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3 years ago

referring to either the CMS file or code book index, what is the cross reference for reduction of a dislocation?

Engineering

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

reposition

Explanation:

if you go to your icd 10 pcs index located on page 1 then look up reduction the subterm is of a dislocation which leads you to the answer reposition

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A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a strip

hram777 [196]

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

$\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}$ ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

$X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167$

and

x₁ = 0.2% = 0.002

$X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}$

Now we have to use the balance equation a

$\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}$ .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

$L_s$ = 300(1 - 0.04) = 288 kmol pure oil

G = $G_s$ = 11.42 kmol

$Y_1$ = 0 , solvent free steam

substitute into the equation a

$\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}$

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ <em>Number of trays = 6</em>

5 0

4 years ago

Read 2 more answers

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

1. If a bolt is size 1/2" or larger, then its corresponding wrench size should be____ larger than the bolt size

Mrac [35]

Answer:

C. 1/4"
B. 3/16"

Explanation:

1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.

2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".

3 0

4 years ago

Why is California a good place for engineers to build suspension bridges?

o-na [289]

Because there a lot of people who live there so they need some engineers

7 0

3 years ago

What is the greatest speedup (in terms of throughput) possible with pipelining? You do not need to worry about the register timi

RUDIKE [14]

Answer:

The greatest speedup possible with pipelining is 2.5 times

Explanation:

As we know

Span of stage = Maximum span of any span

As stage B has the maximum duration of time, So

Duration of time of each stage = Time duration of Stage B = 8 ns

Now calculate the total time without pipeline = Duration of Stage A + Duration of Stage B + Duration of Stage C + Duration of Stage D = 5 ns + 8 ns + 4 ns + 3 ns = 20 ns

So the greatest speedup can be calculated as follow

Greatest speedup = Total time without pipeline / Duration of time of each stage = 20 ns / 8 ns = 2.5 times

3 0

3 years ago