Explanation:
We know that first T-ds equation
Tds=[tex]C_v[/tex]dT+Pdv
For constant volume process dv=0
⇒Tds=[tex]C_v[/tex]dT
So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_v}[/tex] ----(1)
We know that second T-ds equation
Tds=[tex]C_p[/tex]dT-vdP
For constant pressure process dP=0
⇒Tds=[tex]C_p[/tex]dT
So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_p}[/tex] ---(2)
We know that [tex]C_p[/tex] is always greater than [tex]C_v[/tex].
So from equation (1) and (2) we can say that slope of constant volume process will be always greater than constant pressure process on T-s diagram.
Fluid enters a device at 4 m/s and leaves it at 2 m/s. If there is no change in the PE of tihe flow, and there is no heat and (non-flow) work across boundaries of the device, what is the increase in specific enthalpyof the fluid (hg-hi) in kJ/kg? Assume steady state operation of the device.
Answer:
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
Explanation:
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here given Q=0 ,w=0
So [tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]
[tex]V_1=4 m/s,V_2=2 m/s[/tex]
[tex]h_1+\dfrac{4^2}{2000}=h_2+\dfrac{2^2}{2000}[/tex]
[tex]h_2-h_1=6\times 10^{-3}[/tex]
So increase in specific enthalpy
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
An oscillating mechanism has a maximum displacement of 3.2m and a frequency of 50Hz. At timet-0 the displacement is 150cm. Express the displacement in the general form Asin(wt + α).
Given:
max displacement, A = 3.2 m
f= 50 Hz
at t = 0, displacement, d = 150 cm = 1.5 m
Solution:
Displacement in the general form is represented by:
d = Asin(ωt ± α)
d = 3.2sin(2πft ± α)
d = 3.2sin(100πt ± α)
where,
A = 3.2 m,
ω = 2πf = 100π
Now,
at t = 0,
1.5 = 3.2sin(100π(0) ± α )
1.5 = 3.2sinα
sin α = [tex]\frac{1.5}{3.2}[/tex] = 0.4687
α = [tex]sin^{-1}(0.46875)[/tex] = 27.95° = 0.488 radian
Now, we can express displacement in the form of 'Asin(wt + α)' as:
d = 3.2sin(100πt ± 0.488 )
What is Potential flow?
Answer: Potential flow states about the velocity field considering it as gradient of scalar function such as the velocity potential. It is considered same as irrotational flow.
Explanation:In the study of fluid dynamics , potential flow is stated as the term for describing about the velocity field taking it as the gradient of a scalar function. Therefore, potential flow is stated by an irrotational velocity field and there is a irrotation because of the curl of the gradient of scalar quantity is zero.Potential flow is considered same as the irrotational flow that is the particles in the fluids that do not rotate.
Why factor of safety is more than 2 in the gears ? What does effect ?
Answer:
explained
Explanation:
Gear is a mechanical components designed for transfer of torque or power from one shaft to the other. Gear designing is a costly affair. The configuration and geometry of gears are such that its designing is tedious task. A lot of precision is required to design a gear. This is why factor of safety of gears are always kept higher.
The higher factor of safety accounts for durability of gears. Gears once made can used for about 200 hundred years. The wear and tear are less and the failure of gears is avoided. And hence whole machine failure can be avoided.
Consider a falling mass(m) under gravity(9.8m/s). Initial velocity of the mass is 5 m/s upwards. Derive expressions for the velocity and the position of the mass(m) in terms of time and initial velocity/position of the mass. a) -How long will the mass take to reach the maximum height position? b)- What would be the maximum height the mass reach relative to its initial position?
Answer:
a) 0.51 s
b) 1.275m
Explanation:
using equation of linear motion
v=u+gt...........................(1)
[tex]v^{2} -u^{2}=2gh[/tex]...........(2)
[tex]s=ut+\frac{1}{2} gt^2[/tex].......(3)
a) as the ball is thrown upward -ve 'g' will be acting on the body
as the body reaches to the maximum height the final velocity(v) becomes
zero so from equation (1)
0=5-9.8t
[tex]t=\frac{5}{9.8}[/tex]
t=0.51s
b) Now for maximum height calculation using equation (2)
[tex]v^{2} -u^{2}=2gh[/tex]
v=0
[tex]h=\frac{-u^2}{2g}[/tex]
[tex]h=\frac{-5^2}{2\times-9.8}[/tex]
h=1.275m
Which of the following is/are not a cutting tool material (mark all that apply)? a. High-speed steel b. Brass c. Coated carbide d. Diamond
Answer:
The correct option is : b. Brass
Explanation:
The cutting tool materials are materials that are used to make cutting tools. The cutting tools serve a very important roles in the machinery such as milling cutters. The materials used for making the cutting tools must be tougher and harder than the material that is being cut, at all temperatures.
Some of the cutting tool materials are tool steels (carbon tool steel and high speed steel), cemented carbides and super hard materials such as diamond.
Therefore, brass is not used for making cutting tool.
For water the critical temperature is 374 C, water at T- 400 C exist as vapor (____)
Answer:
The statement is False.
Explanation:
Critical temperature is defined as the temperature at and above which the liquid and the gaseous states are indistinguishable. This is the temperature at and above which the liquid cannot be liquefied.
Above critical temperature or pressure, the substance is in supercritical fluid (SCF) state. A supercritical fluid is the one in which liquid and gaseous state can not be distinguished. It can dissolve like liquids and can effuse like solids.
Water above its critical temperature exists in SCF state which is neither vapor nor liquid.
Thus, the statement is false.
An aluminium alloy used for making cans is cold rolled into a strip of thickness 0.3mm and width 1m. It is coiled round a drum of diameter 15cm, and the outer diameter of the coil is 1m. In the cold rolled condition, the dislocation density is approximately 1015 m-2. Estimate: (i) The mass of aluminium on the coil; (ii) The total length of strip on the coil; (iii) The total length of dislocation in the coiled strip.
Answer:
1. Mass = 2070 kg
2.Total length of strip = 2556 m
3. Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
Explanation:
Given:
Aluminium coil thickness, t = 0.3 mm
= 0.3 X [tex]10^{-3}[/tex] m
Width of the coil,w = 1 m
Drum diameter, d = 15 cm
= 0.15 m
Coil outer diameter, d = 1 m
Dislocation density = [tex]10^{15}[/tex] [tex]m^{2}[/tex]
1). Area of the coil, A = [tex]\frac{\pi }{4}\times[/tex] ( [tex]d_{coil} ^{2}[/tex]-[tex]d_{drum} ^{2}[/tex])
A = [tex]\frac{\pi }{4}\times (1^{2}-0.15^{2})[/tex]
A = 0.767 [tex]m^{2}[/tex]
Volume of the coil,V = A X w
= 0.767 X 1
= 0.767 [tex]m^{3}[/tex]
We know density of aluminum at STP = 2.7 X [tex]10^{3}[/tex]
Therefore, mass of the aluminum coil is,
Mass,m = Density of aluminium X Volume
= 2.7 X [tex]10^{3}[/tex] X 0.767
= 2070 kg
Mass = 2070 kg
2). Total length of trip of coil is given by
L = Volume of coil / area of strip
= [tex]\frac{0.767}{1\times 0.3\times 10^{-3}}[/tex]
= 2556 m
Total length of strip = 2556 m
3). Total length of dislocation of the coiled strip = volume X dislocation density
= 0.767 X [tex]10^{15}[/tex]
= 7.67 X [tex]10^{14}[/tex]
Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
How to called crystalline lattice changes at the temperature change a) oxidation; b) primary crystallization c) allotropy d) anisotropy.
Answer: (c)Allotropy
Explanation: Allotropy is defined as the feature of the material to exist in different chemical forms or state while being in the same physical state. So, allotropy responsible for maintenance of the crystalline lattices during the temperature changes and does not alter the physical state of it. It helps elements to exist in different forms by modifying the structure of the chemical state.
An indirect contact heat exchanger operating at steady state contains a shell with a single tube through it. The fluid flowing through the tube enters at 1kg/s with and enthalpy of 100kJ/kg. It exits with an enthalpy of 500kJ/kg. The fluid flowing through the shell enters with a mass flow rate of 4kg/s and an enthalpy of 1000kJ/kg. Determine the enthalpy at the exit of the shell.
Answer:900 KJ/kg
Explanation:
Given data
mass flow rate of fluid through tube is=1 kg/s
Initial enthalpy of fluid through tube=100 KJ/kg
Final enthalpy of fluid through tube=500KJ/kg
mass flow rate of fluid through shell is=4 kg/s
Initial enthalpy of fluid through shell=1000 KJ/kg
Final enthalpy of fluid through shell=[tex]h_2[/tex]
since heat lost by Shell fluid is equal to heat gain by Tube fluid
heat lost by Shell fluid=[tex]4\times c\left ( 1000-h_2\right )[/tex]
Heat gain by tube Fluid=[tex]1\times c\left ( 500-100\right )[/tex]
Equating both heats
[tex]4\times c[/tex][tex]\left ( 1000-h_2\right )[/tex]=[tex]1\times c\left ( 500-100\right )[/tex]
[tex]h_2[/tex]=900 KJ/kg
What is meant by thermodynamic property? How do you classify the property? Explain with specific examples.
Answer:
The condition of very system can be identified by some physical characteristics of the system known as Thermodynamic Property of the system. For example, pressure, temperature, volume, etc.
Explanation:
Thermodynamic properties are nothing but the physical characteristics of any body by which the physical condition of the body can be described. properties describes the different states of a system. They are macroscopic.
Properties are of two types ---
Intensive property
Extensive property
Intensive Property : Those properties of a system which does not depend upon mass of the system are known Intensive property.
For example, pressure, temperature, density,specific volume, etc
Extensive Property : Properties that depends on the mass oh the system is called Extensive properties.
For example, energy, volume, etc.
Air is compressed by a compressor from v1 = 1.0 m3/kg to v2 = 0.71 m3/kg in a reversible, isothermal process. The air temperature is maintained constant at 25 oC during this process as a result of heat transfer to the surroundings. Air moves through the compressor at a rate of m = 1 kg/s. a)- Determine the entropy change of the air per kg of air. b)- What is the power required by the compressor? c)- What is the rate at which entropy leave the compressor?
Answer:
(a)[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K
(b)P= 29.8 KW
(c) [tex]S_{gen}[/tex]= -0.098 KW/K
Explanation:
[tex]V_1=1m^3/kg,V_2=0.71m^3/kg,[/tex] mass flow rate= 1 kg/s.
T=25°C
Air treating as ideal gas
(a)
We know that entropy change for ideal gas between two states
[tex]s_2-s_1=mC_v\ln \frac{T_2}{T_1}+mR\ln \frac{V_2}{V_1}[/tex]
Given that this is isothermal process so
[tex]s_2-s_1=mR\ln \frac{V_2}{V_1}[/tex]
[tex]s_2-s_1=1\times 0.287\ln \frac{0.71}{1}[/tex]
[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K
(b)
Power required
[tex]P=\dot{m}T\Delta S[/tex]
[tex]P=1\times (273+25)(s_2-s_1)[/tex]
[tex]P=1\times (273+25)(-0.098)[/tex]
P= -29.8 KW (Negative sign means it is compression process.)
(c)
Rate of entropy generation [tex]S_{gen}[/tex]
[tex]S_{gen}=\dot{m}T\Delta S[/tex]
[tex]S_{gen}[/tex]=1(-0.098)
[tex]S_{gen}[/tex]= -0.098 KW/K
Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed Mold c)- Preforming d)- Postforming e)- None of the above f)- All the above
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
At any given state the specific internal energy of a gas is always greater its specific enthalpy. a)True b) False
Answer:
(b)False
Explanation:
We know that specific Internal energy of gas u=[tex]C_v[/tex]T
and specific enthalpy of gas h=[tex]C_p[/tex]T
If we take the case of air we know that
[tex]C_v[/tex]=0.707 KJ/Kg=K , [tex]C_p[/tex]=1.005 KJ/Kg=K
If we take A fixed temperature T=300 K
so u=212.1 KJ/ kg ,h=301.5 KJ/kg
So we can say that specific enthaply of gas is always greater than its specific internal energy.
Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C. The carriage and half-nuts D. A1l of the above
Answer:
(d) all of the above
Explanation:
before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial
so option (D) will be correct because we need lubricate in all the given parts
Give two advantages of a four-high rolling mill opposed to a two-high rolling mill for the same output diameter.
Answer:
Four- high rolling mill Two-high rolling mill
1.Small roll radius.That is why required 1.High roll radius.That is required low power. . why required high power.
2. Low roll separating force. 2.High roll separating
force
A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.06 ft*lb b. 19.24 ft*lb ? c)-31.17 ftlb d) 119.07 ftlb e)-9.92 ft*lb
Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
Power: Rate of change of work with respect to time is called power.
We know that P=[tex]Torque\times speed[/tex]
[tex]\omega=\frac{2\pi N}{60}[/tex] rad/sec
So that P=[tex]\dfrac{2\pi NT}{60}[/tex]
So 3281.08=[tex]\dfrac{2\pi \times 2329\times T}{60}[/tex]
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb
A disk brake has two pads which cover 45 degrees of the disk. The outside radius is 6.0 inch and the inside radius is 4.0 inch. Assume a coefficient of friction of 0.4, and a max pressure, pa=100 psi. a) Find the force required to apply one pad. b) Find the torque capacity for both pads.
Answer:
f = 628.32 lb
t = 2513.28 lb-inc
Explanation:
given data:
θ = 45°
outside radius = 6 inch
inside radius = 4 inch
coefficient of friction = 0.4
max pressure = 100 psi
a) determine force required for applying one pad
f = [tex] \frac{\theta }{360}* 2\pi *p_{max}*r_{i}(r_{o}-r_{i})[/tex]
f = [tex] \frac{45 }{360}* 2\pi *100*4(6-4)[/tex]
f = 628.32 lb
b) torque capacity (t)
t = [tex]\mu *f*r_{average}^{}[/tex]
t = 0.4 *628.32*5
torque = 1256.64 lb-inc
for both pad = 2 * 1256.64 =2513.28 lb-inc
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twice of that of loose side, F1= 2F2. Please calculate the F1, Fe, and Fo.
Answer:
F₁ = 1500 N
F₂ = 750 N
[tex]F_{e}[/tex] = 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
[tex]F_{max}[/tex] = 3 x [tex]F_{e}[/tex]
where [tex]F_{e}[/tex] is centrifugal force
[tex]F_{e}[/tex] = [tex]F_{max}[/tex] / 3
= 1500 / 3
= 500 N
Consider a fully developed flow in a circular pipe (negligible entrance effects), how does the convective heat transfer coefficient vary along the flow direction? a) Gradually decrease b) Gradually increase c) Remain constant d) There is not enough information to determine
Answer: A) Gradually decrease
Explanation:
The convection value of heat transfer rate are gradually decreasing with the flow of the heat. Flow in a circular pipe, flow direction does not change in the velocity path. The average of the coefficient of heat transfer and the number of pipes are needed and the effects are get neglected so that is why the flow are fully developed.
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 500C at a rate of 0.065 kg/s and leaves at 40°C. The refrigerant enters the evaporator at 20°C with a quality of 23 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine: (a) The degrees of subcooling of the refrigerant in the condenser, b)-The mass flow rate of the refrigerant . (c) The heating load and the COP of the pump, and d)-The theoretical minimum power input to the compressor for the same heating load.
Answer:2.88
Explanation:
Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s
To hoist a load 6 meters in 1.5 seconds, the cable must be drawn in by the motor at a constant speed of 4 meters per second.
Explanation:The question asks to determine the constant speed at which a cable must be drawn in by a motor to hoist a load to a certain height within a given time frame. This can be solved by understanding the basic concepts of distance, speed, and time.
To find the constant speed, we use the formula:
Speed = Distance / Time. In this case, the distance is 6 meters (the height the load needs to be hoisted) and the time is 1.5 seconds.
Plugging the numbers into the formula gives:
Speed = 6m / 1.5s = 4 m/s.
Therefore, the cable must be drawn in by the motor at a constant speed of 4 meters per second to hoist the load 6 meters in 1.5 seconds.
Convective heal transfer is defined as______
Answer: Convective heat transfer is defined as the transfer of heat from a particular place to another with the help of fluid movements.
Explanation: Convective heat also knows as convection that is transfer of heat from one place to another place by the help of fluid movements around the area having different temperatures. This process has many examples in everyday life to understand the process better.
Ex.- the melting of ice from solid to liquid form as it comes in contact of high temperature as heat surrounds the ice in form of air.
A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- False
Answer:
The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is
b) False
Explanation:
Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.
Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.
Describe ICP/OES in detail.
Answer:
ICP -OES stand for inductively coupled plasma optical emission spectroscopy
Explanation:
It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.
The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.
Answer:
ICP stands for Insane Clown Posse. It is a rap group. The group's members are Violent J and Shaggy 2 Do*e
Explanation:
Air enters a compressor at 100 kPa, 10°C, and 220 m/s through an inlet area of 2 m2. The air exits at 2 MPa and 240°C through an area of 0.5 m2. Including the change in kinetic energy, determine the power consumed by this compressor, in kW.
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]
[tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
[tex]h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W[/tex]
We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]and we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate= [tex]density\times area\times velocity[/tex]
So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
=1.23×200×2 Kg/s
=541.17 Kg/s
[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]
[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
[tex]V_{2}[/tex]=80.07 m/s
Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]
So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]
[tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]
Now by putting the values
[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW
Discuss the importance of dust and fluff removal from spinning mills and how it is realised in air conditioning plants
Answer:
Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.
Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc. Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.
The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.
The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.
Answer:b) Point defects
Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.
The uniform wall thickness that is usually targeted for plastic injection molded parts is roughly: A. 0.5 mm B. 3 mm C. 7 mm D. 12 + mm
Answer:
B. 3 m
Explanation: For plastic injection moulded the thickness is generally between 2 mm to 3 mm
the wall is not too thick because during cooling process there should be defects so thickness of wall is no too high and there is also a problem if we use thicker wall that we need more material for moulding process so the thickness should be in between 2 to 3 mm which is in option B so option B will be the correct option
A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside surface of the tank wall is held constant at 25 oC and the outside surface heat transfer coefficient is 6 W m2 K. Calculate the rate of heat loss from the tank when the outside air temperature is 15°C.
Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W [tex]m^{-1}K^{-1}[/tex]
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 × [tex]10^{-2}[/tex]
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × [tex]10^{-2}[/tex]
temperature = 25°C
surface heat transfer coefficient = 6 6 W [tex]m^{-2}K^{-1}[/tex]
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4[tex]\pi[/tex] × r1 × r2 × K
R1 = ( 5.2 - 5 ) [tex]10^{-2}[/tex] / 4[tex]\pi[/tex] × 5 × 5.2 × 16 × [tex]10^{-4}[/tex]
R1 = 3.825 × [tex]10^{-3}[/tex]
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4 [tex]\pi[/tex] r2²
area = 4 [tex]\pi[/tex] (5.2 × [tex]10^{-2}[/tex])² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 × [tex]10^{-3}[/tex] + 4.90499
heat loss = 2.037152 W