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Galina-37 [17]
2 years ago
14

At the coast on a summer day, the land is hotter than the ocean. Warm air over the land rises and is replaced by cooler air, cau

sing a sea breeze.
At night, the land cools quickly and the ocean is now warmer than the land. What will happen next?

Winds will stop blowing as the convection current ceases to flow.
The convection current will reverse direction, reversing the winds.
Winds will increase as the two convection currents overlap.
The convection current will move seaward, reversing the winds.
Engineering
2 answers:
Alona [7]2 years ago
8 0
B) The convection current will reverse direction, reversing the winds.
wlad13 [49]2 years ago
6 0

Answer:At the coast on a summer day, the land is hotter than the ocean. Warm air over the land rises and is replaced by cooler air, causing a sea breeze.

At night, the land cools quickly and the ocean is now warmer than the land. What will happen next?

1) Winds will stop blowing as the convection current ceases to flow.

CORRECT-->> 2) The convection current will reverse direction, reversing the winds.

3) Winds will increase as the two convection currents overlap.

4) The convection current will move seaward, reversing the winds.

Explanation:

Edge 2021

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A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

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2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

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==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

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3 years ago
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The minimum safe working distance from exposed electrical conductors
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Answer:

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3 years ago
In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie
Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.  

Given:  

Rake angle is 12°.  

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.  

Calculation:  

Step1  

Chip reduction ratio is calculated as follows:  

r=\frac{t}{t_{c}}

r=\frac{0.32}{0.65}

r = 0.4923

Step2  

Shear angle is calculated as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

Here, \phi is shear plane angle, r is chip reduction ratio and \alpha is rake angle.  

Substitute all the values in the above equation as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}

tan\phi=\frac{0.48155}{0.8976}

\phi=28.21^{\circ}

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

\gamma=cot\phi+tan(\phi-\alpha)

\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})\gamma = 2.155.

Thus, the shear strain rate is 2.155.

6 0
3 years ago
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