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galben [10]
3 years ago
15

Ammonia in a piston–cylinder assembly undergoes two processes in series. At the initial state, p1 = 120 lbf/in.2 and the quality

is 100%. Process 1–2 occurs at constant volume until the temperature is 100°F. The second process, from state 2 to state 3, occurs at constant temperature, with Q23 = –98.9 Btu, until the quality is again 100%. Kinetic and potential energy effects are negligible. For 2.2 lb of ammonia, determinea) the heat transfer for Process 1–2, in Btu. b) the work for Process 2–3, in Btu.
Engineering
1 answer:
Kitty [74]3 years ago
4 0

Answer:

The heat transfer \mathbf{\mathbf{Q_{1-2}} = 35.904 \  Btu}

W_{2-3}=  -71.312 \ Btu

Explanation:

At the initial state when P₁ = 10 lbf/in :

We obtain the internal energy u₁ and specific volume v₁.

u₁ = u_g = 574.08 btu/lbm

v₁ = v_g  = 2.4746  ft³/lbm

Process 1–2 occurs at constant volume until the temperature is 100°F.

i.e T₂ = 100⁰ F

At T₂ = 100⁰ F  : v₁ = v₂ = 2.4746  ft³/lbm  

\mathbf{u_2 = 591.28 + \dfrac{2.4746-2.5917}{2.117-2.5917}*(587.68 -591.28)}

\mathbf{u_2 = 591.28 + 0.246682115(-3.6)}

\mathbf{u_2 = 591.28+ (-0.888055614)}

\mathbf{u_2 \approx 590.4 \ btu/lbm}

\mathbf{Q_{1-2}= W+ \Delta U}

\mathbf{Q_{1-2}= W+m( u_2 -u_1)}

\mathbf{\mathbf{Q_{1-2}} = 0+2.2(590.4-574.08)}

\mathbf{\mathbf{Q_{1-2}} = 0+2.2(16.32)}

\mathbf{\mathbf{Q_{1-2}} = 35.904 \  Btu}

b) the work for Process 2–3, in Btu.

At T_3 = 100 ^0 \ F ; u_3 = 577.86 \ Btu/lbm

Q_{2-3} = W_{2-3} + \Delta U

Q_{2-3} = W_{2-3} + m(u_3-u_2)

Q_{2-3} = W_{2-3} +2.2(577.86-590.4)

-98.9 = W_{2-3} +2.2(577.86-590.4)

-W_{2-3}=  2.2(577.86-590.4)+98.9

-W_{2-3}=  -27.588+98.9

-W_{2-3}=  71.312

W_{2-3}=  -71.312 \ Btu

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Naily [24]

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

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Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

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J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

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φdza = 10^-6

z = 0.1

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Where J = Density = -10^6 * z^1.5

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Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

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ρv = -0.015811388300841

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Where Velocity Charge Density = -2000 C/m3

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J = -10^6 * 0.15^1.5

J = -58094.75019311125

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Velocity = 29.05m/s

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