Question

Nuclear power plants often exhaust steam, which you can consider to have the properties of standard air at 5000 ft (since it wants to rise when emitted). Rather than rising, a horizontal wind causes the stack air to move horizontally. What is the horizontal force (in pounds) that the wind imposes on the rising air to cause it to be redirected?

6% 3) (23 pts) Nuclear power plants often exhaust steam, which you' can consider to have the properties of standard air at 5000 ft (since it wants to rise when emitted). Rather than rising, a horizontal wind causes the stack air to move horizontally. What is the horizontal force (in pounds) that the wind imposes on the rising air to cause it to be redirected? , 15%

Answer 1

Wind force is to be calculated for the given elevation of 5000 feet.

As mentioned in the question a horizontal wind causes the stack air to move horizontally – that is there is a fanning plume formation.

Let there is a stack discharge or stack air emission, Q = 1lb/s.

The velocity of emission v_s = is 6ft/s

Force of stack air emission, F_s= Q*v_s

F_s= 6 lbf

The resultant force F_eq =F_s (in magnitude)

And there is a wind velocity, v_w = 15ft/s

F_w = Q_w*v_w

By using vector algebra of forces

F_eq = sqrt (F_w² + F_s² - (2* F_w * F_s))

6 = sqrt (F_w² + 6² - (2* F_w * 6))

36 = F_w² + 36- 12* F_w

F_w = 12 lbf

Horizontal component of wind force is F_{w *} cos 45⁰

F_wH = 8.485 lbf

Wind force is to be calculated for the given elevation of 5000 feet. As mentioned in the question a horizontal wind causes the stack air to move horizontally- that is ere is a fanning plume formation. Let there is a stack discharge or stack air emission, Q-1lb/s e velocity of emission Vs is 6ft/s Force of stack air emission, Fs QV resultant force Fs (in magnitude) And there is a wind velocity, Hx-15ft/s By using vector algebra of forces 36 = Fw2+36-12° Ex Ex= 12 lbf Horizontal component of wind force is Ex cos 45 FeaF 6