Oh my gosh, can't believe I didn't write that! It was late and I was sleepy. Yes, of course! So we substitute that into the mass balance for ##Q## and we're set, right?Chestermiller said:I get $$P_1=P_2+\frac{\rho}{2}\left(\frac{Q}{\pi R^2}\right)^2$$
You mean you substitute that into the momentum balance, not the mass balance. Hopefully, that will lead to a result which matches the given answer.joshmccraney said:Oh my gosh, can't believe I didn't write that! It was late and I was sleepy. Yes, of course! So we substitute that into the mass balance for ##Q## and we're set, right?
Yeeeeees, this is what I meant! Hahahaha evidently I'm still tired. Thanks a ton!Chestermiller said:You mean you substitute that into the momentum balance, not the mass balance. Hopefully, that will lead to a result which matches the given answer.