Answer:
<em>the % recovery of aluminum product is 80.5%</em>
<em>the % purity of the aluminum product is 54.7%</em>
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Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = <em>54.7%</em>
Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
Answer:
a) 60000 psi
b) 1.11*10^6 psi
c) 112000 psi
d) 30.5%
e) 30%
Explanation:
The yield strength is the load applied when yielding behind divided by the section.
yield strength = Fyield / A
A = π/4 * D^2
A = 0.5 in^2
ys = Fy * A
y2 = 30000 * 0.5 = 60000 psi
The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).
k = E * L0 / A
And the stiffness is related to change of length:
Δx = F / k
Then:
Δx = F * A / (E * L0)
E = F * A / (Δx * L0)
When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was
Δx = L - L0 = 1.8075 - 1.8 = 0.0075
Then:
E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi
Tensile strength is the strees at which the material breaks.
The maximum load was 56050 lb, so:
ts = 56050 / 0.5 = 112000 psi
The percent elongation is calculated as:
e = 100 * (L / L0)
e = 100 * (2.35 / 1.8 - 1) = 30.5 %
If it necked with and area of 0.35 in^2 the precent reduction in area was:
100 * (1 - A / A0)
100 * (1 - 0.35 / 0.5) = 30%
