Answer:
(a). 42.1 ft/s, (b). 3366.66 ft^3/s, (c). 0.235, (d). 18.2 ft, (e). 3.8 ft.
Explanation:
The following parameters are given in the question above and they are;
Induced hydraulic jump, j = 80 ft wide channel, and the water depths on either side of the jump are 1 ft and 10 ft. Let k1 and k2 represent each side of the jump respectively.
(a). The velocity of the faster moving flow can be calculated using the formula below;
k1/k2 = 1/2 [ √ (1 + 8g1^2) - 1 ].
Substituting the values into the equation above a s solving it, we have;
g1 = 7.416.
Hence, g1 = V1/ √(L × k1).
Therefore, making V1 the subject of the formula, we have;
V1 = 7.416× √ ( 32.2 × 1).
V1 = 42.1 ft/s.
(b). R = V1 × j × k1.
R = 42.1 × 80 × 1.
R = 3366.66 ft^3/s.
(c). Recall that R = V2 × A.
Where A = 80 × 10.
Therefore, V2 = 3366.66/ 80 × 10.
V2 = 4.21 ft/s.
Hence,
g2 = V1/ √(L × k2).
g2 = 4.21/ √ (32.2 × 10).
g2 = 0.235.
(d). (k2 - k1)^3/ 4 × k1k2.
= (10 - 1)^3/ 4 × 1 × 10.
= 18.2 ft.
(e).The critical depth;
[ (3366.66/80)^2 / 32.2]^ 1/3.
The The critical depth = 3.80 ft.
Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:
[tex] \frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1} [/tex]
[tex] 10*2 = \sqrt{1 + 8f^2 - 1[/tex]
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
[tex] \frac{V_1}{\sqrt{g*y_1}} = 7.416 [/tex]
[tex] V_1 = 7.416 *\sqrt{32.2*1}[/tex]
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
[tex] V_2 = \frac{3666.66}{80*10} [/tex]
= 4.208 ft/sec
Froude number, F2 =
[tex] \frac{V_2}{g*y_2} = \frac{4.208}{32.2*10} [/tex]
F2 = 0.235
d) [tex]El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}[/tex]
[tex] El = \frac{(10-1)^3}{4*1*10} [/tex]
[tex] = \frac{9^3}{40} [/tex]
= 18.225ft
e) for critical depth, we use :
[tex] y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3 [/tex]
= 3.80 ft
) Water flows steadily up the vertical-1-in-diameter pipe and out the nozzle, which is 0.5 in. in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be 30 f t/s. Calculate the minimum gage pressure required at the section (1). If the device were inverted, what would be the required minimum pressure at section (1) to maintain the nozzle exit velocity at 30 f t/s.
Answer:
a) 9.995 psi
b) 1.3725 psi
Explanation:
Given:-
- The diameter of the pipe at inlet, d1 = 1 in
- The diameter of the pipe at exit, de = 0.5 in
- The exit velocity, Ve = 30 ft/s
- The exit discharge pressure Pe = 0 ( gauge )
- The density of water ρ = 1.940 slugs/ft3
Find:-
Calculate the minimum gage pressure required at the section (1)
Solution:-
- The mass flow rate m ( flow ) for the fluid remains constant via the continuity equation applies for all steady state fluid conditions.
m ( flow ) = ρ*An*Vn = constant
Where,
An: the area of nth section
Vn: the velocity at nth section
- Consider the point ( 1 ) and exit point. Determine the velocity at point ( 1 ) via continuity equation.
- The cross sectional area of the pipe at nth point is given by:
An = π*dn^2 / 4
- The continuity equation becomes:
ρ*A1*V1 = ρ*Ae*Ve
Note: Water is assumed as incompressible fluid; hence, density remains constant.
V1 = ( Ae / A1 ) * Ve
V1 = ( (π*de^2 / 4 ) / (π*d1^2 / 4) ) *Ve
V1 = ( de / d1 ) ^2 * Ve
V1 = ( 0.5 / 1 )^2 * 30
V1 =7.5 ft/s
- The required velocity at section ( 1 ) is V1 = 7.5 ft/s.
- Apply the bernoulli's principle for the point ( 1 ) and exit. Assuming the frictional losses are minimal.
P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he
- We will set " h1 " as datum; hence, h1 = 0. The elevation of exit nozzle from point (1) is at he = 10 ft.
- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).
Therefore the simplified equation becomes:
P1 + 0.5*ρ*V1^2 = 0.5*ρ*Ve^2 + ρ*g*he
P1 = 0.5*ρ* ( Ve^2 - V1^2 ) + ρ*g*he
P1 = 0.5*1.940*( 30^2 - 7.5^2 ) + 1.940*32*10
P1 = 818.4375 + 620.8
P1 = 1439.2375 lbf / ft^2 = 9.995 psi
- If the device is inverted then the velocity at the inlet " V1 " would remain same as there is no change in continuity equation - ( Diameters at each section remains same ).
- The only thing that changes is the application of bernoulli's equation as follows:
P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he
- We will set " he " as datum; hence, he = 0. The elevation of point ( 1 ) from exit nozzle is at h1 = 10 ft.
- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).
Therefore the simplified equation becomes:
P1 + 0.5*ρ*V1^2 + ρ*g*h1 = 0.5*ρ*Ve^2
P1 = 0.5*ρ* ( Ve^2 - V1^2 ) - ρ*g*h1
P1 = 0.5*1.940*( 30^2 - 7.5^2 ) - 1.940*32*10
P1 = 818.4375 - 620.8
P1 = 197.6375 lbf / ft^2 = 1.3725 psi
Why does the the diffusion capacitance fall off at high frequencies?
1. At high frequencies the dopants are vibrated across the metallurgical junction and the doping profiles smooths out.
2. Since Gd increases with frequency Cd must decrease with frequency and the parallel combination of Gd and Cd is constant.
3. When the applied bias frequency is higher than the speed which the majority carriers can respond the diode starts to disappear electrically.
4. It takes time to store and remove minority charge, so when the frequency is higher than the inverse of the minority carrier lifetime the carriers can't respond as well.
Answer:
The answer 1 is correct.
Explanation:
When a junction called the p-n is forward biased, the capacitance of diffusion will form or structured across depletion layers.
When at a higher frequency or frequencies, the dopants, are vibrated across the metallurgical junction and the doping profiles smooths out.
An equation used to evaluate vacuum filtration is Q = ΔpA2 α(VRw + ARf) , Where Q ≐ L3/T is the filtrate volume flow rate, Δp ≐ F/L2 the vacuum pressure differential, A ≐ L2 the filter area, α the filtrate "viscosity," V ≐ L3 the filtrate volume, R ≐ L/F the sludge specific resistance, w ≐ F/L3 the weight of dry sludge per unit volume of filtrate, and Rf the specific resistance of the filter medium. What are the dimensions of Rf and and α?
The dimensions of the specific resistance of the filter medium, Rf, are F/L (force per unit length), and the dimensions of the filtrate "viscosity", alpha, are T/L^2 (time per area). This ensures the dimensional consistency of the vacuum filtration equation.
Explanation:To answer the question regarding vacuum filtration, we need to determine the dimensions of Rf and alpha. The equation provided is Q = Delta pA2 alpha(VRw + ARf), where each symbol has an associated dimension. To keep the equation dimensionally consistent, the dimensions of the terms added within the parenthesis must be the same, i.e., the dimensions of VRw must equal the dimensions of ARf.
Let's break down the dimensions:
The dimensions of V (volume) are L3.The dimensions of Rw (specific resistance of the sludge) are L/F.The dimensions of A (area) are L2.Therefore, the dimensions of Rf must be F/L, so when multiplied by A (area), it equals L/F, the same as VRw.Similarly, to ensure that when alpha is multiplied by A2, the result has dimensions of L5/FT to be consistent with Q (Delta pA2) without the VRw + ARf term, the dimensions of alpha must be T/L2.Hence, the dimensions of Rf are F/L (force per unit length), and the dimensions of alpha are T/L2 (time per area).
Thermal energy generated by the electrical resistance of a 5-mm-diameter and 4-m-long bare cable is dissipated to the surrounding air at 20°C. The voltage drop and the electric current across the cable in steady operation are measured to be 60 V and 1.5 A, respectively. Disregarding radiation, estimate the surface temperature of the cable. Evaluate air properties at a film temperature of 60°C and 1 atm pressure. Is this a good assumption?
Answer:
surface temperature = 128.74⁰c
Explanation:
Given data
diameter of cable = 5 mm = 0.005 m
length of cable = 4 m
T∞ ( surrounding temperature ) = 20⁰c
voltage drop across cable ( dv )= 60 V
current across cable = 1.5 A
attached to this answer is the comprehensive analysis and solution to the problem.
The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts
Heat is transferred at a rate of 2 kW from a hot reservoir at 875 K to a cold reservoir at 300 K. Calculate the rate at which the entropy of the two reservoirs changes. (Round the final answer to six decimal places.) The rate at which the entropy of the two reservoirs changes is kW/K. Is the second law satisfied?
Answer:
The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.
Explanation:
Solution
From the question given we recall that,
The transferred heat rate is = 2kW
A reservoir cold at = 300K
The next step is to find the rate at which the entropy of the two reservoirs changes is kW/K
Given that:
Δs = Q/T This is the entropy formula,
Thus
Δs₁ = 2/ 300 = 0.006667 kW/K
Δs₂ = 2 / 875 =0.002285
Therefore,
Δs = 0.006667 - 0.002285
= 0.004382 kW/K
Yes, the second law is satisfied, because it is seen as positive.
HELP!
A roller picture has a front roll of 1.3m diameter by 1.7m long and two rear wheel each of 2.6m diameter
1. From the sizes given for the road roller calculate:
A) The surface area of the front drum (ignore material thickness)
B) The volume of the front roll in litres if the end plates are set in 75mm from each end of the roll and the thickness of the material is 20mm.
C) If one gallon=4.57 litres, calculate in gallons, the amount of water that can be held in the front roll.
D) The ratio in size of the front roll diameter to the rear wheel diameter.
E) The distance travelled by the rear wheel when the front roll turns one revolution, and state this as a percentage of one full revolution.
The statement that is NOT true about the difference between laminar and turbulent boundary layers is:1.the Reynolds number for a turbulent boundary layer is higher than that for a laminar boundary layer. 2.the wall shear stress for a turbulent boundary layer is greater than that for a laminar boundary layer. 3.a turbulent boundary layer is thicker than a laminar boundary layer.4.the velocity gradient at the wall is greater for a laminar boundary layer than a turbulent boundary layer.
Answer:
4. The velocity gradient at the wall is greater for a laminar boundary layer than a turbulent boundary layer.
Explanation:
This is false
Consider the system whose transfer function is given by: 1()(21)(3)Gsss=++. (a)Use the root-locus design methodology to design a lead compensator that will provide a closed-loop damping ζ =0.4and a natural frequency ωn=9 rad/sec.The general transfer function for lead compensation is given by(b)Use MATLAB to plot the root locus of the feed-forward transfer function, D(s)*G(s), and verify if the closed loop system pole is located on a root locus for the calculated value of K
Answer:
transfer function from R(s) to E(s) and determine the steady-state error (ess) for a unit-step reference input (c) Select the system parameters (k, kP, kI) such that the closed-loop system has damping coefficient ζ = 0.707 Figure Control system diagram.
Explanation:
hope this helps
John was a high school teacher earning $ 80,000 per year. He quit his job to start his own business in pizza catering. In order to learn how to run the pizza catering business, John enrolled in a TAFE to acquire catering skills. John’s course was for 3 months. John had to pay $2,000 as tuition for the 3 months.
After the training, John withdrew $110,000 from his savings account. He had been earning 5 percent interest per year for this account. He also borrowed $50,000.00 from his friend whom he pays 6 percent interest per year. Further, to start the business John used his own premises. He was receiving $12,000 from rent per year. Finally, to start the business John uses $50,000 he had been given by his father to go on holiday to USA.
John’s first year of business can be summarised as follows:
Item Amount $
Revenue- Pizza Section 400,000
Revenue- Beverages Section 190,000
2 Cashiers (wages per worker) 55,000
Pizza ingredients 50,000
Manager 75,000
3 Pizza bakers (wages per baker) 60,000
Equipment 10,000
Based on your calculated accounting profit and economic profit, would you advise John to return to his teaching job? Show your work (10 marks)
Answer:
John should continue with his catering business because accounting and economic profits from pizza catering are more than his annual salary from teaching.
Explanation:
John can earn from teaching $80,000 annually
For starting business of pizza catering John will incur following expense
Tuition for catering skills $2,000
Opportunity cost for withdrew of funds ($110,000 * 5%) $5,500
Interest for Borrowing from friend ($50,000 * 6%) $3,000
Opportunity cost of Rent from premises $12,000
Total opportunity cost $22,500
Revenue:
Revenues from Pizza $400,000
Revenue from Beverages $190,000
Total revenue $590,000
Expenses:
Cashier wages ($55,000 * 2) $110,000
Pizza ingredients $50,000
Manager salary $75,000
Pizza bakers wages ($60,000 * 3) $180,000
Equipment $10,000
Total Expense $425,000
Net Income - Accounting Profit ($590,000 - $425,000) = $165,000
Less opportunity cost $22,500
Economic Profit $142,500
g (d) Usually, in the case of two finite-duration signals like in parts (a) and (b), the convolution integralmust be evaluated in 5 different regions: no overlap (on the left side), partial overlap (on the left side),complete overlap, partial overlap (on the right side), and no overlap (on the right side). In this case,there are only 4 regions. Why
you face is A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
the hoop is cast on the rough surface such that it has an angular velocity w=4rad/s and an angular acceleration a=5rad/s^2. also, its center has a velocity v0=5m/s and a deceleration a0=2m/s^2. determine the acceleration of point a at this instant
The acceleration of point A on the hoop can be calculated by considering both the tangential acceleration due to the hoop's angular acceleration and the radial (centripetal) acceleration. These values are vectorially added along with the deceleration of the hoop's center.
Explanation:To determine the acceleration of point A, we must consider both the tangential and radial (centripetal) accelerations since the hoop is experiencing angular acceleration and the center of the hoop has a deceleration. The tangential acceleration (at) is due to the angular acceleration and is calculated by multiplying the angular acceleration (
a) by the radius (r) of the hoop. The radial (centripetal) acceleration (ar) depends on the angular velocity (
w) and the radius of the hoop. The equation for centripetal acceleration is ar = w2
r. The total acceleration (aA) of point A can be found by vectorially adding the tangential and radial accelerations, considering the direction of deceleration of the center as well.