The heat transfer which is in steady state, the heat transfer rate to the wall is equal to the wall.
<u>Explanation:</u>
- The convection transfer of heat to the wall is
- Here,
is the temperature of solid surface, 
is the temperature of moving fluid stream which is adjacent of solid surface, h is the heat transfer coefficient. - The coefficient of convection heat transfers outer surface contains 3 times to the inner surface which experience smaller drop of temperature for 3 times that compares to inner surface.
- Hence, the temperatures outer surface get close to the surroundings of air temperature.
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
Answer:
u will need good car parts and a very well made engine
Explanation:
u need a good engine because if u only work on the outer layer of the car the inner parts will be slow and old and the car will have problems
Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.
