Answer:4050 W
Explanation:
Given
Heat transfer Coefficient(h)=
Air temperature =75 F
surface area(A)=
Temperature of hot tube is 102 F
We know heat transfer due to convection is given by


Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
Answer:
See explanation
Explanation:
The magnetic force is
F = qvB sin θ
We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields
F
=
(
20
×
10
−
9
C
)
(
10
m/s
)
(
5
×
10
−
5
T
)
=
1
×
10
−
11
(
C
⋅
m/s
)
(
N
C
⋅
m/s
)
=
1
×
10
−
11
N
Answer:
DIAMETER = 9.797 m
POWER = 
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as

where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius

v = 0.8409 m^3/ kg
from continuity equation





mass flow rate is given as


the power produced ![\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]](https://tex.z-dn.net/?f=%5Cdot%20W%20%3D%20%5Cdot%20m%20%5Cfrac%7B%20V_1%5E2%20-%20V_2%5E2%7D%7B2%7D%20%3D%20717.3009%20%5B%5Cfrac%7B12%5E2%20-%208%5E2%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2Fs%5E2%7D%5D)

Answer:
a) What is the surface temperature, in °C, after 400 s?
T (0,400 sec) = 800°C
b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s
c) What is the temperature, in °C, 1 mm from the surface after 400 s?
T (1 mm, 400 sec) = 798.35°C
Explanation:
oak initial Temperature = 25°C = 298 K
oak exposed to gas of temp = 800°C = 1073 K
h = 20 W/m².K
From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k
Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.
From energy balance: 
Initial temperature wall = 
Surface temperature = T
Gas exposed temperature = 