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A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).
Correct answer is '135.08'. Can you explain this answer?
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A rigid tank of volume 50 m3 contains a pure substance as a saturated ...
To find the total mass of the liquid-vapour mixture in the tank, we need to use the given information about the composition of the mixture and the specific volumes of the liquid and vapour phases.
Given information:
- Total volume of the tank, V = 50 m^3
- Saturation temperature, Tsat = 143.61°C
- Specific volume of saturated liquid, vf = 0.001084 m^3/kg
- Specific volume of saturated vapour, vg = 0.46242 m^3/kg
- Mass fraction of liquid, Xliquid = 0.20
- Mass fraction of vapour, Xvapour = 0.80
To find the total mass of the mixture, we can use the equation:
V = Xliquid * Vliquid + Xvapour * Vvapour
where Vliquid is the volume occupied by the liquid phase and Vvapour is the volume occupied by the vapour phase.
Let's calculate the values:
1. Calculate the volume of the liquid phase:
Vliquid = Xliquid * V = 0.20 * 50 m^3 = 10 m^3
2. Calculate the volume of the vapour phase:
Vvapour = Xvapour * V = 0.80 * 50 m^3 = 40 m^3
3. Calculate the mass of the liquid phase:
mliquid = Vliquid / vf
Substituting the values, we get:
mliquid = 10 m^3 / 0.001084 m^3/kg = 9224.77 kg
4. Calculate the mass of the vapour phase:
mvapour = Vvapour / vg
Substituting the values, we get:
mvapour = 40 m^3 / 0.46242 m^3/kg = 86.49 kg
5. Calculate the total mass of the mixture:
mtotal = mliquid + mvapour
Substituting the values, we get:
mtotal = 9224.77 kg + 86.49 kg = 9311.26 kg
Since we need to round off to the nearest integer, the total mass of the mixture in the tank is approximately 9311 kg, which rounds off to 9311 kg.
Therefore, the correct answer is 9311 kg.
Given information:
- Total volume of the tank, V = 50 m^3
- Saturation temperature, Tsat = 143.61°C
- Specific volume of saturated liquid, vf = 0.001084 m^3/kg
- Specific volume of saturated vapour, vg = 0.46242 m^3/kg
- Mass fraction of liquid, Xliquid = 0.20
- Mass fraction of vapour, Xvapour = 0.80
To find the total mass of the mixture, we can use the equation:
V = Xliquid * Vliquid + Xvapour * Vvapour
where Vliquid is the volume occupied by the liquid phase and Vvapour is the volume occupied by the vapour phase.
Let's calculate the values:
1. Calculate the volume of the liquid phase:
Vliquid = Xliquid * V = 0.20 * 50 m^3 = 10 m^3
2. Calculate the volume of the vapour phase:
Vvapour = Xvapour * V = 0.80 * 50 m^3 = 40 m^3
3. Calculate the mass of the liquid phase:
mliquid = Vliquid / vf
Substituting the values, we get:
mliquid = 10 m^3 / 0.001084 m^3/kg = 9224.77 kg
4. Calculate the mass of the vapour phase:
mvapour = Vvapour / vg
Substituting the values, we get:
mvapour = 40 m^3 / 0.46242 m^3/kg = 86.49 kg
5. Calculate the total mass of the mixture:
mtotal = mliquid + mvapour
Substituting the values, we get:
mtotal = 9224.77 kg + 86.49 kg = 9311.26 kg
Since we need to round off to the nearest integer, the total mass of the mixture in the tank is approximately 9311 kg, which rounds off to 9311 kg.
Therefore, the correct answer is 9311 kg.
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Community Answer
A rigid tank of volume 50 m3 contains a pure substance as a saturated ...
Given : V = 50 m3 , P = 400 kPa
Liquid = 20 % , Vapor = 80%
vf = 0.001084 m3 / kg
vg = 0.46242 m3 /kg
v = vf + x(vg − vf)
v = 0.001084+ 0.8 (0.46242 − 0.001084)
v = 0.37015 m3 / kg
v = V/m
= 135.8kg
Liquid = 20 % , Vapor = 80%
vf = 0.001084 m3 / kg
vg = 0.46242 m3 /kg
v = vf + x(vg − vf)
v = 0.001084+ 0.8 (0.46242 − 0.001084)
v = 0.37015 m3 / kg
v = V/m
= 135.8kg
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A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. Can you explain this answer?
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A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. 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 A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. 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 A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. Can you explain this answer?.
A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. 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 A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. 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 A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : Saturation temperature, Tsat = 143.61oC; Specific volume of Saturated liquid, vf = 0.001084 m3 /kg; Specific volume of saturated vapour, vg = 0.46242m3/kg . The total mass of liquid vapour mixture in the tank is _______ kg (round off to the nearest integer).Correct answer is '135.08'. Can you explain this answer?.
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