Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*/2*c^4
∅=2TL/G**c^4
c=∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*/2*c^4)]*c
г =[2T/(J**c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
Answer:
Light = A xor B
Explanation:
If switches A and B produce True or False, then Light will be True for ...
Light = A xor B
Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
Answer:
Exit velocity m/s.
Explanation:
Given:
At inlet:
Properties of steam at 100 bar and 600°C
At exit:Lets take exit velocity
We know that if we know only one property inside the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.
Properties of saturated steam at 80 bar
So the enthalpy of steam at the exit of turbine
Now from first law for open system
In the case of adiabatic nozzle Q=0,W=0
m/s
So Exit velocity m/s.