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Rina8888 [55]
3 years ago
10

Mechanical clips are used to close bags and keep things together. Investigate the design of various paper clips, hair clips, and

potato chip bag clips. Choose three (3) different types of clips. Discuss what you think are important design parameters. Discuss the advantages and disadvantages associated with each of the three (3) designs.
Engineering
1 answer:
pochemuha3 years ago
4 0

Answer:

Hair clip: this used to hold human hairs together especially long hairs

The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )

Advantages : very easy to use, holds hairs tight

Disadvantages : Expensive and material used is sometimes brittle

paper clips : This used to hold papers together it is made of a thin elastic wire

The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material

advantages : it is quite cheap and good strength

disadvantages :it can hold a limited amount of papers together

potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight

The design parameters to be considered are : Elasticity of the material to be used , and size

Advantages : it helps to prevent the potato chips from getting exposed to air

disadvantages : Not very popular and its quite expensive

Explanation:

Hair clip: this used to hold human hairs together especially long hairs

The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )

Advantages : very easy to use, holds hairs tight

Disadvantages : Expensive and material used is sometimes brittle

paper clips : This used to hold papers together it is made of a thin elastic wire

The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material

advantages : it is quite cheap and good strength

disadvantages :it can hold a limited amount of papers together

potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight

The design parameters to be considered are : Elasticity of the material to be used , and size

Advantages : it helps to prevent the potato chips from getting exposed to air

disadvantages : Not very popular and its quite expensive

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What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ
swat32

Answer:

1) Dimensions of shear rate is [T^{-1}] .

2)Dimensions of shear stress are [ML^{-1}T^{-2}]

Explanation:

Since the dimensions of velocity 'v' are [LT^{-1}] and the dimensions of distance 'y'  are [L] , thus the dimensions of \frac{dv}{dy} become

\frac{[LT^{-1}]}{[L]}=[T^{-1}] and hence the units become s^{-1}.

Now we know that the dimensions of coefficient of dynamic viscosity \mu are [ML^{-1}T^{-1}] thus the dimensions of shear stress can be obtained from the given formula as

[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]

Now we know that dimensions of momentum are [MLT^{-1}]

The dimensions of Area\times time are [L^{2}T]

Thus the dimensions of \frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]

Which is same as that of shear stress. Hence proved.

4 0
3 years ago
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
3 years ago
Read 2 more answers
Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the
Alex_Xolod [135]

Sorry need.points I'm new

7 0
3 years ago
A certain heat pump produces 200 kW of heating for a 293 K heated zone while only using 75 kW of power and a heat source at 273
vodka [1.7K]

Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

8 0
4 years ago
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