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arlik [135]
3 years ago
6

Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and i

nclude appropriate units.

Engineering
1 answer:
lina2011 [118]3 years ago
3 0

Complete Question

A force of F =460 lb. acts on the curved member as shown. Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and include appropriate units.(Figures Attached)

Answer:

Resultant Normal Force = 55. 02 lb.

Explanation:

In Order to find the resultant normal force we must first resolve the figure to find out the x and y components of the force across the cross-section at point B. Hence after resolving the figure we can find the resultant normal force by finding  the vertical component of the force: ∑ F_y

The normal force acting at point B will be:

∑ F_y :-

N_F + 460(4/5)sin〖30〗^° - 460(3/5) sin〖60〗^° = 0

N_F + 184 - 239.02 = 0

N_F = 55. 02 lb.

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Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

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Mass Balance

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\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

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b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

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