Thermodynamics of a Tire Pump

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1 1 roblem Thermodynamics of a Tire ump Kirk T. McDonald Joseph Henry Laboratories, rinceton University, rinceton, NJ 8544 November 9, 7; updated September 9, 8 lthough a tire pump is a simple device, making a satisfactory, simple thermodynamic model of its behavior is challenging. Suppose the tire has volume V that remains constant as its absolute pressure is increased by the tire pump from,where = atmospheric pressure, to a final pressure at the ambient absolute temperature T. The process of pumping may be assumed to be adiabatic, so the pressure and temperature in the tire reach maximum values of max and T max before the tire cools slowly back to temperature T and the pressure drops to the desired pressure. You may assume that air is a diatomic ideal gas. First discuss how a tire pump at a gas station works, before turning to the slightly more complicated case of a hand pump. t a gas station, a compressor first compresses a lot of air to some desired pressure C and stores this in a tank. Before this air is injected into a tire, it comes back to room temperature T. You may ignore the details of the thermodynamics of the compressor. Then, we connect the storage tank to the tire via a one-way valve. So long as the pressure in the tank is greater than the pressure in the tire, air flows from the tank into the tire. The simple model is that the pressure in the compressor s tank stays constant during this process. Solution ThetireofvolumeV must end up with n = V/RT moles of air assumed to be an ideal gastobeatpressure at ambient temp T,whereR is the gas constant. Suppose the tire has pressure to start with. Then it already contains n = n / = V/RT moles of air. So, we must add n = n n moles from the tank into the tire, which we can also express as n = n. 1.1 The Tire Is umped by a Compressor The volume V of these n moles in the tank at pressure C and temperature T is V = n RT = nrt. C C s air flows from the tank into the tire through the one-way valve, the tank does work W C at constant by assumption pressure C given by W C = C V = nrt 3 1

2 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 W C assuming that the air transfer is adiabatic. Therefore, the temperature of the air in the tire rises to T max = T +ΔT,wherenC V ΔT = ΔU = W C. Here, C V =5R/ is the molar specific heat of air at constant volume. Thus, ΔT = W C = RT nc V C V = 5 and T max = T +ΔT = 1+ 5 s a result, the pressure in the tire rises temporarily to max = nrt max V 1 1 = T max = 1+ T 5 T, 4 ] T. 5 1 ]. 6 fter a while, the air cools down to temperature T, and the pressure will be the desired value. However, the filling process won t work if max > C. Example: =1atm, = 3 atm these are absolute pressures, not gauge pressures, and T = 3K = 3C. Then, max =1..7 =3.8 atm,andt max =1.7T = 38K = 17C. The pressure in the compressor must be greater than 3.8 atm.8 atm gauge pressure to be able to pump the tire up to 3 atm atm gauge pressure quickly.. The Tire Is umped by a Hand ump Now we turn to the case of a hand pump, which has a working volume V p that is small compared to the volume V of the tire. The number n p of moles or air delivered in each stroke of the pump is n p = V p n, 7 RT where = atmospheric pressure. So, a large number N of strokes of the hand pump will be required, N = n = V, 8 n p V p recalling eq. 1, and each stroke will make only a tiny difference in the pressure of the tire. t the beginning of the ith stroke, the tire has pressure i 1,wherefori =1, =initial pressure in the tire. My model is that during the ith stroke the pressure in the pump increases until it reaches pressure i, the pressure in the tire at the end of the stoke, after which the pump pressure is held constant at this value while the air in the pump is transferred into the tire.

3 I will again assume that all this happens quickly i.e., adiabatically. We need to know the volume V p of air in the pump when it first reaches pressure i.for, this, we use the law of adiabatic compression of an ideal gas: V γ = V 7/5 = constant, where γ = C /C V =7R//5R/ = 7/5 for air. Thus, 5/7 V p = V p. 9 i The temperature of the air in the pump has now risen to T i V /7 p i = T = T. 1 V p The temperature rise of the air in the pump is /7 ΔT = T i T = T 1]. 11 To accomplish this, we must have done work W equal to the increase in the internal energy of the air in the pump, namely W = n p C V ΔT = V p 5R /7 RT T i 1] = 5 /7 i V p 1]. 1 Next, we push the hot gas of volume V p into the tire at constant pressure i. This requires that we do additional work 5/7 /7 W = i V i = i V p = V p. 13 i So, the total work required to accomplish the ith stroke is W i = W + W = V p 7 /7 i 5 ]. 14 Not only is the energy W i added to that of the gas in the tire during the ith stroke, but also the initial energy U p = C V n p RT =5/ V p of the gas in the pump. Thus, the total energy added to the gas during the ith stroke is ΔU i = W i + 5 V p = 7 /7 i V p = 7 5/7 /7 i V p. 15 The internal energy U of volume V of air at pressure is related by U = C V nrt = C V V = 5 V, 16 so the change in pressure Δ i in the tire during the ith stroke is related by Δ i = 5 ΔU i V = 7 5 5/7 3 /7 V p i V. 17

4 s Δ i is small, we treat eq. 17 as a differential equation, which integrates to or V p d i = 7 /7 5 5/7 V 5/7 = i = 5/7 + 5/7 V p 5/7 + 5/7 V p di, 18 V i, 19 7/5 V i. fter the N strokes needed, according to eq. 8, to fill the tire so that it has pressure when it returns to temperature T, the maximum pressure in the tire is max = 5/7 + 5/7 7/5 = 1+ 5/7 ] 7/5, 1 which is independent of the volume V p of the hand pump. The maximum temperature of the air in the tire is T max = max T, The final internal energy of the air in the tire at temperature T max isu f =5/ max V, while the initial internal energy of the air in the tire plus the N volumes V p of pumped air is U =5/ V + N V p. Hence, the total work done while pumping is W = U f U = 5 maxv V N V p = 5 5/7 ] 7/5 V Example: = =1atm, =3atm,V =.1 m 3, V p =.1 m 3,andT = 3K. Then, N = strokes, max =4.65 atm, T max = 465K = 19C, and W 41, J. To pump up the tire in 1 minutes requires 115 W of power delivered into the air of the tire. n estimate of the efficiency of the engine that provides this power is beyond the scope of this problem...1 Entropy Change The free expansion during filling of the tire, and the cooling at constant volume after filling, are both irreversible processes, so we expect that the entropy of the system has increased. In classical thermodynamics, the entropy S of an ideal gas can be calculated only up to an additive constant. Namely, n moles of gas at temperature T in a volume V have entropy where is an undetermined constant. S = nr +ln VTC V n 4, 4

5 In the present example, n = V/RT moles of air end up in the tire of volume V at pressure and temperature T. Initially, this air was also at temperature T,withn = n / moles in volume V at pressure,andn = n / moles at pressure where they occupied volume NV p = V /, recalling eqs. 1 and 8. Hence, the change in entropy of the air that ends up inside the tire during the pumping is ΔS tire = nr ln VT5/ n R ln VT5/ n R ln V T 5/ n n n = nr ln 1 n n R ln n R ln = n R ln n R ln n n = nr ln + ln = nr ln + ln, 5 which is negative. However, this is not the total entropy change of the Universe, because as the tire cooled from temperature T max back to T,heatΔQ, givenby ΔQ = C V nrt max T = 5 nrt max 1, 6 was transferred into the environment at temperature T, so the entropy of the environment increased by ΔQ/T. Thus, the total entropy change of the Universe during the pumping is 5 ΔS = nr 1+ 5/7 ] 7/5 5 ln ln. 7 For the example with = 1 =1atmand =3atm,ΔS =.8nR. 5

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