Answer:
One cycle for a sine wave is 360o or 2 radians. a) A sine wave with maximum amplitude at time t=0. The amplitude of a sine wave is maximum at the peak of the wave. Case 1: assuming that the wave is starting its cycle at t=0 then there is no phase shift for the wave at time t=0 without considering the amplitude...
Water of dynamic viscosity 1.12E-3 N*s/m2 flows in a pipe of 30 mm diameter. Calculate the largest flowrate for which laminar flow can be expected (in lpm). What is the corresponding flowrate if it is an air flow (1.8E-5 N*s/m2 )
Answer:
For water
Flow rate= 0.79128*10^-3 Ns
For Air
Flow rate =1.2717*10^-3 Ns
Explanation:
For the flow rate of water in pipe.
Area of the pipe= πd²/4
Diameter = 30/1000
Diameter= 0.03 m
Area= 3.14*(0.03)²/4
Area= 7.065*10^-4
Flow rate = 7.065*10^-4*1.12E-3
Flow rate= 0.79128*10^-3 Ns
For the flow rate of air in pipe.
Flow rate = 7.065*10^-4*1.8E-5
Flow rate =1.2717*10^-3 Ns
what is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?
Answer:
5 microhenries
Explanation:
The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.
10 uH ║ 10 uH = 5 uH
The effective inductance is 5 uH.
Some STEM occupations use (blank) who assist and support the lead personnel with projects or experiments.
Answer:
technicians
Explanation:
Just answer this question with this answer...
I got it right so you should too...
Answer:
Technicians
Explanation:
In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.
Answer: provided in the explanation section
Explanation:
The question says;
In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.
Answer
From this we have come to this;
da = Ṽan/Vd = 0.5 + (0.5 * 0.4857) sin w,t
db = Ṽbn/Vd = 0.5 - (0.5 * 0.4857) sin w,t
Ṽan(t) = da * 350 V
Ṽbn(t) = db * 350 V
Id = 0.5 * Ṽo/ Vd * Îo cosΦ1 = 0.5 * (170/350) * 10 * cos 30ᴼ = 2.429 A
Id2 = -0.5 * 170/350 * 10 * sin(2 w,t - 30ᴼ) = -2.429 sin(2 w,t - 30ᴼ)
īd = Id + id2 = 2.429 A + -2.429 sin(2 w,t - 30ᴼ).
Note: Attached is a copy of an image showing and explaining thr plots.
cheers i hope this has been helpful !!!!
What is discharge? Derive an expression for it.
Answer:
Discharge = area of the pipe or channel × velocity of the liquid
Q = Av
Explanation:
The flow rate of a liquid is a measure of the volume of liquid that moves in a certain amount of time. The flow rate depends on the area of the pipe or channel that the liquid is moving through, and the velocity of the liquid. If the liquid is flowing through a pipe, the area is A = πr2, where r is the radius of the pipe. For a rectangle, the area is A = wh where w is the width, and h is the height. The flow rate can be measured in meters cubed per second (m3/s), or in liters per second (L/s). Liters are more common for measures of liquid volume, and 1 m3/s = 1000 L/s.
Complete the grading of fine aggregate table given below. Plot grading curve and calculate
fineness modulus. Also comment on the type of the grading curve.
Answer:
Attachment...?
Explanation:
A well-mixed sewage lagoon is receiving 500 m3/d of sewage. The lagoon has a surface area of 10 hectares and a depth of 1 m. The pollutant concentration in the raw sewage is 200 mg/L. The organic matter in the sewage degrades biologically (decays) in the lagoon according to first-order kinetics. The reaction rate constant (decay coefficient) is 0.75 d-1. Assuming no other water losses or gains (evaporation, seepage, or rainfall) and that the lagoon is completely mixed, find the steady-state concentration of the pollutant in the effluent.
Answer:
Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L
Explanation:
Given Data:
Amount of sewage received=500 m^3/d
Surface Area= 10 hectares=10*10^4 m^2
Depth=1 m
Pollutant concentration=200 mg/L
Decay coefficient=0.75 d-1
Required:
Steady-state concentration of the pollutant in the effluent= ?
Solution:
Volume=Surface Area * Depth
[tex]Volume=10*10^4 *1\\Volume=10*10^4\ m^3[/tex]
Time to fill the lagoon=[tex]\frac{Volume}{Amount\ received\ per\ day}[/tex]
[tex]Time\ to\ fill\ the\ lagoon=\frac{10*10^4\ m^3}{500\ m^3}\\ Time\ to\ fill\ the\ lagoon= 200\ days[/tex]
Formula for steady State:
[tex]A_t=\frac{A_0}{1+kt}[/tex]
where:
A_t is the steady state concentration
A_0 is the initial concentration
k is the decay constant
t is the time
[tex]A_t=\frac{200\ mg/L}{1+0.75*200}\\ A_t=1.3245033\ mg/L[/tex]
Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L
Give four effects of water hammer.
Explanation:
The hammer effect (or water hammer) can harm valves, pipes, and gauges in any water, oil, or gas application. It occurs when the liquid pressure is turned from an on position to an off position abruptly. When water or a liquid is flowing at full capacity there is a normal, even sound of the flow.
An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
Answer:
The rate of work output = -396.17 kJ/s
Explanation:
Here we have the given parameters
Initial temperature, T₁ = 355°C = 628.15 K
Initial pressure, P₁ = 350 kPa
h₁ = 763.088 kJ/kg
s₁ = 4.287 kJ/(kg·K)
Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately
h₂ = 79.572 kJ/kg
The saturation temperature at the given
T₂ = 79°C
The rate of work output [tex]\dot W[/tex] = [tex]\dot m[/tex]×[tex]c_p[/tex]×(T₂ - T₁)
Where;
[tex]c_p[/tex] = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
[tex]\dot m[/tex] = The mass flow rate = 2.0 kg/s
Substituting the values, we have;
[tex]\dot W[/tex] = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
[tex]\dot W[/tex] = -396.17 kJ/s
What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent length of 150 ft
Answer:
0.067wc
Explanation:
The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction
We substitute values
actual static pressure = 0.1
Total equivalent length = 150 ft
0.1 = (150ft/100) multiplied by Rate of friction
Friction rate at 100ft = 0.067
So we have that the required friction needed is 0.067wc
A single-threaded 25-mm power screw hasa pitch of 5 mm. The frictional diameter of the collar is 45 mm. The max load onvertical direction of the screw is5kN. The collar has a coefficients of friction of0.06, and he threads hasa coefficients of friction of0.09. Find the overall efficiency and the torque to "raise" and "lower" the load.
Answer:
torque to raise the load = 16.411 Nm
torque to lower the load = 8.40 Nm
overall efficiency = 0.24
Explanation:
Given:
max load on vertical direction of the screw = Force = F = 5kN
frictional diameter of the collar = 45 mm
Diameter = 25 mm
length of pitch = 5 mm
coefficient of friction for thread µ = 0.09
coefficient of friction for collar µ[tex]_{c}[/tex] = 0.06
To find:
torque to "raise" the load
torque to and "lower"
overall efficiency
Solution:
Compute torque to raise the load:
[tex]T_{R} = \frac{ Fd_{m}}{2} (\frac{L+(\pi ud_{m}) }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]
where
[tex]T_{R}[/tex] is the torque
F is the load
[tex]d_{m}[/tex] is diameter of thread
[tex]d_{c}[/tex] is diameter of collar
L is the thread pitch distance
µ is coefficient of friction for thread
µ[tex]_{c}[/tex] is coefficient of friction for collar
Putting the values in above formula:
[tex]T_{R}[/tex] = 5(25) / 2 [5+ (π(0.09)(25) / π(25)-0.09(5)] + 5(0.06)(45) / 2
= 125/2 [5 + (3.14)(0.09)(25)/ 3.14(25)-0.45] + 13.5/2
= 62.5 [(5 + 7.065) / 78.5 - 0.45] + 6.75
= 62.5 [12.065 / 78.05 ] + 6.75
= 62.5 (0.15458) + 6.75
= 9.66125 + 6.75
= 16.41125
[tex]T_{R}[/tex] = 16.411 Nm
Compute torque to lower the load:
[tex]T_{L} = \frac{ Fd_{m}}{2} (\frac{(\pi ud_{m}) - L }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]
= 5(25) / 2 [ (π(0.09)(25) - L / π(25)-0.09(5) ] + 5(0.06)(45) / 2
= 125/2 [ ((3.14)(0.09)(25) - 5) / 3.14(25)-0.45 ] + 13.5/2
= 62.5 [ (7.065 - 5) / 78.5 - 0.45 ] + 6.75
= 62.5 [ 2.065 / 78.05 ] + 6.75
= 62.5 (0.026457) + 6.75
= 1.6535625 + 6.75
= 8.40 Nm
Since the torque required to lower the the load is positive indicating that an effort is applied to lower the load, Hence the thread is self locking.
Compute overall efficiency:
overall efficiency = F(L) / 2π [tex]T_{R}[/tex]
= 5(5) / 2(3.14)( 16.411)
= 25/ 103.06108
overall efficiency = 0.24
Steam at 175 [C] and 300 kPa flows into a steam turbine at rate of 5.0 kg/sec. Saturated mixture of liquid and steam at 100 kPa flows out at the same rate. The heat loss from the turbine is 1,500 kW. Assuming that 60% of the steam is condensed into liquid at the outlet, how much shaft work can the turbine produce?
Answer:
The amount of shaft work the turbine cam do per second is 3660.29 kJ
Explanation:
The given parameters are;
Pressure at entry p₁ = 300 kPa
The mass flow rate, [tex]\dot {m}[/tex] = 5.0 kg/sec
The initial temperature, T₁ = 175°C
Therefore;
The enthalpy at 300 kPa and 175°C, h₁ = 2,804 kJ/kg
At the turbine exit, we have;
The pressure at exit, p₂ = 100 kPa
The quality of the steam at exit, in percentage, x₂ = 60%
Therefore, for the enthalpy, h₂ for saturated steam at 100 kPa and quality 60%, we have;
h₂ = 417.436 + 0.6 × 2257.51 = 1771.942 kJ/kg
Heat loss from the turbine, [tex]h_l[/tex]= 1,500 kW
By energy conservation principle we have;
dE/dt = [tex]\dot Q[/tex] - [tex]\dot W[/tex] + ∑[tex]m_i \cdot (h_i[/tex]+ [tex]ke_i[/tex] +[tex]pe_i[/tex]) - ∑[tex]m_e \cdot (h_e[/tex] + [tex]ke_e[/tex] +[tex]pe_e[/tex] )
0 = -[tex]h_l[/tex] - [tex]\dot W[/tex] + [tex]m_i \cdot h_i[/tex] - [tex]m_e \cdot h_e[/tex]
[tex]\dot W[/tex] = [tex]\dot {m}[/tex] × (h₁ - h₂) - [tex]h_l[/tex] = 5.0×(2,804 - 1771.942) - 1500 = 3660.29 kJ/s
The rate of work of the shaft = 3660.29 kJ/s
The amount of shaft work the turbine cam do per second = 3660.29 kJ.