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maw [93]
3 years ago
11

Why do high and low tides happen 1 hour later each day? just answer please and it is a written response

Physics
1 answer:
8_murik_8 [283]3 years ago
6 0
With each<span> passing </span>day<span>, the </span>high tides occur<span> about an </span>hour later<span>. The moon rises about an </span>hour later each day<span>, too (actually, 54 minutes </span>later<span>). Since the moon pulls up the </span>tides<span>, these two delays are connected. As the earth rotates through </span>one day<span>, the moon moves in its orbit.</span>
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Explanation:

Given\\V=3ft^3\\T_{1} =300^oF\\T_{L} =400^oF\\P_{L} =200psia\\V_{f} =0.5V

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a) Taking tank as a system, The energy balance can be define as

E_{in}- E_{out}= E_{sys}

m_{i} h_{L} -Q_{out}=m_{2}  u_{2}-m_{1}  u_{1}

The mass balance could be written as

m_{in}- m_{out}= m_{sys} \\m_{i}= m_{2}- m_{1}

The final pressure in the tank could be defined as following

P_{2} =P_{sat300^oF}

from standard steam table we know at

T_{2}= T_{1}=300^oF\\ P_{2}=67.028psia

b)

From steam table at

T_{1} =300^oF\\v_{1}= v_{g}=6.4663ft^3/lbm\\v_{1}=  v_{g}=1099.8Btu/lbm\\ v_{f} =0.01745ft^3/lbm\\v_{f} =269.51Btu/lbm\\

at\\P_{L} =200psia\\T_{L}=400^oF\\h_{L}=1210.9Btu/lbm

initial mass in the tank could be define as

m_{1}=\frac{V}{v_{1} }  \\m_{1} =\frac{3}{6.4663} =0.464lbm\\

Final mass in the tank could be define as

m_{2}=\frac{V_{f} }{v_{f} }+\frac{V_{g} }{v_{g} }  \\ m_{2} =\frac{1.5}{0.01745} +\frac{1.5}{6.4663} =86.2lbm

The amount of steam that has entered the tank

m_{i}=m_{2}-  m_{1}\\ m_{i}=86.2-0.464=85.74lbm

c)

The internal energy in final state could be defined as following

U_{2}=m_{f}  u_{f}+ m_{g} u_{g}\\ U_{2} =85.96*269.51+0.232*1099.8=23422Btu\\

The heat transfer could be defined as following

Q_{out}=m_{i}  h_{L}+m_{1}  u_{1}-m_{2}  u_{2}\\ Q_{out}=85.74*1210.9+0.464*1099.8-23422=80910Btu

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