Question

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. For air as an ideal gas with Cp = 1.1 Kj/kg * K, determine

(a) the rate power is developed, in kJ per kg of air flowing, and
(b) the rate of entropy production within the turbine, in kJ/kg per kg of air flowing.

Answer 1

a

The rate of work developed is
`(\r W)/(\r m)= 300kJ/kg`

b

The rate of entropy produced within the turbine is
`(\sigma)/(\r m)= 0.0861kJ/kg \cdot K`

Step-by-step explanation:

From the question we are told

The rate at which heat is transferred is
`(\r Q)/(\r m ) = - 30KJ/kg`

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is
`c_p = 1.1KJ/kg \cdot K`

The inlet temperature is
`T_1 = 970K`

The outlet temperature is
`T_2 = 670K`

The pressure at the inlet of the turbine is
`p_1 = 400 kPa`

The pressure at the exist of the turbine is
`p_2 = 100kPa`

The temperature at outer surface is
`T_s = 315K`

The individual gas constant of air R with a constant value
`R = 0.287kJ/kg \cdot K`

The general equation for the turbine operating at steady state is \

`\r Q - \r W + \r m (h_1 - h_2) = 0`

h is the enthalpy of the turbine and it is mathematically represented as

`h = c_p T`

The above equation becomes

`\r Q - \r W + \r m c_p(T_1 - T_2) = 0`

`(\r W)/(\r m) = (\r Q)/(\r m) + c_p (T_1 -T_2)`

Where
`\r Q` is the heat transfer from the turbine

`\r W` is the work output from the turbine

`\r m` is the mass flow rate of air

`(\r W)/(\r m)` is the rate of work developed

Substituting values

`(\r W)/(\r m) = (-30)+1.1(970-670)`

`(\r W)/(\r m)= 300kJ/kg`

The general balance equation for an entropy rate is represented mathematically as

`(\r Q)/(T_s) + \r m (s_1 -s_2) + \sigma = 0`

=>
`(\sigma)/(\r m) = - (\r Q)/(\r m T_s) + (s_1 -s_2)`

generally
`(s_1 -s_2) = \Delta s = c_p\ ln[(T_2)/(T_1) ] + R \ ln[(v_2)/(v_1) ]`

substituting for
`(s_1 -s_2)`

`(\sigma)/(\r m) = (-\r Q)/(\r m) * (1)/(T_s) + c_p\ ln[(T_2)/(T_1) ] - R \ ln[(p_2)/(p_1) ]`

Where
`(\sigma)/(\r m)` is the rate of entropy produced within the turbine

substituting values

`(\sigma)/(\r m) = - (-30) * (1)/(315) + 1.1 * ln(670)/(970) - 0.287 * ln [(100kPa)/(400kPa) ]`

`(\sigma)/(\r m)= 0.0861kJ/kg \cdot K`