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3 years ago

How to solve this question

Engineering

1 answer:

Harrizon [31]3 years ago

3 0

Answer:

Voltage across element X is 49.26 volts

Explanation:

The kirchhoff's voltage law states that the algebraic sum of voltages around a closed loop is equal to zero.

Applying the kirchhoff's voltage law in loop yields,

-160 + V₁ + V₂ + V₃ + Vₓ + 60 + V₄ = 0

-160 + 60 + V₁ + V₂ + V₃ + Vₓ + V₄ = 0

But we know that V = IR

-100 + IR₁ + IR₂ + IR₃ + IRₓ + IR₄ = 0

IR₁ + IR₂ + IR₃ + IRₓ + IR₄ = 100

I(R₁ + R₂ + R₃ + Rₓ + R₄) = 100

I = 100/(R₁ + R₂ + R₃ + Rₓ + R₄)

Substitute the given resistor values

I = 100/(18k + 20k + 45k + 100k + 20k)

I = 100/203k

I = 100/203×10³

I = 0.0004926 A

The voltage drop at element x is

Vₓ = IRₓ

Vₓ = 0.0004926*100×10³

Vₓ = 49.26 V

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Answer:

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Explanation:

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Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

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m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

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3 years ago

A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti

natima [27]

Answer:

Final mass of Argon= 2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

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k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

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Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

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<em>Note: Calculation for T2 is attached below</em>

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