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22 May, 06:49

An automobile tire having a temperature of

-2.6°C (a cold tire on a cold day) is filled to a

gauge pressure of 24 lb/in^2

What would be the gauge pressure in the

tire when its temperature rises to 43◦C? For

simplicity, assume that the volume of the tire

remains constant, that the air does not leak

out and that the atmospheric pressure remains constant at 14.7 lb/in^2

Answer in units of lb/in^2

+3
Answers (1)
  1. 22 May, 08:03
    0
    Psm = 30.66 [Psig]

    Explanation:

    To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.

    P * v = R * T

    where:

    P = pressure [Pa]

    v = specific volume [m^3/kg]

    R = gas constant for air = 0.287 [kJ/kg*K]

    T = temperature [K]

    For the initial state

    P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)

    T1 = - 2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)

    Therefore we can calculate the specific volume:

    v1 = R*T1 / P1

    v1 = (0.287 * 270.4) / 266.8

    v1 = 0.29 [m^3/kg]

    As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.

    V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.

    T2 = 43 + 273 = 316 [K]

    P2 = R*T2 / V2

    P2 = (0.287 * 316) / 0.29

    P2 = 312.73 [kPa]

    Now calculating the manometric pressure

    Psm = 312.73 - 101.325 = 211.4 [kPa]

    And converting this value to Psig

    Psm = 30.66 [Psig]
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