Answer:
In both cases, reynolds number is greater than 2000,so the flows can't be laminar.
Explanation:
A) For flow in a tube of uniform diameter, the reynolds number is defined as;
Re = 2ρvr/η
where;
ρ is the fluid density,
v its speed,
η is viscosity
r is the tube radius.
In this question,
We are given;
r = 0.25cm = 0.25 × 10^(-2) m
η of water has a standard value of 1.005 × 10^(-3)
ρ of water has a standard value of 1000 kg/m³
In the reynolds equation, we don't know the velocity. So let's calculate it from;
Q' = vA
Where; Q' is flow rate = 0.5 L/s = 0.0005 m³/s
Area = πr² = π × (0.25 × 10^(-2))²
Area = 1.963 × 10^(-5) m²
So, v = Q/A = 0.0005/(1.963 × 10^(-5)) = 25.5 m/s
So, Re = 2ρvr/η = (2*1000*25.5*0.25 × 10^(-2))/(1.005 × 10^(-3))
Re = 126865.67
Re > 2000 and so the flow is not laminar
B) Now,
radius = 0.9cm = 0.9 × 10^(-2) m
So, A = πr² = π × (0.9 × 10^(-2))²
A = 2.5447 × 10^(-4) m²
v = Q/A = 0.0005/(2.5447 × 10^(-4))
v = 1.965 m/s
Re = 2ρvr/η = (2*1000*1.965*0.9 × 10^(-2))/(1.005 × 10^(-3))
Re = 35194.03
Re > 2000. So flow is not laminar.
A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.
Answer:
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Explanation:
A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:
[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]
Output power is cleared:
[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]
If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:
[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\dot W_{out} = 1752.825\,kW[/tex]
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Q: Draw shear and bending moment diagram for the beam shown in
the figure. EI= constant
Answer:
Explanation:
Please
To determine the viscosity of a liquid of specific gravity 0.95, you fill to a depth of 12 cm a large container that drains through a 30 cm long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm3 /s. What is the fluid viscosity (assume laminar flow)
Answer:
Fluid viscosity, [tex]\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]
Explanation:
Container depth, D = 12 cm = 0.12 m
Tube length, l = 30 cm = 0.3 m
Specific Gravity, [tex]\rho[/tex] = 0.95
Tube diameter, d = 2 mm = 0.002 m
Rate of flow, Q = 1.9 cm³/s = 1.9 * 10⁻⁶ m³/s
Calculate the velocity at point 2 ( check the diagram attached)
Rate of flow at section 2, [tex]Q = A_2 v_2[/tex]
[tex]Area, A_2 = \pi d^{2} /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^{-6} m^2[/tex]
[tex]v_{2} = Q/A_{2} \\v_{2} =\frac{1.9 * 10^{-6}}{3.14 * 10^{-6}} \\v_{2} = 0.605 m/s[/tex]
Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:
[tex]\frac{p_{1} }{\rho g} + \frac{v_{1}^2 }{2 g} + z_1 = \frac{p_{2} }{\rho g} + \frac{v_{2}^2 }{2 g} + z_2 + h_f\\ z_1 = L + l = 0.12 + 0.3\\z_1 = 0.42\\p_1 = p_{atm}\\v_1 = 0\\z_2 = 0\\\frac{p_{atm} }{\rho g} + \frac{0^2 }{2 g} + 0.42= \frac{p_{atm} }{\rho g} + \frac{0.605^2 }{2 *9.8} +0 + h_f\\h_f = 0.401 m[/tex]
For laminar flow, the head loss is given by the formula:
[tex]h_f = \frac{128 Q \mu l}{\pi \rho g d^4} \\\\0.401 = \frac{128 * 1.9 * 10^{-6} * 0.3 \mu}{\pi *0.95* 9.8* 0.002^4}\\\\\\\mu = \frac{0.401 * \pi *0.95* 9.8* 0.002^4}{128 * 1.9 * 10^{-6} * 0.3} \\\\\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]
Answer:
0.00257 kg / m.s
Explanation:
Given:-
- The specific gravity of a liquid, S.G = 0.95
- The depth of fluid in free container, h = 12 cm
- The length of the vertical tube , L = 30 cm
- The diameter of the tube, D = 2 mm
- The flow-rate of the fluid out of the tube into atmosphere, Q = 1.9 cm^3 / s
Find:-
To determine the viscosity of a liquid
Solution:-
- We will consider the exit point of the fluid through the vertical tube of length ( L ), where the flow rate is measured to be Q = 1.9 cm^3 / s
- The exit velocity ( V2 ) is determined from the relation between flow rate ( Q ) and the velocity at that point.
[tex]Q = A*V_2[/tex]
Where,
A: The cross sectional area of the tube
- The cross sectional area of the tube ( A ) is expressed as:
[tex]A = \pi \frac{D^2}{4} \\\\A = \pi \frac{0.002^2}{4} \\\\A = 3.14159 * 10^-^6 m^2[/tex]
- The velocity at the exit can be determined from the flow rate equation:
[tex]V_2 = \frac{Q}{A} \\\\V_2 = \frac{1.9*10^-^6}{3.14159*10^-^6} \\\\V_2 = 0.605 \frac{m}{s}[/tex]
- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.
[tex]\frac{P_1}{p*g} + \frac{V^2_1}{2*g} + z_t_o_p = \frac{P_2}{p*g} + \frac{V^2_2}{2*g} + z_d_a_t_u_m + h_L[/tex]
- The free surface conditions apply at atmospheric pressure and still ( V1 = 0 ). Similarly, the exit of the fluid is also to atmospheric pressure. Where, z_top is the total change in elevation from free surface to exit of vertical tube.
- The major head losses in a circular pipe are accounted using Poiessel Law:
[tex]h_L = \frac{32*u*L*V}{S.G*p*g*D^2}[/tex]
Where,
μ: The dynamic viscosity of fluid
L: the length of tube
V: the average velocity of fluid in tube
ρ: The density of water
- The average velocity of the fluid in the tube remains the same as the exit velocity ( V2 ) because the cross sectional area ( A ) of the tube remains constant throughout the tube. Hence, the velocity also remains constant.
- The energy balance becomes:
[tex]h + L = \frac{V_2^2}{2*g} + \frac{32*u*L*V_2}{S.G*p*g*D^2} \\\\0.42 = \frac{0.605^2}{2*9.81} + \frac{32*u*(0.3)*(0.605)}{0.95*998*9.81*0.002^2} \\\\u = 0.00257 \frac{kg}{m.s}[/tex]
- Lets check the validity of the Laminar Flow assumption to calculate the major losses:
[tex]Re = \frac{S.G*p*V_2*D}{u} \\\\Re = \frac{0.95*998*0.605*0.002}{0.00257} \\\\Re = 446 < 2100[/tex]( Laminar Flow )
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
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.
==================================
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.
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
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 DesignImplementationTestingDeploymentMaintenanceA very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects calculate the viscosities of the oils.
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = [tex]\frac{A*L}{Vo}[/tex] Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s
since the reaction is in first order
X = 1 - [tex]e^{-kt}[/tex]
[tex]e^{-kt}[/tex] = 1 - X
kt = In [tex]\frac{1}{1-X}[/tex]
k = In [tex]\frac{1}{1-X}[/tex] / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156 [tex]s^{-1}[/tex]
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]
therefore
[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
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
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
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
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.
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.
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 = $
The throttling valve is replaced by an isentropic turbine in the ideal vapor-compression refrigeration cycle to make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.
A. True
B. False
Answer:
False
Explanation:
The given statement is False. In real scenario the throttling valve not replaced by an isentropic turbine in the ideal vapor-compression refrigeration cycle. It is done so that the ideal vapor-compression refrigeration cycle to make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.
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.
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An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal gas at identical pressure but at different temperatures and masses. The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form
T3 = f(m1m2, m2m3, T1, T2)T 3=f( m 2m 1, m 3m 2,T 1,T 2) where m3m 3and T3T 3are the mass and temperature of the final mixture, respectively.
Answer:
Explanation:
Given that
Mass of 1 = [tex]m_1[/tex]
Mass of 2 = [tex]m_2[/tex]
Temperature in 1 = [tex]T_1[/tex]
Temperature in 2 = [tex]T_2[/tex]
Pressure remains i the group apartment
The closed system and energy balance is
[tex]E_{in}-E_{out}=\Delta E_{system}[/tex]
The kinetic energy and potential energy are negligible
since it is insulated tank ,there wont be eat transfer from the system
And there is no work involved
[tex]\Delta U = 0[/tex]
Let the final temperature be final temperature
[tex]m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)[/tex]
Using mass balance
[tex]m_3+m_2+m_1[/tex]
from eqn i
[tex]m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}[/tex]
Therefore the final temperature can be express as
[tex]\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }[/tex]
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, ν = 0.30). Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon, (c) the thickness of portion AB, (d) the cross- sectional area of portion AB.
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
PART (a)For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
PART (b)
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
PART(c)
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
PART (d)
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²
What is the stress and strain in the plate?Let us first find the cross sectional area of AB from the image attached;
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
We are given;
Tensile Load; F = 2.75 kN = 2.75 × 10³ N
Horizontal Stress is calculated from the formula;
σₓ = F/A
σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m
σₓ = 143.23 × 10⁶ Pa
Horizontal Strain is calculated from;
εₓ = σₓ/E
We are given E = 200 GPa = 200 × 10⁹ Pa
Thus;
εₓ = (143.23 × 10⁶)/(200 × 10⁹)
εₓ = 716.15 × 10⁻⁶
Formula for Vertical Strain is;
ε_y = -ν * εₓ
We are given ν = 0.30. Thus;
ε_y = -(0.3) * (716.15 × 10⁻⁶)
ε_y = -214.84 × 10⁻⁶
A) We are given;
Gauge Length; L = 0.05m
Change in gauge length is gotten from;
Δx = L * εₓ
Δx = 0.05 × 716.15 × 10⁻⁶
Δx = 35.808 × 10⁻⁶ m
B) From the attached diagram, the width is;
W = 0.012m
Change in width is;
Δy = W * ε_y
Δy = 0.012 * -214.84 × 10⁻⁶
Δy = -2.5781 × 10⁻⁶m
C) We are given;
Thickness of plate; t = 1.6 mm = 0.0016m
Change in thickness;
Δ_z = t * ε_z
where;
ε_z = ε_y
Thus;
Δ_z = t * ε_y
Δ_z = 0.0016 * -214.84 × 10⁻⁶
Δ_z = -343.74 × 10⁻⁹m
D) The change in cross sectional area is gotten from;
ΔA = A_final - A_initial
From calculating the areas, we have;
A = -8.25 × 10⁻⁹ m²
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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
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 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
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.
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.
Time, budget, and safety are almost always considered to be
1. Efficiency
2. Constraints
3. Trade-offs
4. Criteria
Answer:
The answer is option # 2. (Constraints).
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.
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]
Light acoustical panels in fire rated assemblies generally:
a. Compromise the fire rating by one hour.
b. Require hold down clips
c. Require pressure cleaning.
d. Are not allowed.
Answer:
a. Compromise the fire rating by one hour.
Explanation:
One hour fire rating is given to materials that can resist the fire exposure. The Acoustical panels controls reverberation and they are used for echo controls. The Fire rating is the passive fire protection which can resist standard fire. The test for fire rating also consider normal functioning of the material.
a. Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R.
b. Compute and compare linear density values for these same two directions for iron (Fe).
A) The linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R are;
i) LD_110 = √3/(4R√2)
ii) LD_111 = 1/(2R)
B) The linear density values for these same two directions for iron (Fe) are;
i) LD_110 = 2.4 × 10^(9) m^(-1)
ii) LD_111 = 4 × 10^(9) m^(-1)
Calculating Linear Density of Crystalline StructuresA) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;
√((4R)² - (4R/√3)²)
This reduces to; 4R√(1 - 1/3) = 4R√(2/3)
Now, the expression for the linear density of this direction is;
LD_110 =
Number of atoms centered on (110) direction/vector length of 110 direction
In this case, there is only one atom centered on the 110 direction. Thus;
LD_110 = 1/(4R√(2/3))
LD_110 = √3/(4R√2)
ii) The length of the vector for the direction 111 is equal to 4R, since
all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;
LD_111 = 2/(4R) = 1/(2R)
B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear density for the [110] direction is;
LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))
LD_110 = 2.4 × 10^(9) m^(-1)
ii) for the 111 direction, we have;
LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))
LD_111 = 4 × 10^(9) m^(-1)
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An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.
Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?
Answer:
A.) 7.13 degree north east
B.) 806.23 km/h
C.) 7.44 hours
D.) 0.06 hours
Explanation:
Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction
Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.
A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula
Tan Ø = y/x
Substitute y = 100 km/h and x = 800km/h
Tan Ø = 100/800
Tan Ø = 0.125
Ø = Tan^-1(0. 125)
Ø = 7.13 degrees north east.
Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east
B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem
Speed = sqrt ( 800^2 + 100^2)
Speed = sqrt (650000)
Speed = 806.23 km/h
C.) Let us use the speed formula
Speed = distance / time
Substitute the speed and distance into the formula
806.23 = 6000/ time
Make Time the subject of formula
Time = 6000/806.23
Time = 7.44 hours
D.) If there is no cross wind,
Time = 6000/800
Time = 7.5 hour
Time difference = 7.5 - 7.44
Time difference = 0.06 hours