Exercise:

A star is at Right Ascension 5h 0m and declination +26°20'.
The latitude φ is +56°20'.
Local Sidereal Time is 5h 0m.
Atmospheric pressure is 1050 millibars,
and the temperature is +5°C.
What is the star’s true altitude?

LST = RA, so the star is on the local meridian,
so altitude a = (90°-φ) + δ = 60°.

How much will the star’s image be shifted by atmospheric refraction,
and in which direction?

Zenith distance z = 90°-altitude = 30°.
Angle of refraction R = k tan(z')
where k = 16.27" P/(273+T) = 16.27" x 1050 /278 = 61.45".

However, we don't know z', only z.
For a first, approximate answer, take z = z' = 30°.
This gives R = 35.5",
so z' = z+R = 30° 0' 35.5".

Now recalculate R = k tan(z') = 35.5"
(unchanged, so no need to iterate further).

So the star is raised by 35.5".

What will be the star’s Right Ascension and declination,
corrected for refraction?

Since the star is on the local meridian, the shift is only in declination.
The apparent altitude is increased by 35.5",
so the apparent declination is similarly increased.
So the apparent coordinates are:
Right Ascension 5h 0m and declination +26°20'35.5".

Back to "refraction".