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vladimir1956 [14]
3 years ago
10

Consider a unidirectional continuous fiber-reinforced composite with epoxy as the matrix with 55% by volume fiber.i. Calculate t

he longitudinal and transverse moduli when E-glass fiber is used.ii. Calculate the longitudinal and transverse moduli if carbon fibers are used instead of E-glass. Constants:Epoxy modulus = 2.41GPaE-glass fiber modulus = 72.5 GPaCarbon fiber modulus = 230 GPa
Engineering
1 answer:
ohaa [14]3 years ago
7 0

Answer:

I)E= 40.95 GPa

II)E=5.29 GPa

Explanation:

I)

Given that

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ = 72.5 GPa   ,V₂=0.55

Longitudinal moduli  given as ;

E= E₁V₁+E₂V₂

E= 2.41 x 0.45 + 72.5 x 0.55 GPa

E= 40.95 GPa

II)

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ =230 GPa   ,V₂=0.55

Transverse moduli given as:

\dfrac{1}{E}=\dfrac{V_1}{E_1}+\dfrac{V_2}{E_2}

\dfrac{1}{E}=\dfrac{0.45}{2.41}+\dfrac{0.55}{230}

E=5.29 GPa

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A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of
lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = \frac{v}{3}L

150\times v=P_2\times \frac{v}{3}  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

7 0
3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa
mariarad [96]

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with  water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol  = 0.8 g/cm³

solution

we get here water content that is express as

water content = S_s \times \gamma _w   ....................1

here soil is full saturated so S_s is 100% in both case

so put here value for water

water content = 100 % ×  1

water content = 1 g

and

now we get for alcohol that is

water content  = 100 % × 0.8

water content  = 0.8 g

so here water sample have more water content

5 0
3 years ago
A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol
insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0
3 years ago
The separation and purification process is used for which of the following purposes?
Svetach [21]

The separation and purification process is used for DEVELOPING VACCINES. The development of vaccines must follow strict guidelines.

A vaccine is a biological preparation used to protect against harmful diseases (e.g., viruses) in a simple and safe manner.

Vaccine purification is a strict process consisting of clarification, concentration and chromatography in order to separate the extract that contains the active principle of the vaccine.

The techniques used for vaccine filtration include membrane filtration, depth filtration and centrifugation.

Learn more about vaccines here:

brainly.com/question/6683555

6 0
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