmospheric pressure is 1.01 bar.

bonufazy [111]

Answer:

Given

inlet Pga =40kpa = 40000pa

Patm=1.01bar = 1.01 x 100000pa =101000pa

exit Pab= 6.5 (inlet Pab)

But generally, Pab = Patm + Pga

1. the absolute pressure of the gas at the inlet, inlet Pab?

inlet Pab = Patm + inlet Pga

= 101000pa + 40000pa = 141kpa

the absolute pressure of the gas at the inlet, inlet Pab = 141kpa

2. the gage pressure of the gas at the exit? exit Pga?

exit Pab = Patm + exit Pga

exit Pga = exit Pab - Patm

= (6.5 x 141kpa) - 101kpa

= 815.5kpa

the gage pressure of the gas at the exit exit Pga=815.5kpa

5 0

3 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

3 years ago

One of the disadvantages of the test of the hypothesis is that the final decision cannot be said to be completely correct eviden

vfiekz [6]

Answer:

A. True

Explanation:

4 0

3 years ago

‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?

Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

$E=12\times 9.8\times 25\\\\E=2940\ J$

So, the potential energy of the mass is 2940 J.

3 0

3 years ago

If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit

mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

$\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}$

$\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}$

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

$\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}$

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

$\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}$

choose R2, R1 such that it will maintain required ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0

3 years ago