Answer:
Steady flow
Explanation:
Flows in fluids can be categorized into different classes depending on the type of flow and the variations in their characteristics such as velocity, pressure, density, temperature, e.t.c
When these characteristics do not change when measured over time at a single point, then the flow is said to be steady. For a steady flow, the mathematical expression, amidst other conditions, is given as follows;
[tex]\frac{dV}{dt} = 0, \frac{dP}{dt} = 0, \frac{dT}{dt} = 0[/tex]
Where;
V, P and T are the velocity, pressure and temperature of the fluid.
PS:
Other types of flows include:
i. Unsteady flow
ii. Laminar flow
iii. Turbulent flow
iv. Uniform flow
v Non-uniform flow
vi. Rotational flow
vii. Irrotational flow
Students would like to sell cold drinks to raise money for a field trip. They need to keep the drinks cold for 3 hours at the ball game. Which step of the engineering design process are the students at?
Answer:
Problem definition
Explanation:
The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are
problem definition conceptual designpreliminary designdetailed designdesign communication.The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'
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 !!!!
Why do electricians require critical thinking skills? In order to logically identify alternative solutions to problems in order to understand the implications of new information in order to attend to what others are saying in order to install equipment and wiring to meet specifications
Answer:
In order to logically identify alternative solutions to problems
Explanation:
Electricians are specialized in electrical wiring of buildings, transmission lines, stationary machines, and related equipment. They are either employed in the installations of new electrical components, or to maintain an already installed component. The job of an electrician can be mentally tasking, especially in troubleshooting for fault, and methods of fixing of faults. Some problems might require an out-of-norm approach to solve, and the electrician has to be able to logically identify alternative solutions to problems.
Answer:
in order to logically identify alternative solutions to problems
Explanation:
it makes the most sense
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.
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.
Thermoplastics that exhibit a definite TM point have this kind of crystalline structure:
A. semi-crystalline
B. isotactic
C.amorphous
D. syndiotactic
Answer:
The correct option is A
Explanation:
Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.
There are two main classes of thermoplastics; semi-crystalline and amorphous.
An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A
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
Define water hammer. Give four effects of water hammer.
Answer:
Water Hammer is a knocking sound in an water pipe which occurs when the tap is turned off briskly
Explanation:
A turbine in a simple ideal Rankine cycle with water as the working fluid operates with an inlet temperature of 500°C and pressure of 600 kPa and an exit temperature of 45°С. Find the turbine work in MW if the mass flow rate of the steam is at 11 kg/s.
Answer:
the turbine work in MW is [tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
Explanation:
Given that:
the inlet temperature = 500°C
pressure = 600 kPa
exit temperature = 45°С
mass flow rate = 11 kg/s
The objective is to find the turbine work in MW given that the mass flow rate of the steam is at 11 kg/s.
From the steam tables, the data obtained for the enthalpies at the inlet temperature of 500°C and a pressure of 600 kPa is:
[tex]\mathtt{h_1 : 3482.7 \ kJ/kg}[/tex]
The turbine expansion process is also the isentropic process, as such:
Inlet entropy [tex]\mathbf{s_1}[/tex] = Exit entropy [tex]\mathbf{s_2}[/tex]
The data obtained from the steam table for [tex]\mathbf{s_1}[/tex] = 8.003 kJ/kg.K
[tex]s_2 = s_{f2}+ x_2 *s_{fg2}[/tex]
The data obtained from the steam tables for this entities are as follows:
[tex]\mathsf{s_{f2} = 0.638 \ kJ/Kg/K }[/tex]
[tex]\mathtt{s_{fg2}=7.528 \ kJ/kg/K}[/tex]
since [tex]\mathbf{s_1}[/tex] = [tex]\mathbf{s_2}[/tex] = 8.003 kJ/kg.K
Therefore;
[tex]8.003 = 0.638 + x_2 * 7.528[/tex]
[tex]8.003 -0.638 = 7.528x_2[/tex]
[tex]7.365= 7.528x_2[/tex]
[tex]x_ 2= \dfrac{ 7.365}{7.528}[/tex]
[tex]x_2 = 0.978[/tex]
From the steam tables, the data obtained for the enthalpies at the exit temperature of 45°C and a pressure of 600 kPa is:
[tex]h_{f2}[/tex] = 188.4
[tex]h_{fg2[/tex] =2394.9
Thus;
[tex]h_ 2= 188.4 +0.978*2394.9[/tex]
[tex]h_ 2=2530.61[/tex] kJ/kg
The workdone for the turbine process can be computed as:
[tex]\mathsf{W_t = m(h1-h_2)}[/tex]
[tex]\mathsf{W_t = 11(3482.7-2530.61)}[/tex]
[tex]\mathsf{W_t = 11(952.09)}[/tex]
[tex]\mathsf{W_t = 10472.99}[/tex] kW
To MW, we have
[tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
The spring-held follower AB has weight W and moves back and forth as its end rolls on the contoured surface of the cam, where the radius is r and z = asin(2θ). If the cam is rotating at a constant rate θ', determine the force at the end A of the follower when θ = θ 1. In this position the spring is compressed δ1. Neglect friction at the bearing C.
Answer:
some parts of your question is missing attached below is the missing part
Answer : Fa = 4.46 Ib
Explanation:
use the equation
Z = 0.1 sin2∅
next we will differentiate the equation to get the locus of the velocity
z = 0.2 cos2∅∅
differentiate the equation furthermore to get the locus of acceleration in the horizontal axis
z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅
note : express ∅ as 6 rad.[tex]s^{-1}[/tex] for angular velocity and ∅ = 0 for angular acceleration
equation above becomes :
Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)
= - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )
next calculate The force at the end of A of the follower
Fa - Kx = mz
note: m = w / g hence : Fa - Kx = w/g z ------- (2)
w = weight of the spring-held follower = 0.75 Ib
x = compression of the spring = 0.4
k = spring stiffness = 12 Ib/ft
∅ = 45⁰
g = 32.2 ft/s^2
input these values into equation 2
hence : Fa = 4.46 Ib ( force at the end A of the follower )
Air flow at 15 m/s past a thin flat plate. Estimate the distance x from the leading edge at which the boundary layer thickness will be (a) 10 cm and (b) 1 mm. Use a transition Reynolds number of 5 x 105.
Answer:
a) x = 5.7791 m ( this a turbulent flow )
b) x = 0.04 m ( this is a Laminar flow )
Explanation:
For Air, take
p = 1.2 kg/m³ and u = 1.8×10⁻⁵ kg/m³
So for A , x = 10 cm
Guess turbulent flow;
(Sturb / x) = (0.16 / (p∪x / u)^1/7)
we substitute
(0.1/x = 0.16) / ( 1.2 × 15 × x / 1.8×10⁻⁵)^1/7
x = 5.7791 m
CHECK
Re = (1.2 × 15 × 5.7791) / 1.8×10⁻⁵ = 5.779×10⁶
Therefore this a turbulent flow
for B, x = 1mm
Guess Laminar flow
(S-laminar / x) = ( 5 / (Reₓ)^1/2)
x = (Stutb)²p∪ / 23u
x = ((0.001)² × 1.2 × 15) / (25 × 1.8×10⁻⁵)
x = 0.04 m
CHECK
Re = (1.2 × 15 × 0.04) / (1.8×10⁻⁵)
= 40000
Therefore this is a Laminar flow
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
Why some types of aggregate are susceptible to damage from repeated freezing and thawing? Explain.
Answer:
Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.
When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.
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:
1. At the end of the day, a bakery gives everything that is unsold to food banks for the needy. If it has 12 apple pies left at the end of a given day, in how many different ways can it distribute these pies among six food banks for the needy?
2. In how many different ways can the bakery distribute the 12 apple pies if each of the six food banks is to receive at least one pie?
Answer:
The answer ix below
Explanation:
There are 12 apple pies left.
Given that:
n = number of apple pies left = 12
x = number of food banks = 6
1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:
C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]
12 apple pies can be distributed among 6 food banks in 6188 ways
2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:
C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]
12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways
how many crankshaft is in V8 engine?
Answer:
Explanation:
hola friend!!!!!
there is only one crankshaft in V8 engine
Hope this helps
plz mark as brainliest!!!!!!!!
A three-phase, 60-Hz, completely transposed 345-kV, 200-km line has two 795,000-cmil (403-mm2) 26/2 ACSR conductors per bundle and the following positive-sequence line constants: z 0.032 + 10.35 /km y j4.2 x 10-6 S/km Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at 95% of rated voltage. Assuming a medium-length line, determine the following:
a. ABCD parameters of the nominal π circuit
b. Sending-end voltage Vs, current Is, and real power Ps
c. Percent voltage regulation
d. Transmission-line efficiency at full load
Answer:
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
Explanation:
attached is the detailed solution to the problem
Given data:
l = 200 km
z = 0.032 + j0.35 Ω/km
y = j4.2 * 10^-6 S/km
A) find the total series impedance and shunt admittance
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
a circuit contains four capacitors connected in parallel. The values of the capacitors are 0.1 microfarads, 0.6 microfarads 1.0 microfarads and 0.05 microfarads. what is the total capacitance of this circuit. A.0.46 Microfarads B. 1.75Microfarads C.1.25Microfarads D0.42Microfarads
Answer:
B. 1.75 microfarads
Explanation:
When capacitors are connected in parallel, the total capacitance is equivalent to the sum of each individual capacitor's capacitance.
With this in mind, we have the values, 0.1, 0.6, 1.0, and 0.05, all microfarads. So simply, we need to sum these values.
0.1 + 0.6 + 1.0 + 0.05
= 0.7 + 1.0 + 0.05
= 1.7 + 0.05
= 1.75
So the total capacitance for this circuit would be 1.75 microfarads.
Cheers.
Which 3-input gate outputs a false value only when all inputs are true?
a) AND
b) NAND
c) OR
d) NOR
e) XOR
f) XNOR
Answer:
(b) NAND
(d) NOR
(f) XNOR
Explanation:
This is best explained using a truth table.
A truth table containing 3 hypothetical inputs and their corresponding outputs using the AND, NAND, OR, NOR, XOR, and XNOR gates, has been attached to this response.
From the table, it can be seen that the NAND, NOR and XNOR gates produce a false value (0) when all inputs are true.
Note:
On the table,
For the AND column, a 1 is being output only when all three inputs are 1. Otherwise, a 0 results.
For the NAND column (which is the negation of the AND), a 0 is being output only when all three inputs are 0. Otherwise, a 1 results.
For the OR column, a 1 is being output when at least one of the three inputs is 1. Otherwise, a 0 results.
For the NOR column (which is the negation of the OR), a 0 is being output when at least one of the three inputs is 0. Otherwise, a 1 results.
For the XOR column, a 1 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 0 results.
For the XNOR column (which is the negation of the XOR), a 0 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 1 results.
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
What is the minimum amplitude at time zero sine wave??
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...
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.
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:
Give the general layout of centrifugal pump.
Answer:
A centrifugal pump is a rotating machine that pumps liquid by forcing it through a paddle wheel or propeller called an impeller. It is a most common type of industrial pump. By the effect of the rotation of the impeller, the pumped fluid is drawn axially into the pump, then accelerated radially, and finally discharged tangentially.
Explanation:
Suppliers generally offer charts in the plan, which present the various things at the nominal operating point using two main arrangements: the inductors and the balancing pants.
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
In a voltage-divider biased pnp transistor, there is no base current, but the base voltage is approximately correct. The most likely problem(s) is
Answer:
the base-emitter junction is open and the emitter resistor is open
Explanation:
Because here there will be no base current only when the the base emitter junction is kept open and. Also when emitter resistor is kept open or with thus there will be no voltage drop across the Resistor meaning the base voltage Will be equal to that in the voltage divider circuitry
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
Class hello Sarah{
Pubic static void(“hello Sarah”)
Name=input(“hello Sarah”)
Print(“hello Sarah”).
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
}
}
Explanation:
The question seem incomplete; However, since the program is incorrect, I'll assume the question is to correct the given code;
The program segment was written in Java and the correct program is in the answer section
See bold texts for line by line explanation
This line declares the class name
class helloSarah {
This line declares the main method of the program
public static void main(String [] args) {
This line initializes string variable Name
String Name = "Hello Sarah";
This line prints the initialized variable
System.out.print(Name);
}
}
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
]
Explanation:
what is advantage and disadvantage of flat top sampling in digital signal trasmission?
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.