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4 years ago

14

Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but d

oes not allow either gas to leak into the other side. Initially, one side of the piston contains 1m3 of N2 gas at 500 kPa and 160 degrees C while the other side contains 1m3 of He gas at 500 kPa and 60 degrees C. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move?

1 answer:

inn [45]4 years ago

3 0

Answer:

- the final equilibrium temperature in the cylinder is 85.67 °C

- Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.

Explanation:

Given That;

Compartment Nitrogen:

Volume V1 = 1 m³, Pressure P1 = 500 kPa , Temperature T = 120°C

Compartment of helium:

Volume V1H = 1 m³, Pressure P1H = 500 kPa, Temperature T1H = 40°C

From the ideal specific heat of gases

-Nitrogen

The gas k (constant) and the k (constant) volume specific heats are;

R = 0.2968 kJ

/kg.K

Cv = 0.743 kJ/kg.K

also From the ideal specific heat of gases

-helium

R = 2.0769 kJ /kg.K

Cv = 3.1156 kJ/kg.K

we know that

PV = mRT

Mass of the nitrogen

mN2 = (P1V1 /RT1)_N2

mN2 = (500)(1) / (0.2968)(393)

= 4.29kg

mass of helium

mHe = (P1HVIH /RT1H)_He

mHe = (P1V1 /RT1)_N2

mHe = (500)(1) / (2.0769)(313)

= 0.769kg

Taking the whole contents of the cylinder,

the 1st law relation can be expressed as;

Ein - Eout = ΔEsystem

0 = ΔU = (ΔU)_N2 + (ΔU)_He

0 = [mcV(T2 -T1)]_N2 + [mcV(T2 - Ti)]He

Since T2 =T_F

0 = [mcV(T_F -T1)]_N2 + [mcV(T_F - Ti)]He

we Substitute

(4.29)(0.743)(T_F - 120) + (0.769)(3.1156)(T_F - 40) = 0

T_F = 85.67 °C

therefore the final equilibrium temperature in the cylinder is 85.67 °C

Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.

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Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential

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This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/( $A^{*}$ ₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃ in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403 $e^{j51.34 }$

c) value of A₃ in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937 $e^{j81.34 }$

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁ = 5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( $\frac{5}{4}$ )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A $e^{j\beta }$

so, A₂ in exponential form will be;

A₂ = 6.403 $e^{j51.34 }$

exponential form of A₂ = 6.403 $e^{j51.34 }$

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( $\frac{8 }{9.196}$ )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°

A₃ = 12.19∠41.02°

Therefore, value of A₃ in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/( $A^{*}$ ₂)

giving your answer in exponential form

we know that $A^{*}$ ₂ is the complex conjugate of A₂

so

$A^{*}$ ₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A $e^{j\beta }$

$A^{*}$ ₂ = 6.403 $e^{-j51.34 }$

also

A₁ = 6∠30

we convert to polar form

A₁ = 6 $e^{j30 }$

so A₅ = A₁/( $A^{*}$ ₂)

A₅ = 6 $e^{j30 }$ / 6.403 $e^{-j51.34 }$

A₅ = (6/6.403) $e^{j(30+51.34) }$

A₅ = 0.937 $e^{j81.34 }$

Therefore A₅ in exponential form is 0.937 $e^{j81.34 }$

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3 years ago