![5. 0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076 m3
. What is the
temperature of the steam? If the vessel is cooled, at what temperature will the steam be just dry
saturated? Cooling is continued until the pressure is 11 bar; calculate the dryness fraction of the
steam, and the heat rejected between the initial and the final states.
Solution
Given data
m = 0.05 kg
P1 = 15 bar
V = 0.0076 m3
P2 = 11 bar
The steam is superheated then V = mv1
0.0076 = 0.05 v1
v1 = 0.152m3
/ kg.
From superheated steam table,
corresponding to a steam press of 15 bar and volume 0.152 m3
/kg, the temperature of steam is
250°C. [vg = 0.152 at 15 bar].
Process 1-2. Constant volume cooling (v1 = v2) from dry and saturated steam table,
corresponding to the pressure of 15 bar the saturation temperature of steam is 191.6°C.
From dry and saturated steam table corresponding to the pressure of 11 bar we get.
vg2 = 0.178 m3
/ kg
v1 = v2 = x2 x vg2
0.152 = x2 x 0.178
x2 = 0.854.
Thus the steam will be just dry saturated at the temperature of 191.6°C.
Therefore, the final dryness fraction of steam = 0.854. Ans.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Assignment thermal 2018 . . .
- 1. 2. In the turbine of a gas turbine unit the gases flow through the turbine at 17 kg/s and the power developed by the turbine is 14,000 kW. The specific enthalpies of the gases at inlet and outlet are 1200 kJ/kg and 360 kJ/kg respectively, and the velocity of the gases at the inlet and outlet are 60 m/s and 150 m/s, respectively. (a) Calculate the rate at which heat is rejected from the turbine. (b) Find also the area of the inlet pipe given that the specific volume of the gases at inlet are 0.5 m3 /kg. Solution AE 213/BPE 211:APPLIED THERMODYNAMICS DEC 2018
- 2. 5. 0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076 m3 . What is the temperature of the steam? If the vessel is cooled, at what temperature will the steam be just dry saturated? Cooling is continued until the pressure is 11 bar; calculate the dryness fraction of the steam, and the heat rejected between the initial and the final states. Solution Given data m = 0.05 kg P1 = 15 bar V = 0.0076 m3 P2 = 11 bar The steam is superheated then V = mv1 0.0076 = 0.05 v1 v1 = 0.152m3 / kg. From superheated steam table, corresponding to a steam press of 15 bar and volume 0.152 m3 /kg, the temperature of steam is 250°C. [vg = 0.152 at 15 bar]. Process 1-2. Constant volume cooling (v1 = v2) from dry and saturated steam table, corresponding to the pressure of 15 bar the saturation temperature of steam is 191.6°C. From dry and saturated steam table corresponding to the pressure of 11 bar we get. vg2 = 0.178 m3 / kg v1 = v2 = x2 x vg2 0.152 = x2 x 0.178 x2 = 0.854. Thus the steam will be just dry saturated at the temperature of 191.6°C. Therefore, the final dryness fraction of steam = 0.854. Ans.
- 3. 7. A quantity of a certain perfect gas is compressed from an initial state of 0.085 m3 , 1 bar to a final state of 0.034 m3 , 3.9 bar. The specific heat at constant volume is 0.724 kJ/kgK, and the specific heat at constant pressure is 1.020 kJ/kgK. The observed temperature rise is 146 K. Calculate the specific gas constant, R, the mass of gas present, and the increase of internal energy of the gas. Solution