# An adiabatic compressor compressed steam from 4 MPa and 300 o C to 9 MPa isentropically. What...

## Question:

An adiabatic compressor compressed steam from **4 MPa** and** 300 {eq}^o{/eq}C **to** 9 MPa** isentropically.

What is the final temperature, in {eq}^o{/eq}C, of the steam?

## Answer and Explanation:

Given

Initial temperature of steam {eq}T_1 = 300^{\circ}C {/eq}

Initial pressure {eq}P_1 {/eq} = 4MPa

Final pressure {eq}P_2 {/eq} = 9MPa

From the steam table at {eq}T_1 {/eq} and {eq}P_1 {/eq}

{eq}h_1 = 2961.7 \frac{KJ}{kg} {/eq}

{eq}s_1 = 6.3639 \frac{KJ}{kg-K} {/eq}

as process is isentropic

so, {eq}s_1 = s_2 {/eq}

Thus at {eq}P_2 {/eq} and {eq}s_2 {/eq} , from the steam table chart by interpolation

We get {eq}T_2 = 419.11^{\circ}C {/eq}

Thus the final temperature of the steam is 419.11{eq}^{\circ}C {/eq}

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