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2 years ago

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2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to

a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process.

1 answer:

Korolek [52]2 years ago

8 0

Answer:

Work input =283.47 KJ

Explanation:

Given that

$P_1=150\ KPa$

$P_2=600\ KPa$

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

So now putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=2.5\times 0.287\times 285\ln \dfrac{150}{600}$

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

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Explain why the fundamental software engineering principles of process, dependability, requirements management, and reuse are re

Svet_ta [14]

Answer:

1. Costs and frequency of change.

2. The most important ‘non-functional’ requirements. Different systems have different priorities for non-functional requirements.

3. The software lifetime and delivery schedule.

Explanation:

The process of requirement gathering is required in both generic and customized software.

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2 years ago

Read 2 more answers

. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 r

forsale [732]

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave) = 5.8 in ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = $\frac{w1}{w2}$ = 950 / 250

also since the speed of belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : $\frac{w1}{w2} = \frac{d2}{d1}$ = VR

Appropriate belt size ( d2) can be calculated as

d2 = $\frac{w1d1}{w2}$ = $\frac{950 * 5.8}{250}$ = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

$L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}$

inputting all the values into the above equation including the value of C as calculated below

L ≈ 140 in

Calculating the center distance

we use this equation to get the ideal center distance

$d2< C_{ideal} < 3( d1 +d2)$

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7 but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

$C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}$ -------- equation 3

B = $4L -2\pi (d2 + d1 )$

inputting all the values into (B)

B = 140(4) - 2 $\pi$ ( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * $\frac{d1}{2}$

= 99.48 * 5.8 / 2 = 288.49 in/s

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2 years ago

Assume that an attacker knows that a user’s password is either p1 = abcd or p2 = bedg. Say the user encrypts his password using

denis23 [38]

Answer: I will Show how the attacker can determine the user’s password

Explanation:

If we assume an attacker knows that a user’s password is either abcd or bedg. Say the user encrypts his password using the shift cipher, and the attacker sees the resulting ciphertext. Show how the attacker can determine the user’s password, or explain why this is not possible.

The alphabet{A, B, . . . , Z}is identified with the set Σ = {0, 1, . . . 25} and all additions are implicitly taken mod26.

Then, the possible passwords are p0 = abcd = (0, 1, 2, 3) and p1 = bedg = (1, 4, 3, 6).

All possible encryptions of p0 are C0 = {(k, k + 1, k + 2, k + 3) | k ∈ Σ} and the ones of p1 are C1 = {(k + 1, k + 4, k + 3, k + 6) | k ∈ Σ}.

These two sets are disjoint and so in order to obtain the password it is necessary to check in which set the ciphertext lies.

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2 years ago

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

Vilka [71]

Answer:

24.87 mm

Explanation:

The area of the bolt is given by

$A_b=\pi r^{2}$

Since diameter is 16mm, the radius is 16/2= 8 mm= 0.008 m

Area, $A_b=\pi\times 0.008^{2}=0.000201062 m^{2}$

For safe design of the bolt, we use stress of 201 Mpa

$\sigma_b=\frac {P}{A_b}$ where P is the load and $\sigma_b$ is normal stress.

Making P the subject then

$P=A_b \sigma_b$

Substituting the figures given and already calculated area of bolt

$P=201\times 10^{6}\times 0.000201062 m^{2}=40413.4479 N$

The area of spacer is given by

$\pi (r_o^{2}- r_i^{2})$ where r is radius and the subscripts o and I denote inner and outer respectively

The value of 142 Mpa by default becomes the stress on spacer hence

$\sigma_s=\frac {P}{A_s}$ and making $A_s$ the subject then

$A_s=\frac {P}{\sigma_s}$

$\pi (r_o^{2}- r_i^{2})=\frac {P}{\sigma_s}$

Since we already calculated the value of P as 40413.4479 N and inner radius is 8mm= 0.008 m then

$\pi (r_o^{2}- 0.008^{2})=\frac {40413.4479}{142\times 10^{6}}$

$r_o^{2}=0.000154592$

$r=\sqrt{0.000154592}=0.012433 m$

Therefore, $d=2r=2*0.012433=0.024867 m=24.86697 mm\approx 24.87 mm$

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2 years ago

A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The s

Sonbull [250]

Answer:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A ( here R is the hydraulic Radius)

by substituting the given data in above formula we get:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

here, A is the flow area and P is the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

<u>d = 1.42 feet</u>

<u></u>

Depth of stilling channel before the jump will be = d1 = 8 feet

Depth of stilling channel after the jump will be = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

= 5( 8 - 1.42)

<u>Length of stilling basin = 32.9 feet </u>

Now calculating the height of end sill:

Jump height = (8 - 1.42)

<u>Height of end sill = 6.58 feet</u>

8 0

2 years ago