Answer: true
Explanation:
it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However
Answer:
86701 Micrometers.
Explanation:
Multiply 0.86701 dm by 100,000 to get 86701 um.
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:
The answer is "
"
Explanation:
Air flowing into the![p_1 = 20 \ \frac{lbf}{in^2}](https://tex.z-dn.net/?f=p_1%20%3D%2020%20%5C%20%5Cfrac%7Blbf%7D%7Bin%5E2%7D)
Flow rate of the mass ![m = 230.556 \frac{lbm}{s}](https://tex.z-dn.net/?f=m%20%20%3D%20230.556%20%5Cfrac%7Blbm%7D%7Bs%7D)
inlet temperature ![T_1 = 700^{\circ} F](https://tex.z-dn.net/?f=T_1%20%3D%20700%5E%7B%5Ccirc%7D%20F)
Pipeline![A= 5 \times 4 \ ft](https://tex.z-dn.net/?f=A%3D%205%20%5Ctimes%204%20%5C%20ft)
Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:
![\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}](https://tex.z-dn.net/?f=%5Cto%20%5Cbar%7BR%7D%20%3D%201545%20%5C%20ft%20%5Cfrac%7Blbf%7D%7Blbmol%20%5E%7B%5Ccirc%7D%20R%7D%5C%5C%5C%5C%20%5Cto%20M%3D%2028.97%20%5Cfrac%7Blb%7D%7B%5Cbmol%7D%5C%5C%5C%5C%20%5Cto%20pv%3DRT%20%5C%5C%5C%5C%5Cto%20v%3D%20%5Cfrac%7B%5Cfrac%7B%5Cbar%7BR%7D%7D%7BM%7DT%7D%7Bp%7D)
![= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Cfrac%7B1545%7D%7B28.97%7D%2870%5E%7B%5Ccirc%7DF%2B459.67%29%7D%7B20%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B144%7D%5C%5C%5C%5C%3D9.8%20%5Cfrac%7Bft3%7D%7Blb%7D)
![V= \frac{mv}{A}](https://tex.z-dn.net/?f=V%3D%20%5Cfrac%7Bmv%7D%7BA%7D)
![= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B230.556%20%5Cfrac%7Blbm%7D%7Bs%7D%20%5Ctimes%209.8%20%5Cfrac%7Bft%5E3%7D%7Blb%7D%7D%7B5%20%5Ctimes%204%20%5C%20ft%5E2%7D%5C%5C%5C%5C%3D%20112.97%20%5Cfrac%7Bft%7D%7Bs%7D)
Answer:
Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.
Recovery Point Objective is basically to determine the age of restoration or recovery point.
Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.
Hence, Recovery Time Objective (RTO) is the correct answer.