The following is known about the reals $ \alpha$ and $ \beta$

$ \alpha^{3}-3\alpha^{2}+5\alpha-17=0$ and $ \beta^{3}-3\beta^{2}+5\beta+11=0$

Determine $ \alpha+\beta$
1. We start with the given equations for \(\alpha\) and \(\beta\):
   \[
   \alpha^3 - 3\alpha^2 + 5\alpha - 17 = 0
   \]
   \[
   \beta^3 - 3\beta^2 + 5\beta + 11 = 0
   \]

2. Define the function \( f(x) = x^3 - 3x^2 + 5x \). Notice that:
   \[
   f(\alpha) = \alpha^3 - 3\alpha^2 + 5\alpha
   \]
   \[
   f(\beta) = \beta^3 - 3\beta^2 + 5\beta
   \]

3. From the given equations, we can express \( f(\alpha) \) and \( f(\beta) \) as:
   \[
   f(\alpha) = 17
   \]
   \[
   f(\beta) = -11
   \]

4. Consider the function \( g(x) = f(x) - 3 \). Then:
   \[
   g(x) = x^3 - 3x^2 + 5x - 3
   \]

5. We can rewrite \( g(x) \) in a different form:
   \[
   g(x) = (x-1)^3 + 2(x-1)
   \]

6. Since \( g(x) \) is strictly increasing (as the derivative \( g'(x) = 3(x-1)^2 + 2 \) is always positive), we can use this property to find the relationship between \(\alpha\) and \(\beta\).

7. Given \( g(\alpha) = 14 \) and \( g(\beta) = -14 \), we have:
   \[
   (\alpha-1)^3 + 2(\alpha-1) = 14
   \]
   \[
   (\beta-1)^3 + 2(\beta-1) = -14
   \]

8. Let \( t = \alpha - 1 \) and \( s = \beta - 1 \). Then:
   \[
   t^3 + 2t = 14
   \]
   \[
   s^3 + 2s = -14
   \]

9. Since \( g(x) \) is strictly increasing, the only way for \( t \) and \( s \) to satisfy these equations is if \( t = -s \). Therefore:
   \[
   \alpha - 1 = -( \beta - 1 )
   \]
   \[
   \alpha - 1 = 1 - \beta
   \]

10. Solving for \(\alpha + \beta\):
    \[
    \alpha + \beta - 1 - 1 = 0
    \]
    \[
    \alpha + \beta = 2
    \]

The final answer is \(\boxed{2}\).