Karl Lui Barrus says:
Encrypting with k1 and then k2 leaves you open to the "meet in the middle" attack.
Say I get a copy of the plaintext and ciphertext. I could encrypt the plaintext with 2^56 keys, and decrypt the ciphertext with 2^56 keys. Then by matching results of the above steps, I could figure out k1 and k2.
Tell you what, Karl -- when you build the device that can store 2^56 encryptions, let us know. You'll make a mint in the storage technology business. Also let us know how you'll index and fetch the encryptions in any reasonable time while you are at it, but by comparison thats a tiny problem.
The work for this attack is 2^56 + 2^56 = 2^57, which suggests that double encryption doesn't increase the complexity of breaking your text very much.
Karl, are you sure that you want people to think you believe this? Perry