/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 Gases leaving the propulsion noz... [FREE SOLUTION] | 91Ó°ÊÓ

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Gases leaving the propulsion nozale of a rocket are modeled as flowing radially outward from a point upstream from the nozzle throat. Assume the speed of the exit flow, \(V_{e}\) has constant magnitude. Develop an expression for the axial thrust, \(T_{a}\) developed by flow leaving the nozzle exit plane. Compare your result to the one-dimensional approximation, \(T=m v_{e^{-}}\) Evaluate the percent error for \(\alpha=15^{\circ} .\) Plot the percent error versus a for \(0

Short Answer

Expert verified
For \(\alpha=15^{\circ}\), the percent error calculated between the axial thrust and the one-dimensional approximation of thrust is \(|1-\cos(15^{\circ})|\times 100\%\).

Step by step solution

01

Develop the expression for the axial thrust

The axial thrust, \(T_{a}\), can be calculated using the formula \(T_{a}=m V_{e} \cos(\alpha)\) where m is the mass flow rate of the gases, \(V_{e}\) is the exit flow velocity, and \(\alpha\) is the angle of flow with respect to the axial direction.
02

Compare the derived expression to the one-dimensional approximation

The one-dimensional approximation of thrust is given by the formula \(T=m v_{e^{-}}\). Here, since the speed of the exit flow, \(V_{e}\) is given as constant, \(V_{e}=v_{e^{-}}\). Hence, we can see that the only difference between the two expressions is the \(\cos(\alpha)\) term. The one-dimensional approximation assumes the gas flow is purely axial, i.e., \(\alpha=0\), for which \(\cos(0)=1\). Therefore, the two expressions would be equal at \(\alpha=0\).
03

Evaluate the percent error at \(\alpha=15^{\circ}\)

The percent error can be evaluated using the formula: \(\text{Percent error} = \frac{|T_{a} - T|}{T} \times 100\% = \frac{|m V_{e} \cos(\alpha) - m v_{e^-}|}{m v_{e^-}} \times 100\%\). Substituting \(\alpha=15^{\circ}\) and considering that \(V_{e}=v_{e^-}\), we obtain an error of \(|1-\cos(15^{\circ})|\times 100\%\).
04

Plot the percent error for the range \(0

For each angle \(\alpha\) in the range \(0<a<22.5^{\circ}\), evaluate the percent error using the formula derived in Step 3. The resulting graph will have percent error on the y-axis and \(\alpha\) on the x-axis.

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