Answer:
Taking as a basis of calculation 100 mol of gas leaving the conversion reactor, draw andcompletely label a flowchart of this process. Then calculate the moles of fresh methanol feed,formaldehyde product solution, recycled methanol, and absorber off-gas, the kg of steamgenerated in the waste-heat boiler, and the kg of cooling water fed to the heat exchangerbetween the waste-heat boiler and the absorber. Finally, calculate the heat (kJ) that must beremoved in the distillation column overhead condenser, assuming that methanol enters as asaturated vapor at 1 atm and leaves as a saturated liquid at the same pressure.
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Explanation:
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
ΔQ = 4930.37 BTu
Explanation:
given data
height h = 8ft
Δt = 8 hours
length L = 24 feet
R value = 16.2 hr⋅°F⋅ft² /Btu
inside temperature t1 = 68°F
outside temperature t2 = 16°F
to find out
number of Btu conducted
solution
we get here number of Btu conducted by this expression that s
......................1
here A is area that is = h × L = 8 × 24 = 1492 ft²
put here value we get
solve it we get
ΔQ = 4930.37 BTu
Answer:
Part A:
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
CPI reduced by 
=80%
Part C:
New Execution Time=
Increase in speed=
Explanation:
FP Instructions=50*106=5300
INT Instructions=110*106=11660
L/S Instructions=80*106=8480
Branch Instructions=16*106=1696
Calculating Execution Time:
Execution Time=
Execution Time=
Execution Time=
Part A:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI reduced by =80%
Part C:
New Execution Time=
New Execution Time=
Increase in speed=
