Answer:
93.8 sec
Explanation:
it is given that tab has 350 gallon
we know that 1 gallon = 0.134 cubic foot
350 gallon = 350×0.134=46.9 cubic foot
the delivery pressure is 100 psi which is greater than 80 psi to operate machine
it is given that supply volume is 30 cubic foot per minute
= [tex]\frac{30}{60}=0.5[/tex] [tex]ft^{3}/sec[/tex]
[tex]time\ required\ =\frac{tab\ air }{supply\ volume}[/tex]
[tex]time\ required\= [tex]\frac{46.9}{0.5}[/tex]
=93.8 sec
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
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] .
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.
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.
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:
Radiation heat transfer occurs from any object that is above 0K. a) True b) False
Answer:
True, hope this helps but there no school right now its summer
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
Saturated water vapor at 140°C is compressed in a reversible, steady-flow device to 895 kPa while its specific volume remains constant. Determine the work required.
Answer:
The work required to compress the saturated water vapor to 895 kPa pressure is 130.9540 k J/Kg
Explanation:
Given data in question
temperature = 140°C
pressure (P2) = 895 kPa
To find out
work required for compress saturated water
Solution
We know the equation for reversible work for compress saturated water vapor
i.e.
W = [tex]-\int_{1}^{2}vdP-\Delta ke - \Delta pe[/tex]
w is reversible work, v is specific volume, P is water vapor pressure and
ke is kinetic energy and pe is potential energy
and in question we have given v is constant so ke and pe will be zero
so
W = [tex]-\int_{1}^{2}vdP[/tex]
W = -v( P2 - P1 )
we can given in question temperature = 140°C and use steam table "A-4 saturated water - temperature table"
at this water property P1 will be 361.53 kPa and v will be 0.50850 m³/kg
so put these value in above equation
W = -0.50850( 104 - 361.53 )
W = 130.9540 kJ/Kg
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
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 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 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
Which of the following is a correct formula of Ohm s Law (a) E= R/I (b) E=1+R (c)E=I/R (d) E= IR
Answer: The correct answer is Option d.
Explanation:
Ohm's law is defined as the law which gives us the relationship between voltage and current.
In electronics, the equation used to represent this law is:
[tex]E=IR[/tex]
where,
E = voltage of the circuit. The unit for this is Volts.
I = current of the circuit. The unit for this is Amperes
R = resistance of the circuit. The unit for this is Ohms.
Hence, the correct answer is Option d.
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]
What is the thermal efficiency of this reheat cycle in terms of enthalpies?
Answer:
[tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]
Explanation:
For close gas turbine:
Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.
Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.
In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,
We know that [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]
Now take [tex]h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6}[/tex] represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.
So total heat supplied [tex]Q_S[/tex]=
[tex]\left (h_3-h_2\right )+\left (h_5-h_4\right )[/tex]
Net work out put
[tex]W_{net}[/tex]=[tex]\left (h_5-h_6\right )-\left (h_2-h_1\right )[/tex]
So efficiency [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]
[tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]
What different between 'flow analysis using control volume method' and 'flow analysis using differential method'?
Answer:
control volume
control volume is used to determine the flow characteristics of complex shape like turbine and compressorsdifferential approach
it is carried out by considering infintely small region for fluid analysis.Explanation:
control volume:
control volume is used to determine the flow characteristics of complex shape like turbine and compressorsit is used to determine the flow velocity within in the boundaries of control volume. it can also used for force analysis for flow. one main disadvantage of control volume is that it doesn't provide detail information about stress and pressure variation.differential approach:
it is carried out by considering infintely small region for fluid analysis.solution of the fluid analysis is in the form of differential equationit provide detail information about the flow.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
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.
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.
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
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.
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]
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
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
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
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''
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.
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.
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.
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]