Answer:
The answer is option # 2. (Constraints).
Cathy works in a welding shop. While working one day, a pipe falls from scaffolding above and lands on her head, injuring her. Cathy complains to OSHA, but the company argues that because it has a "watch out for falling pipe" sign in the workplace that it gave fair warning. It also says that if Cathy wasn’t wearing a hardhat that she is responsible for her own injury. Which of the following is true?1. Common law rules could hold Cathy responsible for her own injury.2. Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.3. OSHA rules can hold Cathy’s employer responsible for not maintaining a hazard-free workplace.4. More than one answer is correct.
Answer:1 common law
Explanation:
It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.
What are OSHA rules?In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.
According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.
You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.
Therefore, more than one answer is correct.
Learn more about OSHA, here:
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As an engineer who has just finished taking engineering materials course, your first task is to investigate the causes of an automobile accident. Your findings show that the right rear wheel has broken off at the axle. The axle is bent. The fracture surface reveals a Chevron pattern pointing toward the surface of the axle. Suggest a possible cause for the fracture and why?
Answer is given below
Explanation:
Evidence shows that the axle was not broken before the accident, while the clumsy axle meant that the wheel was still attached when the load was applied. This indicates that the Chevron prototype wheel suffered a severe impact shock, which caused the failure of the transmission to the axle. Preliminary evidence suggests that the driver lost control and crashed. Further examination of the surface, microstructure and structure and characteristics of the fracture can be modified if the axle is properly preparedDescribe how the pair-wise summation computation provided below can be changed to find the maximum element of an array. Array has the following elements (x[0], x[1], x[2], x[3], x[4], x[5], x[6])
t[0] = x[0] + x [1];
t[1] = x[2] + x[3];
t[2] = x[4] + x[5];
t[3] = x[6] + x[7];
t[4] = t[0] + t[1];
t[5] = t[2] + t[3];
sum = t[4] + t[5];
Answer:
To change the pair-wise summation computation provided to find a maximum element of an array we; Take two elements to check max element from that and add to the T ARRAY, the same process is then repeated for the next two elements, again we repeat the same process for the next two elements from T array until we get the max element process going on pair-wise computation
Explanation:
Code written using pair-wise computation to describe how to change the given pair-wise summation computation provided to find the maximum element of an array
The given array element ; (x[0], x[1], x[2], x[3], x[4], x[5], x[6])
IF X[0] > X[1]
T[0]=X[0];
ELSE
T[0]=X[1]
IF X[2] > X[3]
T[1]=X[2];
ELSE
T[1]=X[3];
IF X[4] > X[5]
T[2]=X[4];
ELSE
T[2]=X[5];
T[3]=X[6];
IF T[0] > T[1]
T[4]=T[0];
ELSE
T[4]=T[1]
IF T[2] >T[3]
T[5]=T[2];
ELSE
T[5]=T[3];
IF T[4] > T[5]
MAX=T[4];
ELSE
MAX=T[5]
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa (293400 psi). Assume that the parameter Y has a value of 1.14. (a) If the largest surface crack is 0.2 mm (0.007874 in.) long, determine the critical stress .
Answer:
Explanation:
The formula for critical stress is
[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]
[tex]\sigma_c =\texttt{critical stress}[/tex]
K is the plane strain fracture toughness
Y is dimensionless parameters
We are to Determine the Critical stress
Now replacing the critical stress with 54.8
a with 0.2mm = 0.2 x 10⁻³
Y with 1
[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]
The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa
Therefore, the specimen does not failure for surface crack of 0.2mm
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K
Answer:
The exit temperature of the gas = 32° C
Explanation:
Solution
Given that:
Inlet temperature T₁ = 27°C ≈ 300.15 K
Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa
Volume flow rate , V = 15 m/s³
Diameter of the deduct, D = 500 mm = 0.5 m
Electric heater power, W heater = 130 kW = 130 * 10^3 W
The heat lost Q = 80 kW = 80 * 10^3 W
Now,
From the ideal gas law, density of the air at the inlet is given as :
ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300
=0.6667 kg/m³
The mass flow rate through the duct is computed below:
m = ρ₁ V = 0.6667 * 15 = 10 kg/s
Thus
Applying the first law of thermodynamics to the process is shown below:
Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)
So,
If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:
Q + m (h₁) = m (h₂) + W
or
Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)
Thus
h₂ - h₁ = Cp T₂ - T₁
Now by method of substitution the known values are:
(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)
Note: The heat transfer is taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas
So,
Solving for T₂,
T₂ = 32° C
Therefore the exit temperature of the gas = 32° C
If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the resultant moment produced by all the weights about point A.
Answer:
hello the required diagram is missing attached to the answer is the required diagram
7.9954 kip.ft
Explanation:
AB = 1550-Ib ( weight acting on AB )
BCD = 190 - Ib ( weight of cage )
169-Ib = weight of man inside cage
Attached is the free hand diagram of the question
calculate distance [tex]x![/tex]
= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]
[tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft
calculate distance x
= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]
x = 30 * cos 75⁰ = 7.765 ft
The resultant moment produced by all the weights about point A
∑ Ma = 0
Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )
Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )
= 4014.5 + 1950.35 + 2030.535
= 7995.385 ft. Ib ≈ 7.9954 kip.ft
A 300-ft long section of a steam pipe with an outside diameter of 4 in passes through an open space at 50oF. The average temperature of the outer surface of the pipe is measured to be 280oF, and the average convection heat transfer coefficient between the pipe and surrounding air is estimated to be 6 Btu/h·ft 2·oF. Determine the rate of heat loss from the pipe by convection
Answer:
The rate of heat loss from the pipe by convection is 433113 Btu/h
Explanation:
Given that:
The length of the steam pipe = 300 ft
Temperature of the open space [tex]T_{air[/tex] = 50 ° F
Outside diameter = 4 ㏑ = 4/12 ft = 0.333 ft
Avg, temperature of the surface of the pie [tex]T_s[/tex] = 280° F
Avg. convection of heat transfer h = 6 Btu/h·ft²·°F
We are to Determine the rate of heat loss from the pipe by convection
Let first calculate the area of the heat transfer;
A = πdh
A = πdL
A = π× 0.333 ft ×300 ft
A = 313.85 ft²
The rate of heat loss from the pipe by convection can now be calculated by using the formula:
[tex]Q_{pipe} = hA ( T_s - T_{air})[/tex]
Replacing our values from above ; we have:
[tex]Q_{pipe} = 6* 313.85 ( 280 -50})[/tex]
[tex]Q_{pipe} = 1883.1*( 230})[/tex]
[tex]\mathbf{Q_{pipe} =433113 \ \ Btu/h }[/tex]
Therefore, the rate of heat loss from the pipe by convection is 433113 Btu/h
Q: Draw shear and bending moment diagram for the beam shown in
the figure. EI= constant
Answer:
Explanation:
Please
Consider a refrigerator that consumes 400 W of electric power when it is running. If the refrigerator runs only one-quarter of the time and the unit cost of electricity is $0.13/kWh, what is the electricity cost of this refrigerator per month (30 days)
Answer:
Electricity cost = $9.36
Explanation:
Given:
Electric power = 400 W = 0.4 KW
Unit cost of electricity = $0.13/kWh
Overall time = 1/4 (30 days) (24 hours) = 180 hours
Find:
Electricity cost
Computation:
Electricity cost = Electric power x Unit cost of electricity x Overall time
Electricity cost = 0.4 x $0.13 x 180
Electricity cost = $9.36
Given:
Electric power = 400 W = 0.4 KW
Over all Time = 30(1/4) = 7.5 days
Unit cost of electricity = $0.13/kWh
Find:
Electricity cost.
Computation:
Electricity cost = Electric power x Unit cost of electricity x Over all Time
Electricity cost = 0.4 x 0.13 x 7.5
Electricity cost = $
It is proposed to use water instead of refrigerant-134a as the working fluid in air-conditioning applications where the minimum temperature never falls below the freezing point. Can water be used as the working fluid in air-conditioning applications?
Answer:
No, water can't be used.
Explanation:
No, water cannot be used as the working fluid in air-conditioning applications.
This is because, if we assume the water is maintained at 10°C in the evaporator, the evaporator pressure will now be the saturation pressure that corresponds to this pressure, which in this case would be 1.2 kPa.
So we can conclude that for the refrigerants in the evaporator the temperature of a saturated pressure would be very low and so it's not practical to maintain it with water
Thus, it's is not practical to design refrigeration or air conditioning devices with water as the working fluid because it will involve extremely low pressures.
All circuit conductors between the service equipment, the source of a separately derived system, or other power supply source and the final branch-circuit overcurrent device defines a
Answer: Feeder
Explanation:
The National Electrical Code (NEC) which has been adopted by all 50 stated in the Repubic is a code that governs the installation of electrical devices, circuits and systems. The primary purpose is to safeguard the populace in the use of electricity which can prove to be hazardous if not properly used.
Article 100 of the the NEC defines a Feeder as All circuit conductors between the service equipment, the source of a separately derived system, or other power supply source and the final branch-circuit overcurrent device.
A piston-cylinder device initially at 0.45-m3 contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.2 m3. The work done on the gas during this compression process is _____ kJ.
Answer:
219kJ
Explanation:
The work done (W) on a gas in an isothermal process is given by;
W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex] -----------------(i)
Where;
P₁ = initial pressure of the gas
V₁ = initial volume of the gas
V₂ = final volume of the gas
From the question;
P₁ = 600kPa = 6 x 10⁵Pa
V₁ = 0.45m³
V₂ = 0.2m³
Substitute these values into equation (i) as follows;
W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]
W = -6 x 10⁵ x 0.45 x ln (0.444)
W = -6 x 10⁵ x 0.45 x -0.811
W = 2.19 x 10⁵
W = 219 x 10³
W = 219kJ
Therefore, the work done on the gas during the compression process is 219kJ
An 60-m long wire of 5-mm diameter is made of steel with E = 200 GPa and ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the allowable tension in the wire (b) the corresponding elongation of the wire
Answer:
a) 2.45 KN
b) 0.0375 m
Explanation:
[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]
[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]
[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]
(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]
5. Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm3, respectively. Atomic weight of Ge is 72.64 g/mol
Answer:
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
Explanation:
The masses of silicon and germanium contained in a cubic centimeter of the germanium-silicon alloy by apply the concepts of mass ([tex]m[/tex]), density ([tex]\rho[/tex]) and volume ([tex]V[/tex]), as well as the mass-mass proportion of Germanium ([tex]x[/tex]):
[tex]m_{Ge} = x \cdot \rho_{Ge}\cdot V_{sample}[/tex]
[tex]m_{Ge} = 0.15\cdot \left(5.32\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{3})[/tex]
[tex]m_{Ge} = 0.798\,g[/tex]
The amount of moles of Germanium is obtained after dividing previous outcome by its atomic weight. That is to say:
[tex]n = \frac{m_{Ge}}{M_{Ge}}[/tex]
[tex]n = \frac{0.798\,g}{72.64\,\frac{g}{mol} }[/tex]
[tex]n = 0.011\,mol[/tex]
There are 0.011 moles in a cubic centimeter of the germanium-silicon alloy. According to the Law of Avogadro, there are [tex]6.022 \times 10^{23}\,atoms[/tex] in a mole of Germanium. The quantity of atoms in a cubic centimeter is therefore found by simple rule of three:
[tex]y = \frac{0.011\,mol}{1\,mol}\times \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)[/tex]
[tex]y = 6.624 \times 10^{21}\,atoms[/tex]
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
4. "ABC constriction Inc." company becomes the lowest in the bed process to get a $21
million construction project for "Northern Inc.". Now “ABC construction Inc." planning to
make a formal contract agreement with the "Northern Inc.". What are the main elements of
this agreement to consider it as a legal contract? Explain.
Answer is given below
Explanation:
Agreement is a official contract. It is written form or even in orally form. The Agreement can be written in formal or informal terms or we can use purely verbal language.
Agreement made between two or more party that allow the court to decide.
The main 6 elements are:
1. Offer
2. acceptance
3. consideration
4. intention to create legal relation
5. certainty
6. capacity
1. The first elements of the contract herein are ABC Sanctions Inc., without the offer, as it is not valid under the Contract Act 1950, Contract Act.
2. Once the offer is made in the contract, acceptance must take place. The agreement must be approved by Northern Inc. When Northern Inc is clear with the offer, it will accept it once the terms and conditions of the agreement are clear.
3. Contrarification is the most important aspect of a contract, when considering a contract, the other person will give something in return. It is considered an exchange between ABC Construction Inc and Northern Inc.
4. It is necessary to have these elements in the contract. Contract law 1950 is one of the requirements of a valid contract, although there is silence about the need for a legal relationship.
5. Another important aspect of the contract is of course. The Contract Agreement sets out the terms and conditions that must be clearly understood by both ABC Contracts Inc and Northern Inc.
6. The ability of a contract to have a legal capacity on either side of the contract is more than eighteen years, since the age of 18 years is specified as the age at which the contract is entered into.
for an electromotive force to be induced across a vertical loop from the field of an infinite length line of fixed current in the z axis the loop must be moving to?
Answer:
The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".
Explanation:
Since along that same z-axis none electromagnetic field would be triggered as being in the same orientation loop movement of them across different line portions would allow some caused emf/voltage to be canceled. And the only logical choice seems to be either x or y-axes.The magnetic field of fluctuation should indeed be changed and changed across both X as well as Y directions.So that the above is the appropriate choice.
Eight switches are connected to PORTB and eight LEDs are connected to PORTA. We would like to monitor the first two least significant bits of PORTB (use masking technique). Whenever both of these bits are set, switch all LEDs of Port A on for one second. Assume that the name of the delay subroutine is DELAY. You do not need to write the code for the delay procedure.
Answer:
In this example, the delay procedure is given below in the explanation section
Explanation:
Solution
The delay procedure is given below:
LDS # $4000 // load initial memory
LDAA #$FF
STAA DDRA
LDAA #$00 //load address
STAA DDRB
THERE LDAA PORT B
ANDA #%00000011// port A and port B
CMPA #%0000011
BNE THERE
LDAA #$FF
STAA PORT A
JSR DELAY
LDAA #$00
STAA PORT A
BACK BRA BACK
The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T is that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False
For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger shell (just like a shell and tube heat exchanger). Consider one tube inside such a reactor that is 2.5 m long with an inside diameter of 0.025 m. The catalyst is alumina spheres with a diameter of 0.003 m. The particle density is 1300 kg/m3 and the bed void fraction is 0.38. Compute the pressure drop seen for a superficial mass flux of 4684 kg/m2hr. The feed is methane at a pressure of 5 bar and 400 K. At these conditions the density of the gas is 0.15 mol/dm-3 and the viscosity is 1.429 x 10-5 Pa s.
Answer:
the pressure drop is 0.21159 atm
Explanation:
Given that:
length of the reactor L = 2.5 m
inside diameter of the reactor d= 0.025 m
diameter of alumina sphere [tex]dp[/tex]= 0.003 m
particle density = 1300 kg/m³
the bed void fraction [tex]\in =[/tex] 0.38
superficial mass flux m = 4684 kg/m²hr
The Feed is methane with pressure P = 5 bar and temperature T = 400 K
Density of the methane gas [tex]\rho[/tex] = 0.15 mol/dm ⁻³
viscosity of methane gas [tex]\mu[/tex] = 1.429 x 10⁻⁵ Pas
The objective is to determine the pressure drop.
Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³
SO; we have :
Density = 0.15 mol/dm ⁻³
Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol
Density = [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]
Density = 2400
Density [tex]\rho_f[/tex] = 2.4 kg/m³
Density = mass /volume
Thus;
Volume = mass/density
Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³
Volume of the methane gas = 1951.666 m/hr
To m/sec; we have :
Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec
[tex]Re = \dfrac{dV \rho}{\mu}[/tex]
[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]
[tex]Re=2276.317705[/tex]
For Re > 1000
[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]
[tex]\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}[/tex]
[tex]\Delta P=8575.755212*2.5[/tex]
[tex]\Delta = 21439.38803 \ Pa[/tex]
To atm ; we have
[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]
[tex]\Delta P =0.2115903087 \ atm[/tex]
ΔP ≅ 0.21159 atm
Thus; the pressure drop is 0.21159 atm
When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus _____ percent of the areas of smaller flue collar outlets
Answer:
Fifty (50) percent. [50%]
Explanation:
Water heater is a home appliance that comprises of an electric or gas heating unit as well as a water-tank where water is heated and stored for use.
When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus fifty (50) percent of the areas of smaller flue collar outlets.
A water heater is primarily vented with an approved and standardized plastic or metallic pipe such as flue or chimney, which allows gas to flow out of the water heater into the surrounding environment.
For a draft hood-equipped water heater, both the water heater and the barometric draft regulators must be installed in the same room. Also, the technician should ensure that the vent is through a concealed space such as conduit and should be labeled as Type L or Type B.
The minimum capacity of a water heater should be calculated based on the number of bathrooms, bedrooms and its first hour rating.
Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature of 300 K, and a turbine inlet temperature of 1000 K. The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4. Determine the required mass flow rate of air for a net power output of 70 MW, assuming both the compressor and the turbine have an isentropic efficiency of: a. 100 percent b. 85 percent.
Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)
Answer:
Exit temperature = 32°C
Explanation:
We are given;
Initial Pressure;P1 = 100 KPa
Cp =1000 J/kg.K = 1 KJ/kg.k
R = 500 J/kg.K = 0.5 Kj/Kg.k
Initial temperature;T1 = 27°C = 273 + 27K = 300 K
volume flow rate;V' = 15 m³/s
W = 130 Kw
Q = 80 Kw
Using ideal gas equation,
PV' = m'RT
Where m' is mass flow rate.
Thus;making m' the subject, we have;
m' = PV'/RT
So at inlet,
m' = P1•V1'/(R•T1)
m' = (100 × 15)/(0.5 × 300)
m' = 10 kg/s
From steady flow energy equation, we know that;
m'•h1 + Q = m'h2 + W
Dividing through by m', we have;
h1 + Q/m' = h2 + W/m'
h = Cp•T
Thus,
Cp•T1 + Q/m' = Cp•T2 + W/m'
Plugging in the relevant values, we have;
(1*300) - (80/10) = (1*T2) - (130/10)
Q and M negative because heat is being lost.
300 - 8 + 13 = T2
T2 = 305 K = 305 - 273 °C = 32 °C
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:
a. The oil mean temperature.
b. The centerline temperature.
c. The axial gradient of the mean temperature.
d. The heat transfer coefficient.
Answer:
(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K
Explanation:
Solution
Given that:
The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s
Diameter of the tube, D = 1cm =0.01 m
Electrical heat rate, q =76 W/m
Wall Temperature, Ts = 370 K
Now,
From the properties table of engine oil we can deduce as follows:
thermal conductivity, k =0.139 W/m .K
Density, ρ = 854 kg/m³
Specific heat, cp = 2120 J/kg.K
(a) Thus
The wall heat flux is given as follows:
qs = q/πD
=76/π *0.01
= 2419.16 W/m²
Now
The oil mean temperature is given as follows:
Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)
Tb =370 - 11/24 * (2419.16 * 0.005/0.139)
Tb = 330.12 K
(b) The center line temperature is given below:
Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)
Tc =304.73 K
(c) The flow velocity is given as follows:
V = m/ρ (πR²)
Now,
The The axial gradient of the mean temperature is given below:
dTb/dx = 2 *qs/ρ *V*cp * R
=2 *qs/ρ*[m/ρ (πR²) *cp * R
=2 *qs/[m/(πR)*cp
dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120
dTb/dx = 19.81 K/m
(d) The heat transfer coefficient is given below:
h =48/11 (k/D)
=48/11 (0.139/0.01)
h =60.65 W/m². K
Waste cooking oil is to be stored for processing by pouring it into tank A, which is connected by a manometer to tank B. The manometer is completely filled with water. Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa. To what height h can waste oil be poured into tank A? If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height?
KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.
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Answer:
(1). 1.2 metres.
(2). There is going to be the same pressure.
Explanation:
From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;
" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."
=> Also, the density of oil = 930
That is if Pressure, P in B > 18kpa there will surely be a burst.
The height, h the can waste oil be poured into tank A is;
The maximum pressure = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).
18 × 10^3 = (height, h × 10 × 930) + 10 × (2 - 1.25) × 1000.
When we make height, h the Subject of the formula then;
Approximately, Height, h = 1.2 metres.
(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.
Item110pointseBook HintPrintReferences Check my work Check My Work button is now disabled5Item 1Item 1 10 pointsAn ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rate of heat addition to this cycle when it produces 150 hp of power, the cycle is repeated 1200 times per minute, and the state of the air at the beginning of the compression is 95 kPa and 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.
Answer:
The answer is given below
Explanation:
Given that:
Compression ratio (r) = 18, cut off ratio ([tex]r_c[/tex]) = 1.5, k = 1.4, W = 150 hp, initial temperature (Ti) = 17°C = 290 K
The rate of heat addition (Qin) is given by the equation:
[tex]Q_{in}=\frac{W*r^{k-1}*k(r_c-1)}{r^{k-1}*k(r_c-1)-(r_c^k-1)} \\\\substituting\ values\ into\ the \ equation\ gives:\\\\Q_{in}=\frac{150*18^{1.4-1}*1.4(1.5-1)}{18^{1.4-1}*1.4(1.5-1)-(1.5^{1.4}-1)}=\frac{333.6555}{2.2244-0.7641} =\frac{333.6555}{1.4603} =228.5\ hp[/tex]
Therefore, The rate of heat addition is 228.5 hp
The maximum air temperature is determined using the Compression ratio, cut off ratio and the initial temperature. The maximum air temperature ([tex]T_{max}[/tex]) is given by the formula:
[tex]T_{max}=T_{in}*r^{k-1}*r_c=290*18^{1.4-1}*1.5=1382.3\ K=1109.3^oC[/tex]
The maximum air temperature is 1109.3°C
A cylindrical tank is required to contain a gage pressure 670 kPakPa . The tank is to be made of A516 grade 60 steel with a maximum allowable normal stress of 150 MPaMPa . If the inner diameter of the tank is 2 mm , what is the minimum thickness, tt, of the wall
Answer:
The minimum thickness t of the wall is 0.00446 mm
Explanation:
Solution
Given that
Pressure =670kPa = 0.670
σ allowable normal stress = 150 MPa
Inner diameter = 2mm
Steel = A516 grade 60
Now,
Since the hoop stress is twice the longitudinal stress, the cylindrical tank is more likely to fail from the hoop stress.
Thus
σ allowable = σₙ = pμ/t
=p (d/2)
150 MPa =0.670MPa * 2/2/t
=0.67/t
t=0.67/150
t =0.00446 mm
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at 1 bar, 27 °C. Heat is transferredto the tank by an electric resister at a constant rate for 5 minutes. After heating, the tank pressure is 1 bar and the temperature is 477 °C. Air can be modeled as an ideal gas. Find the power input required, in kW
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
[tex]dQ = m \times c_p \times (T_2 -T_1)[/tex]
For ideal gas, [tex]c_p[/tex] = 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
2. The block is released from rest at the position shown, figure 1. The coefficient of
kinetic friction over length ab is 0.22, and over length bc is 0.16. Using the
principle of work and energy, find the velocity with which the block passes
position c.
Answer:
Velocity = 4.73 m/s.
Explanation:
Work done by friction is;
W_f = frictional force × displacement
So; W_f = Ff * Δs = (μF_n)*Δs
where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24
Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;
mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).
Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;
ΔKE2 + (0.16)(mg cos 24)(2).
Plugging in the relevant values, we have;
1.22mg = ΔKE1 + 0.603mg
ΔKE1 = 1.22mg - 0.603mg
ΔKE1 = 0.617mg
Also,
0.813mg = ΔKE2 + 0.292mg
ΔKE2 = 0.813mg - 0.292mg
ΔKE2 = 0.521mg
Now total increase in Kinetic Energy is ΔKE1 + ΔKE2
Thus,
Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg
Putting 9.81 for g to give;
Total increase in kinetic energy = 11.164m
Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m
m cancels out to give; ½v² = 11.164
v² = 2 × 11.164
v² = 22.328
v = √22.328
v = 4.73 m/s.
Q#1: Provide an example of a software project that would be amenable to the following models. Be specific. a. Waterfall b. Prototype c. Extreme Programming
Answer:
Waterfall model
Explanation:
The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.
The waterfall model is very easy to use. This is the earliest approach of the SDLC.
There are different phase of the waterfall:
Requirement analysisSystem DesignImplementationTestingDeploymentMaintenancePurely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of 3 phase, four-wire system. The line voltage is 415. What is the current in each line conductor ?
Answer:
The phase current in each line conductor are;
[tex]I_{R} = 100.17 < 0A[/tex]
[tex]I_{Y} = 75.13< - 120A[/tex]
[tex]I_{B} = 50.08 <120A[/tex]
Explanation:
Given the following data;
Red phase = 24kW,
Yellow phase = 18kW
Blue phase = 12kW
Line voltage = 415V
For a star connected system, we have;
[tex]Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}[/tex]
[tex]Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}[/tex]
[tex]Phase voltage (V_{p} ) = 239.6V[/tex]
The phase sequence for RYB is given by;
[tex]V_{R} = 239.6<0\\V_{Y} = 239.6<120\\V_{B} = 239.6<-120[/tex]
[tex]Phase current (I) = \frac{Phase power}{Phase voltage}[/tex]
[tex]Hence, I = \frac{P}{V}[/tex]
For the Red phase;
[tex]I_{R} = \frac{24000}{239.6<0}[/tex]
[tex]I_{R} = 100.17 < 0A[/tex]
For the Yellow phase;
[tex]I_{Y} = \frac{18000}{239.6<120}[/tex]
[tex]I_{Y} = 75.13< - 120A[/tex]
For the Blue phase;
[tex]I_{B} = \frac{12000}{239.6<-120}[/tex]
[tex]I_{B} = 50.08 <120A[/tex]
For the line neutral;
[tex]I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}[/tex]
Substituting we have, [tex]I_{N} = 43.29A[/tex]