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3 years ago

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A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

plate remains fixed while the upper plate is subjected to a horizontal force, P. If the top plate displaces 2 mm horizontally, determine:
a. the average shear strain in the material
b. the force P exerted on the upper plate.

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

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Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

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As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

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3 years ago

The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05

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Answer:

$P_m=181.42 KPa$

Explanation:

$P_1=101.325 KPa,V_1=0.05m^3,T_1=32C$ ,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio $r=1+\dfrac{1}{C}$

$r=1+\dfrac{1}{0.08}$

r=13.5

$r=\dfrac{V_1}{V_2}$

$13.5=\dfrac{0.05}{V_2}$

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$\eta =0.56$

$\eta =\dfrac{W}{Q}$

W is the work out put and Q is the heat addition.

$0.56 =\dfrac{W}{15}$

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume $V_s=V_1-V_2$

$V_s=0.05-3.7\times 10^{-3}$

$V_s=0.0463 m^3$

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An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

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This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

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a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

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P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

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the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

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ηth = 1 - 0.2916

ηth = 0.7084

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Qin = W_net.out / ηth

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we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

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c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out / W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

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