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anastassius [24]
3 years ago
10

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where

p is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P-1.2 kg/m and V-60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column
Engineering
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

a. 430.944 pascal

b. 0.0625psi

c. 1.73008inH20

Explanation:

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where p is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P-1.2 kg/m and V-60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column

checking the dimensional consistency

Dp=\frac{kg}{m^3} (\frac{m^2}{s^2} \\\frac{x}{y} \\\\\frac{kg}{ms^2} \\which s the same as\\ \frac{N}{m^2}

Dp=∈(\frac{v^2}{2} )

convert 1 mile to meter

1mile=1609m

1h=3600s

60mile/h=26.8m/s

slotting intpo the relation

Dp=1.2kg/m^3(\frac{26.8^2}{2} )

430.944kg/(ms^2)

which is the same as 430.944N/m^2

expressing in pascal.

We know that

1 pascal=1 N/m^2

430.944 pascal

2. 1 pascal=0.000145psi

answer=0.0625psi

3.1 pascal=0.00401inH20

answer=430.944*0.0040146

1.73008inH20

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Answer:

a) Please see attached copy below

b) 0.39KJ

c)  20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

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b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

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c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

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Which of the following conditions were present in over 80% of paddling fatalities from 1995-2000?
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A sum of $500,000 will be invested by a firm two years from now. If money is worth 12%, what will be the worth of this investmen
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Answer:

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Explanation:

given data

sum = $500,000

rate = 12% =0.12

total time = 10 year

solution

as present value After 2 years from now is $500,000

so time period is now = 8 year  ( 10 - 2 )

so we apply future value formula that is

Future value  = present value × (1+r)^{t}   ............1

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Future value  = $500,000 × (1+0.12)^{8}  

Future value  = $500,000 × 2.476

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8 0
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The entropy change of the air is 0.240kJ/kgK

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T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

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Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

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attached below

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