Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled
hot-water pipe that has an outer diameter of 2.5 cm. The pipe passes through the center of a 14-cm-thick wall filled with fiberglass insulation (k = 0.035 W/m·K). The surfaces of the wall are at 18°C.a. Determine the rate of heat transfer from the pipe to the air in the rooms. b. Determine the temperature drop of the hot water as it flows through the 5-m-long section of the wall where the steady rate of heat transfer from the pipe is 39.153 W. For water, use density (rho) 1000 kg/m^3 and specific capacity of heat (cp) =4180 J/kg.°C.
1 answer:
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Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Explanation:
From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.
Part 1
For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
Part 2
For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
And we can quantify the decrease using the relative change:
of reduction
Part 3
A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Answer:
ordinary bulb total cost is $39.54
fluorescent bulb total cost is $13.05
amount save = 39.54 - 13.05 = $26.49
resistance = 626.1 ohm
Explanation:
in the 1st part
bulb on time = 3 year = 4380 hours
life of bulb = 750 h
so number of bulb required =
number of bulb required = 6
cost of 6 bulb is = 6 × 0.75 = $4.5
so
cost of operation is = 100 × 4380 ×
cost of operation = $35.04
so total cost will be = $4.5 + $35.04 = $39.54
and
when compare with florescent bulb
time = 3 year = 4380 h
life of bulb = 10000 h
so number of bulb required =
number of bulb required = 0.43 = 1
cost of 6 bulb is = 1 × 5 = $5
so
cost of operation is = 23 × 4380 ×
cost of operation = $8.05
so total cost will be = $5 + $8.05 = $13.05
in part 2nd
total amount save while compare bulb is
amount save = 39.54 - 13.05 = $26.49
and in part 3rd
resistance of bulb is
resistance =
resistance =
resistance = 626.1 ohm