Air enters the compressor with: enthalpy of 80 Btu/lb a pressure of 14.5 psia a temperature of 90...

Question:

Air enters the compressor with: enthalpy of 80 Btu/lb a pressure of 14.5 psia a temperature of 90 {eq}^o{/eq}F volumetric flow rate of 10 ft{eq}^3{/eq}/s Heat transfer from the compressor to its surroundings occurs at a rate of 100 Btu/hr.

If the compressor power input is 80 hp, determine the exit enthalpy, in Btu/lb.

Assume that the compressor is operating at steady state.The kinetic energy change across the compressor is negligible and air behaves as an ideal gas.

Compressors:

Compressors are the mechanical device which are used to handle gases. It increases the pressure of the gases and simultaneously the temperature is also increased. It is a work consuming device.

Answer and Explanation:

Given Data

  • The enthalpy at the inlet is: {eq}{h_1} = 80\;{\rm{Btu/lb}} {/eq}
  • The volume flow rate is :{eq}\dot V = 10\;{\rm{f}}{{\rm{t}}^{\rm{3}}}{\rm{/s}} {/eq}
  • The heat rejected by the compressor is :{eq}Q = 100\;{\rm{Btu/hr}} {/eq}
  • The power input to the compressor is: {eq}\dot W = 80\;{\rm{hp}} {/eq}


The mass flow rate is,

{eq}m = \rho \dot V {/eq}

Here, {eq}\rho {/eq} is the density of air.


{eq}\begin{align*} \dot m &= \left( {1.2\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {10\;{\rm{f}}{{\rm{t}}^{\rm{3}}}{\rm{/s}}\left( {\dfrac{{0.0283\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}}}{{{\rm{1}}\;{\rm{f}}{{\rm{t}}^{\rm{3}}}{\rm{/s}}}}} \right)} \right)\\ &= 0.3396\;{\rm{kg/s}} \end{align*} {/eq}

Substitute all the values in the above equation.

Apply the conservation of energy,

{eq}\dot m\left( {{h_1}} \right) + Q = \dot m{h_2} + \dot W {/eq}


Take value of work as negative because work is done on the compressor similarly heat is rejected by the system, so take its values to be negative.


Substitute all the values in the above equation.

{eq}\begin{align*} 0.3396\;{\rm{kg/s}}\left( {{\rm{80}}\;{\rm{Btu/lb}}\left( {\dfrac{{2.326\;{\rm{kJ/kg}}}}{{{\rm{1}}\;{\rm{Btu/lb}}}}} \right)} \right) - 100\;{\rm{Btu/hr}}\left( {\dfrac{{0.000293\;{\rm{kW}}}}{{{\rm{1}}\;{\rm{Btu/hr}}}}} \right) &= 0.3396\;{\rm{kg/s}}\left( {{h_2}} \right) - \left( {80\;{\rm{hp}}\left( {\dfrac{{0.7457\;{\rm{kW}}}}{{{\rm{1}}\;{\rm{hp}}}}} \right)} \right)\\ 63.192\;{\rm{kJ/s}}\left( {\dfrac{{{\rm{1}}\;{\rm{kW}}}}{{{\rm{1}}\;{\rm{kJ/s}}}}} \right) - 0.0293\;{\rm{kJ/s}}\left( {\dfrac{{{\rm{1}}\;{\rm{kW}}}}{{{\rm{1}}\;{\rm{kJ/s}}}}} \right) &= 0.3396\;{\rm{kg/s}}\left( {{h_2}} \right) - 59.656\;{\rm{kW}}\\ {h_2} &= \dfrac{{122.818\;{\rm{kJ/s}}}}{{0.3396\;{\rm{kg/s}}}}\left( {\dfrac{{0.4299\;{\rm{Btu/lb}}}}{{{\rm{1}}\;{\rm{kJ/kg}}}}} \right)\\ {h_2} &= 361.65\;{\rm{kJ/kg}}\left( {\dfrac{{0.4299\;{\rm{Btu/lb}}}}{{{\rm{1}}\;{\rm{kJ/kg}}}}} \right)\\ {h_2} &= 155.47\;{\rm{Btu/lb}} \end{align*} {/eq}


Thus the enthalpy at exit is {eq}155.47\;{\rm{Btu/lb}} {/eq}.


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