A heat powered elevator is to be designed as shown in the figure below. The air

Question

A heat powered elevator is to be designed as shown in the figure below. The air in the tank of volume V= 30 m3 is initially at T1= 25o C and P1= 100 kpa. A piston (mass mp= 230 kg; diameter Dp= 0.75) snugly fits the chute, as shown in the figure. A casting ( mass= 1000kg ) is loaded onto the platform at level 1. Initially, the piston along with the casting rests on the stop at level 1. A bottom chamber of the tank is filled with 1 kg of water at saturated liquid conditions with Tw= 150o C along with a pressure valve to maintain constant pressure inside the chamber. An electric heater ( I= 30 A; V = 240 V ) in the chamber is turned on to heat up the water and in turn, the air in the tank.

Theheating continues until the platform ( piston with the casting ) reaches level 2 that is z = 6 m above level 1, at which point the platform hits a stop at level 2, and the heater is switched off. At this point, the water in the lower chamber is found to be in saturated vapor conditions. Here, the casting can be offloaded from the platform and arrangements are made to cool off the tank air to bring down the platform to level 1. Please answer the questions below.

1) Determine the temperature and pressure of the air in the system when the platform just reaches level 2.

2) Determine the work done by air (kJ) in raising the piston and casting from level 1 to level 2.

3) What is the net heat transfer (kJ) to the air from the water chamber while raising the piston and casting from level 1 to level 2?

4) How long (in minutes) does the electric heater have to run to lift the piston and casting from level 1 to level 2? Air is an ideal gas at conditions specified with R= 0.287 kJ/kg-K and average specific heat is cv=0.718 kJ/kg-K. The outside of the entire system is insulated to prevent heat loss to the surroundings. The piston and casting are exposed to atmospheric pressure of P= 101.325 kPa

atm z=6m casting m, 1000 kg piston m230 kg D, = 0.75 m Level 1 air V 30 m3 .Hea Pressure Valve water Tw = 150° 130 Amps V = 240 Volts

Explanation / Answer

Initila volume V1 = 30 cu.m

area of the piston A= 3.14*(0.75/2)2 = 0.44sq.m

Pressure in the chamber = 101.32kpa + (1000+230)*9.8/0.44 = 128.6 kPa

Total wieght of the Piston and the casting 1230 kg add a force of 1230*9.8 N, it causes an excee pressure of F/A over the atom. pressure.

when the lift reaches level to change in height is 6 m

increase in volume = 6*0.44 = 2.64 cu.m

final volume of air when the lift is at level2 = 30 +2.64 = 32.64 cu.m

There is no change is pressure through out the operation hence

V1/V2 = T1/T2 , T1 = 273 +25 = 298 K

1) final temp T2 = 298*32.6/30 = 323.83 K = 50.83 C

2 ) work done W = Pdv   , P = 128.6 kPa const, dV = 2.64 cu.m

                           = 128.6 *2.64 = 339.504 kJ

3) PV = nRT

P = 128.6 kPa ; V = 30 cu.m T = 298 K , initial conditions

n = 128.6e+3*30/298*8.33 = 1554.18 mol

change in internal energy dU = nCv dT = 1554.18*0.718 *(323.83-298) = 28823.7290292 J

heat transfered Q = W +dU = 339.504 kJ + 28823.7290292 = 368.328 kJ

4) Power dissipated by the heater P = IV = 30*240 = 7200 w

The heater supplies energy at the rate of 7200 J/s

time required   = 368328/7200 = 51.16 s

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