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2020 Sept14 ME004A

The document discusses the ideal gas equation and several gas laws. It provides two example problems: 1) Calculating the mass of air in a room given temperature, pressure, and volume. The mass is found to be 703.51 kgm. 2) Calculating the mass of air in a tire given temperature, pressure, and volume. The mass is found to be 0.50 lbm. It then provides a third problem: given an increased temperature, calculating the resulting percentage increase in gauge pressure for the tire, assuming an inflexible volume. The percentage increase is found to be 57.56%.

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Daniel Manivough
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201 views11 pages

2020 Sept14 ME004A

The document discusses the ideal gas equation and several gas laws. It provides two example problems: 1) Calculating the mass of air in a room given temperature, pressure, and volume. The mass is found to be 703.51 kgm. 2) Calculating the mass of air in a tire given temperature, pressure, and volume. The mass is found to be 0.50 lbm. It then provides a third problem: given an increased temperature, calculating the resulting percentage increase in gauge pressure for the tire, assuming an inflexible volume. The percentage increase is found to be 57.56%.

Uploaded by

Daniel Manivough
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Ideal Gas Equation

𝑃𝑉 = 𝑚𝑅𝑇
𝑤ℎ𝑒𝑟𝑒:
𝑃 = 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔
𝑉 = 𝑉𝑜𝑙𝑢𝑚𝑒 [𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑜𝑟 𝑆𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 ]
𝑉𝑐𝑦𝑙 = 𝜋𝑟 2 ℎ
4 3
𝑉𝑠 = 𝜋𝑟
3
𝑚 = 𝑚𝑎𝑠𝑠
𝑅 = 𝐺𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇 = 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 [𝐾 𝑜𝑟 °𝑅]
𝐾 = ℃ + 273
°𝑅 = ℉ + 460
Gas Constant, R
Ṙ
𝑅=
𝑀𝑊
𝑤ℎ𝑒𝑟𝑒:
Ṙ = 𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝐺𝑎𝑠 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑘𝐽
Ṙ = 8.314
𝑘𝑔𝑚𝑜𝑙 ∙ 𝐾
𝑙𝑏𝑓 ∙ 𝑓𝑡
Ṙ = 1545
𝑙𝑏𝑚𝑜𝑙 ∙ °𝑅
𝑀𝑊 = 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡

Problems:
1. Determine the mass of air in kgm inside
a room with a temperature of 22°C and a
pressure of 101.325 kPa. The dimensions
of the room is 12m x 14m x 3.5m.
Given:
T = 22°C
P = 101.325 kPaa
V = 12m x 14m x 3.5m
Required:
m = ? kgm
Solution:
𝑃𝑉 = 𝑚𝑅𝑇
𝑃𝑉
𝑚=
𝑅𝑇
𝑤ℎ𝑒𝑟𝑒:
𝑃 = 101.325 𝑘𝑃𝑎
𝑉 = [12][14][3.5]𝑚3 = 588𝑚3
𝑘𝐽
𝑅𝑎𝑖𝑟 = 0.28708
𝑘𝑔𝑚 ∙ 𝐾
𝑇 = 22℃ + 273 = 295𝐾
𝑘𝑁
[101.325 2 ] [588𝑚3 ]
𝑚= 𝑚
𝑘𝑁 ∙ 𝑚
[0.28708 ] [295𝐾 ]
𝑘𝑔𝑚 ∙ 𝐾
𝒎 = 𝟕𝟎𝟑. 𝟓𝟏 𝒌𝒈𝒎 (𝑨𝑵𝑺𝑾𝑬𝑹)

2. A tire contains 3730 in3 of air at 32 psig


and 80°F. Determine the mass of air in the
tire in lbm.
Given:
V = 3730 in3
P = 32 psig
T = 80°F
Required:
m = ? lbm
Solution:
𝑃𝑉 = 𝑚𝑅𝑇
𝑃𝑉
𝑚=
𝑅𝑇
𝑤ℎ𝑒𝑟𝑒:
𝑃 = 32 𝑝𝑠𝑖𝑔
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔
𝑃 = 14.7 𝑝𝑠𝑖 + 32 𝑝𝑠𝑖 = 46.7 𝑝𝑠𝑖
𝑉 = 3730𝑖𝑛3
𝑙𝑏𝑓 ∙ 𝑓𝑡
𝑅𝑎𝑖𝑟 = 53.34
𝑙𝑏𝑚 ∙ °𝑅
𝑇 = 80℉ + 460 = 540°𝑅

𝑙𝑏𝑓 3 1 𝑓𝑡
[46.7 2 ] [3730𝑖𝑛 ] [12 𝑖𝑛]
𝑖𝑛
𝑚=
𝑙𝑏𝑓 ∙ 𝑓𝑡
[53.34 ] [540°𝑅 ]
𝑙𝑏𝑚 ∙ °𝑅
𝒎 = 𝟎. 𝟓𝟎 𝒍𝒃𝒎 (𝑨𝑵𝑺𝑾𝑬𝑹)
Condition:
Processes like Compression/Expansion, or
Constant Volume Process, etc.
Initial state >> Final state
Point 1 >> Point 2
If mass is constant for both state;
𝑚1 = 𝑚2
𝑃𝑉
𝑚=
𝑅𝑇
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑅1 𝑇1 𝑅2 𝑇2
𝑅1 = 𝑅2
Combined Gas Law
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
Constant Temperature Process (T=C)
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
𝑇1 = 𝑇2
𝑃1 𝑉1 = 𝑃2 𝑉2 [𝐵𝑜𝑦𝑙𝑒 ′ 𝑠 𝐿𝑎𝑤 ]
Constant Pressure Process (P=C)
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
𝑃1 = 𝑃2
𝑉1 𝑉2
= [𝐶ℎ𝑎𝑟𝑙𝑒𝑠 ′ 𝐿𝑎𝑤 ]
𝑇1 𝑇2
Constant Volume Process (V=C)
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
𝑉1 = 𝑉2
𝑃1 𝑃2
= [𝐺𝑎𝑦 − 𝐿𝑢𝑠𝑠𝑎𝑐 ′ 𝑠 𝐿𝑎𝑤]
𝑇1 𝑇2
Example:
3. From problem number 2, if the air
temperature increases to 145°C, what is the
resulting increase in percentage in gauge
pressure assuming that the tire is inflexible?
Given from problem 2:
V1 = 3730 in3 = V2
P1 = 32 psig
T1 = 80°F
m1 = 0.503961889 lbm
Given from number 3:
T2 = 145°C
Required:
P2 – P1 = ΔP in percentage
Solution:
∆𝑃
%∆𝑃 = × 100%
𝑃1
𝑤ℎ𝑒𝑟𝑒:
∆𝑃 = 𝑃2 − 𝑃1
 Increase in temperature [T2]
 Inflexible [Volume is constant]
From Combined Gas Law:
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
𝑃1 𝑃2
=
𝑇1 𝑇2
𝑃1 𝑇2
𝑃2 =
𝑇1
𝑤ℎ𝑒𝑟𝑒:
𝑃1 = 32 𝑝𝑠𝑖𝑔
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔
𝑃1 = 14.7 𝑝𝑠𝑖 + 32 𝑝𝑠𝑖 = 46.7 𝑝𝑠𝑖
𝑇1 = 80℉ + 460 = 540°𝑅
𝑇2 = 145℃
9
℉ = ℃ + 32
5
9
°𝑅 = [145℃] + 32 + 460
5
𝑇2 = 753°𝑅
[46.7𝑝𝑠𝑖 ][753°𝑅 ]
𝑃2 =
540°𝑅
𝑃2 = 65.12055556 psi [𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒]
𝑃𝑔 = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚
𝑃2 = 65.12055556 𝑝𝑠𝑖 − 14.7𝑝𝑠𝑖
𝑃2 = 50.42055556 psig
𝑃2 − 𝑃1
%∆𝑃 = × 100%
𝑃1
[50.42055556 − 32]psig
%∆𝑃 = × 100%
32𝑝𝑠𝑖𝑔
%∆𝑷 = 𝟓𝟕. 𝟓𝟔% (𝑨𝑵𝑺𝑾𝑬𝑹)

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