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Flauer [41]
3 years ago
11

Act as food engineers or biomedical engineers and write a short paragraph explaining why they need to know about mixtures and so

lutions in creating new food products or medicines.
Engineering
1 answer:
drek231 [11]3 years ago
5 0

Answer:

We use our knowledge of mixtures and solutions when we are designing new synthetic materials. This is especially the case in the biomedical field, where we have to deal with compatibility issues when placing materials made outside the human body into the body.We also design ways to help separate mixtures and solutions in industrial, commercial and environmental processes.

Hope this helps

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A cylindrical tank is required to contain a gage pressure 670 kPakPa . The tank is to be made of A516 grade 60 steel with a maxi
RSB [31]

Answer:

The minimum thickness t of the wall is 0.00446 mm

Explanation:

Solution

Given that

Pressure =670kPa = 0.670

σ allowable normal stress = 150 MPa

Inner diameter = 2mm

Steel = A516 grade 60

Now,

Since the hoop stress is twice the longitudinal stress, the cylindrical tank is more likely to fail from the hoop stress.

Thus

σ allowable = σₙ = pμ/t

=p (d/2)

150 MPa =0.670MPa * 2/2/t

=0.67/t

t=0.67/150

t =0.00446 mm

8 0
4 years ago
Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat
Serggg [28]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

 

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

 

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

 

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11  

Download docx
3 0
3 years ago
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
A student lab group is brainstorming the design of an experiment that uses an ammeter (measures current) and different resistors
givi [52]

Answer:

  Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.

Explanation:

The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.

3 0
3 years ago
Tyuuyiopopiouyttrrtrffrlkl,k;;';'l.l
Viktor [21]

Answer:

Explanation:

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5 0
3 years ago
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