Question

In a steam engine the steam at the beginning of the expansion process is

at 7 bar, dryness fraction 0.98 and expansion follows the law pv1.1 = constant, down to a pressure

of 0.34 bar. Calculate per kg of steam :

(i) The work done during expansion ;

(ii) The heat flow to or from the cylinder walls during the expansion.

Answer 1

Initial pressure of steam,

$p_{1} =7 \;bar = 7 × 10^{5} \; N/m^{2}$

Dryness fraction,

$x_{1} = 0.98$

Law of expansion,

$pv^{1.1} = constant$

Final pressure of steam,

$p_{2} =0.34 \; bar = 0.34 \times 10^{5} \; N/m^{2}$

At 7 bar :

$v_{g} = 0.273 \; m^{3}/kg \\ v_{1}=x_{1}v_{g}= 0.98 \times 0.273 = 0.267 \; m^{3}/kg \\ p_{1}v_{1}^{n}=p_{2}v_{2}^{n} \\ \frac{v_{2}}{v_{1}}=\left ( \frac{p_{1}}{p_{2}} \right )^{\frac{1}{n}} \\ \frac{u_{2}}{0.267}=\left ( \frac{7}{0.34} \right )^{\frac{1}{1.1}}$

or

$v_{2}= 0.267\left ( \frac{7}{0.34} \right )^{\frac{1}{1.1}}= 4.174 \; m^{3}/kg$

(i) Work done by the steam during the process :

$W=\frac{p_{1}v_{1}-p_{2}v_{2}}{n-1} \\ =\frac{7\times 10^{5}\times 0.267-0.34\times 10^{5}\times 4.174}{\left ( 1.1-1 \right )} \\ =\frac{10^{5}}{0.1}(1.869 - 1.419) =10^{5}\times 4.5 \; Nm/kg$

Work done $= \frac{10^{5}\times 4.5}{10^{3}} =450\; kJ/kg.$

At 0.34 bar : $v_{g}= 4.65\; m^{3}/kg$ , therefore, steam is wet at state 2 $(since \;v_{2} < v_{g})$ .

Now, $v_{2}=x_{2}v_{g}$ , where $x_{2} =$ dryness fraction at pressure $p_{2} \; (0.34 \;bar)$

$4.174 = x_{2}\times 4.65$

or

$x_{2} = \frac{4.174}{4.65}= 0.897$

The expansion is shown on a p-v diagram in Fig., the area under 1-2 represents the work done per kg of steam.

(ii) Heat transferred :

Internal energy of steam at initial state 1 per kg,

$u_{1}= (1 - x_{1})u_{f} + x_{1}u_{g} \\ = (1 - 0.98) 696 + 0.98 \times 2573 = 2535.46 \; kJ/kg$

Internal energy of steam at final state 2 per kg,

$u_{2}=\left ( 1-x_{2} \right )u_{f}+x_{2}u_{g} \\ = (1 - 0.897) 302 + 0.897 \times 2472 = 2248.49 \; kJ/kg$

Using the non-flow energy equation,

$Q=\left ( u_{2} -u_{1}\right )+W \\ = (2248.49 - 2535.46) + 450 = 163.03 \; kJ/kg$

Heat supplied = 163.03 kJ/kg.