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3 years ago

5

Consider the series solution, Equation 5.42, for the plane wall with convection. Calculate midplane (x* = 0) and surface (x* = 1

) temperatures θ* for Fo = 0.1 and 1, using Bi = 0.1, 1, and 10. Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations 5.43 and 5.44.

1 answer:

VashaNatasha [74]3 years ago

8 0

Answer:

We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.

Explanation:

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The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c

marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data

str1: .space 20

str2: .space 20

msg1:.asciiz "Please enter string (max 20 characters): "

msg2: .asciiz "\n Please enter string (max 20 chars): "

msg3:.asciiz "\nSAME"

msg4:.asciiz "\nNOT SAME"

.text

.globl main

main:

li $v0,4 #loads msg1

la $a0,msg1

syscall

li $v0,8

la $a0,str1

addi $a1,$zero,20

syscall #got string to manipulate

li $v0,4 #loads msg2

la $a0,msg2

syscall

li $v0,8

la $a0,str2

addi $a1,$zero,20

syscall #got string

la $a0,str1 #pass address of str1

la $a1,str2 #pass address of str2

jal methodComp #call methodComp

beq $v0,$zero,ok #check result

li $v0,4

la $a0,msg4

syscall

j exit

ok:

li $v0,4

la $a0,msg3

syscall

exit:

li $v0,10

syscall

methodComp:

add $t0,$zero,$zero

add $t1,$zero,$a0

add $t2,$zero,$a1

loop:

lb $t3($t1) #load a byte from each string

lb $t4($t2)

beqz $t3,checkt2 #str1 end

beqz $t4,missmatch

slt $t5,$t3,$t4 #compare two bytes

bnez $t5,missmatch

addi $t1,$t1,1 #t1 points to the next byte of str1

addi $t2,$t2,1

j loop

missmatch:

addi $v0,$zero,1

j endfunction

checkt2:

bnez $t4,missmatch

add $v0,$zero,$zero

endfunction:

jr $ra

3 0

3 years ago

You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

Katen [24]

Answer:

The required wall thickness is $1.506 \times 10^{-3}$ m

Explanation:

Given:

Fluid density $\rho = 1200$ $\frac{kg}{m^{3} }$

Diameter of tank $d = 8$ m

Length of tank $l = 4$ m

F.S = 4

For A-36 steel yield stress $\sigma = 250$ MPa,

Allowable stress $\sigma _{allow} = \frac{\sigma}{F.S}$

$\sigma _{allow} = \frac{250}{4} = 62.5$ MPa

Pressure force is given by,

$P = \rho gh$

$P = 1200 \times 9.8 \times 4$

$P = 47088$ Pa

Now for a vertical pipe,

$\sigma _{allow} = \frac{Pd}{4t}$

Where $t =$ required thickness

$t = \frac{Pd}{4 \sigma _{allow} }$

$t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }$

$t = 1.506 \times 10^{-3}$ m

Therefore, the required wall thickness is $1.506 \times 10^{-3}$ m

8 0

3 years ago

You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen

VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

6 0

3 years ago

Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i

stealth61 [152]

Answer:

$\dot m = 2.74 kg/s$

Explanation:

given data:

pressure 1 MPa

diameter of pipe = 30 cm

average velocity = 10 m/s

area of pipe $= \frac[\pi}{4}d^2$

$= \frac{\pi}{4} 0.3^2$

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

$\dot m = \rho A v$

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

$\dot m = 4.068 * 0.070* 10$

$\dot m = 2.74 kg/s$

7 0

3 years ago

Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED

Finger [1]

Answer: LED have the lowest cost of operation.

Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

$Cost=Energy\times cost/unit$

Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.

6 0

3 years ago