Statistical power of tests on unequal and known variances vs equal but unknown variances Suppose you are testing for differences in two normally distributed variables. In the following situations:
-Variables have known but unequal variances
-Variables have variances that are unknown but equal
Which situation would be generally expected to have more statistical power? I believe tests on distributions with unequal variances are expected to have less statistical power, but does knowing the variance allow for more power, even if they are unequal?

The question does not specify all the relevant information:

*

*You do not specify sample sizes. I chose $n_1=n_2=10$ throughout. I would expect the z test
for known, unequal variances to have about the same power
as a roughly comparable pooled t test. (For small sample sizes, maybe a little higher.)


*You do not specify how unequal the variances would be
in the known/unequal case. I chose $\sigma_1=3, \sigma_2=5.$


*You do not specify the difference in means. I have chosen
$\mu_1=50, \mu_2=54$ for both z and t tests.
Below are straightforward simulations in R of the power
in two specific situations:
# known, unequal
set.seed(2022)
sg1 = 3;  sg2 = 5
n = 10
mu1 = 50;  mu2 = 54
se = sqrt(sg1^2/n + sg2^2/n)
z = replicate(10^5, ( mean(rnorm(n,mu1.sg1)) - 
                      mean(rnorm(n,mu2,sg2)) )/se)
mean(abs(z) >= 1.96) # aprx power z test
[1]  0.58432


# unknown, equal
sg = 4
n = 10
mu1 = 50;  mu2 = 54
t = replicate(10^5, t.test(rnorm(n,mu1,sg),
                     rnorm(n,mu2,sg), var.eq=T)$stat)
c = qt(.975, 18); c
[1] 2.100922
mean(abs(t) >= c)
[1] 0.56095

So, for my specific version of your general question,
the z test has power about 58% while the t test has
power about 56%.
Notes: (1) The known/unequal case is a z-test, for which it is not difficult to find the power of a the test against a specific
alternative. The unknown/equal case is a pooled 2-sample
t test, for which the power depends on a non-central t distribution. Also, power and sample size procedures are
widely available in various statistical softwares and online.
So, you should not have difficulty investigating whatever variety of sample sizes, effects, and variances you like.
(2) Here is a partial printout of the power of the
t test from a recent release of Minitab. Results
agree with my second simulation:
Power and Sample Size 

2-Sample t Test

Testing mean 1 = mean 2 (versus ≠)
Calculating power for mean 1 = mean 2 + difference
α = 0.05  Assumed standard deviation = 4


            Sample
Difference    Size     Power
         4      10  0.562007

The sample size is for each group.

(3) If this is a class assignment, I suppose the purpose is for you to look up and use the approximate formula for the power of a two-sided, two-sample z test. Also, to become familiar with methods of finding the power of a pooled two-sample t test.
