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4 years ago

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Why do we study economics

1 answer:

avanturin [10]4 years ago

6 0

Answer:

Because it teaches us about money and how it works

Explanation:

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3 years ago

Air entrainment is used in concrete to: __________.

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Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

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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 = $\frac{23-6}{2.20} = 7.7W/m$

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3 years ago

Read 2 more answers

He engine in Problem 3.8 produces 57kW of brake power at 2000 RPM. Calculate:

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Answer:

Answer is below attached.

Explanation:

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