Question

Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.

Required:
Determine the power output of the turbine, in hp.

Answer 1

Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.

Required:

a)Determine the power output of the turbine, in hp.

b) The Flow rate

Answer:

a)
`w=74.26hp`

b)
`m=0.22`

Step-by-step explanation:

From the question we are told that:

Initial Pressure
`p_1= 145 psi`

Initial Temperature
`T_1 =2700oR=>2240.33^oF`

Final Pressure
`p_2= 29 psi`

Final Temperature
`t_2=1974oR=>1514.33^oF`

Output Power
`w=74.26hp`

Heat transfer Rate
`Q=14BTU/s`

Generally the equation for Steady flow energy is mathematically given by

`Q-w=m(h_2-h_1)`

Where

`m=Flow\ rate`

From Steam table

`h_1=704btu/ib (at\ p_1= 145\ psi,\ T_1 =2700oR=>2240.33^oF )`

`h_2=401btu/ib (at\ p_2= 29psi\ t_2=1974oR=>1514.33^oF )`

Therefore

`-14-74.26=m(401-704)`

`m=(-14-74.26)/((401-704))`

`m=0.22`