Select the answer that shows how the recognition of depreciation expense

A sample of cast iron with .35 wt%C is slow cooled from 1500C to room temperature. What is the fraction of proeutectoid Cementit

Marizza181 [45]

Answer:

So the fraction of proeutectoid cementite is 44.3%

Explanation:

Given that

cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.

We know that

Fraction of proeutectoid cementite phase gievn as

${w_p}'=\dfrac{{C_o}'-0.022}{0.74}$

Now by putting the values

${w_p}'=\dfrac{0.35-0.022}{0.74}$

${w_p}'=0.443$

So the fraction of proeutectoid cementite is 44.3%

6 0

2 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:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
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

Given vectors A = xˆ2−yˆ +zˆ3 and B = xˆ3−zˆ2, find a vector C whose magnitude is 9 and whose direction is perpendicular to both

natka813 [3]

Answer:

Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z

Explanation:

The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.

A detailed step by step explanation is attached below.

7 0

3 years ago

Read 2 more answers

How to tie shoe please

Genrish500 [490]

Step 1: Unknot Shoelaces. Make sure that the two ends of the shoelace are completely untangled and free of knots past the tongue. ...
Step 2: Create Overhand Knot. ...
Step 3: Finish the Basic Shoe Knot. ...
Step 4: Create Double Knot. ...
Step 5: Finished!

4 0

2 years ago

Work done by a system during a process can be considered as a property of the system. a)True b) False

SVEN [57.7K]

Answer:

b) False

Explanation:

Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.

6 0

3 years ago