A positive slope on a position-time graph suggests

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago

Based on the graphs of stress-strain from the V-MSE site, how would you characterize the general differences between polymers an

Pepsi [2]

Answer:

Option A

Explanation:

Alloys are metal compounds with two or more metals or non metals to create new compounds that exhibit superior structural properties. Alloys have high level of hardness that resists deformation thereby making it less ductile compared to polymers. This is due to the varying difference in the chemical and physical characteristics of the constituent metals in the alloy.

6 0

3 years ago

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0

3 years ago

(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific

lina2011 [118]

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content = 66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

5 0

4 years ago