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skelet666 [1.2K]

3 years ago

8

R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its

specific entropy remains constant. Determine the change in the enthalpy of R-134a as it passes through the turbine.

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant

hence ; s1 = s2

by interpolation ; h2 = 106.95

Δ enthalpy = h2 - h1

= ( 106.95 - 129.85 ) = -23 Btu/Ibm

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In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

5 0

3 years ago

Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should w

Setler79 [48]

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

$Re_{5}$ = $Re_{12}$

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

$Re_{5}$ = Reynold number of water pipe

$Re_{12}$ = Reynold number of oil pipe

$V_{5}$ = Velocity of water 5 diameter pipe = ?

$V_{12}$ = Velocity of oil 12 diameter pipe = 2.30

$v_{5}$ = Kinetic Viscosity of water = 1 x $10^{-6}$ $m^{2}$ /s

$v_{12}$ = Kinetic Viscosity of oil = 4 x $10^{-6}$ $m^{2}$ /s

$D_{5}$ = Diameter of pipe used for water = 0.05 m

$D_{12}$ = Diameter of pipe used for oil = 0.12 m

Use the formula

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

By Removing square rots on both sides

${\frac{V_{5}XD_{5} }{v_{5}}}$ = ${\frac{V_{12}XD_{12} }{v_{12}}}$

${V_{5}$ = ${\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}$ x $v_{5}$

${V_{5}$ = [ (0.23 x 0.12m ) / (4 x $10^{-6}$ $m^{2}$ /s) x 0.05 ] 1 x $10^{-6}$ $m^{2}$ /s

${V_{5}$ = 1.38 m/s

4 0

3 years ago

Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address

Arada [10]

Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

5 0

3 years ago

Two steel plates of 15 mm thickness each are clamped together with an M14 x 2 hexagonal head bolt, a nut, and one 14R metric pla

Blababa [14]

Answer:

a) 50 mm

b) 808.24 MN/m

Explanation:

Given:

Thickness of each steel plate = 15mm

a) To find suitable length of bolt, we'll use:

Length of bolt = grip length + height of nut

To find the grip length since there is a washer, we'll use:

Grip length = plate thickness + washer thickness

Since we have 2 plates of 15mm thickness,

Plate thickness = 15 + 15 = 30mm

Using the table, a metric plain washer has a thickness of 3.5mm

Grip length = 30 + 3.5 =33.5 mm

Height of nut: Using table A-31, height of hexagonal nut is 12.8 mm

Therefore,

Length of bolt = 33.5 + 12.8 = 46.3

Rounde up to the nearest 5mm, we'll get 50mm

Length of bolt = 50mm

b) Bolt stiffness:

Threaded length for L ≤ 125mm

LT = 2*d + 6

Where d = 14mm, from table(8-7)

= 2*14 + 6

= 34 mm

Area of unthreaded portion:

Ad= $\frac{\pi}{4} d^2 = \frac{pi}{4} * 14^2 = 153.94 mm^2$

Length of unthreaded portion in grip:

Ld = 50 - 34 = 16mm

Length of threaded portion in grip:

Lt = 33.5 - 16 = 17.5mm

Bolt stiffness = $\frac{A_d A_t E}{A_d l_t + A_t l_d}$

$= \frac{153.94 * 115 * 207}{(153.94*17.5)+(115*16)} = 808.24$

Bolt stiffness = 808.24 MN/m

7 0

3 years ago

Thomas Edison modification of the filament of the light bulb is an example of the "identify the problem" step of the engineering

777dan777 [17]

True , i’m pretty sure

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3 years ago