Thermodynamics Question and My Answer

  • Thread starter Thread starter Altairs
  • Start date Start date
  • Tags Tags
    Thermodynamics
Click For Summary
SUMMARY

The discussion addresses a thermodynamics problem involving the inflation of car tires under varying temperatures. The driver inflates the tire to 38 PSI (Gauge) at 77°F, and after a temperature increase of 36°F, the final pressure is calculated to be 41.54 PSI (Gauge). Using the ideal gas law and the relationship between pressure and temperature, it is confirmed that the tire will not burst, as the pressure remains below the maximum allowed limit of 44 PSI (Gauge).

PREREQUISITES
  • Understanding of the Ideal Gas Law
  • Knowledge of pressure conversions from PSI to lbf/ft²
  • Familiarity with temperature conversions from Fahrenheit to Rankine
  • Basic principles of thermodynamics
NEXT STEPS
  • Study the Ideal Gas Law in detail, focusing on its applications in real-world scenarios.
  • Learn about pressure measurement units and conversion techniques, specifically between PSI and lbf/ft².
  • Explore the effects of temperature changes on gas behavior, particularly in closed systems.
  • Investigate safety standards and maximum pressure limits for various tire types.
USEFUL FOR

Students studying thermodynamics, automotive engineers, and anyone involved in tire safety and pressure management.

Altairs
Messages
125
Reaction score
0
[SOLVED] Thermodynamics Question and My Answer

Homework Statement



During a cloudy day of the summer season, a driver inflates his car tyres (max allowed pressure is 44PSI(Gauge)) to 38PSI(Gauge) at a petrol station. the temperature of the air inside one of the tires is [tex]77^oF[/tex] and volume of the tire is [tex]1ft^3[/tex]. Later, the sun comes out and the air inside the tire experience a temperature rise of [tex]36^oF[/tex]. Assuming the volume to be constant determine whether the tire will burst or not.

Homework Equations



P(Absolute) = P(Ambient) + P (Gauge)
Conversion from F to R
Conversion from PSI to [tex]\frac{lbf}{ft^2}[/tex]

The Attempt at a Solution



First did the relevant conversions.

Then used :-

[tex]P_{1}v = RT_{1}[/tex]

[tex]P_{2}v = RT_{2}[/tex]

Where [tex]R[/tex] = Individual Gas Constant

Where [tex]v[/tex] = Specific Volume

Took the ratios, because the volume remains constant and so does the value of R :-

[tex]\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}[/tex]

Made relevant calculations and then did reverse conversions. Got the answer to be 41.54 PSI (Gauge). Which indicates that tire doesn't bursts. Is this the correct answer ?
 
Last edited:
Physics news on Phys.org
Yes.
 

Similar threads

Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 67 ·
3
Replies
67
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K