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TEA [102]
3 years ago
12

What are the benefits of using the engineering design process? combining scientific knowledge and creativity in a rigid and stru

ctured approach to problem-solving combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving combining scientific knowledge and logic in a rigid and structured approach to problem-solving combining careful reasoning and artistic invention in a flexible approach to problem-solving
Engineering
1 answer:
Trava [24]3 years ago
7 0

Answer:

combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving

Explanation:

You might be interested in
Could I please get help with this​
alex41 [277]

Answer:

1.I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = 36.661458\overline 3 in.⁴

Iₓ = 28.6458\overline 3 in.⁴

I_y = 138.6548\overline 3 in.⁴

2. I_{xc} = 114.\overline 3 in.⁴

I_{yc} = 37.\overline 3 in.⁴

Iₓ = 457.\overline 3 in.⁴

I_y = 149.\overline 3 in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458\overline 3

I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458\overline 3

I_{yc} = 36.661458\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458\overline 3

Iₓ = 28.6458\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548\overline 3

I_y = 138.6548\overline 3 in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 4 × 7³/12 = 114.\overline 3

I_{xc} = 114.\overline 3 in.⁴

I_{yc} = h·b³/12 = 7 × 4³/12 = 37.\overline 3

I_{yc} = 37.\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457.\overline 3

Iₓ = 457.\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 149.\overline 3

I_y = 149.\overline 3 in.⁴

3. The deflection, \delta _{max}, of a simply supported beam having a point load at the center is given as follows;

\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, I_{xc} = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552

The maximum deflection of the beam, \delta _{max} = 2.55552 inches

5 0
3 years ago
the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i
Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

V(t) = 5(1-e^{-0.5t})

The current flowing through the inductor is given by

i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\

i(t) = 1000t +2000e^{-0.5t} -2000\\

So the current at t = 2 seconds is

i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A

The energy stored in the inductor at t = 2 seconds is

W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J

4 0
2 years ago
Can you reduce energy use without compromising people's basic needs (such as opening a car, cooking food, home lighting, electri
Liono4ka [1.6K]

Answer:

   yes

  • electric and/or hybrid cars
  • microwave ovens
  • LED lighting
  • low-power electronics

Explanation:

Advances in technology and changes in social organization have brought about reductions in energy use on many fronts.

  • hybrid/electric vehicles have reduced transportation energy needs
  • microwave ovens have reduced cooking energy needs
  • LED lighting has reduced lighting energy needs
  • low-power electronics have reduced the energy cost of technology and entertainment
  • heat pumps and insulated windows have reduced energy needs for home heating and cooling
  • zoning laws have reduced the need for travel to work and shopping areas
4 0
3 years ago
A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
Read 2 more answers
An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred b
irinina [24]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx
4 0
3 years ago
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