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

Initial pressure of steam, p₁ = 5.5 bar = 5.5 × 10⁵ N/m² 

Initial condition of steam, x₁ = 1

Final pressure of steam, p₂ = 0.75 bar = 0.75 bar × 10⁵ N/m²

Since v₂ > v_g (at 0.75 bar), therefore, the steam is superheated at state 2. Interpolating from superheat tables at 0.75 bar, we have

For dry saturated steam at 5.5 bar 

u₁ = u_g = 2565 kJ/kg 

Hence, gain in internal energy 

= u₂ – u₁ = 2567.25 – 2565 = 2.25 kJ/kg 

The process is shown on a p-v diagram in  the shaded area representing the work done

The constant is either p₁v₁ or p₂v₂

Using non-flow energy equation, we get

Q = (u₂ – u₁) + W

= 2.25 + \(\cfrac{375543}{10^3}\) = 378 kJ/kg

Heat supplied = 378 kJ/kg.

Answer 2

Initial pressure of steam,   p_{1}= 5.5 bar = 5.5 ×10^{5}N/m^{2}p1​=5.5bar=5.5×105N/m2

Initial condition of steam,     x_{1}=1x1​=1

Final pressure of steam,     p_{2}= 0.75p2​=0.75 bar = 0.75 bar × 10^{5} N/m^{2}105N/m2

At 5.5 bar,   v_{1}=v_{g}=0.3427m^{3}/kgv1​=vg​=0.3427m3/kg

Also   p_{1}v_{1}=p_{2}v_{2}p1​v1​=p2​v2​

\therefore∴     v_{2}=\frac{p_{1}v_{1}}{p_{2}}=\frac{5.5\times 0.3427}{0.75}=2.513 m^{3}/kgv2​=p2​p1​v1​​=0.755.5×0.3427​=2.513m3/kg

At 0.75 bar,   v_{g}=2.217 m^{3}/kgvg​=2.217m3/kg .

Since    v_{2}> v_{g}v2​>vg​ (at 0.75 bar),therefore, the steam is superheated at state 2.Interpolating from superheat tables at 0.75 bar,we have

u_{2}=2510+\left ( \frac{2.513-2.271}{2.588-2.271} \right )(2585 – 2510)u2​=2510+(2.588−2.2712.513−2.271​)(2585–2510)

= 2510 +\frac{0.242}{0.317}\times 75=2567.25=2510+0.3170.242​×75=2567.25 kJ/kg.

For dry saturated steam at 5.5 bar

u_{1}=u_{g}= 2565u1​=ug​=2565 kJ/kg

Hence, gain in internal energy

=u_{2}-u_{1}= 2567.25 – 2565 = 2.25=u2​−u1​=2567.25–2565=2.25 kJ/kg

The process is shown on a p-v diagram in Fig. 2.61, the shaded area representing the work done.

Now,     W=\int_{v_{1}}^{v_{2}} p dvW=∫v1​v2​​pdv

=\int_{v_{1}}^{v_{2}}\left ( \frac{constant}{v} \right )dv=∫v1​v2​​(vconstant​)dv      \left [ \because pv = constant      and     p=\frac{constant}{v}\right ][∵pv=constant   and  p=vconstant​]

=constant \left [ \log_{e} v\right ]_{v_{1}}^{v_{2}}[loge​v]v1​v2​​

The constant is either   p_{1}v_{1}p1​v1​   or   p_{2}v_{2}p2​v2​

i.e.,   W = 5.5 × 10^{5} × 0.3427 × \log_{e}\frac{p_{1}}{p_{2}}W=5.5×105×0.3427×loge​p2​p1​​     \left [ \because p_{1} v_{1}=p_{2}v_{2}    or     \frac{v_{2}}{v_{1}}=\frac{p_{1}}{p_{2}}\right ][∵p1​v1​=p2​v2​  or  v1​v2​​=p2​p1​​]

= 5.5 × 10^{5}\times 0.3427 ×\log_{e}\left ( \frac{5.5}{0.75} \right )= 375543=5.5×105×0.3427×loge​(0.755.5​)=375543 Nm/kg.

Using non-flow energy equation, we get

Q=\left ( u_{2}-u_{1} \right )+WQ=(u2​−u1​)+W

=2.25+\frac{375543}{10^{3}}=377.79=2.25+103375543​=377.79  say  378 kJ/kg

i.e.,      Heat supplied = 378  kJ/kg