Question

You are pumping up a road bicycle tire with a compressor. The inner tube is initially completely deflated, and you inflate it to a final pressure of 120psig and final volume confined by the tire tread and rim (Vtire= 500mL). The compressor supplies air steadily at 24°C and 190 psig. Assume air is an ideal gas, with CP= (7/2)Rand R= 8.314 J/mol·K.Clearly state all assumptions.Final answers should be appropriately rounded.

a.Define an opensystem.

b.Calculate the work of expansion ofthe inner tube(against the atmosphere).Assume negligibletensile force from the inner tube

.c.Starting with ageneral energy balance, calculate the final temperature inside the tire. Assume thatthe time to inflate the tire is much less than the time for heat dissipation through the tire and rim.

d.For this process, it is possible for the final temperature to be less than the compressor supply temperature? Explain briefly.e.After having inflated the tire,you let the bicycle stand outside, where it is heated under the sun. At what temperature will the tire burst?(The tire is designedfor 180 psig).

Answer 1

a) Open system is Compressor inner tube in tire of bicycle.

b) Work

Pressure (P) is 190 psig = 1.31e+6

Initial volume (V1) is 0 ml. Final volume is (V2) is 500 ml.

c)

The tire of volume V must end up with n = P V/RT₀ moles of air (assumed to be an ideal gas) to be at pressure P at ambient temp T₀.

Suppose the tire has pressure P₀ to start with, then it already contains n₀ = n(P₀/P) = P₀V/RT₀ moles of air.

So, we must add n' = n − n₀ moles from the tank into the tire, which we can also express as

n' = n(P − P₀)/P

The volume V' of these n' moles in the tank at pressure P_c and temperature T₀ is,

V = n'RT₀/P_c = nRT₀*[(P − P₀)/PP_c]

As air flows from the tank into the tire through the “one-way” valve, the tank does work W_c at constant (by assumption) pressure P_c given by

W_c = P_cV' = nRT₀[(P − P₀)/P],

on the n' moles of air. This work appears as internal energy of the air that is injected into the tire.

Given the assumption that air is an ideal gas, the motion of air through the “one-way” valve into the tire is a free expansion, and no work is done in the latter process. So, the internal energy U of the air in the tire rises by amount WC (assuming that the air transfer is adiabatic).

Therefore, the temperature of the air in the tire rises to Tmax = T0+ΔT , where nC_vΔT = ΔU = W_c. Here, C_v is the molar specific heat of air at constant volume. Thus,

ΔT = W_c/nC_v = (RT₀/C_v)[(P − P₀)/P]

In the given case,

P₀ = P_atm = 14.7 psia (since the tire is completely deflated)

P = 190 psig = 190 + P_atm psia = 204.7 psia

C_p = 7R/2 => C_v = C_p - R = 5R/2

T₀ = 24⁰C = 297 K

Therefore,

ΔT = (2/5)[1 − (P₀/P)]T₀,

ΔT = 110.27 K

Therefore, the temperature in the tire rises to

T_max = T₀ + ΔT = 407.27 K

After some time the air cools down to T₀ and pressure will be the desired value.

d)

No, final temperature will always be greater than T₀(compressor supply temperature)

e)

Final pressure will be 180 psig. After that tire will be burst.