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Chapter 4: Problem 84

At rated thrust, a liquid-fueled rocket motor consumes \(80 \mathrm{kg} / \mathrm{s}\) of nitric acid as oxidizer and \(32 \mathrm{kg} / \mathrm{s}\) of aniline as fuel. Flow leaves axially at \(180 \mathrm{m} / \mathrm{s}\) relative to the nozzle and at \(110 \mathrm{kPa}\). The nozzle exit diameter is \(D=0.6 \mathrm{m}\). Calculate the thrust produced by the motor on a test stand at standard sealevel pressure.

Short Answer

Expert verified
The thrust produced by the rocket motor on a test stand at standard sealevel pressure is 22596 N.

Step by step solution

01

Gather the given data

From the exercise, the following data is given:\[ \text{Mass flow rate of oxidizer } (m_1) = 80 \, \text{kg/s} \] \[ \text{Mass flow rate of fuel } (m_2) = 32 \, \text{kg/s} \] \[ \text{Ejection velocity } (V_e) = 180 \, \text{m/s} \] \[ \text{Nozzle exit pressure } (P_e) = 110 \, \text{kPa} \] \[ \text{Nozzle exit diameter } (D) = 0.6 \, \text{m} \] \[ \text{Standard sealevel pressure } (P_a) = 101.3 \, \text{kPa} \]
02

Calculate total mass flow rate

The total mass flow rate is given by the sum of the mass flow rates of the fuel and oxidizer. It is calculated as follows: \[ m = m_1 + m_2 = 80 \, \text{kg/s} + 32 \, \text{kg/s} = 112 \, \text{kg/s} \]
03

Convert pressures into same units

Before carrying out the thrust calculations, ensure that the pressures (nozzle exit pressure and standard sealevel pressure) are in the same units. Given that the sealevel pressure is in kPa, convert 110 kPa to Pa. \[ P_e = 110 \times 10^3 \, \text{Pa} = 110000 \, \text{Pa} \] \[ P_a = 101.3 \times 10^3 \, \text{Pa} = 101300 \, \text{Pa} \]
04

Calculate the area of rocket nozzle

The area of the rocket nozzle is calculated using the equation for the area of a circle. The resulting value is necessary for the thrust calculation. \[ A = \frac{蟺D^2}{4} = \frac{蟺 \times (0.6)^2}{4} = 0.28 \, \text{m}^2 \]
05

Calculate the thrust

Using the rocket thrust equation: Thrust \( T = mV_e +(P_e - P_a)A \), where \( m \) is total mass flow rate, \( V_e \) is ejection velocity, \(P_e\) is nozzle exit pressure, \(P_a\) is atmospheric pressure and \( A \) is the nozzle area. Therefore, \[ T = (112 \, \text{kg/s} \times 180 \, \text{m/s}) + [(110000 \, \text{Pa} - 101300 \, \text{Pa}) \times 0.28 \, \text{m}^2] = 20160 \, \text{N} + 2436 \, \text{N} = 22596 \, \text{N} \]

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