Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on forward speed, angular speed, size, and viscosity and density ofthe fluid. How would the expression change if gravity were a relevant variable in the case of a ship propeller?
Answers
Answer:
Find the given attachment
Related Questions
a turbine operating at steady state at 500 kPa, 860 K and exists at 100 kPa. A temperature sensor indicates that the exit air temperature is 460 K. Stray heat transfer and kinetic and potential energy effects are negligible. The air can be modeled as an ideal gas. Determine if the exit temperature reading can be correct. If yes, determine the power developed by the turbine for an expansion between these states,in kJ/kg of air flowing. If no, provide an explanation with supporting calculations
Answers
Answer:
Given:
P₁ = 500 kPa
T₁ = 860 K
P₂= 100 kPa
T₂ = 460 K
Let's take entropy properties of T1 and T2 from ideal properties of air,
at T = 860K, s(T₁) = 2.79783 kJ/kg.K
at T = 460K, s(T₂) = 2.13407 kJ/kg.K
using entropy balance equation:
[tex] \frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}] [/tex]
[tex] \frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}] [/tex]
= - 0.2018 kJ/kg. K
In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).
A rigid tank contains 5 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. (a) Sketchthe process with respect to the saturation lines on a T-vdiagram. Indicatepressure valuesat each state. (5pt) (b) Determinethe entropy change of the steam, in kJ/K. (10 pt) (c) Determinethe totalentropy change associated with this process, in kJ/K. (15pt)
Answers
Answer:
Explanation:
Part c:
(Q)out = m(u1 - u2) = 5(2506-193.67) = 11562KJ
Total Entropy change = Δs + (Qout/Tsurrounding) = -33.36 + (11562/393) = 5.44KJ/K
In a given rocket engine, a mass flow of propellants equal to 87.6 lbm/s is pumped into the combustion chamber, where the stagnation temperature after combustion is 6000°R. The combustion products have mixture values R = 2400 ft·lb/(slug·°R) and γ = 1.21. If the throat area is 0.5 ft2 , calculate the stagnation pressure in the combustion chamber in lb/ft2 . Hint: use your knowledge about choked flow.
Answers
Answer:
15 atm
Explanation:
Solution
Given that:
The first step is to convert mass flow rate unit
m = 87.6 lbm/s = 2.72 slug/sec
Now,
We express the mass flow rate, which is
m = p₀ A* /√T₀√y/R ( 2/ y+1)^(y+1) (y-1)
so,
2.72 = p₀ ^(0.5)/√6000√1.21/2400 (2/2.21)^10.52
Thus,
p₀ = 31736 lb/ft^2
Then,
We convert the pressure in terms of atmosphere
p₀ = 31736/ 2116 = 15 atm
A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 10.1 Ω resistor to create a closed circuit. During a time interval of 0.163 s, the magnetic field strength decreases uniformly from 0.573 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.
Answers
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
[tex]emf= N\frac{d \phi}{dt}[/tex]
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
[tex]emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V[/tex]
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermal conductivity k = 0.78 W/m·K, and emissivity ε = 0.9. The room and the walls that face the window are maintained at 25°C, and the average temperature of the inner surface of the window is measured to be 5°C. If the temperature of the outdoors is −5°C, determine (a) the convection heat transfer coefficient on the inner surface of the window, (b) the rate of total heat transfer through the window, and (c) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case?
Answers
Answer:
Explanation:
The solutions to this question can be seen in the following screenshots taken from the solution manual.
Technician A says inspecting the operation of the automatic emergency brake system helps determine whether the spring brakes will apply during a loss of system pressure. Technician B says the inspection also helps identify whether several controlled applications of the brake pedal can be used to help slow a vehicle with a sudden severe drop in air pressure. Who is correct?
Answers
Answer: Both Technicians
Explanation:
When testing a spring break it is advisable to Step on and off the brake, with the engine off, the parking brake knob is expected to pop out when air pressure falls between 20-40 psi.
Go under the vehicle and pull the spring brakes.
Turn on the engine back and pump the brake pedal down to the floor. To test the spring breaks
The cross section at right is made of 2024-T3 (Ftu=62 ksi, E=10.5 Msi,Ec=10.7 Msi) aluminum clad sheet. Dimensions of the angle areb1=1.25", b2=1.75", t1=0.050", & t2=0.080". If the angle is 20" long, and the ends can be considered pinned, determine the critical buckling allowable Pcr for the angle & the allowable buckling stress σcr.
Answers
Answer:
See explaination
Explanation:
Given that
the cross section at right is mode 2024-T3
Ftu=62 ksi
E=10.5 Msi
Ec=10.7 Msi
the angle b1=1.25"
b2=1.75"
t1=0.050"
t2=0.080"
Kindly check attachment for the step by step solution of the given problem.
Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is to be heated by electric resistance heaters. If the house is to be maintained at 25 0C, determine the reversible work input for this process and the irreversibility.
Answers
Answer:
a) [tex]\dot W = 0.978\,kW[/tex], b) [tex]I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW[/tex]
Explanation:
a) The ideal Coefficient of Performance for the heat pump is:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
[tex]COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}[/tex]
[tex]COP_{HP} = 14.198[/tex]
The reversible work input is:
[tex]\dot W = \frac{\dot Q_{H}}{COP_{HP}}[/tex]
[tex]\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]
[tex]\dot W = 0.978\,kW[/tex]
b) The irreversibility is given by the difference between real work and ideal work inputs:
[tex]I = \dot W_{real} - \dot W_{ideal}[/tex]
[tex]I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW[/tex]
Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific volume is one-tenth of the specific volume at state 1. Applying the ideal gas model and assuming variations in specific heat, determine (a) T2, in °C, and (b) the work, in kJ/kg.
Answers
(a)[tex]\( T_2 = 87.92°C \), (b) \( W = 8089.91 \text{ kJ/kg} \)[/tex] for the isentropic compression of air in a piston-cylinder assembly.
To solve this problem, we'll use the isentropic compression process for an ideal gas. The process being isentropic means that entropy remains constant during the compression, which implies [tex]\( S_1 = S_2 \). Given that it's an ideal gas, we can use the specific gas constant \( R \) and the specific heats \( C_{p} \) and \( C_{v} \) to solve the problem. The relationship between these variables is \( C_{p} - C_{v} = R \).[/tex]
First, let's find
[tex]\( T_2 \) using the given information:\( T_1 = 35°C \)\( v_2 = \frac{1}{10} v_1 \)Since it's an isentropic process, we have:\[ \frac{T_2}{T_1} = \left( \frac{v_1}{v_2} \right)^{k-1} \][/tex]
[tex]where \( k = \frac{C_p}{C_v} \) is the ratio of specific heats. We need to express \( T_1 \) and \( v_2 \) in terms of \( v_1 \) to solve for \( T_2 \):\[ v_2 = \frac{1}{10} v_1 \]\[ v_1 = \frac{RT_1}{P_1} \][/tex]
Since it's an ideal gas, we can use the ideal gas law [tex]\( Pv = RT \):[/tex]
[tex]\[ P_1v_1 = RT_1 \]\[ P_1 = \frac{RT_1}{v_1} \]Now substitute \( v_1 \) into the equation for \( P_1 \):\[ P_1 = \frac{RT_1}{v_1} = \frac{RT_1}{\frac{RT_1}{P_1}} = P_1 \]So, \( P_1 \) cancels out. This means pressure remains constant during the process. Now, let's substitute \( v_1 \) and \( v_2 \) into the equation for \( T_2 \):[/tex]
[tex]\[ T_2 = T_1 \left( \frac{v_1}{v_2} \right)^{k-1} \]Now, calculate \( T_2 \):\[ T_2 = 35 \left( \frac{RT_1}{\frac{RT_1}{10}} \right)^{k-1} \]\[ T_2 = 35 \left( 10 \right)^{k-1} \]We know \( k = \frac{C_p}{C_v} \), and since \( C_p - C_v = R \), we can rewrite \( k \) as \( k = 1 + \frac{R}{C_v} \). Substituting this into the equation:\[ T_2 = 35 \left( 10 \right)^{\left( 1 + \frac{R}{C_v} \right) - 1} \]\[ T_2 = 35 \left( 10 \right)^{\frac{R}{C_v}} \][/tex]
Now, let's find the specific heat ratio[tex]\( k \):\[ k = 1 + \frac{R}{C_v} \]\[ k - 1 = \frac{R}{C_v} \]\[ C_v = \frac{R}{k - 1} \]Given that \( C_p - C_v = R \), we can also express \( C_p \) in terms of \( k \):\[ C_p = C_v + R = \frac{R}{k - 1} + R = \frac{Rk}{k - 1} \][/tex]
Now substitute
[tex]\( C_v \) and \( C_p \) into the equation for \( T_2 \):\[ T_2 = 35 \left( 10 \right)^{\frac{R}{\frac{R}{k - 1}}} \]\[ T_2 = 35 \left( 10 \right)^{k - 1} \]Since \( k = \frac{C_p}{C_v} \):\[ T_2 = 35 \left( 10 \right)^{\frac{C_p}{C_v} - 1} \]\[ T_2 = 35 \left( 10 \right)^{\frac{\frac{Rk}{k - 1}}{\frac{R}{k - 1}} - 1} \]\[ T_2 = 35 \left( 10 \right)^{k - 1} \][/tex]
Now we can calculate
[tex]\( T_2 \):\[ T_2 = 35 \times 10^{0.4} \]\[ T_2 = 35 \times 2.5119 \]\[ T_2 = 87.9185°C \][/tex]
Now that we have [tex]\( T_2 \), we can calculate the work done during the process using the first law of thermodynamics:\[ W = C_v (T_1 - T_2) \]Given that \( C_v = \frac{R}{k - 1} \), we can substitute this expression for \( C_v \):\[ W = \frac{R}{k - 1} (T_1 - T_2) \][/tex]
Substitute the known values:
[tex]\[ W = \frac{8.314 \text{ kJ/kg-K}}{\frac{1.4 - 1}{1.4}} (308.15 - 87.9185) \]\[ W = \frac{8.314 \times 1.4 \text{ kJ/kg-K}}{0.4} \times 220.2315 \]\[ W = 36.779 \times 220.2315 \text{ kJ/kg} \]\[ W = 8089.91 \text{ kJ/kg} \]So, the final answers are:(a) \( T_2 = 87.9185°C \)(b) \( W = 8089.91 \text{ kJ/kg} \)[/tex]
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.
Answers
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
a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in from the top of the sprue for design purposes. If a flow rate of 40 in3/s is to be achieved, what should be the diameter at the bottom of the sprue
Answers
Answer:
See explaination
Explanation:
We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.
See the attached file for detailed solution of the given problem.
Answer:
The diameter at the bottom of the sprue is =0.725 in, and the sprue will not occur when 0.021<0.447.
Explanation:
Solution
The first step to take is to define the Bernoulli's eqaution
h effective = v²top/2g + ptop /ρg = hbottom + v² bottom/2g + p bottom/ ρg
h effective + 0 +0= 0 +v² bottom/2g + 0
Thus,
v bottom = √ 2gh total
=√ 2 (32. 6 ft/ s²) + (12/12 ft)
Which is = 8.074 ft/s
We now, express the relation for flow rate.
Q =π/4 D² bottom v bottom
= 40 in 3/s = π/4 D²₃ ( 8.074 ft/s) (12 in/ ft)
so,
D bottom = 0.725 in.
Then,
We express the relation to avoid aspiration
A₃/A₂ < √ h top /h total
= π/4 D²₃/ π/4 D²₂ < √3/15
= 0.725²/5² < √3/15
=0.021<0.447
Therefore, the aspiration will not happen or occur
Natural Convection from an Oven Wall. The oven wall in Example 4.7-1 is insulated so that the surface temperature is 366.5 K instead of 505.4 K. Calculate the natural convection heat-transfer coefficient and the heat-transfer rate per m of width. Use both Eq. (4.7-4) and the simplified equation. (Note: Radiation is being neglected in this calculation.) Use both SI and English units.
Answers
Answer:
i) Heat transfer coefficient (h) = 7 w/m²k
ii) Heat transfer per meter width of wall
= h x L x 1 x (Ts - T₆₀)
= 7 x 0.3048 x (505.4 - 322) = 414.747 w/m
Explanation:
see attached image
1. A solar concentrator produces a heat flux of 2500 W/m2 on the projected area of a tube of diameter 50 mm. Water flows through the tube at a rate of 0.015 kg/s. If the water temperature at the inlet is 15°C, what length of pipe is required to produce water at a temperature of 85°C?
(Water at 50 degree C has: rho = 990 kg m^-3, k = 0.64 Wm^-1 K^-1, c = 4180 Jkg^-1 K^-1.)
Answers
Answer:
[tex]L = 0.319\,m[/tex]
Explanation:
Let suppose that temperature of air is 15°C. The heating process of the solar concentrator is modelled after the First Law of Thermodynamics:
[tex]\dot Q = h\cdot \pi\cdot D\cdot L\cdot (\bar T-T_{\infty})[/tex]
The required length is:
[tex]L = \frac{\dot Q}{h\cdot \pi\cdot D\cdot (\bar T-T_{\infty})}[/tex]
But,
[tex]\dot Q = \dot m \cdot c_{w}\cdot (T_{o}-T_{i})[/tex]
[tex]\dot Q = \left(0.015\,\frac{kg}{s}\right)\cdot \left(4180\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (85^{\circ}C-15^{\circ}C)[/tex]
[tex]\dot Q = 4389\,W[/tex]
[tex]\bar T = \frac{15^{\circ}C + 85^{\circ}C}{2}[/tex]
[tex]\bar T = 50^{\circ}C[/tex]
[tex]L = \frac{4389\,W}{\left(2500\,\frac{W}{m^{2}} \right)\cdot \pi \cdot (0.05\,m)\cdot (50^{\circ}C-15^{\circ}C)}[/tex]
[tex]L = 0.319\,m[/tex]
g A heat exchanger is designed to is to heat 2,500 kg/h of water from 15 to 80 °C by engine oil. The configuration of the heat exchanger is in shell-and tube and the water is flowing through the tubes. The oil makes a single shell pass, entering at 160 °C and leaving at 90 °C, with an averaged heat transfer coefficient from the oil to the outer wall of the tubes equals 400 W/m2·K. The water flows through 11 brass tubes of 30 mm diameter with each tube making four passes through the shell. Assuming fully developed flow for the water inside the tubes, determine the tube length per pass to achieve these specified output temperatures. You may assume the tube wall thickness is negligible in the determination of overall heat transfer coefficient. (
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file gave a detailed solution of the problem.
ou want to amplify a bio-potential signal that varies between 2.5 V and 2.6 V. Design an amplifier circuit for this signal such that the output spans 0 V to +10 V. The signal cannot be inverted. You can use any number of op amps and any number of resistors (with any values). But you can use only one +10 V DC voltage source (for powering the op amps as well as for any other needs). Clearly draw the complete circuit and show all component values.
Answers
Answer:
See attachment
Explanation:
Gain= Vo/Vin
If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7
Using above value of gain, let's design non-inverting op-amp configuration
Gain= 1+Rf/Rin
3.7= 1= Rf/Rin
2.7= Rf/Rin
If Rin=100Ω then Rf= 270Ω
A 0.91 m diameter corrugated metal pipe culvert (n = 0.024) has a length of 90 m and a slope of 0.0067. The entrance has a square edge in a headwall. At the design discharge of 1.2 m3 /s, the tailwater is 0.45 m above the outlet invert. Determine the head on the culvert at the design discharge. Repeat the calculation for head if the culvert is concrete.
Answers
Answer:
HW=1.71m
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The average starting salary for this year's graduates at a large university (LU) is $20,000 with
a standard deviation of $8,000. Furthermore, it is known that the starting salaries are normally
distributed.
a. What is the probability that a randomly selected LU graduate will have a starting salary of at least $30,400?
b. What is the probability that a randomly selected LU graduate will have a salary of
exactly $30,400?
c. Individuals with starting salaries of less than $15600 receive a low income tax break. What percentage of the graduates will receive the tax break?
d. If 189 of the recent graduates have salaries of at least $32240, how many students
graduated this year from this university?
Answers
Answer:
(a) 0.0968 (b) the probability that a randomly selected LU graduate will have a salary of exactly $30,400 is 0.0000 (c) 29.12% (d) 3000 students graduates this year from this university
Explanation:
Solution
For the problem given,
The average salary starting for this year's graduates at a large university (LU) is = 20,000
So,
The mean μ = $ 20,000
Standard deviation is б = $ 8000
Note: kindly find the complete steps taken to get the solution to this questions attached below.
the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the inductor at t=2 seconds
Answers
Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
[tex]V(t) = 5(1-e^{-0.5t})[/tex]
The current flowing through the inductor is given by
[tex]i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)[/tex]
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
[tex]i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\[/tex]
[tex]i(t) = 1000t +2000e^{-0.5t} -2000\\[/tex]
So the current at t = 2 seconds is
[tex]i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A[/tex]
The energy stored in the inductor at t = 2 seconds is
[tex]W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J[/tex]
Consider a large vertical plate with a uniform surface temperature of 100°C suspended in quiescent air at 25°C and atmospheric pressure. (a) Estimate the boundary layer thickness at a location 0.28 m measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and at what position in the boundary layer does the maximum occur? (c) Using the similarity solution result, Equation 9.19, determine the heat transfer coefficient 0.25 m from the lower edge. (d) At what location on the plate measured from the lower edge will the boundary layer become turbulent?
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
The vertical force P acts on the bottom of the plate having a negligible weight. Determine the shortest distance d to the edge of the plate at which it can be applied so that it produces no compressive stresses on the plate at section a-a. The plate has a thickness of 10 mm and P acts along the center line of this thickness.
200mm is the width and d is the distance from the right edge to the force P.
Answers
Answer:
The shortest distance d to the edge of the plate is 66.67 mm
Concepts and reason
Moment of a force:
Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.
Moment(M) = Force(F)×distance(d)
Moment of inertia ( I )
It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.
First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress
Solution and Explanation:
[Find the given attachments]
The shortest distance (d) to the edge of the rectangular-plate is equal to 66.65 mm.
Given the followin data:
Thickness (length) of plate = 10 mm to m = 0.001 m.
Width of plate = 200 mm to m = 0.2 m.
c = [tex]\frac{0.2}{2}[/tex] = 0.1 m.
How to calculate the shortest distance.First of all, we would determine the area of the rectangular-plate and its moment of inertia.
For area:
[tex]A=LW\\\\A=0.001 \times 0.2\\\\A= 0.002 \;m^2[/tex]
For moment of inertia:
Mathematically, the moment of inertia of a rectangular-plate is given by this formula:
[tex]I=\frac{b^3d}{12} \\\\I=\frac{0.2^3 \times 0.01}{12} \\\\I=\frac{0.008 \times 0.01}{12}\\\\I=\frac{0.0008 }{12}\\\\I=6.67 \times 10^{-6}\;m^4[/tex]
The compressive stress of a rectangular-plate with respect to axial and bending stress is given by this formula:
[tex]\sigma = \frac{P}{A} -\frac{Mc}{I} \\\\\sigma = \frac{P}{0.002} -\frac{P(0.1-d)\times 0.1}{6.67 \times 10^{-6}} \\\\\sigma = \frac{P}{0.002} -\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}} \\\\\frac{P}{0.002}=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\500P=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\3.335 \times 10^{-3}P=0.01P-0.1Pd\\\\[/tex]
[tex]0=0.01P-0.1Pd-3.335 \times 10^{-3}P\\\\0=(0.01-0.1d-3.335 \times 10^{-3})P\\\\0.01-0.1d-3.335 \times 10^{-3}=0\\\\0.1d=0.01-3.335 \times 10^{-3}\\\\d=\frac{6.665\times 10^{-3}}{0.1} \\\\d=6.665\times 10^{-2}\;m\\\\[/tex]
d = 66.65 mm.
Read more on moment of inertia here: https://brainly.com/question/3406242
A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to the x ?or y ?axis. Find the induced emf across the open-circuited ends of the coil if the magnetic field is given by (a) B = z 20 e ^-3t (T) (b) B = z 20 cos x cos 10^3 t (T) (c) B = z 20 cos x sin 2y cos 10^3 t (T)
Answers
Find solution in attachments below
10.34 Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Spheroidite to tempered martensite (b) Tempered martensite to pearlite (c) Bainite to martensite (d) Martensite to pearlite (e) Pearlite to tempered martensite (f) Tempered martensite to pearlite (g) Bainite to tempered martensite (h) Tempered martensite to spheroidite
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Answer:
Explanation:
The solutions to this question can be seen in the screenshot taken from the solution manual.
For all the problems describe all pieces to the equations. 1.What is the equation for normal stress? 2.What is the equation for shear stress? 3.What is the equation for cross sectional area of a beam? 4.What is the equation for cross sectional area of a shaft? 5.What is the equation for shear stress at an angle to the axis of the member? 6.What is the equation for normal stress at an angle to the axis of the member? 7.What is the equation for the factor of safety? 8.What is the equation for strain under axial loading?
Answers
Answer:
stress equation : [tex]\frac{p}{A}[/tex] Shear stress equation : [tex]\frac{Qv}{Ib}[/tex] cross sectional area of a beam equation : b*d cross sectional area of a shaft equation : [tex]\frac{\pi }{4} (d)^{2}[/tex] shear stress at an angle to the axis of the member equation: [tex]\frac{P}{A}[/tex] sin∅cos∅. Normal stress at an angle to the axis of the member equation: [tex]\frac{P}{A} cos^{2}[/tex]∅factor of safety equation : [tex]\frac{ultimate stress}{actual stress}[/tex] strain under axial loading equation: [tex]\frac{PL}{2AE}[/tex]Explanation:
The description of all the pieces to the equations
stress equation : [tex]\frac{p}{A}[/tex] p = axial force, A = cross sectional areaShear stress equation : [tex]\frac{Qv}{Ib}[/tex] Q = calculated statistical moment, I = moment of inertia, v = calculated shear, b = width of beamcross sectional area of a beam equation : b*d b=width of beam, d =depth of beamcross sectional area of a shaft equation : [tex]\frac{\pi }{4} (d)^{2}[/tex] d = shaft diametershear stress at an angle to the axis of the member equation: [tex]\frac{P}{A}[/tex] sin∅cos∅. P = axial force, A = cross sectional area ∅ = given angleNormal stress at an angle to the axis of the member equation: [tex]\frac{P}{A} cos^{2}[/tex]∅ p = axial force , A = cross sectional area, ∅ = given anglefactor of safety equation : [tex]\frac{ultimate stress}{actual stress}[/tex] strain under axial loading equation: [tex]\frac{PL}{2AE}[/tex] P = axial force, L = length, A = cross sectional area, E = young's modulusA flat-plate solar collector is 2 m long and 1 m wide and is inclined 60o to the horizontal. The cover plate is separated from the absorber plate by an air gap of 2 cm thick. If the temperatures of the cover plate and the absorber plate are 305 K and 335 K, respectively. Calculate the convective heat loss. Take the pressure at 1 atm.
Answers
Answer:
164.4 W
Explanation:
We can define Heat loss as a measure of the total transfer of heat through the fabric of a building from inside to the outside, either from conduction, convection, radiation, or any combination of the these.
Please kindly check attachment for the solution of the heat loss as asked in the question.
The attached file have the solved problem.
A tachometer has an analog display dial graduated in 5-revolutions-per-minute (rpm) increments. The user manual states an accuracy of 1% of reading. Estimate the design-stage uncertainty in the reading at 50, 500, and 5,000 rpm.
Answers
To estimate the uncertainty of a tachometer with 1% accuracy, you calculate 1% of the rpm readings at 50, 500, and 5,000 rpm, resulting in uncertainties of 0.5 rpm, 5 rpm, and 50 rpm respectively.
Explanation:To estimate the design-stage uncertainty in a tachometer reading with an accuracy of 1% of reading, we calculate the uncertainty for different values of revolutions per minute (rpm). Given readings at 50 rpm, 500 rpm, and 5,000 rpm, the uncertainty can be calculated by taking 1% of each reading:
At 50 rpm: Uncertainty = 1% of 50 rpm = 0.5 rpmAt 500 rpm: Uncertainty = 1% of 500 rpm = 5 rpmAt 5,000 rpm: Uncertainty = 1% of 5,000 rpm = 50 rpmThis means that the display of the tachometer can vary by the calculated uncertainty for each of these readings, which represents the design-stage accuracy of the instrument.
Q2: The average water height of an ocean area is 2.5 m high and each wave lasts for an average period of 7 s. Determine (a) the energy and power of the wave per unit area and (b) the work and average power output of a wave power plant on this ite with a plant efficiency of 35% and a toltal ocean wave area of 1 km2 . Take the density of the seawater to be 1025 kg/m3 .
Answers
Answer:
(a) 561.12 W/ m² (b) 196.39 MW
Explanation:
Solution
(a) Determine the energy and power of the wave per unit area
The energy per unit are of the wave is defined as:
E = 1 /16ρgH²
= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²
=3927. 83 J/m²
Thus,
The power of the wave per unit area is,
P = E/ t
= 3927. 83 J/m² / 7 s = 561.12 W/ m²
(b) The average and work power output of a wave power plant
W = E * л * A
= 3927. 83 J/m² * 0.35 * 1 *10^6 m²
= 1374.74 MJ
Then,
The power produced by the wave for one km²
P = P * л * A
= 5612.12 W/m² * 0.35 * 1* 10^6 m²
=196.39 MW
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. Assuming the air is modeled as an ideal gas with variations in specific heat, determine a) the rate power is developed, in kJ per kg of air flowing.b) the rate of entropy production within the turbine, in kJ/K per kg of air flowing.
Answers
Answer:
a) [tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex], b) [tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]
Explanation:
a) The process within the turbine is modelled after the First Law of Thermodynamics:
[tex]-q_{out} - w_{out} + h_{in}-h_{out} = 0[/tex]
[tex]w_{out} = h_{in} - h_{out}-q_{out}[/tex]
[tex]w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}[/tex]
[tex]w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}[/tex]
[tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex]
b) The entropy production is determined after the Second Law of Thermodynamics:
[tex]-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)[/tex]
[tex]s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)[/tex]
[tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]
On a cold winter day, wind at 55 km/hr is blowing parallel to a 4-m high and 10-m long wall of a house. If the air outside is at 5o C and the surface temperature of the wall is 12o C, find the rate of heat loss from the wall by convection
Answers
The student's question regarding heat loss by convection requires the convective heat transfer coefficient, which is not provided, making it impossible to calculate the rate of heat loss without additional data or empirical formulas.
Explanation:The question pertains to heat loss from the wall of a house by convection on a cold winter day when the wind is blowing parallel to the wall. However, to answer the question accurately, we need additional information related to the convective heat transfer coefficient for the scenario described, which typically requires the use of empirical relationships that take into account the flow of air over the surface (such as the wind speed) and the properties of the air (temperature, viscosity, etc.). Unfortunately, the student's question does not provide the necessary details or equations to calculate the convective heat transfer coefficient, which is crucial to determining the rate of heat loss.
To calculate the rate of heat loss through convection, the formula Q = hA(T_s - T_{∞}) is typically used, where Q is the heat transfer rate, h is the convective heat transfer coefficient, A is the surface area, T_s is the surface temperature, and T_{∞} is the ambient temperature. Without the convective heat transfer coefficient, we cannot complete this calculation.
a) A coil connected to a 250-V, 50-Hz sinusoidal supply takes a current of 10 A at a
phase angle of 30°. Calculate the resistance and inductance of, and the power
taken by the coil.
b) A voltage v = 0.95 sin754t is applied to a series RLC circuit, which has L = 22.0
mH, R = 23.2 k2, and C = 0.30 uF.
(i) What is the impedance and phase angle7 element?
Answers
The question involves calculations within AC circuits, focusing on impedance, phase angles, power in RLC circuits, and the effects of frequency on circuit behavior.
Explanation:The student has presented a scenario involving alternating current (AC) circuits with various components like coils, resistors, inductors, and capacitors. These types of problems require an understanding of impedance, phase angles, power calculations, and resonance in RLC circuits. Calculating these values involves the use of formulas derived from AC circuit theory which take into account the sinusoidal nature of the voltage and current.
To illustrate, in an AC circuit with a resistor (R) and an inductor (L), the total impedance (Z) can be found using the formula Z = \(\sqrt{R^2 + (\omega L - 1/\omega C)^2}\), where \(\omega\) is the angular frequency of the source. The phase angle (\(\phi\)) between the voltage and the current can be determined using the arctangent of the reactive component over the resistive component, \(\phi = arctan((\omega L - 1/\omega C)/R)\). Power calculations involve both the voltage and current, with the power factor being the cosine of the phase angle.
Consider viscous flow over a flat plate a. Write the definition of the Rex , based on distance, x, from the leading edge of a flat plate, and explain its significance b. Sketch the development of the velocity boundary layer height, d, beginning at the plate’s leading edge, and extending into the turbulent region, and indicate the range of the Reynolds number in the relevant regions c. How does d change with distance x in the laminar region.
Answers
Answer:
Explanation:
Solution:-
- To categorize the flow conditions of any fluid we utilize a dimensionless number, called Reynold's number ( Re ) to study the behaviour of the fluid.
- Reynold's number is proportional to the ratio of inertial forces ( forces that resist any change in motion of a unit mass ) and viscous forces ( forces that resist any iner-plane deformations between layers of fluid ).
- Considering 2-Dimensional viscous flow over a flat plate, the Reynold number (Re) is mathematically expressed as a function of distance "x" denoted from leading edge and along the length of the plate:
[tex]Re_x = \frac{U*x}{v}[/tex]
Where,
U: The free stream velocity of the fluid
ν: The kinematic viscosity of the fluid
- The distance "x" along the length of the plate is substituted in the above formula and the corresponding Reynold number is evaluated. This gives a highly localized value about "x".
- The purpose of the Reynold number is the substitution of dynamically similar fluids i.e Fluid with the same Reynold's number when testing models to see how they would behave in a specific environment.
- The Reynolds number has a set of ranges above and below the critical range defined by the critical length "xc" along the plate. The range below the critical has laminar flow characteristics, whereas the range above the critical has turbulent flow. The laminar region has flow along smooth streamlines, while turbulent region is characterized by 3 - dimensional random eddy.
- The critical length " xc " is determined from the critical Reynold number i.e ( 5 x 10^5 ) which is a small region that has mixed characteristics of laminar and turbulent conditions.
- The flow at the boundaries has zero velocity, there is a steep velocity gradient from the boundary into the flow. This velocity gradient in a real fluid sets up shear forces near the boundary that reduce the flow speed to that of the boundary. That fluid layer which has had its velocity affected by the boundary shear is called the "boundary layer"
- For smooth upstream boundaries, " x << xc " or " Re_x < 5 x 10^5 " , the boundary layer starts out as a laminar boundary layer in which the fluid particles move in smooth layer.
- As the laminar boundary layer increases in thickness, it becomes unstable as changes in motion become more predominant ( inertial ) than viscous effect of fluid layers. This leads to a transformation of laminar boundary layer into turbulent boundary layer in which fluid particles move in haphazard paths. " x > xc " or " Re_x > 5 x 10^5 "
- The boundary layer thickness/height d increases as x increases. The relationships for laminar and turbulent regions of boundary layer are given as follows:
[tex]\frac{d}{x} = \left \{ {{\frac{5}{\sqrt{Re_x} } , 10^3 < Re_x < 10^6} \\\\\atop {\frac{0.16}{\frac{1}{7} \sqrt{Re_x} } , 10^6 < Re_x}} \right.[/tex]
- To construct a function of boundary layer thickness " d " and length from leading edge of the plate " x ". We use the Re_x relation and substitute, we get the following proportionalities for our sketch:
d ∝ √x .... Laminar region
d ∝ [tex]x^\frac{6}{7}[/tex] .... Turbulent region
- Use the above relation to develop sketch for the boundary layer along the length "x" from leading edge.
- The sketch is given as an attachment.