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o-na [289]
3 years ago
8

2.4 kg of nitrogen at an initial state of 285K and 150 kPa is compressed slowly in an isothermal process to a final pressure of

600 kPa. Determine the work.
Engineering
1 answer:
Neko [114]3 years ago
8 0

Answer:

W=-280.67 KJ

Explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

W=mRT\ ln\dfrac{P_1}{P_2}

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

W=mRT\ ln\dfrac{P_1}{P_2}

W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.

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Multiple Choice
mote1985 [20]
I think it’s manufacturing
7 0
3 years ago
Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture
alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

b) 2.0 MΩ, +/- 20%

Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

1)  Red = 1st digit = 2

2) Black = 2nd digit = 0

3) Green = multiplier = 10e5

4) Nothing = tolerance = +/- 20%

This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.

7 0
3 years ago
Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
motikmotik

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

8 0
3 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
15 POINTS! Help.
IRISSAK [1]

Answer: it would  overload

Explanation:

4 0
3 years ago
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