Answer:
a
The rate of work developed is [tex]\frac{\r W}{\r m}= 300kJ/kg[/tex]
b
The rate of entropy produced within the turbine is [tex]\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K[/tex]
Explanation:
From the question we are told
The rate at which heat is transferred is [tex]\frac{\r Q}{\r m } = - 30KJ/kg[/tex]
the negative sign because the heat is transferred from the turbine
The specific heat capacity of air is [tex]c_p = 1.1KJ/kg \cdot K[/tex]
The inlet temperature is [tex]T_1 = 970K[/tex]
The outlet temperature is [tex]T_2 = 670K[/tex]
The pressure at the inlet of the turbine is [tex]p_1 = 400 kPa[/tex]
The pressure at the exist of the turbine is [tex]p_2 = 100kPa[/tex]
The temperature at outer surface is [tex]T_s = 315K[/tex]
The individual gas constant of air R with a constant value [tex]R = 0.287kJ/kg \cdot K[/tex]
The general equation for the turbine operating at steady state is \
[tex]\r Q - \r W + \r m (h_1 - h_2) = 0[/tex]
h is the enthalpy of the turbine and it is mathematically represented as
[tex]h = c_p T[/tex]
The above equation becomes
[tex]\r Q - \r W + \r m c_p(T_1 - T_2) = 0[/tex]
[tex]\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)[/tex]
Where [tex]\r Q[/tex] is the heat transfer from the turbine
[tex]\r W[/tex] is the work output from the turbine
[tex]\r m[/tex] is the mass flow rate of air
[tex]\frac{\r W}{\r m}[/tex] is the rate of work developed
Substituting values
[tex]\frac{\r W}{\r m} = (-30)+1.1(970-670)[/tex]
[tex]\frac{\r W}{\r m}= 300kJ/kg[/tex]
The general balance equation for an entropy rate is represented mathematically as
[tex]\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0[/tex]
=> [tex]\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)[/tex]
generally [tex](s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ][/tex]
substituting for [tex](s_1 -s_2)[/tex]
[tex]\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ][/tex]
Where [tex]\frac{\sigma}{\r m}[/tex] is the rate of entropy produced within the turbine
substituting values
[tex]\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ][/tex]
[tex]\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K[/tex]
