3) An insulated piston-cylinder device contains steam at 300 kPa, 200oC, occupyi

Question

3) An insulated piston-cylinder device contains steam at 300 kPa, 200oC, occupying a volume of 1 m3 , and having a specific volume of 0.716 m3 /kg. It is heated by an internal electrical heater until the volume of steam doubles due to an increase in temperature. (a) Determine the final specific volume of steam. (b) If the diameter of the piston is 20 cm and the outside pressure is 100 kPa, determine the mass of the weight placed on the piston to maintain a 300 kPa internal pressure. (c) Calculate the WB (magnitude only) done by the steam in kJ. (d) Calculate the amount of WM (magnitude only) transferred into the weight in kJ. (e) If you are asked to choose one of the three values - 1300 kJ, 300 kJ, 200 kJ - as the magnitude of Wel, which one will be your educated guess? Why?

Explanation / Answer

The Question mentions a thermally insulated cylinder so it seems it is an adiabatic system but it isn't as there is an electric heater which obviously draws power from outside. Therefore it is not insulated.

Given that the Initial Volume of steam is 0.716 m3/Kg and volume is 1 m3 the mass is (1/0.716) Kg. Given that the final volume is double the initial volume, the mass of steam being the same the Final specific volume is (2m3/(1/0.716Kg)) = 2*0.716 m3/Kg. = 1.432 m3/Kg.

Therefore For Part(a) Final specific volume is 1.432 m3/Kg.

b) Area of Piston = Pi*r2 = 3.14159*(0.1)2 = 0.0314159m2. Force Exerted by External Air Pressure = Area of Piston * Pressure = 0.0314159*(100*1000) N = 0.0314159*105 = 3141.59 N.

Force Exerted by Internal Pressure = 3*Force Exerted by External Air Pressure(Because the internal Pressure is 3 times external pressure) = 3*3141.59 N. = 9424.77 N.

= Force Exerted by External Pressure + Weight on Piston

So Weight on Piston = 9424.77 N - 3141.59 N. = 6283.18 N.

Assuming the mass is m then weight mg = 6283.18 N.

Assuming g=9.8m/s2 m = 6283.18/9.8 Kg = 641.1408 Kg

Therefore Mass on Piston = 641.1408 Kg.

c) The change in Volume occurs due to change in Temperature and the Final Pressure is still 300 KPA as the External Pressure and Weight of Piston are still the same as calculated above. Given the Initial and Final Pressures are the same and assuming Ideal Conditions the process can be assumed to be an Isobaric Process.

Therefore for Isobaric Process the work done by Steam = P*(Change in Volume) = 300kPa*1m3 = 300KJ

d) The Magnitude of Work Done on the Mass on Piston = Weight of Piston*Displacement of Piston Vertically = 6283.18 N*Displacement of Piston Vertically because it moves against gravity. This is Because we are assuming Ideal Conditions therefore the mass gain only potential energy and no kinetic energy.

To calculate Displacement of Piston apply formula of Volume of Cylinder = Pi*r2L. We already know area of Piston = Pi*r2 = 0.0314159m2. Now since the Gas expands by 1 m3 therefore this cylindrical Volume should be equal to Pi*r2L. Therefore L = 1m3/0.0314159m2 = 31.8310155049 m. Therefore the Work Done on Piston = 31.8310155049m * 6283.18 N = 200000 J = 200 KJ.

e) Taking an Educated Guess I would estimate the WEl to be 1300 KJ because this includes both the work done to expand the gas against external forces and also raising the temperature of the gas. Since the work done by the steam itself is 300 KJ the WEL is obviously more.

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