Problem 1.13

In a proposed jet propulsion system for an automobile, air is drawn in vertically through a large intake in the roof at a rate of 3 kg/s, the velocity through this intake being small. Ambient pressure and temperature are 100 kPa and 30^oC, respectively. This air is compressed and heated and then discharged horizontally out of a nozzle at the rear of the automobile at a velocity of 500 m/s and a pressure of 140 kPa. If the rate of heat addition to the air stream is 600 kW, find the nozzle discharge area and the thrust developed by the system.

Given:
Air is drawn vertically, $\dot{m} = 3 kg/s$ , $V_{1} \approx 0 m/s$ , $p_{amb} = p_{0} = 100 kPa$ , $T_{amb} = T_{0} = 30^{\circ} C = 303 K$ .
Air is compressed and heated and discharged horizontally, $V_{e} = 500 m/s$ , $p_{e} = 140 kPa$ , $\dot{q} = 600 kW$ .

To calculate: $A_{e} = ?$ , $Thrust = ?$ .

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

Applying the conservation of energy through the system,

\begin{aligned} Rate of energy entering & + Rate of heat addition \\ = Rate of energy leaving + Work done by the system \end{aligned}

\dot{m} c_{p} T_{0} + \dot{q} = \dot{m} c_{p} T_{e} + \dot{m} \frac{V_{e}^{2}}{2} + W

3 \times 1005 \times 303 + 600 \times 10^{3} = 3 \times 1005 \times T_{e} + 3 \times \frac{500^{2}}{2} + 0

T_{e} = \frac{3 \times 1005 \times 303 + 600 \times 10^{3} - 3 \times \frac{500^{2}}{2}}{3 \times 1005} = 377.627 K .

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The density at the exit can be calculated using the ideal gas equation,

ρ_{e} = \frac{p_{e}}{R T_{e}} = \frac{140 \times 10^{3}}{287 \times 377.6} = 1.29176 {kg/m}^{3}

Applying the conservation of mass equation,

A_{e} = \frac{\dot{m}}{ρ_{e} V_{e}} = \frac{3}{1.29176 \times 500}

A_{e} = 0.0046448 m^{2} .

Applying the conservation of horizontal momentum,

Thrust = {\dot{m}}_{e} V_{e} + (p_{e} - p_{amb}) A_{e}

Thrust = 3 \times 500 + (140 \times 10^{3} - 100 \times 10^{3}) \times 0.0046448

Thrust = 1685.79 N .

Problem 1.13:

Solution: