Describe what V1-V4 is

Brazing, Soldering and Adhesive Bonding are the types of • Liquid Solid System Welding Solid State Welding • Fusion Welding . No

olga_2 [115]

Answer: Solid state welding

Explanation: Solid state welding is the welding procedure which is based on the temperatures and pressure but without any liquid or vapor as aid for welding.This process is carried mainly cohesive forces and considering forces as less important. Brazing,soldering and adhesive is the process in which material are joint which the help of solid welding agent.Thus solid state welding is the correct option.

6 0

2 years ago

Fill in the blank to correctly complete the statement below.

frutty [35]

Answer:

The invention of the pendulum-driven ___<u>clocks</u>___ in the 1600s paved the way for a new industrial era.

4 0

3 years ago

If your accelerator pedal gets stuck, what is the first thing you should do?

Anna35 [415]

If your accelerator gets stuck down, do the following: Shift to neutral. Apply the brakes. Keep your eyes on the road and look for a way out.If your accelerator gets stuck down, do the following:

Shift to neutral.

Apply the brakes.

Keep your eyes on the road and look for a way out.

Warn other drivers by blinking and flashing your hazard lights.

Try to drive the car safely off the road.

Turn off the ignition when you no longer need to change direction.

8 0

2 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr

Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, $\dot m$ = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, $\dot m$ ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, $\dot m$ ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, $P_w$ = $\dot m$ ·g·h

∴ $P_w$ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

$P_w$ ≈ 14.1130725 Hp

The power output of the pump, $P_{out}$ = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, $P_f$ = $P_{out}$ - $P_w$

∴ $P_f$ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction, $P_f$ ≈ 1.9 hp

7 0

2 years ago

A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh

sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

$T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C$

to kelvin = (5 + 273)K = 278 K

From the " thermophysical properties of gases at atmospheric pressure" table; At $T_f$ = 278 K ; by interpolation; we have the following

$\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}$ → v 13.93 (10⁻⁶) m²/s

$\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}$ → k = 0.0245 W/m.K

$\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}$ → ∝ = 19.6(10⁻⁶)m²/s

$\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}$ → Pr = 0.713

$\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}$

The Rayleigh number for vertical cavity

$Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}$

= $\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}$

= $8.38*10^5$

$\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24$

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

$Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}$

$NU_L= \frac{hL}{k}= 4.878$

$\frac{hL}{k}= 4.878$

$\frac{h*0.06}{0.0245}= 4.878$

$h = \frac{4.878*0.0245}{0.06}$

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0

3 years ago