STP stands for standard temperature pressure and NTP stands for normal temperature pressure
Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature
= 300 F = 422.04 K
Duct temperature
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant
= 5.67 ×

Heat transfer per foot length is given by
Q = ∈
A (
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 ×
× 3.14 × 0.0508 × 1 × (
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
If your accelerator gets stuck down, do the following: Shift to neutral. Apply the brakes. Keep your eyes on the road and look for a way out.If your accelerator gets stuck down, do the following:
Shift to neutral.
Apply the brakes.
Keep your eyes on the road and look for a way out.
Warn other drivers by blinking and flashing your hazard lights.
Try to drive the car safely off the road.
Turn off the ignition when you no longer need to change direction.
Answer:
98°C
Explanation:
Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7
22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²
Temperature change, t = (50 - 25)°C = 25°C = 298K
Hence, Temperature = 150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =
∴ Temperature change = 2.00K
But temperature, T= (373 - 2)K = 371 K
In °C = (371 - 273)K = 98°C