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4 years ago

What are the three most common metals used in die casting?

Engineering

1 answer:

Anit [1.1K]4 years ago

3 0

Answer:

Aluminium,Copper,Magnesium

Explanation:

The three most common metal of die casting are as follows

1.Aluminium

2.Copper

3.Magnesium

Die casting is the process in which metal is forced in the die to produces the desired casting product.Generally two type of machines are used like cold chamber and hot chamber machining,it depends on the metals.Die casting produces simple shape of casting ,it can not use for complex casting.

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Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water

sveticcg [70]

Answer:

$\dot Q = -2.341\,kJ$

Explanation:

The rate of heat transfer from the balls to the air is:

$\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)$ $\dot Q = -2.341\,kJ$

3 0

4 years ago

Read 2 more answers

Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158

ahrayia [7]

Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter = $d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

$= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$

8 0

3 years ago

Why are plastics known as synthetic materials?

GarryVolchara [31]

Ans: Synthetic polymers are made up of long chains of atoms, arranged in repeating units, often much longer than those found in nature. It is the length of these chains, and the patterns in which they are arrayed, that make polymers strong, lightweight, and flexible. In other words, it's what makes them so plastic.

3 0

2 years ago

A rectangular steel alloy A-36 (structural steel) plate is hanging vertically and supporting a hanging weight of 90 kN. The plat

KengaRu [80]

Answer:

a) Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b)Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm\

c) % Reduction in area = (450-449.7391/450 ) = 0.58 %.

Explanation:

a) Change in length = Pl /AE = 90*1000*0.5*1000/75*207*6*10^3

= 0.4838mm Expansion.

Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b )Change in width = - μpt/AE = -(0.3*90*1000*6/ 75*207*6*10^3)

= -1.739 * 10^-3 mm

Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm

c )

New C/s area = 74.97827 *5.998260 = 449.7391 mm^2

% Reduction in area = (450-449.7391/450 ) = 0.58 %.

3 0

4 years ago

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0

3 years ago