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3 years ago

Represent the following sentence by a Boolean expression:

Engineering

1 answer:

Sphinxa [80]3 years ago

5 0

I’m really sorry That I can’t help

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Why is logging done during drilling?

Solnce55 [7]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

8 0

2 years ago

The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa

Elanso [62]

Answer:

load = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height = 13 ft

solution

we get here load so first we get wood stud load and that is

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster = 12 psf

and we have given height 13 ft

so load will be = 12 psf × 13 ft

load = 156 lb/ft

7 0

3 years ago

Safety measures to be taken during technical drawing<br>

Pavlova-9 [17]

Explanation:

Couldn't you just leave it in centimeters?

3 0

2 years ago

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit

Natali [406]

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$

3 0

3 years ago

The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made

Rom4ik [11]

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

δd/5 = δb/3

δd = (5/3) * 0.025

δd = 0.0417 in

- Compute εde i.e strain in DE:

εde = δd / Lde

εde = 0.0417 / 3*12

εde = 0.00116

- Compute stress in DE, σde:

σde = E*εde

σde = 29000*0.00116

σde = 33.56 ksi

- Compute the Force F_de:

F_de = σde *A_de

F_de = 33.56*0.002

F_de = 0.0672 kips

- Equilibrium conditions apply:

(M)_a = 0

3*W - 5*F_de = 0

W = (5/3)*F_de

W = (5/3)* 0.0672 = 112 lb

4 0

3 years ago