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Bingel [31]
3 years ago
12

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced press

ure PR approaches zero. How do you explain this behavior?
Engineering
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

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What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-
Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}

Where \sigma_{m} is the magnitude of he maximum stress at the tip of the crack, \sigma_{p} is the magnitude of the tensile stress, l_{c} is 1/2 the length of the internal crack, and r_{c} is the radius of curvature of the crack.

We have:

r_{c}=1.9*10^{-4} [mm]

l_{c}=3.8*10^{-2} [mm]

\sigma_{c}=140 [MPa]

We replace:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}

We get:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]

5 0
3 years ago
use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square
Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25  

 = 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000  

 = 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives  

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh  

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as  

E = 968.75* 103 / 30000  

E = 32.291 KBtu/sf.

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4 years ago
How to find Catenary length in sag
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Answer:

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2 years ago
Pls help!!! will give brainly!!!
gogolik [260]
Short-term goals are planned for the near future while long-term goals will take longer to complete. An example of a short-term goal is trying to get into a certain college. A long-term goal is wanting to have a big family.
3 0
3 years ago
An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into
olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}

For the 9 mm:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391

For the 0.44 caliber:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460

The ratio is 460 / 391 = 1.176

6 0
3 years ago
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