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3 years ago

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Using the data given below, calculate the drop in the temperature of molten aluminum as it flows in a 15 cm long round channel o

f 2 cm diameter in a water-cooled die and what will be the temperature drop if the metal were to flow through a 15 cm long channel having a square cross-section (side = 2 cm) in an identical mold? Assume that Al enters the channel at 800 C. The average flow velocity is 4 cm/s. Given: density of Al=2.2 g/cm3, specific heat of Al=0.28 cal/g.K, mold-metal interface heat transfer coefficient=0.01 cal/cm2.s.K, and the die temperature=27 C (assumed constant).

1 answer:

Luda [366]3 years ago

8 0

Answer:

99.12 degree centigrade

Explanation:

Please kindly check attachment for the detailed step by step solution to the given problem.

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A force of 250 N is acting on an area of 2 m2. The pressure is therefore:

lianna [129]

Answer:

125 Pascal

Explanation:

We know that

$Pressure=\frac{Force}{Area}$

Applying given values we get

$Pressure=\frac{250N}{2m^{2}}$

$Pressure=125\frac{N}{m^{2}}$

Pressure = 125 Pascal

7 0

3 years ago

Nitrogen (N2) contained in a piston–cylinder arrangement, initially at 10 bar and 405 K, undergoes an expansion to a final tempe

yKpoI14uk [10]

Answer:28.21 kJ/kg

Explanation:

Given

$P_1=10\ bar$

$T_1=405\ K$

$T_2=300\ K$

Process $PV^{1.3}=constant$

Work done for Polytropic process

$W=\dfrac{P_1V_1-P_2V_2}{n-1}$

where n=Polytropic index

$W=\dfrac{R(T_1-T_2)}{n-1}$

$W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]$

$W=103.6\ kJ\kg$

Now Calculating change in Internal energy

$\Delta U=c_v(T_2-T_1)$

$\Delta U=0.718\times (300-405)$

$\Delta U=-75.39\ kJ/kg$

Now applying First law concept

$\Delta U=Q-W$

$Q=W+\Delta U$

$Q=103.6-75.392$

$Q=28.21\ kJ/kg$

6 0

4 years ago

Grace [21]

I need help my self lol XD

5 0

3 years ago

Read 2 more answers

The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large

SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

$exp(-kt^n)=1-y$

taking the natural logarithm of both sides:

$-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}$

Substituting values:

$k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}$

Also

$t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}$

Substituting values and y = 99% = 0.99

$t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s$

∴ t ≅ 305 s

It would take approximately 305 s to go to 99% completion

8 0

3 years ago

Read 2 more answers