Ammonia
exists as a saturated mixture at 240.21 kPa and -14.6oC in a rigid
vessel with a volume of 1.0
m3. The specific volume of the saturated liquid and saturated vapor are 1.5195 L/kg and 0.50063 m3/kg, respectively. |
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The quality of the ammonia is 0.275 kg vap/kg. What is the total mass of Ammonia
inside the vessel in kg? |
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Read : |
There are two keys to
this problem. The first is the relationship between the total mass, the total
volume and the overall, average or mixture specific volume. The other key is
how to use quality and specific properties of saturated liquid and saturated vapor
to determine a specific property of a saturated mixture. |
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Given: |
P |
240.21 |
kPa |
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Vsat liq |
1.5195 |
L/kg |
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T |
-14.6 |
oC |
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Vsat vap |
0.50063 |
m3/kg |
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Vtotal |
1 |
m3 |
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x |
0.275 |
kg vap/kg |
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Find: |
Mtotal |
??? |
kg |
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Assumptions: |
None. |
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Equations
/ Data / Solve: |
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Let's begin with the
relationship between mass, volume and specific volume for the entire system. |
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Eqn 1 |
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We want to determine Mtotal and we know Vtotal, so all we need to
do is determine Vmix and we will be able to use Eqn 1 to solve this problem. |
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The specific volume of
the two-phase mixture is
related to the quality
and the specific volumes of the saturated liquid and saturated vapor by the following equation. |
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Eqn 2 |
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We know the values of
all of the variables on the right-hand side of Eqn 2, so we can plug-in values to determine Vmix. |
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Vmix |
0.1388 |
m3/kg |
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Be careful with the
units in Eqn 2. You must
convert L to m3 in Vsat
liq to make all of the units work out
properly. |
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Now, we can plug
values into Eqn 1 to complete
this problem. |
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Mtotal |
7.2059 |
kg |
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Verify: |
There are no assumptions
to verify in this problem. |
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Answers : |
Mtotal |
7.206 |
kg |
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