Answer Explanation :
We know that entropy is the measure of molecular randomness it is denoted by S It is a thermodynamic property of gas it is an extensive property means its depend on the mass of the the system and specific entropy is defined as the entropy per unit mass its unit is KJ/kg.k
by using standard table the specific entropy for the subcooled liquid is 1.057 KJ/kg.k standard table is used for knowing the specific entropy of different liquid and gaes
Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show your work until you start computing x1, etc. Then just write down what your calculator gives you.
Answer:
x3=0.100167
Explanation:
Let's find the answer.
Because we are going to find the solution for sin(Ф)=0.1 then:
f(x)=sin(Ф)-0.1 and:
f'(x)=cos(Ф)
Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:
x1=x0-f(x0)/f'(0)
x1=0.001-(sin(0.001)-0.1)/cos(0.001)
x1= 0.100000
x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)
x2=0.100167
x3=0.100167
In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle temperature is 2031K. Determine the thermal efficiency.
Answer:
[tex]\eta[/tex]=0.60
Explanation:
Given :Take [tex]\gamma[/tex]=1.4 for air
[tex]P_1=100 KPa ,T_1=300K[/tex]
[tex]\frac{V_1}{V_2}[/tex]=r ⇒ r=16
As we know that
[tex]T_2=T_1(r^{\gamma-1})[/tex]
So [tex]T_2=300\times 16^{\gamma-1}[/tex]
[tex]T_2[/tex]=909.42K
Now find the cut off ration [tex]\rho[/tex]
[tex]\rho=\frac{V_3}{V_2}[/tex]
[tex]\frac{V_3}{V_2}=\frac{T_3}{T_2}[/tex]
[tex]\rho=\frac{2031}{909.42}[/tex]
[tex]\rho=2.23[/tex]
So efficiency of diesel engine
[tex]\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}[/tex]
Now by putting the all values
[tex]\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}[/tex]
So [tex]\eta[/tex]=0.60
So the efficiency of diesel engine=0.60
What is the most common type of pump?
Answer:
The most common type of pumps are Positive displacement and Non positive displacement pumps.
Explanation:
Pumps are two types:
(A) Positive displacement pump
(a)Gear pump
(1) Ge rotor pumps
(2)Internal gear pumps
(3)Lobe pumps
(4) External gear pumps
(b)Piston pump
(1)Radial piston
(2)Axial piston
(c)vane pump
(B) Non positive displacement pump
(a) Centrifugal pump
A horizontal pipe is fitted with a nozzle. The inlet diameter of the nozzle is 40 mm and the outlet diameter is 20 mm. The flow rate in the pipe is 1.2 m3 /min and water density is 1000 kg/m3 . Determine the force exerted by the nozzle on the water.
Answer:
969.68N
Explanation:
d₁=0.04 m A₁=[tex]\frac{\pi d^2_{1} }{4}[/tex]
[tex]A_{1} =\frac{\pi \times .04^2}{4}= 0.00125m^{2} \[/tex]
d₂=0.02 m A₂=[tex]\frac{\pi d^2_{2} }{4}[/tex]
[tex]A_{2} =\frac{\pi \times .02^2}{4}= 0.00031m^{2} \[/tex]
Q=1.2m³/min Q=1.2/60=0.02m³/s
using continuity equation
Q₁=A₁v₁
v₁=Q₁/A₁=0.02/0.00125=16m/s
Q₂=A₂v₂
v₂=Q₂/A₂=0.02/0.00031=64.5m/s
[tex]F_{inlet}=\rho A_{1}v_1^{2}[/tex]
[tex]F_{inlet}=1000\times 0.00125\times16^{2}=320N[/tex]
[tex]F_{outlet}=\rho A_{2}v_2^{2}[/tex]
[tex]F_{outlet}=1000\times 0.00031\times64.5^{2}=1289.68N[/tex]
Force on the nozzle=F_{outlet}-F_{inlet}
= 1289.68-320
=969.68N
The two stroke engines has greater cooling and lubrication requirements than four stroke engine. Explain why?
Answer:
A two stroke engine produces twice the power compared to a four stroke engine of same weight and size.
Explanation:
In a two stroke engine, all the four processes namely, intake stroke, compression stroke, power stroke and exhaust stroke takes place in one revolution of crankshaft or two strokes of the piston. While in a four stroke engine, all the four processes namely intake stroke, compression stroke, power stroke and exhaust stroke take place in two revolution of crankshaft or four strokes of the piston.
Therefore, there is one power stroke in one revolutions of the crankshaft in case of a two stroke engine as compared to the four stoke engine where there is one power stroke for two revolutions of the crank shaft.
So the power developed in a two stroke engine is more ( nearly twice ) as compared to a four stroke engine of the same capacity. When power produced is more, the heat dissipation is also more in case of a two stroke engine. So greater cooling is required to dissipate heat from a two stroke engine as compared to a four stroke engine.
Also in a two stroke engine, the lubricating oil is used with the oil whereas a four stroke engine has a separate tank for lubricating oil. So the lubricating oil gets burnt quickly in a two stroke engine.
Thus, to dissipate more heat, a two stroke engines has greater cooling and lubrication requirements than a four stroke engines as power produce in a two stroke engine is more than a four stoke engine with same weight or size.
The drilling pipe on an oil rig is made from steel pipe which having thickness of 5-mm and an outside diameter of 90-mm. Calculate the maximum shear stress occur in the pipe if the pipe is turning at 650 rev/min while being powered using 12kW motor.
Answer:
[tex]\tau_{max}= 3.28 \ MPa[/tex]
Explanation:
outside diameter = 90 mm
inside diameter = 90- 2× t= 90- 2× 5 = 80mm
where t is thickness of pipe.
power (P) = 12 kW
Revolution (N)= 650 rev/min
we
Power = torque × angular velocity
P= T× ω
ω = [tex]\frac{2 \pi N}{60}[/tex]
[tex]P=T \times \frac{2\pi N}{60}\\12 \times 10^3=T\times \frac{2\pi \times 650}{60}[/tex]
T= 176.3 Nm
for maximum shear stress
[tex]\frac{\tau_{max}}{y_{max}}=\frac{T}{I_p}[/tex]
where ymax is maximum distance from neutral axis.
[tex]y_{max}=\frac{d_0}{2}= \frac{90}{2}[/tex]= 45 mm
[tex]I_p[/tex]= polar moment area
= [tex]\frac{\pi}{32} (d_o^4-d_i^4)=\frac{\pi}{32} (90^4-80^4)[/tex]
= 2,420,008 mm⁴
[tex]\dfrac{\tau_{max}}{45}=\dfrac{176.3 \times 10^3}{2,420,008}[/tex]
[tex]\tau_{max}= 3.28 \ MPa[/tex]
The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is linear with depth what is the surface velocity of flow in the channel?
Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
[tex]\dot{m}=\rho AV[/tex]
Putting all the value to get the velocity of the flow
[tex]\frac{\dot{m}}{\rho A} = V[/tex]
[tex]V = \frac{4000}{1000*4*2}[/tex]
V = 0.5 m/s
Air is heated from 50 F to 200 F in a rigid container with a heat transfer of 500 Btu. Assume that the air behaves as an ideal gas. Determine the volume of air [ft3] if the initial pressure is 2 atm. Also show the process on a P-v state diagram. Use the following temperature conversion: T[R] = T[F] + 460.
Answer:
[tex]V=68.86ft^3[/tex]
Explanation:
[tex]T_1[/tex] =10°C,[tex]T_2[/tex] =93.33°C
Q=500 btu=527.58 KJ
[tex]P_1= 2atm[/tex]
If we assume that air is ideal gas PV=mRT, ΔU=[tex]mC_v(T_2-T_1)[/tex]
Actually this is closed system so work will be zero.
Now fro first law
Q=ΔU=[tex]mC_v(T_2-T_1)[/tex]+W
⇒Q=[tex]mC_v(T_2-T_1)[/tex]
527.58 =[tex]m\times 0.71(200-50)[/tex]
m=4.9kg
PV=mRT
[tex]200V=4.9\times 0.287\times (10+273)[/tex]
[tex]V=1.95m^3[/tex] ([tex]V=1m^3=35.31ft^3[/tex])
[tex]V=68.86ft^3[/tex]
What is the overall transfer function for a closed-loop system having a forward-path transfer function of 5/(s + 3) and a negative feedback-path transfer function of 10?
Answer:
transfer function T(S)=[tex]\frac{5}{S+53}[/tex]Explanation:
NEGATIVE FEEDBACK TRANSFER FUNCTIONnegative feedback control of the amplifier is achieved by applying output voltage signal back to inverting input terminal by feedback
transfer function is T=[tex]\frac{g}{1+Gh}[/tex]
where G=forward Path gain
H=negative feedback gain
here G=[tex]\frac{5}{S+3}[/tex]
H=10
T(S)=[tex]\frac{G}{1+GH}[/tex]
=[tex]\frac{5}{S+53}[/tex]
What is considered a method for inducing heat transfer? (1) -heat power and convection (2)-preseribed temperature (3)-radiation (4)-thermal insulation (5)-prescribed strain
Answer: (1) heat power and convection
(3)radiation
Explanation: Heat can be transferred in many different ways such as conduction,radiation form and convection etc.
Convection is a method of transferring of the heat from a particular surface by the help of fluids .E.g.- air
Radiation is the method of transfer of heat by the emission or absorption process in the other surface.E.g.- earth getting warm due to sun.
Therefore the answer to the question is option (1) and (3).
Which of the following is not a method of heat transfer? A. Conduction B. Convection C. Injection D. Radiation is desirable.
Answer:
C) Injection
Explanation:
Injection is a molding process, not a heat transfer mechanism.
A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False
Answer:
True
Explanation:
For point in xz plane the stress tensor is given by[tex]\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right][/tex]
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane
Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False
Answer: False
Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.
How to convert a friction to decimal ?
Answer:
To convert a fraction to a decimal, divide the numerator by the denominator.
A heat engine operates between a hot reservoir at 2000°C and the atmosphere (cold reservoir) at 25°C. it produces 50 MW of power while rejecting 40 MW of waste heat. Determine the maximum possible thermal efficiency of the engine in percent.
Answer:
55.56%
Explanation:
Given data
Temprature of hot reservior =2000°c=2273k
Temprature of Cold reservior=25°c=298k
Power produced by engine=50MW
Heat rejected =40MW
we know that Effeciency(η) of heat engine=[tex]\frac{Work produced}{heat supplied}[/tex]
Also we know that
heat supplied[tex]\left ({Q_s} \right )=work produced{W}+Heat rejected{Q_r}[/tex]
Q_s=50+40=90MW
η=[tex]\frac{W}{Q_s}[/tex]
η=[tex]\frac{50}{90}[/tex]
η=55.55%
An aluminum electrical cable is 20 mm in diameter is covered by a plastic insulation (k = 1 W/m-k) of critical thickness. This wire is placed in a room with an air flow heat transfer coefficient of 50 W/m^2-K. Compared to the bare aluminum wire, the heat loss from this insulated wire will be a) LESS b) MORE c) SAME d) ZERO
Answer:
the heat loss from this insulated wire is less
Explanation:
Given data in question
diameter of cable (d) = 20 mm
( K ) = 1 W/m-k
heat transfer coefficient (h) = 50 W/m²-K
To find out
the heat loss from this insulated wire
solution
we will find out thickness of wire
heat loss is depend on wire thickness also
we have given dia 20 mm
so radius will be d/2 = 20/ 2 = 10 mm
Now we find the critical thickness i.e.
critical thickness = K / heat transfer coefficient
critical thickness = 1 / 50 = 0.02 m i.e. 20 mm
now we can see that critical thickness is greater than radius 10 mm
so our rate of heat loss will be decreasing
so we can say our correct option is (a) less
The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03
Answer:
Answer is c 0.500
Explanation:
[tex]SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}[/tex]
We know that [tex]\rho_{water}=62.42lb/ft^{3}[/tex]
Applying values we get
[tex]SpecificGravity=\frac{31.2}{62.4}=0.5[/tex]
A solid 0.75 in diameter steel shaft transmits 7 hp at 3,200 rpm. Determine the maximum shear stress magnitude produced in the shaft. Hint: Use P=Tω and convert hp to ft-lbf/s. Find τ by using Tc/J. Recall max shear stress will be on the outer most surface.
Answer:[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Explanation:
Given data
[tex]power[/tex][tex]\left ( P\right )[/tex]=7 hp=5220 W
N=3200rpm
[tex]\omega [/tex]=[tex]\frac{2\pi\times N}{60}[/tex]=335.14 rad/s
diameter[tex]\left ( d\right )[/tex]=0.75in=19.05mm
we know
P=[tex]Torque\left ( T\right )\omega [/tex]
5220=[tex]T\times 335.14[/tex]
T=15.57 N-m
And
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Polar\ modulus}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Z_{P}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{16\times T}{\pi d^{3}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Answer:
Maximum shear stress is 11.47 MPa.
Explanation:
Given:
D=.75 in⇒D=19.05 mm
P=7 hp⇒ P=5219.9 W
N=3200 rpm
We know that
P=T[tex]\times \omega[/tex]
Where T is the torque and [tex]\omega[/tex] is the speed of shaft.
P=[tex]\frac{2\pi N\times T}{60}[/tex]
So 5219.9=[tex]\frac{2\pi \times 3200\times T}{60}[/tex]
T=15.57 N-m
We know that maximum shear stress in shaft
[tex]\tau _{max}=\dfrac{16T}{\pi \times D^3}[/tex]
[tex]\tau _{max}=\dfrac{16\times 15.57}{\pi \times 0.01905^3}[/tex]
[tex]\tau _{max}[/tex]=11.47 MPa
So maximum shear stress is 11.47 MPa.
Fluid power is a. The technology that deals with the generation, control, and transmission of power-using pressurized fluids b. muscle that moves industry. c. used to push, pull, regulate, or drive virtually all the machine of modern industry d. probably as old as civilization itself e. all of the above
Answer: a) The technology that deals with the generation, control and transmission of power using pressurized fluids
Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.
How is heat transfer defined in an internally reversible process
Answer:
Heat transfer for a internally reversible process.
Explanation:
Internally reversible means that there is entropy generation ' with in ' the system.
Heat transfer of a process is considered to be reversible if it occurs because of any minute temperature difference between the surrounding and the system .
Let us consider an example ,
Transferring of the heat across the difference in temperature of 0.0001 °C appears as more reversible than for the difference in temperature of 100 °C .
Hence ,
By heating or cooling a system for a number of infinitesimally small steps , we can approximate a reversible process.
The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion. a)-True b)-False
Answer:
a)-True
Explanation:
The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion.
This statement is true.
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=[tex]\frac{s}{(s+3)^2}[/tex]
U(s)=[tex]\frac{1}{s}[/tex] for unit step function
output =H(s)×U(s)
=[tex]\frac{s}{(s+3)^2}[/tex]×[tex]\frac{1}{s}[/tex]
=[tex]\frac{1}{(s+3)^2}[/tex]
taking inverse laplace of output
output=t×[tex]e^{-3t}[/tex]
at t=0 putting the value of t=0 in output
output =0
A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, determine the magnitude of the particle's acceleration when t= 2 s.
Answer:
4.83m/[tex]s^{2}[/tex]
Explanation:
For a particle moving in a circular path the resultant acceleration at any point is the vector sum of radial and the tangential acceleration
Radial acceleration is given by [tex]a_{radial}=w^{2}[/tex]r
Applying values we get [tex]a_{radial}=(2t)^{2}[/tex]X0.3m
Thus [tex]a_{radial}=1.2t^{2}[/tex]
At time = 2seconds [tex]a_{radial}= 4.8m/s^{2}[/tex]
The tangential acceleration is given by [tex]a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}[/tex]
[tex]a_{tangential}=\frac{d(2tr)}{dt}[/tex]
[tex]a_{tangential}= 2r[/tex]
[tex]a_{tangential}=0.6m/s^{2}[/tex]
Thus the resultant acceleration is given by
[tex]a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}[/tex]
[tex]a_{res} =\sqrt{4.8^{2}+0.6^{2} } =4.83m/s^{2}[/tex]
Internal flow is one in which the flow is not bounded. a) True b) False
False it is External
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T that may be applied to the shaft if the allowable shear stress is equal to 100 MPa.
Answer:
Explanation:
Using equation of pure torsion
[tex]\frac{T}{I_{polar} }=\frac{t}{r}[/tex]
where
T is the applied Torque
[tex]I_{polar}[/tex] is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
[tex]D_{0} =[/tex] Outer Diameter
[tex]D_{i} =[/tex] Inner Diameter
[tex]I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}[/tex]
Applying values in the above equation we get
[tex]I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\
I_{polar}= 1.74[/tex] x [tex]10^{-7} m^{4}[/tex]
Thus from the equation of torsion we get
[tex]T=\frac{I_{polar} t}{r}[/tex]
Applying values we get
[tex]T=\frac{1.74X10^{-7}X100X10^{6} }{.021}[/tex]
T =829.97Nm
If I add 30J of heat to a system so that the final temperature of the system is 300K, what is the change in entropy of the system? a)-1 J/K b)- 3 J/K c)- 1 J/K d)- 9 J/K e)- 10 J/K
Answer:
0.1 J/K
Explanation:
entropy change equation is as followed:
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
where ΔS=entropy change
Q=Heat transfer
T= temperature
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
[tex]\Delta S=\frac{30}{300}[/tex]
[tex]\Delta S=0.1 J/K[/tex]
hence the change in entropy of system which is [tex]\Delta S[/tex]is equal to 0.1 J/K
A Mariner vessel, floating at a draft of 23'6", has a GM of 1.5 feet which does not meet the required GM standard. How far above the keel must 1,400 tons be loaded to increase the GM to 2.0 feet?
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 hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of 300 KNm, as shown. The modulus of rigidity G for the steel is 80 GPa. Determine: (a) The maximum shearing stress in the shaft. (b) The shearing stress on a traverse cross section at the inside surface of the shaft (c) The magnitude of the angle of twist for a (L) -2.5 m length.
Answer:
a. [tex]\tau=51.55 MPa[/tex]
b.[tex]\tau=42.95MPa[/tex]
c.[tex]\theta=7.67\times 10^{-3}[/tex] rad.
Explanation:
Given: [tex]D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa [/tex]
We know that
[tex]\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}[/tex]
J for hollow shaft [tex]J=\dfrac{\pi (D_o^4-D_i^4)}{64}[/tex]
(a)
Maximum shear stress [tex]\tau =\dfrac{16T}{\pi Do^3(1-K^4)}[/tex]
[tex]K=\dfrac{D_i}{D_o}[/tex]⇒K=0.83
[tex]\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}[/tex]
[tex]\tau=51.55 MPa[/tex]
(b)
We know that [tex]\tau \alpha r[/tex]
So [tex]\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}[/tex]
[tex]\dfrac{51.55}{\tau}=\dfrac{210}{175}[/tex]
[tex]\tau=42.95MPa[/tex]
(c)
[tex]\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}[/tex]
[tex]\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}[/tex]
[tex]\theta=7.67\times 10^{-3}[/tex] rad.
What is the principle of operation of a mechanical dynamometer?
The mechanical dynamometer is an instrument used to measure forces or to calculate the weight of objects. The traditional dynamometer, invented by Isaac Newton, bases its operation on the stretching of a spring that follows the law of elasticity of Hooke in the measurement range. Like a scale with elastic spring, it is a spring scale, but it should not be confused with a scale of saucers (instrument used to compare masses).
These instruments consist of a spring, generally contained in a cylinder that in turn can be inserted into another cylinder. The device has two hooks or rings, one at each end. The dynamometers have a scale marked on the hollow cylinder that surrounds the spring. When hanging weights or exerting a force on the outer hook, the cursor of that end moves on the external scale, indicating the value of the force.
The dynamometer works thanks to a spring or spiral that has inside, which can be lengthened when a force is applied on it. A point or indicator usually shows, in parallel, the force.
Define Viscosity. What are the main differences between viscous and inviscid flows?
1. Define Viscosity
In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.
What are the main differences between viscous and inviscid flows?
Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density
Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.