Piston-Cylinder Device

Page 1 4-33 A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa and 25oC. The mass of the piston is such that 500 kPa of pressure is required to raise it. How much heat must be added to the helium before the piston starts rising. In this problem, we are given the mass, m = 0.5 kg, which we assume is constant. We are also given the initial pressure and temperature, P1 = 100 kPa and T1 = 25oC = 298.15 K. The fact that the piston is resting on stops as heat is added means that the volume is constant. Because we stop adding heat when the pressure is just enough to raise the piston, we know that the final pressure, P2 = 500 kPa, the pressure required to raise the piston. We also know that V2 = V1 because of the constant volume process. We will assume that helium is an ideal gas at these low pressures. We can find the heat from the first law: Q = ∆U + W = m∫cvdT + W for an ideal gas. In this constant-volume process the work, W = 0. From Figure 2-68 on page 102, we see that the heat capacity of helium is essentially constant over a wide range of temperatures. Thus we can use the constant heat capacity of helium from Table A-2 on page 825: cv = 3.1156 kJ/(kg∙K). The same table gives the gas constant for helium; R = 2.0769 kJ/(kg∙K). From the equation that Q = m∫cvdT + W, we see that Q = mcv(T2 – T1) in this (W = 0) case. We do not know the final temperature, T2, but we can find it from the ideal gas law (and V2 = V1). T2 = P2V2 P2V1 = = mR mR P2 mRT1 P1 PT (500 kPa )( 298.15 K ) = 2 1 = = 1490 K mR P1 (100 kPa ) We can now find the heat transfer.  2.0769 kJ Q = mcv (T2 − T1 ) = (0.5 kg )  kg ⋅ K 4-36  (1490 K − 298.15 K ) = 1,857 kJ .  A mass of 15 kg of air in a piston-cylinder device is heated from 25 to 77oC by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process and a heat loss of 60 kJ occurs. Determine the electrical energy supply in kWh. The first law for this problem can be written as Q = ∆U + W = ∆U + We,out + ∫PdV, where We,out is the electrical work done by the system. We expect that this will be negative since we see tghat there is an electrical energy input to the system. We can say that We,in = -We,out = ∆U + ∫PdV – Q. For this constant pressure process, ∫PdV = P(V2 – V1) where P = P1 = P2 is the constant pressure of 300 kPa. With this work term we can write our first law as follows. We,in = ∆U + P(V2 – V1) – Q = U2 – U1 + P2V2 – P1V1 – Q = H2 – H1 – Q = m(h2 – h1) – Q We can find the specific enthalpy change, h2 – h1 = ∫cpdT, where cp can be found from Table A2(b) on page 826. At the average temperature of (25oC + 77oC)/2 = 51oC = 324.15 K, we can interpolate to find cp = 1.0065 kJ/(kg∙K). With this value of cp (and using Q = -60 kJ because it is a heat loss) we can find the electrical work input as follows. We,in = h2 − h1 − Q = mc p (T2 − T1 ) − Q =   1.0065 kJ (15 kg )  kg ⋅ K   1 kWh  (350.15 K − 298.15 K ) − ( −60 kJ )   3600 kJ We,in = 0.235 kWh . Page 2 4-40 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 6,500 kJ/h, the air temperature in the room remains constant, even though the heater operates continuously. Determine the power rating of the heater in kW. Since the temperature of the air is constant, the internal energy is constant. Thus the heat loss through the wall is balanced by the heat input from the electric heater. This input must be 6500 kJ/h = (6500 kJ/h)[(1 kW∙h) / (3600 kJ) ] = 1.81 kW . 4-41E A piston-cylinder device contains 3 ft3 of air at 60 psia and 150oF. Heat is transferred to the air in the amount of 40 Btu as the air expands isothermally. Determine the amount of boundary work during this process. From the first law, Q = ∆U + W, with ∆U = m∫cvdT for ideal gases. We see that ∆U for this process is zero, since the temperature is constant. Thus, we have W = Q = 40 Btu . A piston-cylinder device, with a set of stops on the top, initially contains 3 kg of air at 200 kPa and 27oC. Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a P-v diagram. From the problem description, we see that the path is a two step path. The first part is constant pressure as the piston rises to the stops. During this part, the volume doubles. The second part occurs at constant volume after the piston reaches the stops. (The actual data points plotted on the diagram are found in the calculations below.). Problem 4-44 450 Final point 400 350 Pressure (kPa) 4-44 300 250 Initial point At stops 200 150 Work = Area under path 100 50 0 0 1 2 3 Volume (m3) If we denote the initial point as P1 = 200 kPa and T1 = 27oC = 300.15 K, and the second point as P2 = 2 P1 = 400 kPa and V2 = 2 V1, the intermediate point, where the piston reaches the stops has V = V2 and P = P1. The mass m = 3 kg. All the work is done during the constant pressure part of the path. This work is simply found as the rectangular area under the path: W = P1(V2 – V1). The heat transfer is then found as Q = ∆U + W = m∆u + P1(V2 – V1). If we assume that air behaves as an ideal gas at these pressures, we can compute the initial volume from the ideal gas law. Page 3  0.2870 kJ  (3 kg ) ( 27 + 273.15 K ) kg ⋅ K  mRT1  = V1 = = 1.292 m 3 1 kJ P1   ( 200 kPa ) 3   kPa ⋅ m  Since we are given V2 = 2 V1, we know that V2 = 2.584 m3. We can now find the work as follows. W = P1(V2 – V1) = (200 kPa)( 2.584 m3 – 1.292 m3) = 258.4 kPa∙m3 = 258.4 kJ To compute the change in internal energy, we need the final temperature. We can find this final temperature from the ideal gas law. T2 = P2V2 P2 ( 2V1 ) = = mR mR P2 2mRT1 P1 2P T 2( 400 kPa )(300.15 K ) = 2 1 = = 1200.6 K mR P1 ( 200 kPa ) For this large temperature difference we should consider the temperature dependence of the heat capacity. We can use the ideal gas tables for air on pages 849-50. Interpolating in these tables we find the following values for the ideal gas internal energy: u1 = u(300.15 K) = 214.18 kJ/kg u2 = u(1200.6 K) = 933.86 kJ/kg We can now find the heat transfer, Q = ∆U + W = m(u2 – u1) + W = (3 kg)( 933.86 kJ/kg – 214.18 kJ/kg) + 258.4 kJ. This gives Q = 2,417.5 kJ .