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Galina-37 [17]
3 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]3 years ago
8 0
B) The convection current will reverse direction, reversing the winds.
wlad13 [49]3 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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The self weight of dam.

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An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s
nignag [31]

Given:

X_{L} = 50.0 \ohm

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

V_rms} = 85.0 V

Solution:

To calculate max current in inductor, I_{L(max):

At f = 60.0 Hz

X_{L} = 2\pi fL

50.0 = 2\pi\times 60.0\times L

L = 0.1326 H

Now, reactance X_{L} at f' = 45.0 Hz:

X'_{L} = 2\pi f'L

X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm

Now, I_{L(max) is given by:

I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}

I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A

Therefore,  max current in the inductor, I_{L(max) = 2.13 A

7 0
3 years ago
What impact does modulus elasticity have on the structural behavior of a mechanical design?
devlian [24]

Answer with Explanation:

The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:

1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.

2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.

3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity  the resistance it offer's to shear forces and the torques is more.

While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.

8 0
3 years ago
A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg
lakkis [162]

Answer:

L=107.6m

Explanation:

Cold water in: m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C

Hot water in: m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C

D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m

Step 1: Determine the rate of heat transfer in the heat exchanger

Q=m_{c}C_{c}(T_{c,out}-T_{c,in})

Q=1.2*4.18*(80-20)

Q=1.2*4.18*(80-20)

Q=300.96kW

Step 2: Determine outlet temperature of hot water

Q=m_{h}C_{h}(T_{h,in}-T_{h,out})

300.96=2*4.18*(160-T_{h,out})

T_{h,out}=124\°C

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

dT_{1}=T_{h,in}-T_{c,out}

dT_{1}=160-80

dT_{1}=80\°C

dT_{2}=T_{h,out}-T_{c,in}

dT_{2}=124-20

dT_{2}=104\°C

LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}

LMTD = \frac{104-80}{ln(\frac{104}{80})}

LMTD = \frac{24}{ln(1.3)}

LMTD = 91.48\°C

Step 4: Determine required surface area of heat exchanger

Q=UA_{s}LMTD

300.96*10^{3}=649*A_{s}*91.48

A_{s}=5.07m^{2}

Step 5: Determine length of heat exchanger

A_{s}=piDL

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L=107.57m

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