Answer to Question #231113 in Molecular Physics | Thermodynamics for Unknown346307

Question

In a steam engine cylinder the steam expands from 5.5 bar to 0.75 bar

according to a hyperbolic law, pv = constant. If the steam is initially dry and saturated, calculate

per kg of steam :

(i) Work done ;

(ii) Heat flow to or from the cylinder walls

Answer 1

Solution;

Given;

Initial pressure of steam,$p_1=5.5$ bar

Final pressure of steam,$p_2=0.75$ bar

At 5.5 bar;

$v_1=v_g=0.3427m^3/kg$

Also;

$p_1v_1=p_2v_2$

$v_2=\frac{p_1v_1}{p_2}=\frac{5.5×0.3427}{0.75}$ =2.513m³/kg

At 0.75 bar;

$v_g=2.217m^3/kg$

Since $v_2>v_g$ (At 0.75 bar),the steam is superheated at state 2.

From tables, interpolating,at 0.75 bar,we have;

$u_2=2510+(\frac{2.513-2.271}{2.588-2.271})(2585-2510)=2567.25kJ/kg$

From tables,at 5.5 bar;

$u_1=u_g=2565kJ/kg$

Gain in internal energy is;

$\Delta u=u_2-u_1$ =2567.25-2565=2.25kJ/kg

(i)Work done;

$W=\int_{v_1}^{v_2}pdv$

But pv=c

$W=c [log_ev]_{v_1}^{v_2}$

In which ;

$c=p_1v_1=p_2v_2$

And;

$\frac{p_1}{p_2}=\frac{v_2}{v_1}$

Hence ;

$W=5.5×10^5×0.3427×log_e(\frac{5.5}{0.75})$

W=375.543kN-m/kg

(ii)Heat flow;

Using non flow energy equation;

$Q=(u_2-u_1)+W$

$Q=2.25+375.543$

$Q=377.79kJ/kg$