Question

find v if T=427 degrees celcius and P=39MPa and no qaulity is given fluid is water. clear explaination and work should be shown

Answer 1

$\textbf{Given data}\rightarrow$

T= 427 °C

P= 39 MPa

$\textbf{Solution}\rightarrow$

From superheated steam table properties 35 MPa and 427 °C

$T\textup{ in }^{o}\textup{C}$	$v\textup{ in m}^{3}\textup{/kg}$
	0.00308380
427
430	0.00378000

$\textup{By interpolation}$

$\frac{430-420}{0.0037800-0.00308380}=\frac{430-427}{0.0037800-v_{1}}$

$v_{1}=0.00357114\textup{ m}^{3}\textup{/kg}$

$\textup{From superheated steam table properties 40 MPa and 427 }^{o}\textup{C}$

$T\textup{ in }^{o}\textup{C}$	$v\textup{ in m}^{3}\textup{/kg}$
	0.00236010
427
430	0.00274370

$\textup{By interpolation}$

$\frac{430-420}{0.00274370-0.00236010}=\frac{430-427}{0.00274370-v_{2}}$

$v_{2}=0.00262862\textup{ m}^{3}\textup{/kg}$

$\textup{From superheated steam table properties 39 MPa and 427 }^{o}\textup{C}$

$P\textup{ in MPa}$	$v\textup{ in m}^{3}\textup{/kg}$
35	$v_{1}=0.00357114\textup{ m}^{3}\textup{/kg}$
39	v
40	$v_{2}=0.00262862\textup{ m}^{3}\textup{/kg}$

$\textup{By interpolation}$

$\frac{40-35}{0.00262862-0.00357114}=\frac{40-39}{0.00262862-v}$

$v=0.00281712\textup{ m}^{3}\textup{/kg}$