Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho
1 answer:
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b)
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10
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Answer:
Inlet :
Outlet:
Explanation:
1) Notation and important concepts
Flow of mass represent "the mass of a substance which passes per unit of time".
Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"
Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."
Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".
We know that the flow of mass is given by the following expression
, where
represent the flow rate and
the specific volume at the pressure and temperature given.
is the inlet area
pressure at the inlet area
temperature at the inlet area
is the outlet area
pressure at the outlet area
temperature at the outlet area
represent the flow of mass
If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is
and the entropy is
With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy
Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)
Our interest value would be given using interpolation like this:
2) Solution to the problem
Now since we have all the info required to solve the problem we can find the velocities on this way.
We know from the definition of flow of mass that , but since
we have this:
If we solve from the velocity v we have this:
(*)
And now we just need to replace the values into equation (*)
For the inlet case:
For the oulet case:
Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α =
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹