Answer:
c Waste stack
Explanation:
A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit
Answer:
T = 438.87 N.m
Explanation:
The power required to raise the 4961 N load in 10 meters for 2 minutes is:
[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]
P = Torque × W
[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]
[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]
[tex]T = \dfrac{413.42}{0.942}[/tex]
T = 438.87 N.m
A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?
Answer:
10.007
Explanation:
Assuming we have to find out the compression ratio of the engine
Given information
Cubic capacity of the engine, V = 245 cc
Clearance volume, V_c = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]
plugging values we get
245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]
Solving we get D =7 cm
therefore, L= 7/1.1 =6.36 cm
Now,
the compression ratio is given as:
r =(V+V_c)/V_c
on substituting the values, we get
r = (245+27.2)/27.2 =10.007
Hence, Compression ratio = 10.007
How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including the requirements needed?
Answer:
The application of regulations in locations are very important.
Explanation:
The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.
i need some help with this
Java programing
def digit_sum(number):
sumOfDigits = 0
while number > 0:
sumOfDigits += number % 10
number = number // 10
return sumOfDigits
or you can use,
In [2]: digit_sum(10)
Out[2]: 1
In [3]: digit_sum(434)
Out[3]: 11
or you can use,
digits = "12345678901234567890"
digit_sum = sum(map(int, digits))
print("The equation is: ", '+'.join(digits))
print("The sum is:", digit_sum)
Have a great day <3
Car pool vehicles are those with
Answer:
The pavement of the carpool lanes is marked with the diamond symbol. These carpool lanes are reserved for buses and vehicles with a minimum of two or three people and that includes the driver
Explanation:
hope this helped
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.