# Thrust and specific impulse of a rocket

A picture of the question this originates from is given below.

Note: I normally work with SI units and am mostly unfamiliar with those used here.

The specific impulse of a rocket can be calculated by dividing the thrust force by the weight flow rate of propellants.

To calculate the thrust I take the product of mass flow rate and exhaust velocity. This gives 7000*280=1,960,000lbf. Dividing this by 280*32.174 (propellant weight flow rate) gives the correct specific impulse of 217.5s.

The total impulse should therefore be 1960000*65, the integral of the force over the burn time, however, this result is out by a factor of 32.174.

Evidently the thrust force is wrong. The thrust I have calculated is far higher than any liquid engine I know of including the Rocketdyne F-1 (which is the largest...?). Producing that much thrust with only 280lbs/sec of propellant is highly unlikely. But the bit I can't get my head around is, if my thrust is wrong, why did that still give me the correct answer for specific impulse?

Also, why is my thrust wrong? Is this because ambient pressure cannot be assumed to be equal to exit pressure? (Is it normally close?) The thrust for a rocket (can be demonstrated with global conservation of momentum on control volume set around the rocket) is, in general: $$\mathbb{T}=\dot{m}u_e + A_e\left( p_e - p_a \right)$$ where $\dot{m}$ is the mass flow rate $\left(\frac{M_p}{t_b} \right)$ (being $M_p$ the mass of propellant and $t_b$ the burning time) and $u_e$ the gases exit velocity from the nozzle.

Sometimes the thrust is also expressed as: $$\mathbb{T} = \dot{m}c$$ being $c = u_e + A_e\frac{\left( p_e - p_a \right)}{\dot{m}}$ the effective velocity.

The specific impulse is calculated as: $$I_s=\frac{\mathbb{T}}{\dot{m}g_0} = \frac{c}{g_0}$$

being $g_0 = 9.81... \left[\frac{\text{N}}{\text{kg}}\right]$

The total impulse is: $$I_{\text{tot}}=\overline{T}\cdot t_b$$

Since I'm not used to British units I've run the calculation converting to SI units:

$$c = 2133.6 \left[\frac{\text{m}}{\text{s}}\right]$$ $$\dot{m} = 126.98 \left[\frac{\text{kg}}{\text{s}}\right]$$ $$I_s =\frac{c}{g_0}= 217.4 \left[s\right]$$ $$\mathbb{T}=\dot{m}c=270924.528 \left[ N\right]$$ $$I_{\text{tot}}=\overline{T}\cdot t_b=17610094 \left[\text{Ns}\right]=3958747.936 \left[\text{lbf s}\right]$$ $$M_p = \dot{m}\,t_b=8253.7 \left[\text{kg}\right] = 18200 \left[\text{lbm}\right]$$

Anyway, generally, you try to size the nozzle, on the base of operative condition, in the way that $\left( p_e - p_a \right) = 0$. This, can be demonstrated, is the optimal expansion.

• Thanks for the answer, I feel I would have gotten it right if the units had been in SI. However, I still don't see why T=m˙c works for SI units but not British units? SI units give the correct thrust of 126.98*2133.6=270924N=60687lbf But using the British units gives 280*7000=1960000lbf Feb 5, 2015 at 22:30
• @DylanCleaver for what I see here $1 \text{lbf} = 1 \text{lbm} \cdot 32.174049 \left[\frac{\text{ft}\cdot\text{lbm}}{\text{s}^2}\right]$. So $\frac{1960000}{32.174049}$ should work as expected because $\left[1960000\right]=\left[\frac{\text{lbm}\cdot \text{ft}}{\text{s}^2}\right]$ Feb 5, 2015 at 22:46
• Now this is beginning to make more sense. However, shouldn't this mean that the answer should be in lbm-sec rather than lbf-sec, since when we use 60687*65 we are using the lbm value? Feb 5, 2015 at 23:16
• @DylanCleaver Well, in the SI system the total impulse is expressed as "Force x Time". [lbf] is the measure of a force (something like [kgf]), while [lbm] is the unit of mass, so I think it is correct to express it as [lbf s] Feb 5, 2015 at 23:23