1. A cylinder with piston contains 1.00 kg of water at 100°C and 1.00 atm. Half

Question

1. A cylinder with piston contains 1.00 kg of water at 100°C and 1.00 atm. Half of the water is in the liquid state and half in the vapor state. The combined volumes of the two phases is 0.837 m. With the temperature held constant, the piston is pressed into the cylinder reducing the volume to 0.600 m Find the work 2. A cylinder with piston contains 1.00 mol of an ideal gas at 27.5 and 1.35 atm. The gas is heated at constant pressure to 88.7 . Find the work. lf Cr = 2.94.k//mol. K , find the heat. 3. A cylinder with piston contains 1.00 mol of an ideal gas at 27.5°C and 1.35 atm. The gas is compressed at constant temperature to one half of its initial volume. Find the work Find the heat. . Water is pumped through a 0.225 m diameter pipe from ground level to a height of 25.8 m. The velocity of flow in the pipe is 1.55 m/s. Assume the pump and pipe are adiabatic Find the rate at which the pump performs shaft work 5. Steam at 9000 kPa and 600 enters an adiabatic turbine at a rate of 1.00 kgs. It leaves at 2400 kPa and 400 ·The turbine performs shaft work at a rate of 360 kW. Find the change in enthalpy of the steam from inlet to outlet

Explanation / Answer

Answer to question number 1

Total mass of water = 1 Kg

T = 100 oC

P = 1 atm

Now half of the water is steam and the other half is vapor

mwater = Mvapor =0.5

combined volume V =0.837

Now using this we will first find quality of this mixture

From steam table

Vv = 1.676 m3/Kg = 1.676*0.5 =0.838 m3

Vl =0.001 m3/Kg = 0.001*0.5 =0.00052 m3

V =x*Vv + (1-x)*Vl

0.837 =x*0.838 + (1-x)*0.00052

x =0.99999

which is approximately 1

Hence although by the statement of half of water being liquid and other half being steam it seems steam of quality 0.5 , but as total volume is provided we conclude that the mixture is actually saturated steam

Hence initially the mixture must be saturated Vapor

Hv =2675.57 KJ/Kg

Now the we have an isothermal process where in the total final volume is 0.6 m3

as container consists of 1 Kg of mixture

V = 0.6 m3/Kg

now we will have to look in the steam table and find the Pressure corresponding to this volume and temperature of 100 oC

From steam table we come to the conclusion that we have two phase mixture

hence again

V= x*Vv + (1-x)*Vl

0.6 = x*0.838+ (1-x)*0.00052

x =0.7158

Now at 1 from steam table

Hv =2675.572 KJ/Kg

Hl =419.09 KJ/Kg

H =x*Hv +(-x)*Hl

H = 0.7158*2675.572 + (1-0.7158)*419.09 =2034.2798 KJ/Kg

Net work done = (2034.2798 -2675.572) =-641.29 KJ/Kg = - 641.29 KJ (since mass of total mixture is 1 Kg)

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