The internal energy of wet steam is 2000KJ/kg. If the pressure is 1.18 MPa, what is...

The internal energy of wet steam is 2000KJ/kg. If the pressure is 1.18 MPa, what is the quality of the mixture

Solutions

Expert Solution

Given specific internal energy of steam = u=2000KJ/kg, pressure p=1.18MPa=1180KPa.

In steam tables, the data of steam at 1200.5 KPa and at 1148.3KPa are given.

we will find out the data of steam at 1180KPa by interpolation

The data is as follows

T ()	(KPa)	(KJ/kg)	(KJ/kg)	(KJ/kg)
186	1148.30	788.060	2585.74	1797.68
188	1200.50	796.899	2586.91	1790.01

By interpolation

at 1180Kpa = at 1148.3KPa + ( at 1200.5KPa - at 1148.3KPa)*(1180-1148.3)/(1200.5-1148.3)

= 788.06 + (796.899-788.06)*(1180-1148.3)/(1200.5-1148.3)

=793.427KJ/kg

at 1180Kpa = at 1148.3KPa + ( at 1200.5KPa - at 1148.3KPa)*(1180-1148.3)/(1200.5-1148.3)

= 2585.74 + (2586.91-2585.74)*(1180-1148.3)/(1200.5-1148.3)

=2586.45KJ/kg

Now, the formula to calculate quality of steam is

Substituting the values, we get

2000=793.427 + x(2586.45-793.427)

=> x = 0.673 (The quality of steam)

samet mamat answered 1 year ago

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