Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + 
..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 + 
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - 
) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - 
)
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = 
..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 = 
solve it we get
e = 2%
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) 
b) 
Explanation:
The explanation is shown on the second and third uploaded image
It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.
<h3>What is
Maclaurin Expansion?</h3>
The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
<h3>
What is the explanation for the above?</h3>
as indicated above, the Maclaurin infinite series expansion is given as:
F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
If F(0) = Cot 0
F(0) = ∝ = 1/0
This is not definitive,
Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.
Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
#SPJ1
Answer:
peak flow and any engineering considerations related thereto
Explanation:
It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.
Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.
It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)
