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3 years ago

6

Unlike expendable molds, permanent molds do not collapse, so the mold must be opened before appreciable cooling contraction occu

rs in order to prevent cracks from developing in the casting: True or False

1 answer:

Pani-rosa [81]3 years ago

8 0

Answer: True

Explanation:

Permanent molds do not collapse, unlike expendable molds so the mold must be opened before appreciable cooling contraction occurs in order to prevent cracks from developing in the casting.

The metal casting becomes solid inside the mold after it has been poured. But during the process of manufacture, before the would cools any further, they usually remove the metal cast in order to stop excess contractions of the solid metal casting in the mold. This is done to prevent prevent cracks from developing in the casting since permanent mold do not collapse.

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Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom,

djyliett [7]

Answer:

Following are the solution to this question:

Explanation:

The number of vacancies by the cubic meter is determined.

$N_V =N exp(\frac{Q_v}{kT})$

$= \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})$

$= \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\= \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\= 0.846215158 \times 10^{23} \exp(-30.290856)\\\\$

$=1.57 \times 10^{25} \ cm^{-3}$

7 0

3 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

Stells [14]

Answer:

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

Explanation:

Given that

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

We know that

Rex=ρvx/μ

So

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

$Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}$

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

$Nu_x=C.x^{0.8}=C.x^{4/5}$

We also know that

Nux=hx/K

$C.x^{4/5}=\dfrac{hx}{k}$

m is the constant

$h=mx^{-1/5}$

This is local heat transfer coefficient

The average value of h given as

$\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}$

$\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}$

$\bar{h}=\dfrac{5m}{4}L^{-1/5}$ ---------1

The value of local heat transfer coefficient at x=L

$h=mx^{-1/5}$

$h=mL^{-1/5}$ -----------2

From 1 and 2 we can say that

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

8 0

4 years ago

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$

$Volume= 84.95m^{3}$

$Since 1ft^{3} =0.028m^{3}$

Now the weight of water displaced = $Weight =\rho \times Volumewhererho$ is density of water = 1000kg/ $m^{3}$

Thus weight of liquid displaced = $\frac{84.95X1000}{1000}tonnes=84.95 tonnes$ ..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0

3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit

8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, $A_1 = 0.0044 m^2$

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

$KE = \frac{V_2^2 - V_1^2}{2} \\$ .........(1)

Where Inlet velocity, V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

$KE = \frac{50^2 - 80^2}{2} \\$

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a steady state flow.

$q - w = h_2 - h_1 + KE + g(z_2 - z_1)$

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, $W = \dot{m}w$ ..........(3)

Mass flow rate, $\dot{m} = 12 kg/s$

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, $h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg$ , x₂ = 0.92

specific enthalpy at the outlet, h₂ = $h_1 + x_2 h_{fg}$

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

$A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2$

3 0

4 years ago

\o you sell them?” "Fivepence farthing for one—Twopence for two,” the Sheep replied. "Then two are cheaper than one?” Alice said

NemiM [27]

Answer:

unusual and strange

Explanation:

3 0

3 years ago