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4 years ago

An example of a robot functionality would be:

Engineering

2 answers:

SSSSS [86.1K]4 years ago

5 0

It should be able to move atleast in one direction

Ronch [10]4 years ago

4 0

Poop!!!!!!!!!!!!!!!!!!!

You might be interested in

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

a) Determine the yielding factor of safety

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

4 0

3 years ago

“We’re late for homeroom,” said Bonnie, surprised to hear herself say “we.” “EARL is a tool, Bonnie’s mother kept reminding her,

bogdanovich [222]

Answer:

Hi there

Your answer is:

Explanation:

The "metal contraption", as the text goes on to say, is treated like a friend to Bonnie. Her mom comments on this by contrasting the metal contraption to a puppy.

Hope this helps

8 0

3 years ago

Read 2 more answers

A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure meth

GuDViN [60]

Answer:

mass of LNG: 129501.3388 kg

quality: 0.005048662

Explanation:

Volume occupied by liquid:

400 m^3*0.9 = 360 m^3

Volume occupied by vapor

400 m^3*0.1 = 40 m^3

A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present

For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg

From specific volume definition:

vf = liquid volume/liquid mass

liquid mass = liquid volume/vf

liquid mass = 360 m^3/0.002794 m^3/kg

liquid mass = 128847.5304 kg

vg = vapor volume/vapor mass

vapor mass = liquid volume/vg

vapor mass = 40 m^3/0.06118 m^3/kg

vapor mass = 653.8084341 kg

total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg

Quality is defined as the ratio between vapor mass and total mass

quality = 653.8084341 kg/129501.3388 kg = 0.005048662

4 0

3 years ago

State the mathematical expression to define the availability of equipment over a specified time and operational availability?

Gelneren [198K]

Answer:

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Explanation:

Availability:

It define as the probability of system which perform desired task before showing any failure .

The availability can be define as follows

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Or we can say that

$Availability=\dfrac{Up\ time}{total\ time}$

Availability can also be express as

$Availability=\dfrac{MTBF}{MTBF+MTTR}$

Where MTBF is the mean time between two failure.

MTTR is the mean time to repair.

7 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago