- #1
Baartzy89
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Hi everybody,
I've got a thermodynamics exam coming in a few weeks. I couldn't sit the exam when I took the subject 12 months ago because of work commitments so I'm sitting a deferred exam this semester. Trouble is I have forgotten most of it, and having difficulty getting help from lecturers and working it out myself. Below is a question with my answers, but I know I've made errors...just needing a bit of direction...thanks!
Question 1: A 1.8m3 rigid tank contains steam at 220°C. One third of the volume is in the liquid phase and the rest is in the vapor form. Determine:
a)the pressure of the steam, b)the quality of the saturated mixture, and c) the density of the mixture.
a)From steam table A-4, saturated vapor at 220°C has a pressure of 2319.6
kPa, or 2.312 MPa.
b)If volume of the tank is 1.8m3 and one third is liquid, two thirds are vapor:
1.2m3 is vapor 0.6m3 is liquid
from the steam tables the specific volume for the two states are: vf = 0.001190, vg = 0.086094
Therefore the mass of each phase must equal the volume of the phase " multiplied " by its specific volume:
mf = 0.6 x 0.001190 = 7.14 x 10-4, mg = 1.2 x 0.086094 = 0.1033128
Quality
= mg / mtotal, where mtotal = mg + mf
=7.14x10-4 +0.1033128
=0.1040268 = 0.1033128 / 0.1040268
= 0.993136384 or .99 or 99%
c) Using density formula: Density = Mass / Volume, and mtotal = 0.1040268 and V = 1.8= 0.1040268 / 1.8 = 0.057792666
= 0.058 kg/m3
I've got a thermodynamics exam coming in a few weeks. I couldn't sit the exam when I took the subject 12 months ago because of work commitments so I'm sitting a deferred exam this semester. Trouble is I have forgotten most of it, and having difficulty getting help from lecturers and working it out myself. Below is a question with my answers, but I know I've made errors...just needing a bit of direction...thanks!
Question 1: A 1.8m3 rigid tank contains steam at 220°C. One third of the volume is in the liquid phase and the rest is in the vapor form. Determine:
a)the pressure of the steam, b)the quality of the saturated mixture, and c) the density of the mixture.
a)From steam table A-4, saturated vapor at 220°C has a pressure of 2319.6
kPa, or 2.312 MPa.
b)If volume of the tank is 1.8m3 and one third is liquid, two thirds are vapor:
1.2m3 is vapor 0.6m3 is liquid
from the steam tables the specific volume for the two states are: vf = 0.001190, vg = 0.086094
Therefore the mass of each phase must equal the volume of the phase " multiplied " by its specific volume:
mf = 0.6 x 0.001190 = 7.14 x 10-4, mg = 1.2 x 0.086094 = 0.1033128
Quality
= mg / mtotal, where mtotal = mg + mf
=7.14x10-4 +0.1033128
=0.1040268 = 0.1033128 / 0.1040268
= 0.993136384 or .99 or 99%
c) Using density formula: Density = Mass / Volume, and mtotal = 0.1040268 and V = 1.8= 0.1040268 / 1.8 = 0.057792666
= 0.058 kg/m3