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) $m=0.22$

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

 $-14-74.26=m(401-704)$

 $m=\frac{-14-74.26}{(401-704)}$

 $m=0.22$