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horrorfan [7]
2 years ago
15

In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict

the output). BR there Here: .Word 9 there: DECO here, d stop . end.
Engineering
1 answer:
allsm [11]2 years ago
4 0

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
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Answer:

86 mm

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\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

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R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

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R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

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\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

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L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
2 years ago
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