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3 years ago

Which of the following is NOT true about drawings and specifications?

Engineering

1 answer:

hoa [83]3 years ago

3 0

Answer:

sorry but we can't tell the answer because your not showing the drawings and specifications.

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10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

$\delta = \frac{wa^2b^2}{3EI}$

$= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}$

$=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}$

= 20.83\E ft

at mid span

$\delta = \frac{wl^3}{48EI}$

$= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}$

$\delta = 1.23\E ft$

shear stress

$\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi$

6 0

3 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

3 years ago

Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an

Svet_ta [14]

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

6 0

3 years ago

Read 2 more answers

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

A) each cycle receives same amount of energy by heat transfer

( Given that ; Л1 = 1/3 ЛR )

1) develops greater bet work 

WR develops greater work ( i.e. Wr > WI )

2) discharges greater energy by heat transfer

Qc' > Qc

solution attached below

B) If Each cycle develops the same net work

1) Receives greater net energy by heat transfer from hot reservoir

QH' > QH ( solution is attached below )

2) discharges greater energy by heat transfer to the cold reservoir

Qc' > Qc

solution attached below

4 0

2 years ago

Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?

Firdavs [7]

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

6 0

3 years ago