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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?
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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.
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
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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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? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 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? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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