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3 years ago

Which step in the engineering design process likely broke down in the following scenario?

Engineering

1 answer:

Mademuasel [1]3 years ago

6 0

Answer:

C. Design and build a prototype

Explanation:

The correct answer is - C. Design and build a prototype

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Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel f

Morgarella [4.7K]

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

$T_{ \infty} = 27^0C= 273+27 = 300K$

$T_s= 127^0C = (127+273)= 400K$

P = 1atm

Film temperature

$T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K$

dynamic viscosity =

$\mu =20.9096\times 10^{-6} m^2/sec$

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

$Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}$

$=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043$

we have,

$Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k$

we have,

heat transfer rate from top plate

$\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w$

7 0

3 years ago

A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus

iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}\\\\$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5 $\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

E = 8.83 kips

7 0

3 years ago

A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an

konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

$=1953.83 \ KJ/Kg$

and,

$P_3=15 \ mPa$

$h_3=hg$

$=2610.8 \ KJ/Kg$

$s_3=sg$

$=5.3108 \ KJ/Kgh$

At 45,

⇒ $x_{45} = \frac{5.3108-1.0912}{6.5019}$

$=0.66$

At $P_4=50 \ Kpa$ ,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

$=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

$=\frac{2610.8-1953.83}{2610.8-1836.26}$

$=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

$=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}$

$=28.45$ (%)

4 0

3 years ago

A cylindrical aluminum core is surrounded by a titanium sleeve, and both are attached at each end to a rigid end-plate. Set up t

NeX [460]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The elongation of the composite bar is given as δ = 0.072 in

b) The axial stress induced in each material is = 5485.7 psi

Explanation:

The explanation to the answer above is shown on the second uploaded image

3 0

3 years ago

Why were disc brakes invented?

kolezko [41]

Answer:

The answer is disc brake first evolved as a motorcycle braking system pattern just to become a smaller and remarkably lighter version of them.

Explanation:

It consisted of a metal disc called rotor and callipers that were attached to the fork or frame and equipped with pads to squeeze the rotor for braking.

6 0

2 years ago