Answer:
The infiltration rate is of 0.042 m^3/s.
Explanation:
The total volume of the room is:
V = 3 * 10 * 20 = 600 m^3
If the air infiltration rate is of 1/4 volume per hour:
v = V/4 * 1 hour /3600 seconds
Replacing:
v = 600/4 * 1/3600 = 0.042 m^3/s
The infiltration rate would then be of 0.042 m^3/s.
Do a summary what happen to titanic in the aspect of material(body) and the ductile brittle temperature (DBT) of the material.
Explanation:
A ductile material can convert into brittle material due to following reasons
1.At very low temperature
2.Due to presence of notch
In titanic ,the base of ship strike to the large ice cube and lower part of titanic ship material was made of steel .We know that steel is ductile material and when steel came with very low temperature of ice due to this ductile material converted in to brittle material and titanic ship failed.Brittle material does not show any indication before failure.
In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain is 0.226. Determine strain hardening exponent, n, in the flow curve for the plastic region of this steel.
Answer:n=0.973
Explanation:
Given
When True strain[tex]\left ( \epsilon _T_1\right )=0.171[/tex]
at [tex]\sigma _1=263.8 MPa[/tex]
When True stress[tex]\left ( \sigma _2\right )[/tex]=346.2 MPa
true strain [tex]\left ( \epsilon _T_2\right )[/tex]=0.226
We know
[tex]\sigma =k\epsilon ^n [/tex]
where [tex]\sigma [/tex]=True stress
[tex]\epsilon [/tex]=true strain
n=strain hardening exponent
k=constant
Substituting value
[tex]263.8=k\left ( 0.171\right )^n------1[/tex]
[tex]346.2=k\left ( 0.226\right )^n-----2[/tex]
Divide 1 & 2 to get
[tex]\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n[/tex]
[tex]1.312=\left ( 1.3216\right )^n[/tex]
Taking Log both side
[tex]ln\left ( 1.312\right )=nln\left ( 1.3216\right )[/tex]
n=0.973
Determine the factor of safety for a 9 foot long hollow steel
column 3.5 inches on a side that has a wall thickness of 0.225
inches and is loaded with a 22 kip load. Use the steel E of 29 *10
^6 psi and assume the column is pin connected at each end.
Answer:
factor of safety for A36 structural steel is 0.82
Explanation:
given data:
side of column = 3.5 inches
wall thickness = 0.225 inches
load P = 22 kip
Length od column = 9 ft
we know that critical stress is given as
[tex]\sigma_{cr} = \frac{\pi^2 E}{(l/r)^2}[/tex]
where
r is radius of gyration[tex] = \sqrt{\fra{I}{A}}[/tex]
Here I is moment od inertia [tex]= \frac{b_1^2}{12} - \frac{b_2^2}{12}[/tex]
[tex] I == \frac{3.5^2}{12} - \frac{3.05^2}{12} = 5.294 in^4[/tex]
For hollow steel area is given as [tex]A = b_1^2 -b_2^2[/tex]
[tex]A = 3.5^2 -3.05^2 = 2.948 in^2[/tex]
critical stress [tex]\sigma_{cr} = = \frac{\pi^2\times 29\times 10^6}{((9\times12)/(1.34))^2}[/tex]
[tex]\sigma_{cr} = 44061.56 lbs/inc^2[/tex]
considering Structural steel A36
so A36[tex] \sigma_y = 36ksi[/tex]
factor of safety [tex]= \frac{yield\ stress}{critical\ stress}[/tex]
factor of safety =[tex]\frac{36\times10^3}{44061.56} = 0.82[/tex]
factor of safety for A36 structural steel is 0.82
The student's query involves calculating the factor of safety for a hollow steel column, but without complete material strength data, the calculation cannot be precisely done.
Explanation:The student's question pertains to determining the factor of safety for a hollow steel column with specified dimensions, material properties (steel E), and loading conditions. To calculate the factor of safety for the column, we would need to compare the column's actual stress under load to its maximum allowable stress. However, due to insufficient data about material yield strength or ultimate strength and considering the provided information does not match the examples given, this calculation cannot be accurately completed without further specifics on the steel's properties.
The absolute pressure of an automobile tire is measured to be 320 kPa before a trip and 349 kPa after the trip. Assuming the volume of the tire remains constant at 0.022 m^3, determine the percent increase in the absolute temperature of the air in the tire. The percent increase in the absolute temperature of the air in the tire is_____ %.
Answer:
9%
Explanation:
An ideal gas is one that has its molecules widely dispersed and does not interact with each other, studies have shown that air behaves like an ideal gas, so the state change equation for ideal gases can be applied.
P1V1T2 = P2V2T1
where 1 corresponds to state 1 = 320kPa
and 2 is state 2 = 349kPa.
Given that the volume remains constant the equation is:
P1T2=P2T1
SOLVING for T2/T1
[tex]\frac{T2}{T1} =\frac{P2}{P1} =\frac{349}{320} =1.09\\\\[/tex]
The equation to calculate the percentage increase is as follows
%ΔT=[tex]\frac{(T2-T1)100}{T1} =(\frac{T2}{T1} -1)100=(1.09-1)100=(0.09)100=9%[/tex]
In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm. Roll diameter is 600mm and coefficient of friction between rolls and strip 0.15. The specification is that the draft is to be equal on each pass. Determine a) Minimum number of passes required? b) Draft for each pass?
Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip ([tex]\nu [/tex])=0.15
maximum draft[tex](d_{max})=\nu ^2R[/tex]
[tex]=0.15^2\times 300=6.75 mm[/tex]
Minimum no of passes[tex]=\frac{45-20}{6.75}=3.70\approx 4[/tex]
(b)draft per each pass
[tex]d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}[/tex]
[tex]d=\frac{45-20}{4}=6.25 mm[/tex]
How does fouling affects the performance of a heat exchanger?
Answer:
Fouling :
When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.
Effect of heat exchanger are as follows"
1.It decreases the heat transfer.
2.It increases the thermal resistance.
3.It decreases the overall heat transfer coefficient.
4.It leads to increase in pressure drop.
5.It increases the possibility of corrosion.
You are hitting a nail with a hammer (mass of hammer =1.8lb) the initial velocity of the hammer is 50 mph (73.33 ft/sec). The time of impact is .023 sec. Assuming the nail heads directly in the -j direction, what is the magnitude of the force exerted on the hammer from the nail?
Answer:
The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.
Explanation:
Since we know that the force is the rate at which the momentum of an object changes.
Mathematically [tex]\overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}[/tex]
The momentum of any body is defines as [tex]\overrightarrow{p}=mass\times \overrightarrow{v}[/tex]
In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as
[tex]\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}[/tex]
[tex]\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds[/tex]
A car enters a circular off ramp (radius of 180 m) at 30 m/s. The car's on - bond accelerometers sense a total acceleration magnitude of 7.07 m/s2. What is the car's acceleration magnitude a2 (the component tangent to its path)?
Answer:
Horizontal acceleration will be [tex]2.07m/sec^2[/tex]
Explanation:
We have given radius of track = 180 m
velocity = 30 m/sec
Total acceleration [tex]a_t=7.07m/sec^2[/tex]
The centripetal;l acceleration which is always normal to the velocity also known as normal acceleration is given by [tex]a_n=\frac{v^2}{r}=\frac{30^2}{180}=5m/sec^2[/tex]
So the tangent acceleration will be [tex]a_t=7.07-5=2.07m/sec^2[/tex]
The velocity of a particle along the s-axis is given by v = 14s^7/6 where s is in millimeters and v is in millimeters per second. Determine the acceleration when s is 5.5 millimeters.
Answer:
The acceleration is [tex]2220.00m/s^{2}[/tex]
Explanation:
We know that the acceleration is given by
[tex]a=v\frac{dv}{ds}........................(i)[/tex]
The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]
Thus the acceleration as obtained from equation 'i' becomes
[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]
Hence acceleration at s = 5.5 equals
[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]
The boiler pressure is 38bar and the condenser pressure 0.032 bar.The saturated steam is superheated to 420 oC before entering the turbine. a) Calculate the cycle efficiency of the Rankine cycle. b) Calculate the work ratio of the Rankine cycle.
Answer:
a)38.65%
b)1221KJ/kg
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
first we calculate the enthalpies in all states
h1=3264kJ/Kg
h2=2043kJ/Kg
h2=h3=105.4kJ/Kg
a)we use the ecuation for efficiency
Efficiency = (h1-h2) / (h1-h4)
Efficiency = (3264-2043) / (3264-105.4)
=0.3865=38.65%
b)we use the ecuation for Wout
Wout = m (h1-h2)
for work ratio=
w = (h1-h2)
w=(3264-2043)=1221KJ/kg
Four kilograms of gas were heated at a constant pressure of 12 MPa. The gas volumes were 0.005 m^3 and 0.006 m^3 in the initial and final states, respectively, and 3.9 kJ of heat was transferred to the gas. What is the change in specific internal energy between the initial and final states?
Answer:
Specific change in internal energy is - 2.025 kj/kg.
Explanation:
The process is constant pressure expansion. Apply first law of thermodynamic to calculate the change in internal energy.
Given:
Mass of gas is 4 kg.
Initial volume is 0.005 m³.
Final volume is 0.006 m³.
Pressure is 12 Mpa.
Heat is transfer to the gas. So it must be positive 3.9 kj.
Calculation:
Step1
Work of expansion is calculated as follows:
[tex]W=P(V_{f}-V_{i})[/tex]
[tex]W=12\times10^{6}(0.006-0.005)[/tex]
W=12000 j.
Or,
W=12 Kj.
Step2
Apply first Law of thermodynamic as follows:
Q=W+dU
3.9=12+dU
dU = - 8.1 kj.
Step3
Specific change in internal energy is calculated as follows:
[tex]u=\frac{U}{m}[/tex]
[tex]u=\frac{-8.1}{4}[/tex]
u= - 2.025 kj/kg.
Thus, the specific change in internal energy is - 2.025 kj/kg.
If the local atmospheric pressure is 14.6 psia, find the absolute pressure (in psia) in a column of glycerin (rho = 74.9 lbm/ft^3) at depth of 27in.
Answer:
52.2538 psia
Explanation:
The absolute pressure at depth of 27 inches can be calculated by:
Pressure = Local pressure + Gauge pressure
Also,
[tex]P_{gauge}=\rho\times g\times h[/tex]
Where,
[tex]\rho[/tex] is the density of glycerin ([tex]\rho=74.9\ lbm/ft^3[/tex])
g is the gravitational acceleration = 32.1741 ft/s²
h = 27 in
Also, 1 in = 1/12 ft
So,
h = 27 / 12 ft = 2.25 ft
So,
[tex]P_{gauge}=74.9\times 32.1741\times 2.25\ lbf/ft^2=5422.1402\ lbf/ft^2[/tex]
Also,
1 ft = 12 inch
1 ft² = 144 in²
So,
[tex]P_{gauge}=5422.1402\ lbf/ft^2=\frac {5422.1402\ lbf}{144\ in^2}=37.6538\ lbf/in^2=37.6538\ psia[/tex]
Local pressure = 14.6 psia
So,
Absolute pressure = 14.6 psia + 37.6538 psia=52.2538 psia
We discover a nearby star with two planets. The first planet has an orbit period of 10 years and is in a circular orbit with radius 106 km. The second planet has an orbit period of 15 years. What is its orbit radius? You may assume it is also in a circular orbit.
Answer: 139 Km.
Explanation:
The question tells us that a planet A has an orbit period of 10 years and its circular orbit has a radius of 106 Km, whilst a planet B has an orbit period of 15 years (also assuming a circular orbit), both orbiting a nearby star.
This information allow us to use the Kepler's 3rd law, for the special case in which the orbit is circular.
Kepler's 3rd law, tells that there exist a direct proportionality between the square of the orbit period, and the cube of the orbit radius (in the more general case, with the cube of the semi-major axis of the elipse), for celestial bodies orbiting a same star.
(like Earth and Mars orbiting Sun).
So, for planet A and planet B (orbiting a same star), we can write the following:
(TA)²/ (TB)² = (rA)³ / (rB)³
Replacing by TA= 10 years, TB= 15 years, rA= 106 Km, and solving for Rb, we get RB= 139 Km.
There is 200 kg. of saturated liquid water in a steel tank at 97 C. What is the pressure and volume of the tank?
Answer:
[tex]V = 0.208 m^3[/tex]
saturated pressure = 0.91 bar 0r 91 kPa
Explanation:
Given data:
Mass of liquid water 200 kg
Steel tank temperature is 97 degree celcius = 97°C
At T = 97 degree celcius, saturated pressure = 0.91 bar 0r 91 kPa
And also for saturated liquid
Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg
Volume of tank is given as
[tex]V = \nu * m[/tex]
[tex]V = 0.00104 * 200[/tex]
[tex]V = 0.208 m^3[/tex]
Convert each of the following to three significant figures: (a) 20 lb.ft to N.m, (b) 450 Ib/ft^3 to kN/m^3, and (c) 15 ft/h to mm/s.
Answer:
(a)27.12 N-m (b) 69.96 [tex]KN/m^3[/tex] (c) 1.27 mm/sec
Explanation:
We have to convert
(a) 20 lb-ft to N-m
We know that 1 lb = 4.45 N
So 20 lb = 20×4.45 = 89 N
1 feet = 0.3048 m
So [tex]20lb-ft=20\times 4.45N\times 0.3048N=27.12N-m[/tex]
(b) 450 [tex]lb/ft^3[/tex] to KN/[tex]m^3[/tex]
We know that 1 lb = 4.45 N = 0.0044 KN
[tex]1ft^3=0.0283m^3[/tex]
So [tex]450lb/ft^3=\frac{450\times 0.0044KN}{0.0283m^3}=69.96KN/m^3[/tex]
(c) 15 ft/hr to mm/sec
We know that 1 feet = 0.3048 m = 304.8 mm
And 1 hour = 60×60=3600 sec
So [tex]15ft/h=\frac{15\times 304.8mm}{3600sec}=1.27 mm/sec[/tex]
Answer:
a)20 lb.ft=27.2 N.m
b)[tex]450\dfrac{lb}{ft^3}=70.65\dfrac{KN}{m^3}[/tex]
c)15 ft/h = 1.26 mm/s
Explanation:
a)
As we know that
1 ft.lb= 1.36 N.m
So 20 lb.ft = 1.36 x 20 N.m
20 lb.ft=27.2 N.m
b)
We know that
[tex]1\dfrac{lb}{ft^3}=0.157\dfrac{KN}{m^3}[/tex]
So
[tex]450\dfrac{lb}{ft^3}=450\times 0.157\dfrac{KN}{m^3}[/tex]
[tex]450\dfrac{lb}{ft^3}=70.65\dfrac{KN}{m^3}[/tex]
c)
As we know that
1 ft/h=0.084 mm/s
So
15 ft/h = 0.084 x 15 mm/s
15 ft/h = 1.26 mm/s
The primary heat transfer mechanism that warms me while I stand next to a campfire is: a)- Conduction b)- Impeadance c)- Convection d)- Radiation
Answer:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Explanation:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.
A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60kJ/min. Determine: (a) The electric power consumed by the refrigerator, and (b) The rate of heat transfer to the kitchen air.
Answer:
a) Power =50 KJ/min
b)Rate of heat transfer = 110 KJ/min
Explanation:
Given that
COP = 1.2
Heat removed from space Q = 60 KJ/min
As we know that COP of refrigerator is the ratio of heat removed to work input.
Lets take power consume by refrigerator is W
So
COP= Q/W
1.2=60/W
W=50 KJ/min
So the power consume is 50 KJ/min.
From first law of thermodynamic
Heat removed from the kitchen = 50 + 60 KJ/min
Heat removed from the kitchen =110 KJ/min
Describe the physics associated with the concept of thermal resistance.
Answer:
Thermal resistance:
Thermal resistance is the property which oppose the the heat transfer.It is the property for measurement of heat flow.
As we know that heat transfer take place from the high temperature to low temperature.Heat transfer Q given as
[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]
Where Δt is the temperature difference
[tex]{R_{th}}[/tex] is the thermal resistance.
Connection:
1. Series connection
[tex]R_{total}={R_1}+{R_2}+{R_1}----{R_n}[/tex]
2. Parallel connection
[tex]\dfrac{1}{R_{total}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}-----\dfrac{1}{R_n}[/tex]
Water at room temperature of 20.0°C is poured into an aluminum cylinder which has graduation markings etched on the inside. The reading in the graduations is 300.0 cc. The cylinder with the water in it is then immersed in a constant temperature bath at a temperature of 100°C. What is the reading for the level of water on the graduations of the cylinder after the water and the cylinder reach thermal equilibrium with the bath? The volume coefficient of expansion of water is 2.07 × 10 -4 K -1, and the linear coefficient of expansion of aluminum is 23.0 × 10 -6 K -1. 305.0 cc 303.5 cc 304.5 cc 304.0 cc 303.3 cc
Answer:
304.42 cc
Explanation:
When the aluminum expands the markings will be further apart. If the 300 cc mark was at a distance l0 of the origin at 20 C, at 100 C it will be
l = l0 * (1 + a * (t - t0))
l = l0 * (1 + 23*10^-6 * (100 - 20))
l = l0 * 1.0018
The volume of water would have expanded by
V = V0 (1 + a * (t - t0))
V = 300 (1 + 2.07*10^-4 * (100 - 20))
V = 304.968 cc
Since the markings expanded they would measure
304.968/1.0018 = 304.42 cc
The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2250 K. Assume constant specific heats evaluated at 300 K. Determine the cut-off ratio. There is a +/- 5% tolerance.
Answer:
Cut off ratio=2.38
Explanation:
Given that
[tex]T_1=300K[/tex]
[tex]P_1=100KPa[/tex]
[tex]P_2=P_3=7200KPa[/tex]
[tex]T_3=2250K[/tex]
Lets take [tex]T_1[/tex] is the temperature at the end of compression process
For air γ=1.4
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]
[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]
[tex]T_2=1070K[/tex]
At constant pressure
[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{V_3}{V_2}=2.83[/tex]
So cut off ratio
[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]
Cut off ratio=2.38
Evaluate each of the following and express with an appropriate prefix: (a) (430 kg)^2 (b) (0.002 mg)^2, and (c) (230 m)^3
The expressions squared or cubed with appropriate metric prefixes are: (a) 184.9 Mg^2, (b) 4 x 10^-18 kg^2, and (c) 12.167 km^3. Metric prefixes are used for clarity in representing large quantities.
Explanation:The question deals with the operation of squaring and cubing given quantities and expressing them with the appropriate metric prefixes. Let's evaluate each expression step by step:
(a)It is important to use the proper metric prefixes when
expressing large numbers to ensure clarity and avoid confusion.
A(n)_____is an interconnected collection of computers.
Answer:
Network is a collection of computers
Explanation:
A network is a gathering of at least two gadgets that can convey. Practically speaking, a system is included various distinctive PC frameworks associated by physical or potentially remote associations.
The scale can run from a solitary PC sharing out fundamental peripherals to huge server farms situated far and wide, to the Internet itself. Despite extension, all systems permit PCs or potentially people to share data and assets.
How much power would you need to cool down a closed, 1 Liter container of water from 100°C to 20°C in 5 minutes? (a) 1.1W (b)1.1kW (c)67kW (d)334 kJ
Answer:
The power required to cool the water is 1.11Kw.
Hence the correct option is (b).
Explanation:
Power needed to cool down is equal to heat extract from the water.
Given:
Volume of water is 1 liter.
Initial temperature is 100C.
Final temperature is 20C.
Time is 5 minutes.
Take density of water as 100 kg/m3.
Specific heat of water is 4.186 kj/kgK.
Calculation:
Step1
Mass of the water is calculated as follows:
[tex]\rho=\frac{m}{V}[/tex]
[tex]1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}[/tex]
m=1kg
Step2
Amount of heat extraction is calculated as follows:
[tex]Q=mc\bigtriangleup T[/tex]
[tex]Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)[/tex]
Q=334880 j.
Step3
Power to cool the water is calculated as follows:
[tex]P=\frac{Q}{t}[/tex]
[tex]P=\frac{334880}{(5min)(\frac{60s}{1min})}[/tex]
P=1116.26W
or
[tex]P=(1116.26W)(\frac{1Kw}{1000 W})[/tex]
P=1.11 Kw.
Thus, the power required to cool the water is 1.11Kw.
Hence the correct option is (b).
A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine the final pressure of the water.
Answer:
final pressure is 6847.41 kPa
Explanation:
given data:
[tex]P_1 = 10,000 kPa[/tex]
[tex]T_1 =520\ degree\ celcius = 793 K[/tex]
[tex]T_2 = 270 degree celcius = 543 K[/tex]
as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream
for ischoric process we know that
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
[tex]\frac{10*10^6}{793} = \frac{P_2}{543}[/tex]
[tex]P_2 = 6.84741*10^6 Pa[/tex]
final pressure is 6847.41 kPa
A water jet that leaves a nozzle at 60 m/s at a certain flow rate generating power of 250 kW by striking the buckets located on the perimeter of a wheel. Determine the mass flow rate of the jet.
Answer:
The mass flow rate of jet =69.44 kg/s
Explanation:
Given that
velocity of jet v= 60 m/s
Power P=250 KW
As we know that force offered by water F
[tex]F=\rho\ A \ v^2[/tex]
Power P= F.v
So now power given as
[tex]P=\rho\ A \ v^3[/tex]
We know that mass flow rate = ρAv
[tex]P=mass\ flow\ rate\ \times v^2[/tex]
250 x 1000 = mass flow rate x 3600
mass flow rate = 69.44 kg/s
So the mass flow rate of jet =69.44 kg/s
Is refrigerator with an ice-maker an open or a closed system? Explain your answer
Answer:
Open system
Explanation:
In refrigerator there is a interaction between refrigerator and environment.
In refrigerator heat moves from the system to the outer environment means there is transfer of heat from one system to environment.
We know that whenever there is transfer of energy or mass from system then it is known as pen system
So refrigerator is a open system
Please define the specific heat of material?
Answer and Explanation:
SPECIFIC HEAT :
Specific heat is denoted by [tex]c_v[/tex]It is the heat required for increasing the temperature of a substance which has mass of 1 kg.Its SI unit is joule/kelvinIt is physical property It can be calculated by [tex]c_v=\frac{Q}{m\Delta T}[/tex], here Q is heat energy m is mass of gas and [tex]\Delta T[/tex] is change in temperature.Is CO, an air pollutant? How does it differ from other emissions resulting from the combustion of fossil fuels?
Answer:
Explanation:
CO, carbon monoxide is a toxic gas. It casues asphixiation on people and animals by interfering with hemoglobin, not allowing blood to transport oxygen to the cells in the body.
The normal emissions resulting from the combustion of fussil fuels are CO2 (carbon dioxide) and H2O (water). Carbon monoxide is formed by an incomplete combustion of fossil fuels or carbon containing fuels in general, this not only produces toxic gas, but also is an inefficient combustion that wastes energy.
A round steel bar, 0.02 m in diameter and 0.40 m long, is subjected to a tensile force of 33,000 kg. Y=E= 2E10 kg/m^2. (modulus).Calculate the elongation in meters.
The elongation of the steel bar under the given tensile force is approximately 2.10 millimeters.
The cross-sectional area (A) of the steel bar can be calculated using the formula for the area of a circle:
A = π * r²
Now, let's calculate A:
A = π * (0.01 m)²
≈ π * (0.0001 m²)
≈ 0.000314 m²
Now, we can calculate the elongation (ΔL) using Hooke's Law:
ΔL = (F * L) / (A * E)
Given:
F = 33,000 kg
L = 0.40 m
E = 2 * 10¹⁰ kg/m²
Now, plug in the values:
ΔL = (33,000 kg * 0.40 m) / (0.000314 m² * 2 * 10₈⁰ kg/m²)
Now, perform the calculation
ΔL ≈ (13,200 kg*m) / (6.28 * 10 kg/m²)
ΔL ≈ 2.10 * 10⁻₀ meters
Describe the differences between convection and thermal radiation.
Answer:
Explanation:
Convection needs a fluid to transport the heat, while radiation doesn't.
Generally convection transports a lot more heat than radiation.
Since fluids tend to expand, when in a gravitational field such as that of Earth convection tends to move heat upwards while radiation is indifferent to gravity.