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dalvyx [7]
3 years ago
10

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)

1.25 kg of Sn and 14 kg Pb at 200 o C
Engineering
1 answer:
Novosadov [1.4K]3 years ago
6 0

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%

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3 years ago
A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in
DochEvi [55]

Answer:

a=5.515\frac{m}{s^{2} }

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

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Performing the operations

60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}

Now, we will use the acceleration formula

a=\frac{v}{t}

Where v = speed and t = time

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a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }

7 0
3 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

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v2 = 6.017 m/s

7 0
4 years ago
A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
jeyben [28]

Answer:

  2.728 kg

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The units help you keep the calculation straight.

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4 0
3 years ago
The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump h
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Answer:500,551.02

Explanation:

Given

Initial enthaly of pump \left ( h_1\right )=500KJ/kg

Final  enthaly of pump \left ( h_2\right )=550KJ/kg

Final  enthaly of pump when efficiency is 100%=h_2^{'}

Now pump efficiency is 98%

\eta=\frac{h_2-h_1}{h_2^{'}-h_1}

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therefore initial and final enthalpy of pump for 100 % efficiency

initial=500KJ/kg

Final=551.02KJ/kg

3 0
3 years ago
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