Question

A turbine in a simple ideal Rankine cycle with water as the working fluid operates with an inlet temperature of 500°C and pressure of 600 kPa and an exit temperature of 45°С. Find the turbine work in MW if the mass flow rate of the steam is at 11 kg/s.

Answer 1

the turbine work in MW is
`\mathbf{W_t = 10.47299 \ kW}`

Step-by-step explanation:

Given that:

the inlet temperature = 500°C

pressure = 600 kPa

exit temperature = 45°С

mass flow rate = 11 kg/s

The objective is to find the turbine work in MW given that the mass flow rate of the steam is at 11 kg/s.

From the steam tables, the data obtained for the enthalpies at the inlet temperature of 500°C and a pressure of 600 kPa is:

`\mathtt{h_1 : 3482.7 \ kJ/kg}`

The turbine expansion process is also the isentropic process, as such:

Inlet entropy
`\mathbf{s_1}` = Exit entropy
`\mathbf{s_2}`

The data obtained from the steam table for
`\mathbf{s_1}` = 8.003 kJ/kg.K

`s_2 = s_(f2)+ x_2 *s_(fg2)`

The data obtained from the steam tables for this entities are as follows:

`\mathsf{s_(f2) = 0.638 \ kJ/Kg/K }`

`\mathtt{s_(fg2)=7.528 \ kJ/kg/K}`

since
`\mathbf{s_1}` =
`\mathbf{s_2}` = 8.003 kJ/kg.K

Therefore;

`8.003 = 0.638 + x_2 * 7.528`

`8.003 -0.638 = 7.528x_2`

`7.365= 7.528x_2`

`x_ 2= ( 7.365)/(7.528)`

`x_2 = 0.978`

From the steam tables, the data obtained for the enthalpies at the exit temperature of 45°C and a pressure of 600 kPa is:

`h_(f2)` = 188.4

`h_{fg2` =2394.9

Thus;

`h_ 2= 188.4 +0.978*2394.9`

`h_ 2=2530.61` kJ/kg

The workdone for the turbine process can be computed as:

`\mathsf{W_t = m(h1-h_2)}`

`\mathsf{W_t = 11(3482.7-2530.61)}`

`\mathsf{W_t = 11(952.09)}`

`\mathsf{W_t = 10472.99}` kW

To MW, we have

`\mathbf{W_t = 10.47299 \ kW}`