How to create ground responce analysis using matlab

A 2-cm-diameter vertical water jet is injected upward by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat p

Ede4ka [16]

Answer:58.28 N

Explanation:

Given data

$dia. of nozzle \left ( d\right )=2 cm$

$initial velocity\left ( u\right )=15 m/s$

$height\left ( h\right )=2m$

Now velocity of jet at height of 2m

$v^2-u^2=2gh$

$v^2=15^2-2\left ( 9.81\right )\left ( 2\right )$

$v=\sqrt{185.76}=13.62 m/s$

$Now\ forces\ on\ plate\ are\ weight\left ( Downward\right ) and jet\ force\left ( upward\right )$

equating them

W= $\left ( \rho Av\right )v$

W= $10^{3}\times \frac{\pi}{4}\left ( 0.02\right )^2\times 13.62^2$

W=58.28 N

7 0

3 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

For the pipe 1, the flow velocity is:

$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

8 0

2 years ago

The inspector should inspect insulation in unfinished spaces, including attics, _____ and foundation areas.

RoseWind [281]

Answer:

basements

Explanation:

6 0

1 year ago

Does the Diesel engine have engine knock or detonation problem? Why?

Luda [366]

Explanation:

Yes Diesel engine have problem of knocking.

We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.

Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.

6 0

2 years ago

After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st

Mazyrski [523]

B evaluate the solution. Sorry if I'm wrong.

4 0

3 years ago