# Efficiency of a nuclear thermal rocket

I am trying to calculate the specific impulse and thrust of a nuclear thermal engine, given the thermal power of the reactor, the maximum working temperature and the propellant.

I'm calculating the exhaust velocity using the formulas from https://www.grc.nasa.gov/www/k-12/airplane/rktthsum.html and the mass flow using the heat capacity of the propellant and the temperature difference.

The issue is that the thrust power I get is about equal to my thermal power or slightly larger, which means this engine is 100% efficient. What am I missing in my calculation?

• Welcome. It would be helpful if you could include your actual calculations in your question Mar 13 '19 at 23:33

The equations given at the link you give are for an isentropic, work-free flow of gas. So, in the limit of infinite expansion into a vacuum ($$p_0=0$$, $$M_e\rightarrow\infty$$, $$p_e\rightarrow 0$$) you will find that the thermal efficiency does approach 100%, meaning that almost all of the thermal energy originally in the hot gas has been converted to the kinetic energy of the exhaust. This mathematical model leaves it with no other place to go.
Working from the equations given, you can calculate the energy balance (really, enthalpy balance) in terms of $$M_e$$, the Mach number of the exhaust. Assuming, as the equations do, an ideal gas with constant specific heat, a unit mass of gas will have thermal enthalpy $$\gamma R T_t/(\gamma-1)$$ to begin with. When it has expanded to a state where the Mach number is $$M=M_e$$, the total enthalpy in a unit mass of gas will be partitioned into a thermal enthalpy of $$\gamma R T_e/(\gamma-1)$$ and a kinetic energy of $$\frac{1}{2} V_e^2$$. Re-expressed as fractions of the initial thermal enthalpy, the thermal fraction is $$\frac{2}{2+(\gamma-1)M_e^2}$$ and, after some simplification, the kinetic fraction is $$\frac{(\gamma-1)M_e^2}{2+(\gamma-1)M_e^2}.$$ These two fractions sum to 1.