How to calculate expected revenue in reverse auctions (procurement)
Hint: In second price sealed bid, it is optimal to bid your own valuation for the good. So in a reverse auction, it is optimal for each person to bid their lowest possible price they are willing to supply at. Intuitively, if one individual deviates from that strategy, it doesn't increase their chance of winning the auction or increase their chance at getting a higher priced contract, since the price they supply is based on other contractors' bids.

So the expected revenue is going to be the average value of the second lowest bidder amongst ten bidders. Try to set up an integral to describe the distribution and go from there.

For a normal second price auction (not reverse):

$F(x)$ is the probability that the price will be less than or equal to some given $x$, and $f(x)$ is the corresponding density function.

$$F(x) = \frac{x-50}{50}; \quad f(x) = F'(x) = \frac{1}{50}$$

Probability that second highest bid is $x$, with $n$ bidders:

$$h(x,n) = n(n−1)f(x)(1−F(x))F(x)^{n−2}$$

Expected value of $x$ (the price):

$$\int^{100}_{50} x h(x, n) dx$$

So you can easily apply a similar logic to your reverse auction. Set up $F(x)$ so that it represents the probability that the price will be greater than or equal to some given $x$.