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Saturated liquid at a higher pressure P1 having h11 = 1000 kJ/kg is throttled to a lower pressure P2 . The enthalpy of saturated liquid and saturated vapour are 800 kJ/kg and 2800 kJ/kg respectively. Find the dryness fraction of vapour after throttling.
  • a)
    0.1
  • b)
    0.2
  • c)
    0.8
  • d)
    0.9
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Saturated liquid at a higher pressure P1 having h11 = 1000 kJ/kg is th...
In throttling process enthalpy remains constant.
h1 = h2
1000 = 800 + x(2800 — 800)
x = 0.1.
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Most Upvoted Answer
Saturated liquid at a higher pressure P1 having h11 = 1000 kJ/kg is th...
To find the dryness fraction of the vapor after throttling, we can use the First Law of Thermodynamics for a control volume. The equation is given as:

Δh = h2 - h1 = (v2^2 - v1^2) / 2 + g(z2 - z1) + Q - W

Where:
Δh = change in enthalpy
h2 = enthalpy of the saturated vapor
h1 = enthalpy of the saturated liquid
v2 = specific volume of the saturated vapor
v1 = specific volume of the saturated liquid
g = acceleration due to gravity
z2 = height of the control volume at P2
z1 = height of the control volume at P1
Q = heat transfer
W = work done

In this case, the system is undergoing throttling, which means there is no heat transfer or work done. Therefore, the equation simplifies to:

Δh = h2 - h1 = (v2^2 - v1^2) / 2 + g(z2 - z1)

Since the system is at a higher pressure P1 initially, the saturated liquid at P1 is throttled to a lower pressure P2. This means that the specific volume of the saturated liquid remains the same, but the specific volume of the saturated vapor changes.

Given:
h1 = 1000 kJ/kg
v1 = specific volume of saturated liquid = constant
h2 = 2800 kJ/kg
v2 = specific volume of saturated vapor after throttling

Substituting the given values into the equation:

2800 - 1000 = (v2^2 - v1^2) / 2

Rearranging the equation:

v2^2 = 2 * (2800 - 1000) + v1^2

v2^2 = 3600 + v1^2

Now, we can use the definition of dryness fraction (x) to find the specific volume of the mixture:

v = x * v_g + (1 - x) * v_f

Where:
v = specific volume of the mixture
x = dryness fraction
v_g = specific volume of the saturated vapor
v_f = specific volume of the saturated liquid

Given:
v_f = constant
v_g = v2 (specific volume of saturated vapor after throttling)

Substituting the given values into the equation:

v = x * v2 + (1 - x) * v1

Since v1 is constant, the equation simplifies to:

v = x * v2 + v1 - x * v1

Now, we can substitute this equation into the previous equation:

v2^2 = 3600 + (x * v2 + v1 - x * v1)^2

Expanding and rearranging the equation:

v2^2 = 3600 + x^2 * (v2 - v1)^2 + v1^2 - 2 * x * v1 * (v2 - v1)

Simplifying the equation:

v2^2 - x^2 * (v2 - v1)^2 - v1^2 + 2 * x * v1 * (v2 - v1) - 3600 = 0

This equation is a quadratic equation in terms of x. We
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Saturated liquid at a higher pressure P1 having h11 = 1000 kJ/kg is throttled to a lower pressure P2 . The enthalpy of saturated liquid and saturated vapour are 800 kJ/kg and 2800 kJ/kg respectively. Find the dryness fraction of vapour after throttling.a)0.1b)0.2c)0.8d)0.9Correct answer is option 'A'. Can you explain this answer?
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