Question

1. A rocket turbine is driven by gases having specific heat ratio of 1.3 and specific heat of 1070 J/kg-K. The turbine entry stagnation conditions are 810 K and 2 MPa. If the discharge pressure from the turbine is 0.14 MPa, calculate its specific work. h_ts = 0.70.                  {Ans.: 2.78 (10⁵) J/kg}

Answer 1

Data for rocket turbine :

Specific heat ratio Y = 1.3

Specific heat Cp = 1070 J/kg.k

Turbine entry stagnation condition T1 = 810 k and P1 =2MPa

And discharge pressure P2 = 0.14 MPa

Isentropic efficiency hts = 0.70

calculation :

Stagnation condition shows isentropic process.

Then by isentropic process,

T2 = T1*(P2/P1)^(Y-1)/Y

T2 = 810*(0.14/2)^(1.3-1)/1.3

T2 = 438.5 K

Discharge temperature of gas from turbine T2 = 438.5

Isentropic efficiency = real turbine work /isentropic turbine work

0.7 = real turbine work/Cp(T1 - T2)

Real turbine work = 0.7*1070J/kg.k * (810 - 438.5)k

= 278253.5 J/kg

= 2.78*10^5 J/kg

required specific work for turbine w = 2.78*10^5 J/kg

ans : 2.78*10^5 J/kg

Extra work for turbine : if discharge real temperature is T2' and we calculated T2 = 438.5 K is the isentropic discharge temperature.

also isentropic efficiency = Cp(T1 - T2') /Cp(T1 - T2)

0.7 =( 810 - T2') /(810 - 438.5)

T2' = 549.95 K

Actual discharge temperature from turbine T2'= 549.95 K