A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
ter and is situated in a surrounding ambient temperature of 15°C where the natural convection heat transfer coefficient is 12 W/m^2·K.
If the efficiency of the electrical heater to transfer heat to the plate is 80%, determine the electric power that the heater needs to keep the surface temperature of the plate at 192°C.
If the efficiency of the electrical heater to transfer heat to the plate is 80%, determine the electric power that the heater needs to keep the surface temperature of the plate at 192°C.
1 answer:
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Answer:
The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.
Explanation:
From table 3.5 of Traffic Engineering by Mannering
R_v=835
R=835+(10ft/2)= 840 ft.
Now T is given as
T=R tan(Δ/2)
Here Δ is the central angle of curve given as 35°
So
T=R tan(Δ/2)
T=840 x tan(35/2)
T=840 x tan(17.5)
T=264.85
Now
STA PC=482+72-(2+64.85)=480+07.15
Also L is given as
L=(π/180)RΔ
Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.
L=(π/180)RΔ
L=(π/180)x840 x35
L=512.87 ft
STA PT=480+07.15+5+12.87=485+20.02
Now Ms is the minimum distance which is given as
Here R_v is given as 835
SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering
So Ms is
Now for the clearance from the inside lane
Ms=Ms-lane length
Ms=26.92-5= 21.92 ft.
So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.
Answer:
M10 × 1.5
least number of threads is 4.3
Explanation:
given data
static tensile load = 4 kN = 4000 N
safety factor = 5
coarse‐thread metric = 5.8
solution
we know that proof load for 5.8 class is Sp = 380 MPa
so area of cross section is express as
area = \frac{force \times FOS}{Sp} ......................1
area = \frac{4000 \times }{380}
area = 52.6 mm²
so by table at 58 mm² we can say we resist M10 × 1.5
and
bolt tensile strength is express as
bolt tensile strength = nut shear strength ......2
so here bolt tensile strength = At × Sy (bolt) ...............3
and nut shear strength = πd ( 0.75 t ) Sys
nut shear strength = π × 10 × ( 0.75 t ) × 0.58 × × Sy(bolt) ............4
so from equation 3 and 4 we get
t = 6.37 mm
and
for the pitch is = 1.5 mm
least number of threads is 4.3