Roku internet service providet

Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)

Novosadov [1.4K]

Answer:

a) ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b)

The phase is; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0

3 years ago

Explain by Research how a basic generator works ? using diagram<br>

natulia [17]

Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry

8 0

3 years ago

Supercharging is the process of (a) Supplying the intake of an engine with air at a density greater than the density of the surr

iVinArrow [24]

Answer:

a)supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere

Explanation:

Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere.

By doing this , it increases the power out put and increases the brake thermal efficiency of the engine.It also increases the volumetric efficiency of the engine.

So the our option a is right.

4 0

3 years ago

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy + flow work

$E = \dot m ( \frac{v^2}{2] + zg + p\nu)$

$E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})$

$57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})$

solving for flow rate

$\dot m = 99.977we know that [tex]\dot m = \rho AV$

$\dot m = 780 \frac{\pi}{4} D^2\times 16$

solving for d

$99.97 = 780 \times \frac{\pi}{4} D^2\times 16$

d = 0.090 m

so radius = 0.045 m

3 0

4 years ago

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar

Saturation temperature corresponding to saturation pressure =40°C

$h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}$

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

$Total\ heat\ =C_p\Delta T+\Delta h$

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

$h_1=\dfrac{\Delta h}{\Delta T}$

$h_1=\dfrac{2406}{20}$

$h_1=120.3\frac{KJ}{kg-m^2K}$

Now film coefficient after inclusion of sensible heat

$h_2=\dfrac{total\ heat}{\Delta T}$

$h_2=\dfrac{2,544}{20}$

$h_2=127.2\frac{KJ}{kg-m^2K}$

$So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100$

$=\dfrac{127.2-120.3}{120.3}\times 100$

=5.75 %

So Percentage change 5.75 %.

3 0

3 years ago

Roku internet service providet​

Roku internet service providet