Engineering, 16.11.2020 16:40 bellamore

Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water and the work required, in kJ/kg, for this compression.

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Answer from: elijahjacksonrp6z2o7

The final temperature of the water is 676.164 °C and the specific work required is 1171.384 kilojoules per kilogram.

Let suppose that compression occurs quasi-statically, work is done on the closed system and enthalpy is increased. By First Law of Thermodynamics, we model compression process as following:

$W_{in} + (U_{1} - U_{2}) + (P_{1}\cdot V_{1} - P_{2}\cdot V_{2}) = 0$ (1)

Where:

$W_{in}$ - Compression work, in kilojoules. $U_{1}$ , $U_{2}$ - Initial and final internal energies of the system, in kilojoules. $P_{1}$ , $P_{2}$ - Initial and final pressures of the system, in kilopascals. $V_{1}$ , $V_{2}$ - Initial and final volumes of the system, in cubic meters.

By definition of enthalpy, in kilojoules per kilogram, and by dividing the resulting expression by the mass of the entire system, we have the following expression:

$w_{in} = h_{2}-h_{1}$ (2)

Where:

$w_{in}$ - Specific compression work, in kilojoules per kilogram. $h_{1}$ , $h_{2}$ - Initial and final specific enthalpies, in kilojoules per kilogram.

From steam tables we find that initial and final states of the water are represented by the following data:

Initial state

$P = 70\,kPa$ , $T = 100\,^{\circ}C$ , $h = 2679.76\,\frac{kJ}{kg}$ , $s = 7.56162\,\frac{kJ}{kg\cdot K}$ (Superheated steam)

Final state

$P = 4000\,kPa$ , $T = 676.164\,^{\circ}C$ , $h = 3851.144\,\frac{kJ}{kg}$ , $s = 7.56162\,\frac{kJ}{kg}$ (Superheated steam)

By (1) we have that the specific work required is:

$w_{in} = 3851.144\,\frac{kJ}{kg} - 2679.76\,\frac{kJ}{kg}$

$w_{in} = 1171.384\,\frac{kJ}{kg}$

The final temperature of the water is 676.164 °C and the specific work required is 1171.384 kilojoules per kilogram.

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Answer from: aletadaboss

The answer is "909.3928 KJ".

Explanation:

$70 \ kPa \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\$

The method is isentropic since the cylinders are shielded.

Calculating the work:

$w= u_2-u_i \\\\$

$= 3418.7728-2509.38 \\\\=909.3928 \ KJ$

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Answer from: Quest

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Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the fi...

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