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blagie [28]
3 years ago
10

How would you describe the stability of the atmosphere if you noted a dry adiabatic rate of 10ºC/1000 meters, a wet adiabatic ra

te of 6.5ºC/1000 meters, and an environmental lapse rate of 7.8ºC/1000 meters?
Engineering
2 answers:
Nady [450]3 years ago
8 0

Answer:

conditionally unstable

Explanation:

Given environmental lapse rate (ELR) = 7.8°C/1000 m

Wet adiabatic lapse rate (WALR) = 6.5°C/1000 m  

Dry adiabatic lapse rate (DALR) = 10°C/1000 m  

The stability of the atmosphere can be described in three ways which are;  

a.      Absolutely stable atmosphere: Here the lifted air whether saturated or unsaturated is colder and hence heavier than the surrounding air and it would return to its initial potion when released. In this case the lifted air will descent back to its original position. It occurs when WALR is greater than ELR.  

b.     Absolutely unstable: here both saturated (and moist) and unsaturated (dry) air parcels are warmer than the surrounding and hence lighter. This means both moist and dry air parcels will rise continuously when released. This occurs when DALR is less than ELR.  

c.      Conditionally unstable atmosphere: This occurs where moist air is unstable while dry air is stable. That is moist or saturated air is warmer than the surrounding and hence it will continuously rise when released while dry air is colder than the surrounding and hence it will descent back to its original position when released.  It occurs when DALR is greater than ELR and ELR is greater than WALR (DALR>ELR>WALR).  

The situation given in the question is "c". 10°C/1000 m > 7.8°C/1000 m > 6.5°C/1000 m

muminat3 years ago
3 0

Answer:

conditional instability (Γd >  Γe > Γw)

Explanation:

Given;

dry adiabatic rate, Γd = 10ºC/1000 meters

wet adiabatic rate, Γw= 6.5ºC/1000 meters

environmental lapse rate, Γe = 7.8ºC/1000 meters

Stability of the atmosphere can be described as Absolute stability, Absolute instability or conditional instability.

Conditions for Absolute stability:

Γd > Γw > Γe

Conditions for Absolute instability:

Γe > Γd > Γw

Conditions for conditional instability:

Γd >  Γe > Γw

Thus, conditional instability satisfies the given values of the atmospheric condition: Γd (10) > Γe (7.8) >  Γw (6.5)

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Rolling and Shearing are the types of a)-Bulk Deformation Process b)- Sheet Metal Process c)- Machining Process d)- Both a &
Margarita [4]

Answer:

a)Bulk deformation process  

Explanation:

<u>Rolling</u>

Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.

<u>Shearing</u>

In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.

So from above we can say that option a is right.

5 0
3 years ago
Helium gas expands in a piston-cylinder in a polytropic process with n=1.67. Is the work positive, negative or zero?
IRINA_888 [86]

Answer:

work will be positive when it is under polytropic expansion process

Explanation:

It states a polytropic  process with n equal to 1.67. there is a polytropic expansion that mean work is positive and if it was polytropic compression then it would   be negative

PV^n = const

P_1V_1 = P_2V_2

Also work during the process of polytropic is given as

W_{1-2} =\frac{P_1V_1 -P_2V_2}{n-1}

the work will be positive when it is under the polytropic expansion process

3 0
2 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D
Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ  F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

                                         $\Delta T = -2.43-87.745$

                                           $\Delta T=-90.175^\circ  F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

   $=11.401+0.32198(82.898)$

   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

5 0
2 years ago
What are the horizontal structures beneath a slab that help transfer the load from the slab to the columns?
Delicious77 [7]

Answer: D) Beams

Explanation:

3 0
2 years ago
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Calculate the tensile modulus of elasticity for a laminated composite consisting of 62 percent by volume of unidirectional carbo
kompoz [17]

Answer:

4.30 gp

Explanation:

''.''

5 0
2 years ago
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