Answer:
True.
Explanation:
According the engineering flow they don not possess flow energy when they are in rest.
When they are in motion they show a translation energy.
The features if fluids may be different according the variables of pressure and temperature.
A long homogeneous resistance wire of radius ro = 5 mm is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of g=5'107 W/m as a result of resistance heating. If the temperature of the outer surface of the wire remains at 180°C, determine the temperature at r = 2 mm after steady operation conditions are reached. Take the thermal conductivity of the wire to be k = 8 W/m x °C.
Answer:
T = 212.8125°C
Explanation:
Given
radius of the wire, [tex]r_{0}[/tex] = 5 mm 0.005 m
heat generated, g = 5 x [tex]10^{7}[/tex] W/[tex]m^{3}[/tex]
outer surface temperature, [tex]T_{S}[/tex] = 180°C
Thermal conductivity, k = 8 W / m-k
Now maximum temperature occurs at the center of the wire
that is at r=0,
Therefore, [tex]T_{o}=T_{S}+\frac{g\times r_{o}^{2}}{4\times k}[/tex]
[tex]T_{o}=180+\frac{5\times 10^{7}\times 0.005^{2}}{4\times 8}[/tex]
[tex]T_{o}=219.0625[/tex]°C
Therefore, temperature at r = 2 mm
[tex]\frac{T-T_{S}}{T_{O}-T_{S}}= 1-\left (\frac{r}{r_{O}} \right )^{2}[/tex]
[tex]\frac{T-180}{219.0625-180}= 1-\left (\frac{2}{5} \right )^{2}[/tex]
Therefore, T = 212.8125°C
The velocity of flow over a flat plate is doubled. Assuming the flow remains laminar over the entire plate, what is the ratio of the new thermal boundary layer thickness to the original boundary layer thickness?
Answer:
Given:
laminar flow
and since velocity of flow is doubled, we consider [tex]v_{n}[/tex] as new velocity and [tex]v_{o}[/tex] as original velocity
Explanation:
As per laminar flow, thickness, t is given by
t = [tex]\frac{4.91x}{\sqrt( R_{ex}) }[/tex]
t = [tex]\frac{4.91x}{\sqrt{\frac{\rho vx}{\mu }}}[/tex]
t = [tex]\frac{4.91x\mu }{\sqrt{\rho vx}}[/tex]
where,
[tex]R_{ex}[/tex] = Reynold's no.
therefore,
t ∝ [tex]\frac{1}{\sqrt{v} }[/tex]
Now,
[tex]\frac{t_{n} }{t_{o} }[/tex] = [tex]\sqrt{(\frac{v_{o} }{v_{n} })}[/tex]
[tex]\frac{t_{n} }{t_{o} }[/tex] = [tex]\sqrt{(\frac{v_{o} }{2v_{o} } )} =\frac{1}{\sqrt{2} }[/tex]
therefore,
[tex]t_{n}:t_{o} = 1:\sqrt{2}[/tex]
What is (a) body forces (b) surface forces?
Answer:
A) Body forces-
Body forces is the forces which acts throughout the volume of the body. It is basically distributed over the volume and mass of the element of the body. In a body force other body exerts a force without being contract.
For example : Gravity forces, electromagnetic forces, centrifugal forces.
B) Surface forces-
Surface forces is the forces which are distributed all over the free surface of the body. Surfaces forces can be further divided into two perpendicular components as: normal forces and shear forces.
For example : Pressure forces and viscous forces.
A metal rod, 20 mm diameter, is tested in tension (force applied axially). The total extension over a length of 80 mm is 3.04 x 102 mm for a pull of 25 kN. Calculate the normal stress, normal strain and modulus of elasticity (Young's modulus), assuming the rod is linear elastic over the load range.
Answer:stress=79.56MPa
strain=[tex]3.8\times 10^{-4}[/tex]
Young Modulus=209.36 GPa
Explanation:
Given data
d=20 mm
Length[tex]\left ( L\right )[/tex]=80mm
[tex]\Delta {L}[/tex]=[tex]3.04\times 10^{-2}[/tex]mm
Load=[tex]25\times 10^{3}[/tex]N
[tex]\left ( i\right )[/tex]
Stress=[tex]\frac{Load\ applied}{cross-section}[/tex]
Stress=[tex]\frac{25\times 10^{3}}{314.2}[/tex]
Stress=79.56MPa
[tex]\left ( ii\right )[/tex]
Strain=[tex]\frac{Change\ in\ length}{Length}[/tex]
Strain=[tex]\frac{3.04\times 10^{-3}}{80}[/tex]
Strain=[tex]3.8\times 10^{-4}[/tex]
[tex]\left ( iii\right )[/tex]
young modulus[tex]\left ( E\right )[/tex]=[tex]\frac{Stress}{Strain}[/tex]
E=[tex]\frac{79.56\times 10^{6}}{3.8\times 10^{-4}}[/tex]
E=209.36GPa
Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 seconds.
Answer:
work done= 2.12 kJ
Explanation:
Given
N=2500 rpm
T=4.5 N.m
Period ,t= 30 s
[tex]torque =\frac{power}{2\pi N}[/tex]
[tex]power=2\pi N\times T[/tex]
P=[tex]2\times \pi \times2500 \times 4.5[/tex]
P=70,685W
P=70.685 KW
power=[tex]\frac{work done}{time}[/tex]
work done = power * time
= 70.685*30=2120.55J
= 2.12 kJ
Convection is a function of temperature to the fourth power. a)-True b)-False
Answer:
The given statement for temperature and convection is False.
Explanation:
Convection is not a function of temperature to the fourth power but it depends linearly on temperature. the below equation shows the linear relation of heat transfer due to convection and temperature:
Q = [tex]H_{c}A(T_{hot} - T_{cold} )[/tex]
Whereas, radiation is a function of temperature to the fourth power.
The Stefan-Boltzmann law gives the relationship between an object's temperature and the amount of radiation it emits. The law is given by:
[tex]Q=\sigma T^{4}[/tex]
The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False
Answer:
(b)False
Explanation:
[tex]I_{xy}[/tex] defined as
[tex]I_{xy}[/tex] =[tex]\int \left (x\cdot y\right )dA[/tex]
Where x is the distance from centroidal x-axis
y is the distance from centroidal y-axis
dA is the elemental area.
The product of x and y can be positive or negative ,so the value of [tex]I_{xy}[/tex] can be positive as well as negative .
So from the above expressions we can say that the product of [tex]I_{x},I_y[/tex] is different from [tex]I_{xy}[/tex] .
Refectories are one of the types of ceramics that have low melting temperature. a)-True b)-False
Answer:
b). False
Explanation:
A refractory material is a type of material that can withstand high temperatures without loosing its strength. They are used in reactors, furnaces, kilns, etc.
Refractory materials are certain super alloys and ceramics materials.
Properties of refractory materials :
1. Refractory materials have high melting point.
2.They acts barriers between high heat zone and low heat zone.
3. The specific heat of refractory material is very low.
4. Refractories that have high bulk densities are better in quality.
Hence, Refractory materials have a very high melting temperature.
Shear strain can be expressed in units of either degrees or radians. a)True b)- False
Answer:
true
Explanation:
shear strain is define as the ratio of change in deformation to the original length perpendicular to the axes of member due to shear stress.
ε = deformation/original length
strain is a unit less quantity but shear stain is generally expressed in radians but it can also be expressed in degree.
Which of the following is/are FALSE about refining aluminum from the ore state (mark all that apply) a)- A blast furnace is used b)-The ore is called bauxite c)-The process uses a lot of electricity d)-Coke is used to produce the heat
Answer:
The options a)- A blast furnace is used and d)-Coke is used to produce the heat are FALSE.
Explanation:
Aluminium is a chemical element and the most abundant metal present in the Earth's crust. An aluminium ore is called bauxite. Aluminium is extracted from its ore by the process of electrolysis, called the Hall–Héroult process. The extraction of aluminium is an expensive process as it requires large amount of electricity. The bauxite is purified to produce aluminium oxide. Then, aluminium is extracted from the aluminium oxide.
Therefore, the refining of aluminum from its ore does not involve the use of a blast furnace and coke to produce heat.
Describe how the Rotary Engine works.
Answer:
Rotary engine was early known by the name of internal combustion engine. It convert heat from a high pressure of combustion. The main advantage of rotary engine is that it can be operate with less number of vibration. It works on the principle of converting pressure into rotating motion. In rotary engine the expansion pressure is applied on the flank rotor.
Answer: The rotary engine works on the same basic principle as the piston engine: combustion in the power plant releases energy to power the vehicle. However, the delivery system in the rotary engine is wholly unique. The piston engine performs four key operations: intake, compression, combustion, and exhaust.
Explanation:
Briefly describe the function of the thermostatic expansion valve in a vapour compression refrigeration system
Answer:
Explanation:
Thermostatic expansion valve is mainly a throttling device commonly used in air conditioning systems and refrigerators.
It is an automatic valve that maintains proper flow of refrigerant in the evaporator according to the load inside the evaporator. When the load in the evaporator is higher the valve opens and allows the increase in flow of refrigerant and when the load reduces the valve closes a bit and reduces the flow of refrigerant. This process leads to higher efficiency of compressor as well as the whole refrigeration system. Thus TEV works to reduce the pressure of refrigerant from higher condenser pressure to the lower evaporator pressure. It also keeps the evaporator active.
A solar panel measures 80 cm by 50 cm. In direct sunlight, the panel delivers 4.8 A at 15 V. If the intensity of sunlight is 1 000 W/m2, what is the efficiency of the solar panel in converting solar energy into electrical energy?
Answer:
The solar panel efficiency in converting solar energy in electrical energy is 18%.
Explanation:
The solar panel's efficiency can be defined as:
[tex] n= \frac{Pe}{Ps}*100\%[/tex]
Where Pe and Ps are the output electrical power and input solar power respectively. The electrical is computing in terms of the voltage and current delivered:
[tex]Pe=I.V[/tex]
[tex]Pe=4.8 A * 15 V[/tex]
[tex]Pe= 72 AV = 72 W [/tex]
The net solar power of panel is found by multiplying the solar intensity by the panel area in square meters:
[tex]Ps = Is.Ap[/tex]
[tex]Ps = 1000 W/m^2 *(0.8 m * 0.5 m)[/tex]
[tex]Ps = 1000 W/m^2 *(0.40 m^2)[/tex]
[tex]Ps = 400 W [/tex]
Finally the panel efficience n is:
[tex] n= \frac{72 W}{400 W}*100\%[/tex]
[tex] n= 0.18*100\% = 18\%[/tex]
_______On what basis composites are classified a)- shape of dispersed phase b)-matrix materials c)-chemistry of dispersed phase d)-a & b
Answer: d) a & b
Explanation: Composite materials are made up of two or more different types of phases which include dispersed phase and matrix phase as most important phases.
Matrix phase is a types of continuous phase which is responsible for holding of the dispersed phase.It shows good property of ductility.Dispersed phase is also known as the secondary phase which is harder in nature than matrix phase.Water flovs in a pipe of diameter 150 mm. The velocity of the water is measured at a certain spot which reflects the average flow velocity. A pitot static tube has a meter coefficient of C = 1,05 and is joined to a mercury manometer indicating a reading of 167 mm. Determine the flow rate of the water.
Answer:
Q = 0.118 [tex]m^{3}[/tex]/s
Explanation:
Given :
diameter of the pipe, d = 150 mm
= 0.15 m
Pitot tube co efficient, [tex]C_{v}[/tex] = 1.05
manometer reading is given, x = 167 mm
= 0.167 m
From manometer reading,we can find the difference between the manometer height, h
[tex]h =x\times\left [ \frac{S_{m}}{S_{w}}-1 \right ][/tex]
[tex]h =0.167\times\left [ \frac{13.6}{1}-1 \right ][/tex]
h = 2.1042 m
Now, average velocity is v = [tex]C_{v}[/tex][tex]\sqrt{2.g.h}[/tex]
= [tex]1.05\times \sqrt{2\times 9.81\times 2.1042}[/tex]
= 6.74 m/s
Area of the pipe, A = [tex]\frac{\pi }{4}\times d^{2}[/tex]
= [tex]\frac{\pi }{4}\times 0.15^{2}[/tex]
= 0.0176 [tex]m^{2}[/tex]
Therefore, flow rate is given by, Q = A.v
= 0.0176 X 6.74
= 0.118[tex]m^{3}[/tex]/s
A strip ofmetal is originally 1.2m long. Itis stretched in three steps: first to a length of 1.6m, then to 2.2 m, and finally to 2.5 m. Compute the true strain after each step, and the true strain for the entire process (i.e. for stretching from 1.2 m to 2.5 m).
Answer:
strains for the respective cases are
0.287
0.318
0.127
and for the entire process 0.733
Explanation:
The formula for the true strain is given as:
[tex]\epsilon =\ln \frac{l}{l_{o}}[/tex]
Where
[tex]\epsilon =[/tex] True strain
l= length of the member after deformation
[tex]l_{o} = [/tex] original length of the member
Now for the first case we have
l= 1.6m
[tex]l_{o} = 1.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{1.6}{1.2}[/tex]
[tex]\epsilon =0.287[/tex]
similarly for the second case we have
l= 2.2m
[tex]l_{o} = 1.6m[/tex] (as the length is changing from 1.6m in this case)
thus,
[tex]\epsilon =\ln \frac{2.2}{1.6}[/tex]
[tex]\epsilon =0.318[/tex]
Now for the third case
l= 2.5m
[tex]l_{o} = 2.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{2.5}{2.2}[/tex]
[tex]\epsilon =0.127[/tex]
Now the true strain for the entire process
l=2.5m
[tex]l_{o} = 1.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{2.5}{1.2}[/tex]
[tex]\epsilon =0.733[/tex]
A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air compressor to the drill has a 40 mm diameter and is smooth. The maximum compressor discharge gauge pressure is 690 kPa. Neglect changes in air density and any effects of hose curvature. Air leaves the compressor at 40° C. What is the longest hose that can be used?
Answer:
L = 46.35 m
Explanation:
GIVEN DATA
\dot m = 0.25 kg/s
D = 40 mm
P_1 = 690 kPa
P_2 = 650 kPa
T_1 = 40° = 313 K
head loss equation
[tex][\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m[/tex]
where[tex] h_l = \frac{ flv^2}{2D}[/tex]
[tex]h_m minor loss [/tex]
density is constant
[tex]v_1 = v_2[/tex]
head is same so,[tex] z_1 = z_2 [/tex]
curvature is constant so[tex] \alpha = constant[/tex]
neglecting minor losses
[tex]\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}[/tex]
we know[tex] \dot m[/tex] is given as[tex] = \rho VA[/tex]
[tex]\rho =\frac{P_1}{RT_1}[/tex]
[tex]\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3[/tex]
therefore
[tex]v = \frac{\dot m}{\rho A}[/tex]
[tex]V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}[/tex]
V = 25.90 m/s
[tex]Re = \frac{\rho VD}{\mu}[/tex]
for T = 40 Degree, [tex]\mu = 1.91*10^{-5}[/tex]
[tex]Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}[/tex]
Re = 4.16*10^5 > 2300 therefore turbulent flow
for Re =4.16*10^5 , f = 0.0134
Therefore
[tex]\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}[/tex]
[tex]L = \frac{(P_1-P_2) 2D}{\rho f v^2}[/tex]
[tex]L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}[/tex]
L = 46.35 m
In a photonic material, signal transmission occurs by which of the following? a)- Electrons b)- Photons
B. Photons.
In a photonic material, signal transmission occurs by photons which are light particles.
In a vapour absorption refrigeration system, the compressor of the vapour compression system is replaced by a a)- absorber, generator and liquid pump. b)-absorber and generator. c)- liquid pump. d)-generator.
Answer:
a). absorber, generator and liquid pump
Explanation:
The Vapour absorption system consists of compression, expansion, condensation and evapouration processes. This system uses ammonia, lithium bromide or water as refrigerant.
An absorber, pump and generator is used in place of a compressor in the vapour compression refrigeration system. The operation is smooth in vapour absorption system since all the moving elements are in the pump only. This system make use of low energy like heat and can work on lower evapourator pressure. It has low Coefficient of performance.
Convert 30.12345 degrees into degrees, minutes and seconds.
Answer:
30.12345° can be written as : 30°7'20.42''
Or,
30 degrees 7 minutes and 20.42 seconds.
Explanation:
1 degree consists of 60 arc minutes.
1 arc minutes consists of 60 arc seconds.
Thus, 30.12345° can be written as:
30.12345°= 30° + 0.12345°
1° = 60'
So,
0.12345° = 0.12345*60' = 7.407'
Thus, 7.407' can be written as:
7.407' = 7' + 0.407'
1' = 60''
So,
0.407' = 0.407*60'' = 20.42''
Thus,
30.12345° can be written as : 30°7'20.42''
In general, this of the following methods yields the most conservative fatigue strength proof (a) Saderberg method (b)-Goodman method (c)-Gerber line (d)-The ASME elliptic curve.
Answer:
a). Soderberg method
Explanation:
A straight line joining the endurance limit, [tex]S_{e}[/tex] on the ordinate and to the yield strength,[tex]S_{yt}[/tex] on the abscissa is know as Soderberg line.
The Soderberg line is the most conservative failure criteria and in this there is no need to consider yielding point in this case.
The equation for Soderberg is given by
[tex]\frac{\sigma _{m}}{S_{yt}}+\frac{\sigma _{a}}{S_{e}}=1[/tex]
where [tex]\sigma _{m}[/tex] is mean stress
[tex]\sigma _{a}[/tex] is amplitude stress
A typical aircraft fuselage structure would be capable of carrying torsion moment. a)True b)- False
Answer:
True
Explanation:
An aircraft is subject to 3 primary rotations
1) About longitudinal axis known as rolling
2) About lateral axis known as known as pitching
3) About the vertical axis known as yawing
The rolling of the aircraft induces torsion in the body of the aircraft thus the fuselage structure should be capable of carrying torsion
Calculate the change of entropy of 2 kg of air when its temperature increases from 400 K to 500 K at constant pressure equal to 300 kPa.
Answer:
0.45516
Explanation:
ENTROPY : Entropy is a measure of molecular disorder it is denoted by S. Entropy is also measured in terms of thermal energy and temperature it is equal to thermal energy per unit temperature.
from the table S₁=1.99194 KJ/kg.k (at 400k)
from the table S₂=2.21952 KJ/kg.k (at 500k)
so total entropy change is given by =m (S₂-S₁)
=2(2.21952-1.99194)
=0.45516
Crystal lattice can be characterized with a) angle, geometry and coordination number b) the color, size, lattice type; c) hardness, geometry d) atomic size, plasticity
Answer: a) angle, geometry and coordination number
Explanation: Crystal lattice is described as arrangement of groups of atoms inside three dimensional structure of crystal. It has a particular geometry in which atoms are placed in a symmetry. The also have angle in which placement of atoms are done. Co-ordination number also determines the crystal lattice by counting the atoms with which it is bonded. Thus, option (a) is the correct option.
In a flow over a flat plate, the Stanton number is 0.005: What is the approximate friction factor for this flow a)- 0.01 b)- 0.02 c)- 0.001 d)- 0.1
Answer:
(a) .01
Explanation:
stanton number is a dimensionless quantity stanton is expressed as [tex]\frac{heat transer}{thermal capacity}[/tex]stanton number is discovered by Thomas edward stantonthere is relation between friction factor and stanton number and friction factor that is stanton number is half of friction factor
stanton number =[tex]\frac{friction factor}{2}[/tex]
.005=[tex]\frac{friction factor}{2}[/tex]
friction factor =2×.005
friction factor=.01
Critical Reynolds number for internal turbulent flow is 500,000. a) True b) False
Answer:
FALSE
Explanation:
REYNOLDS NUMBER :Reynolds number is used to indicate whether the fluid flow past a body or turbulent. it is a dimensionless number
REYNOLDS NUMBER OF A INTERNAL TURBULENT FLOW: For a flow in a pipe experimental observation show that the critical reynolds number is about 2300 for the practical purpose . so the reynolds number can not be so high as 500000
A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)
Answer:
Force on the bolt = 24.525 kN
Force on the 1st hinge = 8.35 kN
Force on the 2nd hinge = 16.17 kN
Explanation:
Given:
height = 2 m
width =1 m
depth of the door from the water surface = 1.5 m
Therefore,
[tex]\bar{y}[/tex] =1.5+1 = 2.5 m
Hydrostatic force acting on the door is
[tex]F= \rho \times g\times \bar{y}\times A[/tex]
[tex]F= 1000 \times 9.81\times 2.5\times 2\times 1[/tex]
= 49050 N
= 49.05 kN
Now finding the Moment of inertia of the door about x axis
[tex]I_{xx}=\frac{1}{12}\times b\times h^{3}[/tex]
[tex]I_{xx}=\frac{1}{12}\times1\times 2^{3}[/tex]
= 0.67
Now location of force, [tex]y^{*}[/tex]
[tex]y^{*}=\bar{y}+\frac{I_{xx}}{A\times \bar{y}}[/tex]
[tex]y^{*}=2.5+\frac{0.67}{2\times 1\times 2.5}[/tex]
= 2.634
Therefore, calculating the unknown forces
[tex]F=F_{A}+R_{B}+R_{C} = 49.05[/tex] ------------------(1)
Now since [tex]\sum M_{R_{A}}=0[/tex]
∴ [tex]R_{B}\times L+R_{C}\times L-F\times \frac{1}{2}=0[/tex]
[tex]R_{B}+R_{C}-F\times \frac{1}{2}=0[/tex]
[tex]R_{B}+R_{C}=\frac{F}{2}[/tex]
[tex]R_{B}+R_{C}=24.525[/tex] -----------------------(2)
From (1) and (2), we get
[tex]R_{A} = 49.05-24.525[/tex]
= 24.525 kN
This is the force on the Sliding bolt
Taking [tex]\sum M_{R_{C}}=0[/tex]
[tex]F\times 0.706-R_{A}\times 0.84-R_{B}\times 1.68 = 0[/tex]
[tex]49.05\times 0.706-24.525\times 0.84-R_{B}\times 1.68 = 0[/tex]
[tex]R_{B}[/tex] =8.35 kN
This is the reaction force on the 1st hinge.
Now from (1), we get
[tex]R_{C}[/tex] =16.17 kN
This is the force on the 2nd hinge.
A pipe which is on a slope, transports water downwards. A doubling of cross sectional area takes place 6 above the reference level. The pressure in the smaller pipe, just before the enlargement, is 860 kPa. The flow velocity in the large pipe is 2,4 m/s. Determine the pressure in kPa at a point 1,5 m above the reference level. Ignore friction losses.
Answer:
P₂ = 830.75 kPa
Explanation:
Given:
Pressure in the smaller pipe,P₁ = 860 kPa
Velocity in the larger pipe, v₂ = 2.4 m/s
Therefore velocity in the smaller pipe, v₁ = 4.8 m/s ( velocity gets doubled since area is reduced to half )
Height at section where the area is doubled, z₁ = 6 m
Height at the section where pressure is to be calculated, z₂ = 1.5 m
Now apply Bernouli Equation between the section of enlargement and at section where pressure is to be calculated,
[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}[/tex]
[tex]\frac{860}{1000 \times 9.81}+\frac{4.8^{2}}{2\times 9.81}+6 = \frac{P_{2}}{1000 \times 9.81}+\frac{2.4^{2}}{2\times 9.81}+1.5[/tex]
P₂ = 830.75 kPa
Therefore, pressure at the section 1.5 m above datum is 830.75 kPa
Explain the following terms; i.Water content in air ii. Relative humidity iii. Enthalpy
Answer:
Explanation:
WATER CONTENT IN AIR-the water content of the air varies from place to place and from time to time because water content in air is dependent on temperature if temperature is change then water content also change water exist in air as a solid liquid and gas
RELATIVE HUMIDITY-Relative humidity is the ratio of partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature relative humidity depends on temperature and pressure of the system
Enthalpy-when a substance changes at constant pressure enthalpy tells how much heat and work was added or removed from the substance
enthalpy is equal to the sum of system internal energy and product of its pressure and volume.it is denoted H
It is true about polymers: a)-They are light-weight materials b)-There are three general classes: thermosets, thermoplastics and thermoset c)-They present long term instability under load d)-All the above
Answer: d) All of the above
Explanation: Polymers are the substances that have molecular structure with having same bonds in the entire molecule together.There are light weight substance which occur natural as well as artificially. They are also categorized as thermoplastics ,thermosets, and elastomers. They also have the property of being stretching and bending under the pressure or load they are also instable. Therefore, all the options are correct statement about polymers.