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Zielflug [23.3K]
4 years ago
12

A strip of AISI 304 stainless steel, 2mm thick by 3cm wide, at 550°C, continuously enters a cooling chamber that removes heat at

a rate of 120 kW. How fast must it move to achieve a temperature of 140 °C at the exit?
Engineering
1 answer:
Zanzabum4 years ago
6 0

Answer:

V = 1.23 m/s

Explanation:

Given data:

AISI 304 steel

thickness of steel = 2 mm

width  =  3 cm

Temperature =  550 degree celcius

wer know that heat is given as Q  

Q = m\times Cp \Delta T

\Delta T = 550 - 140 = 410 \degree\ celcius

for AISI 304 steel Cp is 502 J/kg . K

We know that

\dot mass  =\rho A V

\rho = 7.9\times 10^3 kg/m^3

A = 2\times 10^{-3} \times 3\times 10^{-2} m^2

therefore VELOCITY V  will be

120\times 10^{3} = 7.9\times 10^{3} \times 6\times 10^{-5} \times V\times 502\times 410

solving for V we get

V = 1.23 m/s

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A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False
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Answer:

True

Explanation:

For point in xz plane the stress tensor is given by\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]

where Dx is the direct stress along x ; Dz is direct stress along z ;  tzx and txz are the  shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz  ( obtained by moment equlibrium )

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3 years ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
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Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

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d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

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1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

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AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

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