Question

2-lbm of water at 500 psia intially fill the 1.5-ft3 left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300°F. Determine the final pressure of water, in psia, and the total internal energy, in Btu, at the final state.

Answer 1

Step-by-step explanation:

Formula for final volume of chamber if the partition is ruptured will be as follows.

`V_(2)` = 1.5 + 1.5

= 3.0
`ft^(3)`

As mass remains constant then the specific volume at this state will be as follows.

`\\u_(2) = (V_(2))/(m)`

=
`(3.0)/(2)`

= 1.5
`ft^(3)/lbm`

Now, at final temperature
`T_(2)` = 300 F according to saturated water tables.

`\\u_(f) = 0.01745 ft^(3)/lbm`

`\\u_(fg) = 6.4537 ft^(3)/lbm`

`\\u_(g) = 6.47115 ft^(3)/lbm`

Hence, we obtained
`\\u_(f) < \\u_(2) < \\u_(g)` and the state is in wet condition.

`\\u_(2) = \\u_(f) + x_(2)\\u_(fg)`

1.5 =
`0.01745 + x_(2) * 6.4537`

`x_(2)` = 0.229

Now, the final pressure will be the saturation pressure at
`T_(2)` = 300 F

and,
`P_(2)` =
`P_(sat)` = 66.985 psia

Formula to calculate internal energy at the final state is as follows.

`U_(2) = m(u_(f)_(300 F) + x_(2)u_{fg_(300 F)}`

=
`2(269.51 + 0.229 * 830.45)`

= 920.56 Btu

Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.