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Alekssandra [29.7K]
3 years ago
9

Air as an ideal gas in a closed system undergoes a reversible process between temperatures of 1000 K and 400 K. The beginning pr

essure is 200 bar. Determinc the highest possible ending pressure for this process.If the ending pressure is 3 bar, determine the heat transfer and work per unit mass, if the boundary of the system is in constant contact with a reservoir at 400 K.

Engineering
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

highest possible ending pressure for this process is 8.0954 bar

Explanation:

We can say that Heat transfer is any or all of several kinds of phenomena, considered as mechanisms, that convey energy and entropy from one location to another. The specific mechanisms are usually referred to as convection, thermal radiation, and conduction.

Please see attachment for the solution.

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No

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3 years ago
A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.0
harina [27]

Answer:

Load carried by shaft=9.92 ft-lb

Explanation:

Given:    Power P=4.4  HP

                    P=3281.08 W

<u><em>Power:  </em></u>Rate of change of work with respect to time is called power.

We know that P=Torque\times speed

     \omega=\frac{2\pi N}{60} rad/sec

So that P=\dfrac{2\pi NT}{60}

So   3281.08=\dfrac{2\pi \times 2329\times T}{60}

      T=13.45 N-m         (1 N-m=0.737 ft-lb)

 So T=9.92 ft-lb.

Load carried by shaft=9.92 ft-lb

3 0
3 years ago
A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-4
Nat2105 [25]

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

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5 0
2 years ago
A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo
Finger [1]

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

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using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

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m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric  property table

specific heat content initial  100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final  600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

3 0
3 years ago
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