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Tatiana [17]
3 years ago
12

Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ind profile you identified, calculate the 0-1 km and 0-6 km bulk wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at 1 km and 5 km. The atmospheric pressure at 1 km above sea level is typically very close to 850 mb. The pressure at 6 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-6 km wind shear values in knots (kts). For simplicity, assume that the surface winds are due south easterly, the 850 mb winds are due southerly, and the 500 mb winds are due westerly. Show your work.

Engineering
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

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Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by
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Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

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Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
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Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

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Answer:

T_{P}=(2.6667)(10^{-3})h

Explanation:

Let's write the equation of the production rate for the assembly machine :

T_{P}=T_{C}+(n).(m).(p).(T_{D})

Where T_{P} is the production rate for the assembly machine.

Where T_{C} is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where T_{D} is the average downtime per breakdown

We are looking for the hourly production rate ⇒

1h=60min\\1min=60s ⇒

1h=3600s ⇒

6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h

60min=1h ⇒

1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h

T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h

m = 1.0 in the equation.

3 0
3 years ago
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