Answer:
0.5 feet
Explanation:
it is given that the martin floats at draft of 23'6"
GM=1.5
The load is given as follows
1400 tons is loaded as 2 feet above keel
1400 tons-----kg----2 feet
final kg = [tex]\frac{final moment }{final dispacement}[/tex]
[tex]\frac{weight}{1400 kg}[/tex] = [tex]\frac{kg}{2 feet}[/tex] = [tex]\frac{moment of keel }{2800}[/tex]
final kg = [tex]\frac{2800}{1400}[/tex]=2 feet
final GM =2 feet-1.5 feet
=0.5 feet
A 9-cm diameter, 11-cm high hollow metal can floats in water vertically with 8 cm of its height under water. What is the weight of the can?
Answer:
Weight of the can is 5 N
Explanation:
The weight of the can is a downward force. When an object floats in a fluid, it is acted upon by an upward force which balances the weight of the body that acts downwards. This upward force is know as Buoyant force. This buoyant force is also the weight of the fluid displaced by the object.
Therefore, we know that
Total buoyant force = total weight of the body
And total weight of the body is nothing but the weight of the fluid displaced by the body.
So, weight of the body = weight of the fluid displaced by the body.
= [tex]\rho g\times[/tex]volume of water displaced by the body.
= [tex]\rho g\times[/tex]volume of the body submerged in the water
Now we know that,
density of water, [tex]\rho[/tex] = 1000 kg/[tex]m^{3}[/tex]
acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]
Volume of the body submerged in water, V = [tex]\frac{\prod }{4}\times d^{2}\times h[/tex]
= [tex]\frac{\prod }{4}\times 0.09^{2}\times 0.08[/tex]
= 5.05[tex]\times 10^{-4}[/tex] [tex]m^{3}[/tex]
Therefore, weight of can = [tex]\rho g\times[/tex]volume of the body submerged in the water
= 1000[tex]\times 9.81\times 5.05\times 10^{-4}[/tex]
= 4.95 N
[tex]\simeq[/tex] 5 N
Therefore the weight of the can is 5 N.
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the inlet. The volumetric flow rate is 0.2 m3/s, and the inlet and exit areas are 0.06 m2 and 0.03 m3, respectively. (a) Assuming the temperature to remain constant and neglecting any heat transfer, determine the power input to the pump in kW. (b) What-if Scenario: What would the necessary power input be if the change in KE were neglected in the analysis??
Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation
[tex]\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m[/tex]
Work done by pump
[tex]W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W[/tex]
∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy
[tex]h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\[/tex]
Work done by pump
[tex]W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W[/tex]
∴ Power input to the pump 20.7088 kW
It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)-They present high modulus of elasticity d)-a and c are correct
Answer:
(d) a and c are correct
Explanation:
METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity
for example : iron, gold ,silver, copper
ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property
for example ; bronze and brass
so from above discussion it is clear that option (d) will be the correct option
Answer:
d)-a and c are correct
Explanation:
Hello,
As long as metals have specific molecular arrangements (closely assembled molecules) they have a high capacity to transfer both electrical and thermal energy. On the other hand, the modulus of stability is considered as a measure of material's stiffness or resistance to elastic deformation, thus, due to the very same aforesaid molecular arrangement of metals, they are hard to deform so that modulus is considered as high.
Best regards.
A closed-loop system has a forward path having two series elements with transfer functions 5 and 1/(s + 1). If the feedback path has a transfer function 2/s, what is the overall transfer function of the system?
Answer:
Transfer function for feedback path is given by:
[tex]\frac{C(s)}{R(s)}[/tex]=[tex]\frac{G(s)}{1+G(s)H(s)}[/tex] (1)
Explanation:
with reference to fig1:
two blocks in series are multiplied:
[tex]\frac{5}{s(s+1)}[/tex]
for feedback function:
1+G(s)H(s)=[tex]1+\frac{5}{s+1}.\frac{2}{s}[/tex]
Now from eqn (1):
[tex]\frac{C(s)}{R(s)} = \frac{5}{s(s+1)+10}[/tex]
A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa
Answer: c) 450 kPa
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 150 kPa
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = v L
[tex]V_2[/tex] = final volume of gas = [tex]\frac{v}{3}L[/tex]
[tex]150\times v=P_2\times \frac{v}{3}[/tex]
[tex]P_2=450kPa[/tex]
Therefore, the new pressure of the gas will be 450 kPa.
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air properties find the cutoff ratio, the expansion ratio v4/v3, and the exhaust temperature.
Answer:
Cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]
Cxpansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex]
The exhaust temperature[tex]T_4=1997.5R[/tex]
Explanation:
Compression ratio CR(r)=20
[tex]\dfrac{V_1}{V_2}=20[/tex]
[tex]P_2=P_3=920 psia[/tex]
[tex]T_1=520 R ,T_{max}=T_3,T_3=3200 R[/tex]
We know that for air γ=1.4
If we assume that in diesel engine all process is adiabatic then
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{520}=20^{1.4 -1}[/tex]
[tex]T_2=1723.28R[/tex]
[tex]\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}[/tex]
[tex]\dfrac{V_3}{V_2}=\dfrac{3200}{520}[/tex]
So cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]
[tex]\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}[/tex]
Now putting the values in above equation
[tex]\dfrac20=\dfrac{V_4}{V_3}\times 6.15[/tex]
[tex]\dfrac{V_4}{V_3}=3.25[/tex]
So expansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex].
[tex]\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}[/tex]
[tex]\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}[/tex]
[tex]T_4=1997.5R[/tex]
So the exhaust temperature[tex]T_4=1997.5R[/tex]
A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank (m3 ). Later, a pump is used to extract 100 kg of water from the tank. The water remaining in the tank eventually reaches thermal equilibrium with the surroundings at 30 °C). Determine: b) the final pressure (kPa).
Given:
mass of water, m = 2000 kg
temperature, T = [tex]30^{\circ}C[/tex] = 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = m[tex]\bar{R}[/tex]T (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
[tex]\bar{R} = \frac{R}{M} [/tex]
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
[tex]V = \frac{m\bar{R}T}{P}[/tex]
[tex]V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}[/tex]
[tex]V = 2762.44 m^{3}[/tex]
Therefore, the volume of the tank is [tex]V = 2762.44 m^{3}[/tex]
b) After extracting 100 kg of water, amount of water left, m' = m - 100
m' = 2000 - 100 = 1900 kg
The remaining water reaches thermal equilibrium with surrounding temperature at T' = [tex]30^{\circ}C[/tex] = 303 K
At equilibrium, volume remain same
So,
P'V = m'[tex]\bar{R}[/tex]T'
[tex]P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}[/tex]
Therefore, the final pressure is P' = 96.258 kPa
What is the temperature dependency of the electrical conductivity for metals and semiconductors??
Answer and Explanation:
TEMPERATURE DEPENDENCY ON ELECTRICAL CONDUCTIVITY OF METALS : Metals are good conductors of electricity but when we increase the temperature the free electrons of metals collide with each other due to heat.There collision become very fast and so the resistance increases and so the electrical conductivity of metals decreases on increasing temperature.
TEMPERATURE DEPENDENCY ON ELECTRICAL CONDUCTIVITY OF SEMICONDUCTOR : The electrical conductivity of semiconductor is mainly sue to presence of impurities and defects as the temperature increases the impurities and defects also increases so the electrical conductivity of semiconductor increases on increasing temperature.
Define the various properties and constraints that ensure weldability/joinability.
Answer:
Listed Below
Explanation:
Weldability which is also known as joinability is defined as the ease with which a material can be welded without producing any other effect .
Some factors which affect weldability are
1.Melting point of metal
Higher the melting point lower will be the weldability.2.Thermal conductivity
Materials with higher thermal conductivity are difficult to join.3.Reactivity
If the material reacts with surroundings like water or air, it is difficult to weld.4.Coefficient of thermal expansion(α)
Higher the thermal expansion lower will be the weldability.5.Surface condition
The material with dirty surface is difficult to weld for example if a surface has oxide layer on it it become difficult to weld it.Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K. Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.
Solution:
Given:
[tex]T_{H}[/tex] = 1200 K
[tex]T_{L}[/tex] = 600 K
Q = 100 kJ
The Entropy change of the two reservoirs is given by the sum of entropy change of each reservoir system and is given by the formula:
[tex]\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}[/tex]
[tex]\Delta s = \frac{Q(T_{L}-T_{_{H}})}{T_{H}T_{L}}[/tex]
[tex]\Delta s = \frac{-100(600-1200)}{1200\times 600}[/tex]
[tex]\Delta s = 0.0833kJ/K
Since, the change in entropy is positive and according to the Increase in entropy principle, for any process the total change in entropy of a system is always greater than or equal to zero (with its enclosing adiabatic surrounding).
Therefore, the entropy principle is satisfied.
What do you understand by the term redundant work?
Answer:
Redundant work refers to the work done during the process of deformation due to friction. It happens during the wire drawing. Redundant work per unit volume increases when the radial position becomes higher. The redundant work factor is defined as increased strain of the deformation to the stress. It is basically related to the deformation area geometry.
Consider three identical items in parallel. What is the system reliability if each item has a reliability of 98%?
Answer:
Reliability of system = 99.9992%
Explanation:
Given Reliability of item = 98%
therefore probability of failure = 100%-98% =2%
For system in parallel the probability of failure is product of individual probabilities of items
thus
[tex]Probabilityfail_{system}=p_{1}\times p_{2}\times p_{3}\\\\Probabilityfail_{system}=0.02^{3}=0.000008\\\\Realibility_{system}=100- Probabilityfail_{system}\\\\Realibility_{system}=99.9992%[/tex]
A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the inlets are 1kg/s, 1.5kg/s and 2 kg/s. The enthalpy of those inlets are the 100kJ/kg, 120kJ/kg, and 500kJ/kg, respectively. What is the enthalpy at the outlet?
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:
[tex]m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s[/tex]
[tex]h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg[/tex]
We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation
[tex]m_1+m_2+m_3=m[/tex]
m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation
[tex]m_1h_1+m_2h_2+m_3h_3=mh[/tex]
By putting the values
[tex]1\times 100+1.5\times 120+2\times 500=4.5\times h[/tex]
So h=284.44 KJ
Proper grain stress/strain analysis is required in solid motor design: A. To ensure good motor structure B. To avoid surface cracks C. To avoid bond separation D. All of the above
Answer:
The correct answer to the given statement is
option D. All of the above
Explanation:
Analysis of proper grain stress/strain is important as it ensures good mechanical motor structure.
With the help of analysis surface cracks can be checked and proper maintenance can be provided. It further helps in keeping check in order to avoid separation of bonds.
Therefore, qualitative analysis and in depth analysis can reduce errors and helps to maintain the qualitative parameters.
The thermal efficiency of two reversible power cycles operating between the same thermal reservoirs will a)- depend on the mechanisms being used b)- be equal regardless of the mechanisms being used c)- be less than the efficiency of an irreversible power cycle
In thermodynamicsedependent properties means?
Answer:
Explanation:
Thermodynamics properties are the properties which defined the state of any system.
some of the thermodynamics properties are pressure, temperature etc
thermodynamics are broadly divided into two type
1)intensive and
2)extensive properties
Dependent properties are the properties that are dependent on other properties. Extensive property are those which are dependent on the extent of system. Example volume. if size of the system increase or decrease then volume also have same effect according to the changes
What is the typical bonding in a conductor and a semiconductor??
Answer:
The typical bonding in conductors is defined as which contain free valence electrons and free ions as, it is typically known as metallic bonding. In a group of free electrons metallic ions are made up from lattice and to conduct the electricity the free electron are the main reason for ability of the metals.
On the other hand, semiconductors can be arranged as structure of lattice and also there is a covalent bonds are present.
When designing solid rockets, thrust and mass flow must be considered time dependent. a) True b) False
Answer:
the answer is true when designing sold rockets thrust and mass flow
A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a polytropic process during which PV^1.49 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process.
Answer:-0.4199 J/k
Explanation:
Given data
mass of nitrogen(m)=1.329 Kg
Initial pressure[tex]\left ( P_1\right )[/tex]=120KPa
Initial temperature[tex]\left ( T_1\right )=27\degree \approx[/tex] 300k
Final volume is half of initial
R=particular gas constant
Therefore initial volume of gas is given by
PV=mRT
V=0.986\times 10^{-3}
Using [tex]PV^{1.49}[/tex]=constant
[tex]P_{1}V^{1.49}[/tex]=[tex]P_2\left (\frac{V}{2}\right )[/tex]
[tex]P_2[/tex]=337.066KPa
[tex]V_2[/tex]=[tex]0.493\times 10^{-3} m^{3}[/tex]
and entropy is given by
[tex]\Delta s[/tex]=[tex]C_v \ln \left (\frac{P_2}{P_1}\right )[/tex]+[tex]C_p \ln \left (\frac{V_2}{V_1}\right )[/tex]
Where, [tex]C_v[/tex]=[tex]\frac{R}{\gamma-1}[/tex]=0.6059
[tex]C_p[/tex]=[tex]\frac{\gamma R}{\gamma -1}[/tex]=0.9027
Substituting values we get
[tex]\Delta s[/tex]=[tex]0.6059\times\ln \left (\frac{337.066}{120}\right )[/tex]+[tex]0.9027 \ln \left (\frac{1}{2}\right )[/tex]
[tex]\Delta s[/tex]=-0.4199 J/k
The melting point of W (tungsten) is 3380°C, is the processing at 1100 C hot working or cold working?
Answer:
cold working
Explanation:
Given data in question
melting point tungsten (W) = 3380°C = 3653 K
processing temperature = 1100°C = 1373 K
To find out
process is hot or cold working
solution
we know hot working and cold working process is depend upon the Recrystallization and Recrystallization is a process in which deformed grains of crystal structures are replace with stress-free grains that nucleate and grow till actual grains have been consumed fully
and we know that ratio of processing temperature and melting point tungsten is greater than 60% than the process is start of Recrystallization so we check ratio
ratio = processing temperature / melting point tungsten
ratio = 1373 / 3653
ratio = 0.3758 = 37.58 %
we can see this is less than 60 % so our process is cold working
A dielectric is an insulating material or a very poor conductor of electric current. (True , False )
True.
Dielectric is a material with a low electrical conductivity (σ << 1); that is, an insulator, which has the property of forming electric dipoles inside it under the action of an electric field. Thus, all dielectric materials are insulators but not all insulating materials are dielectric.
Some examples of this type of materials are glass, ceramics, rubber, mica, wax, paper, dry wood, porcelain, some fats for industrial and electronic use and bakelite. As for the gases, the air, nitrogen and sulfur hexafluoride are used as dielectrics.
Just-in-time delivery is not an important component of Lean manufacturing. a)- True b)- False
Answer:
False
Explanation:
Just-in-time manufacturing means that production is done when there is demand. It is a pull system which means that demand dictates production. Lean manufacturing is method of production which aims to reduce the seven types of waste which are
Logistics
Inventory
Motion of objects during manufacturing
Unnecessary delay
Producing more than demand
Over Processing of goods
Malfunctioning items.
Here it can be seen that Producing more than demand is a waste. This can be reduced by Just in Time manufacturing.
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the piston until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.
Answer:
n=2.32
w= -213.9 KW
Explanation:
[tex]V_1=0.3m^3,T_1=298 K[/tex]
[tex]V_2=0.1m^3,T_1=1273 K[/tex]
Mass of air=1 kg
For polytropic process [tex]pv^n=C[/tex] ,n is the polytropic constant.
[tex]Tv^{n-1}=C[/tex]
[tex]T_1v^{n-1}_1=T_2v^{n-1}_2[/tex]
[tex]298\times .3^{n-1}_1=1273\times .1^{n-1}_2[/tex]
n=2.32
Work in polytropic process given as
w=[tex]\dfrac{P_1V_1-P_2V_2}{n-1}[/tex]
w=[tex]mR\dfrac{T_1-T_2}{n-1}[/tex]
Now by putting the values
w=[tex]1\times 0.287\dfrac{289-1273}{2.32-1}[/tex]
w= -213.9 KW
Negative sign indicates that work is given to the system or work is done on the system.
For T_V diagram
We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.
Define the Static Balancing.
Answer: Static Balancing is when a stationary object has the the ability to balance because its core centre of gravity is on the axis of rotation.
A flywheel accelerates for 5 seconds at 2 rad/s2 from a speed of 20 rpm. Determine the total number of revolutions of the flywheel during the period of its acceleration. a.5.65 b.8.43 c. 723 d.6.86
Answer:
option (a)
Explanation:
t = 5 sec, α = 2 rad/s^2, f0 = 20 rpm = 20 / 60 rps
Use second equation of motion for rotational motion
θ = ω0 x t + 1/2 α t^2
θ = 2 x 3.14 x 5 x 20 / 60 + 0.5 x 2 x 5 x 5
θ = 10.47 + 25 = 35.47 rad
Number of revolution = 35.47 / (2 x 3.14) = 5.65
A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027). Determine the terminal speed.
Given:
diameter of sphere, d = 6 inches
radius of sphere, r = [tex]\frac{d}{2}[/tex] = 3 inches
density, [tex]\rho}[/tex] = 493 lbm/ [tex]ft^{3}[/tex]
S.G = 1.0027
g = 9.8 m/ [tex]m^{2}[/tex] = 386.22 inch/ [tex]s^{2}[/tex]
Solution:
Using the formula for terminal velocity,
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2V\rho g}{A \rho C_{d}}}[/tex] (1)
[tex](Since, m = V\times \rho)[/tex]
where,
V = volume of sphere
[tex]C_{d}[/tex] = drag coefficient
Now,
Surface area of sphere, A = [tex]4\pi r^{2}[/tex]
Volume of sphere, V = [tex]\frac{4}{3} \pi r^{3}[/tex]
Using the above formulae in eqn (1):
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}[/tex]
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2gr}{3C_{d}}}[/tex]
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}[/tex]
Therefore, terminal velcity is given by:
[tex]v_{T}[/tex] = [tex]\frac{27.79}{\sqrt{C_d}}[/tex] inch/sec
With increases in magnification, which of the following occur? a. The field of view decreases. b. The ambient illumination decreases. c. The larger parts can be measured. d. The eyepiece must be raised.
By increasing magnification you decrease the field of view.
The answer is A.
Hope this helps.
r3t40
The interactions between a closed system and its surroundings include energy transfer by heat, boundary work and flow work. a)True b) False
Answer:
b) False
Explanation:
In close system only energy transfer take place and mass transfer is zero.But on the other hand in open system energy as well mass transfer take place.
Energy transfer means work as well as heat transfer.But we know that in close system there is no any flow of mass so there will not be any flow work,only boundary work will associated with close system.But in open system flow work take place.
The two windings of transformer is: a)- Conductively linked. b)- Not linked at all. c)- Inductively linked d)- Electrically linked.
The two windings of transformer is c)- Inductively linked
Hope this helped!
Answer:
The two windings of transformer is Inductively linked -c)
At winter design conditions, a house is projected to lose heat at a rate of 60,000 Btu/h. The internal heat gairn from people, lights, and appliances is estimated to be 6000 Btuh Ifthis house is to be heated by electric resistance heaters, determine the required rated power of these heaters in kW to maintain the house at constant temperature.
Answer:
15.8529 kW
Explanation:
Rate of heat loss = 60000 Btu/h
Internal heat gain = 6000 Btu/h
Rate of heat required to be supplied
[tex]P_{Sup}=\text{Rate of heat loss}-\text{Internal heat gain}\\\Rightarrow P_{Sup}=60000-6000\\\Rightarrow P_{Sup}=54000\ Btu/h[/tex]
Converting 54000 Btu/h to kW (kJ/s)
1 Btu = 1.05506 kJ
1 h = 3600 s
[tex]P_{Sup}=54000\times \frac{1.05506}{3600}\\\Rightarrow P_{Sup}=15.8529\ kW[/tex]
∴ Required rated power of these heaters is 15.8529 kW
Answer:
Q = 15.8 kW
Explanation:
Given data:
Heat loss rate is 60,000 Btu/h
Heat gain is 6000 Btu/h
Rate of heat required is computed as
Q = (60000 - 6000) Btu/h
Q = 54000 Btu/h
change Btu/h to Kilo Watts
[tex]Q = 54000 Btu/h (\frac{1W}{3.412142\ Btu/h})[/tex]
[tex]Q = 15825.8 W(\frac{1 kW}{1000 W})[/tex]
Q = 15.8 kW