¿Cuál de los siguientes factores reduciría el riesgo de roturas por un choque térmico a. Alto coeficiente de dilatación térmico
1 answer:
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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
Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:
Given that the temperature of the window is below the dew point it will condensate.
Explanation:
A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.
On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.
