Answer:
Explanation:
Mean temperature is given by
Tmean = (Ti + T∞)/2
Tmean = 107.5⁰C
Tmean = 107.5 + 273 = 380.5K
Properties of air at mean temperature
v = 24.2689 × 10⁻⁶m²/s
α = 35.024 × 10⁻⁶m²/s
= 221.6 × 10⁻⁷N.s/m²
= 0.0323 W/m.K
Cp = 1012 J/kg.K
Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶
= 0.693
Reynold's number, Re
Pv = 4m/πD²
where Re = (Pv * D)/
Substituting for Pv
Re = 4m/(πD)
= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)
= 28728.3
Since Re > 2000, the flow is turbulent
For turbulent flows, Use
Dittus - Doeltr correlation with n = 0.03
Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k
(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³
(h₁ × 0.006)/0.0323 = 75.962
h₁ = (75.962 × 0.0323)/0.006
h₁ = 408.93 W/m².K
The best step for the engineers to make next is option D. Begin to design an airplane using this metal.
<h3>What is the metallic is plane parts?</h3>
Aluminum and its alloys are nevertheless very famous uncooked substances for the production of business planes, because of their excessive electricity at exceedingly low density. Currently, excessive-electricity alloy 7075, which includes copper, magnesium and zinc, is the only used predominantly withinside the plane industry.
The solution is D, due to the fact even as it's far crucial to marketplace the fabric and ensure humans are inquisitive about buying, they first want to attempt to layout aircraft the usage of this fabric. There isn't anyt any use promoting an aircraft constituted of this material_ if a aircraft can not be built.
Read more about the aircraft:
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Answer:
See explanation
Explanation:
Solution:-
- The shell and tube heat exchanger are designated by the order of tube and shell passes.
- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.
- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.
- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.
- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:
U ∝ v^( 0.8 ) .... ( turbulence )
- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.
Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).
The best answer would be
D. Int calculateCost(int count);
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%
