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ruslelena [56]
3 years ago
13

The grade is a measure of quality and it captures concentration levels (i.e., how pure a certain fraction is). If grade captures

quality, then Recovery is a measure of quantity: it describes how much of certain material in the input stream is made available for reprocessing. A recovery of R% means that (100 - R)% of the material going into the process is lost, ending up either in mine tailings or as a contaminant in one or more fractions. By weight, copper wire contains about 70% copper and 30% PVC. Suppose we process 1 ton of wire per hour into a copper fraction of 0.74 ton/hour, of which 0.69 is copper and 0.05 is PVC. What is the weight and composition of the tailing?

Engineering
1 answer:
Leokris [45]3 years ago
3 0

Answer:

Check the explanation

Explanation:

To calculate weight when the mass is already known, make use of the formula weight = mass times gravitational acceleration. But kindly note that on the surface of the earth, gravitational acceleration is always 9.8 m/s^2, so simply plug in the mass and multiply it by 9.8 to get the weight in newtons.

Kindly check the attached image below to know the step by step explanation.

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A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
3 years ago
Identify parts of the E-Cig that constitute voltage, current, and resistance. Discuss the role each plays in the E-Cig and typic
Leno4ka [110]

Answer: c

Explanation:

7 0
2 years ago
Which type of software is LibreOffice an example? open source freeware O beta O firmware​
Fittoniya [83]

Answer:

LibreOffice is an integrated suite of software applications used to perform office tasks such as word processing, presentation preparation and spreadsheet calculations

4 0
3 years ago
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Fill in the blank to correctly complete the statement below.
lapo4ka [179]

Explanation:

The invention of the pendulums driver ____ ao in the 1600s paved the way for a new industrial era. Add answer.

6 0
3 years ago
A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness
Bingel [31]

Answer:

a) heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

Explanation:

Assumptions:

  1. Constant properties
  2. Steady state conditions
  3. Negligible effect of radiation
  4. Negligible constant resistance between tube and insulation
  5. one dimensional radial conduction

a) What is the heat gain per unit tube length

R_{conv,i}'=\frac{1}{2\pi r_1h_i}

d_1=36mm Therefore r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}

r_2=2mm=2*10^{-3}m

k_{st}=14.2W/m.k

h_o=6W/m^2

h_i=400W/m^2

R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W

R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W

R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W

R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W

heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

r_3=r_1+r_2+10mm=30mm=0.03m

R_{conv,i}' and R_{cond,st}' are the same, but R_{conv,o}' changes.

Therefore:

R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W

R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W

The total resistance R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W

heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

8 0
3 years ago
Read 2 more answers
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