Given: \( p_{1}=p_{2} \)
case 1 - without afterburner: , , .
case 2 - with afterburner: .
To calculate: static thrust for case 1 and case 2.

The schematic diagram of the problem description is shown in Fig. 1.

Case1: without afterburner

Using the conservation of mass,

Case1: without afterburner

Assuming \( V_{1}\sim 0 \) (since engine is stationary on ground).

The conservation of momentum gives,

$$ \begin{aligned} \text{Thrust} =&\ \text{rate of momentum exiting }-\text{ rate of momentum entering}\\ &\ +\text{ pressure force at exit }-\text{ pressure force at inlet}\\ \\ \text{Thrust} =&\ \dot{m}_{2}V_{2}-\dot{m}_{1}V_{1}+\left(p_{2}-p_{1}\right)A_{\text{exit}} \end{aligned} $$

Case2: with afterburner

Using the conservation of mass,

Assuming \( V_{1}\sim 0 \) (since engine is stationary on ground).

The conservation of momentum gives,

$$ \begin{aligned} \text{Thrust} =&\ \text{rate of momentum exiting }-\text{ rate of momentum entering}\\ &\ +\text{ pressure force at exit }-\text{ pressure force at inlet}\\ \\ \text{Thrust} =&\ \dot{m}_{2}V_{2}-\dot{m}_{1}V_{1}+\left(p_{2}-p_{1}\right)A_{\text{exit}} \end{aligned} $$