From 9199e95568faaf8166e384f1b466f9bf4d9b25b0 Mon Sep 17 00:00:00 2001
From: Alfin Johny <alfin.johny@tum.de>
Date: Mon, 10 Feb 2025 10:33:18 +0100
Subject: [PATCH] Align math equations to the middle
---
.../sizing/propulsion_design/engineering_principles.md | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/docs/documentation/sizing/propulsion_design/engineering_principles.md b/docs/documentation/sizing/propulsion_design/engineering_principles.md
index 51bfec5..1c15b66 100644
--- a/docs/documentation/sizing/propulsion_design/engineering_principles.md
+++ b/docs/documentation/sizing/propulsion_design/engineering_principles.md
@@ -47,15 +47,15 @@ The _scale factor_ is necessary because (as conceptual aircraft designer), we us
So, the scaling is based on continuity principle assuming that the operating condition is constant (commonly known station numbering; assuming no pressure drop).
-$ \textcolor{white}{T = \dot{m} \cdot (V_9 - V_0)} $
+$$ \textcolor{white}{T = \dot{m} \cdot (V_9 - V_0)} $$
Therefore, thrust $T$ is proportional to the mass flow $\textcolor{white}{\dot{m}}$, which is related to the cross-sectional area $A$ of the engine.
-$ \textcolor{white}{\dot{m}} = \rho \cdot V \cdot A = \rho \cdot V \cdot \pi \left(\frac{d}{2}\right)^2 $
+$$ \textcolor{white}{\dot{m}} = \rho \cdot V \cdot A = \rho \cdot V \cdot \pi \left(\frac{d}{2}\right)^2 $$
Because area $A$ is proportional to the square of the diameter $d$ , it follows that the diameter should be proportional to the square root of the scale factor.
-$ \textcolor{white} d_{new} = d_{ref} \cdot ( \frac{T_{new}}{T_{ref}} )^{0.5} $
+$$ \textcolor{white} d_{new} = d_{ref} \cdot ( \frac{T_{new}}{T_{ref}} )^{0.5} $$
An exemplary simplified calculation (data from the V2527-A5): the current engine provides $127.27~kN$ as sea level static thrust, but for the design only $100~kN$ are needed. The scaling factor would be $0.7857$. Assuming an initial diameter $2~m$, the new diameter would be $1.773~m$ with the scaling factor of $(0.7857)^{0.5} = 0.8864$.
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