A piston-cylinder contains 3 kg of wet steam at 1.4 bar. The initial v

1.	A piston-cylinder contains 3 kg of wet steam at 1.4 bar. The initial volume is 2.25 m3. The steam is heated until its temperature reaches 400°C . The piston is free to move up or down unless it reaches the stops at the top. When the piston is up against the stops the cylinder volume is 4.65 m3. Determine the amount of work and heat transfer to or from steam.
Answer» Initial volume per kg of steam = \(\cfrac{2.25}3\) = 0.75 m³/kg Specific volume of steam at 1.4 bar = 1.2363 m³/kg Dryness fraction of initial steam = \(\cfrac{0.75}{1.2363}\) = 0.607 At 1.4 bar, the enthalpy of 3 kg of steam = 3[h_f + xh_fg] = 3 [ 458.4 + 0.607 × 2231.9] = 5439.5 kJ At 400°C, volume of steam per kg = \(\cfrac{4.65}3\) = 1.55 m³/kg At 400°C, when v_sup = 1.55 m³/kg, from steam tables, Pressure of steam = 2.0 bar Saturation temperature = 120.2°C, h = 3276.6 kJ/kg Degree of superheat = t_sup – t_s = 400 – 120.2 = 279.8°C Enthalpy of superheated steam at 2.0 bar, 400°C = 3 × 3276.6 = 9829.8 kg Heat added during the process = 9829.8 – 5439.5 = 4390.3 kJ. Internal energy of 0.607 dry steam at 1.4 bar = 3 × 1708 = 5124 kJ. Internal energy of superheated steam at 2 bar, 400°C = 3(h_sup – pv) = 3(3276.6 – 2 × 10² × 1.55) = 8899.8 kJ (\(\because\) 1 bar = 10² kPa) Change in internal energy = 8899.8 – 5124 = 3775.8 kJ Hence, work done = 4390.3 – 3775.8 = 614.5 kJ.(\(\because\) W = Q – ∆U)

A piston-cylinder contains 3 kg of wet steam at 1.4 bar. The initial volume is 2.25 m3. The steam is heated until its temperature reaches 400°C . The piston is free to move up or down unless it reaches the stops at the top. When the piston is up against the stops the cylinder volume is 4.65 m3. Determine the amount of work and heat transfer to or from steam.

Answer»

Initial volume per kg of steam = \(\cfrac{2.25}3\) = 0.75 m³/kg

Specific volume of steam at 1.4 bar = 1.2363 m³/kg

Dryness fraction of initial steam = \(\cfrac{0.75}{1.2363}\) = 0.607

At 1.4 bar, the enthalpy of 3 kg of steam

= 3[h_f + xh_fg] = 3 [ 458.4 + 0.607 × 2231.9] = 5439.5 kJ

At 400°C, volume of steam per kg = \(\cfrac{4.65}3\) = 1.55 m³/kg

At 400°C, when v_sup = 1.55 m³/kg, from steam tables,

Pressure of steam = 2.0 bar

Saturation temperature

= 120.2°C, h = 3276.6 kJ/kg

Degree of superheat

= t_sup – t_s = 400 – 120.2 = 279.8°C

Enthalpy of superheated steam at 2.0 bar,

400°C = 3 × 3276.6 = 9829.8 kg

Heat added during the process

= 9829.8 – 5439.5 = 4390.3 kJ.

Internal energy of 0.607 dry steam at 1.4 bar

= 3 × 1708 = 5124 kJ.

Internal energy of superheated steam at 2 bar, 400°C

= 3(h_sup – pv) = 3(3276.6 – 2 × 10² × 1.55) = 8899.8 kJ (\(\because\) 1 bar = 10² kPa)

Change in internal energy = 8899.8 – 5124 = 3775.8 kJ

Hence, work done = 4390.3 – 3775.8 = 614.5 kJ.(\(\because\) W = Q – ∆U)