Let’s first be less vague. We’ll say pressure, there’s an intensive property. Also, let’s identify a material. We’ll say water as its vapor behavior is so well understood.
So, if what you are asking is would the pressure of saturated water vapor that boiled off of pure water be different than the pressure of saturated water vapor that was generated from a solution (let’s not require it be saturated) of, say, water and sugar that was mostly sugar, if the temperatures are equal then the answer is yes. Why? Because the boiling points are different.
For example, if you were to find saturated vapor in a system at atmospheric pressure above water, then that steam would be at 212 °F and 0 psig (14.7 psia). Water vapor above a 50%+ DS (dissolved solid to water by mass percentage) solution of dextrose and water would have a partial pressure lower than atmospheric in the same setting. Why? Because we still have a ways to go before that solution will actually boil.
Now, when you do reach its boiling point you will have a vapor that is at 0 psig but it will not be saturated. It will instead be superheated, as the dextrose molecules have a negligible vapor pressure. The vapor you get is basically all water (with some tiny entrained but not vaporized sugar droplets), and as the boiling point was greater than the saturated temperature for that pressure it’s superheated out of the gate. If you were to take it out of that system and allow it to desuperheat on its own (without the addition of a water spray) it would lose pressure to achieve saturation.
I hope that helped.
An injector pressure drop test to see if the injector is restricted with deposits can be done using an integer test.
<h3>What is injector strain drop?</h3>
Fuel injectors have a glide charge. This glide charge of the injector is rated at a sure strain drop throughout the injector. Meaning the injector glide is say 30#/hr at 42.five PSI. This approach that that injector will glide 30#/hr so long as the strain on the deliver side, minus the strain withinside the manifold is 42.five psiListening or Clicking Test.
Start the engine and permit it to idle. Keep the engine strolling and contact the give-up of a protracted steel screwdriver towards the gas injector. Put your ear on the alternative give up of the screwdriver. A clicking sound approaches the injector's working.
Read more about the injector pressure drop:
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Answer: blue
Explanation: blue cuz you look like somebody who likes blue
Answer:
isentropic efficiency = 0.818
Explanation:
given data
pressure P1 = 95 kPa
temperature = 27°C
pressure P2 = 600 kPa
temperature = 277°C
to find out
isentropic efficiency of the compressor and exit temperature of the air
solution
we know from ideal gas of properties of air is
Pr1 at 27°C = 1.3860
and h1 at 300 K = 300.19 kJ/kg
and h2 at 550 K = 555.74 kJ/kg
and
we know equation for isentropic process that is
.........................1
put here value we get
solve we get Pr2
Pr2 = 8.75
by ideal gas of properties of air will be at Pr2
h2s = 508.66
T2s = 505.5 K
'so
isentropic efficiency will be here as
isentropic efficiency =
isentropic efficiency =
isentropic efficiency = 0.818
Answer:
no, it seems that everyone is having the same issue. If you use the app you can still find the answers and see them.
Explanation: