Having problems trying to solve part b of this problem.

Rocket Propulsion Elements 8th - Chapter 2

  1. A jet of fluid hits a stationary flat plate in the manner shown below.

(a) If there is 50 kg of fluid flowing per minute at an absolute velocity of 200 m/sec, what will be the force on the plate?

(b) What will this force be when the plate moves in the direction of flow at u = 50 km/h? Explain the methodology.

Part A just involves plugging in mass flow rate, changing units to seconds, multiplying it by characteristic velocity to get force.

Part B, I have zero idea what's going on. I thought it was a conservation of momentum problem, so I set up the problem as F = dm * v^2, where v is 200-13.8 (13.8 being the change in speed in m/s as the plate moves away). The problem is, I'm getting 155 N, whereas the correct answer is 144 N.

  • $\begingroup$ Not an answer: $F = dp/dt$. If $p = mv$ then $F= d(mv)/dt$, if $v$ ($dx/dt$) is constant then it's $F = v \ dm/dt$ Step 1 is figure out your flow in kg/m ($dm/dx$, kilograms per meter) in the initial case, then choose your velocity carefully, then use $dm/dt = dm/dx \times dx/dt$. hmm... I get (200 - 13.899) * 0.8333 = 155.08 N also! I get 144 only if the block moves at 100 km/hr. $\endgroup$ – uhoh Nov 11 '18 at 18:32
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    $\begingroup$ The answer might be misprinted as well. Check with your classmates solution as well to verify the answer. $\endgroup$ – Amar Nov 11 '18 at 20:05
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    $\begingroup$ Just thinking of it another way: if the plate was moving 200 m/s the force would be zero N, when it's moving 0 m/s the force is 167 N. I see nothing to indicate the variation isn't linear so 155 N at a speed of 13.8 m/s makes sense to me. Incidentally in my tattered 4th edition it's 50 lb/s, 200 ft/s, and 50 mph. Plus they don't give the answers :) $\endgroup$ – Organic Marble Nov 11 '18 at 22:07
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    $\begingroup$ @amar actually not reading this book for a class: My school doesn't have an aeronautics or astronautics program, so I'm going through the process of self-teaching myself. Good to know I wasn't completely crazy in my calculation, thanks for the help everyone! $\endgroup$ – Colin Warn Nov 17 '18 at 16:25

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