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Ket [755]
3 years ago
13

Describe how to mix and apply body filler?

Engineering
1 answer:
malfutka [58]3 years ago
8 0
This is all you need to mix body filler.
Start with a golf-ball size glob of filler.
Squeeze a ribbon of hardner across the filler.
Stir the mixture quickly, but do not "whip" it. ...
Mix until overall color is even.
Pick up an appropriate amount of filler for the task at hand.
You might be interested in
A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump.
larisa86 [58]

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

7 0
3 years ago
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati
Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

\dot m = 5000 kg/s

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy  are taken as zero.

change in mechanical energy is givrn as

e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}

                         = 0.491 kJ/kg

\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW

\eta_{OVERALL}  = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76

B) \eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80

c) \dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)

\dot W_{shaft} = 1964 kW

7 0
3 years ago
Do the coil resistances have any effect on the plots?
PolarNik [594]
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
8 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
3 years ago
Ages have been identified by the materials developed and used in those eras.
Westkost [7]
Ok what is the question
6 0
2 years ago
Read 2 more answers
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