Answer:
The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.
Explanation:
The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform. In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.
An open tank contains ethylene glycol at 25°C. Compute the pressure at a depth of 3.0m.
Answer:
pressure at depth 3 m is 33.255 kPa
Explanation:
given data
temperature = 25°C
depth = 3 m
to find out
pressure
solution
we know pressure formula that is
pressure = ρ g h ................1
and we know specific gravity of Ethylene glycol is 1.13
so
1.13 = [tex]\frac{\rho (e)}{\rho (water)}[/tex]
ρ (Ethylene glyco) = 1130 Kg/m³
so
pressure will be by equation 1
pressure = 1130 × 9.81 × 3
pressure = 33255.9 Pa
so pressure at depth 3 m is 33.255 kPa
In crash tests, a shock absorber is used to slow the test car. The shock absorber consists of a piston with small holes that moves in a cylinder containing water. Viscous dissipation in the water transforms work into heat. How much heat will be transferred from the water after a 2000 kg car crashes into the shock absorber at a speed of 40 m/s?
Answer:
The heat transferred to water equals 1600 kJ
Explanation:
By the conservation of energy we have
All the kinetic energy of the moving vehicle is converted into thermal energy
We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by
[tex]K.E=\frac{1}{2}mv^{2}[/tex]
Thus
[tex]K.E_{car}=\frac{1}{2}\times 2000\times 40^{2}=1600\times 10^{3}Joules[/tex]
Thus the heat transferred to water equals [tex]1600kJ[/tex]
The head difference between the inlet and outlet of a 1km long pipe discharging 0.1 m^3/s of water is 0.53 m. If the diameter is 0.6m, what is the friction factor? Is a) 0.01 b) 0.02 c) 0.03 d) 0.04 e) 0.05
Answer:
The correct option is 'e': f = 0.05.
Explanation:
The head loss as given by Darcy Weisbach Equation is as
[tex]h_{l}=\frac{flv^{2}}{2gD}[/tex]
where
[tex]h_{l}[/tex] is head loss in the pipe
'f' is the friction factor
'l' is the length of pile
'v' is the velocity of flow in pipe
'D' is diameter of pipe
From equation of contuinity we have [tex]v=\frac{Q}{A}[/tex]
Thus using this in darcy's equation we get
[tex]h_{l}=\frac{flQ^{2}}{2gDA}[/tex]
where
'Q' is discharge in the pipe
'A' is area of the pipe [tex]A=\frac{\piD^2}{4}[/tex]
Applying the given values we get
[tex]h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}[/tex]
Solving for 'f' we get
[tex]f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05[/tex]
An inventor proposes an engine that operates between the 27 deg C warm surface layer of the ocean and a 10 deg C layer a few meters down. The claim is that this engine can produce 100 kW at a flow of 20 kg/s. Is this possible?
Answer:
Engine not possible
Explanation:
source temperature T1 = 300 K
sink temperature T2= 283 K
therefore, carnot efficiency of the heat engine
η= 1- T_2/T_1
[tex]\eta= 1-\frac{T_1}{T_2}[/tex]
[tex]\eta= 1-\frac{283}{300}[/tex]
= 0.0566
= 5.66%
claims of work produce W = 100 kW, mass flow rate = 20 kg/s
[tex]Q=mc_p(T_1-T_2)[/tex]
[tex]Q=20\times4.18(300-283)[/tex]
= 1421.2 kW
now [tex]\eta= \frac{W}Q}[/tex]
now [tex]\eta= \frac{100}{1421.2}[/tex]
=7%
clearly, efficiency is greater than carnot efficiency hence the engine is not possible.
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to convert the ice to water at 200°F?
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice [tex]C_p=2.03\ KJ/kg.K[/tex]
Latent heat for ice H=336 KJ/kg
Specific heat for ice [tex]C_p=4.187\ KJ/kg.K[/tex]
We know that sensible heat given as
[tex]Q=mC_p\Delta T[/tex]
Heat for -15F to 32 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 2.03(32+15) KJ[/tex]
Q=858.69 KJ
Heat for 32 Fto 200 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 4.187(200-32) KJ[/tex]
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1.3, find the final temperature. charge of air at 1 bar & 20°C and a) T2= 1093 K b) T2=151.8 K c) T2=983.6 K d) T2 =710.9 K e) T2= 424.8 K
Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
[tex]\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }[/tex] ...........1
put here all these value and we get t2
[tex]\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }[/tex]
t2 = 424.8
final temperature is 424.8 K
so correct option is e
How much heat (Btu) is prod uced by a 150-W light bulb that is on for 20-hours?
Answer:
heat produced by 150 watt bulb is 510 Btu
Explanation:
Given data:
Power of bulb is 150 W
Duration for light is 20 hr
We know that 1 Watt = 3.4 Btu
hence using above conversion value we can calculate the power in Btu unit
so for [tex]150 W\ bulb = 150\times 3.4 = 510 Btu[/tex]
therefore, 150 W bulb consist of 510 Btu power for 20 hr
output heat by 150 watt bulb is 510 Btu
What is the definition of diameter pitch?
Answer:
Diameter pitch is the parallel thread of the imaginary cylinder of the diameter that basically intersect in the surface of thread. It is also known as effective diameter. The diameter pitch is basically used to determine the threated parts.
The pitch diameter is the essential component for determining the basic compatibility in the externally and internally thread parts. The diameter pitch is the sensitive measuring tool for determine the specific measurements.
What is specific gravity? How is it related to density?
Answer:
Specific gravity is defined as the ratio of the Densities of the two substances.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
Explanation:
Specific gravity is defined as the ratio of the Densities of the two substances.
For the standardization, the density ration of densities is calculated with respect to the density of water i.e the denominator is the density of water.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water}}[/tex]
Also,
At STP density of water is 1000 Kg/m³
Therefore the relation between the specific gravity and density is,
The Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water at STP}}[/tex]
or
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
Discuss the differences between conduction and convective heat transfer.
Answer:
Basically there are two principal differences between the convection and conduction heat transfer
Explanation:
The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.
We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.
A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. What is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?
Answer:
0.2 m/s
Explanation:
The velocity of a point on the edge of a disk rotating disk can be calculated as:
[tex]v=\omega*r[/tex]
Where [tex]\omega[/tex] is the angular velocity and r the radius of the disk. This leads to:
[tex]v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s[/tex]
Answer:
0.2 m/s
Explanation:
Step 1: identify the given parameters
angular velocity, ω = 0.5 rad/s
radius of the disk, r = 0.4m
Note: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.
Step 2: calculate the velocity of the disk at the outer edge in m/s
Velocity = Angular velocity (rad/s) X radius (m)
Velocity = 0.5 rad/s X 0.4 m
Velocity = 0.2 m/s
NASA SPACE SHUTTLE QUESTION:
What lessons have we learned from the shuttle program and space travel?
-Be detailed
-full sentences
-2-3 sentence response
Explanation:
The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.
A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enthalpy is then raised by 52kJ (note the unit) by heating at constant pressure. Assuming that all processes occur reversibly, sketch them on a p-v chart and calculate the total work done and the total heat transfer.
Answer:
Total work: -5.25 kJ
Total Heat: 52 kJ
Explanation:
V0 = 0.15
P0 = 350 kPa
t0 = 150 C = 423 K
P1 = 105 kPa (isentropical transformation)
Δh1-2 = 52 kJ (at constant pressure)
Ideal gas equation:
P * V = m * R * T
m = (R * T) / (P * V)
R is 0.287 kJ/kg for air
m = (0.287 * 423) / (350 * 0.15) = 2.25 kg
The specifiv volume is
v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg
Now we calculate the parameters at point 1
T1/T0 = (P1/P0)^((k-1)/k)
k for air is 1.4
T1 = T0 * (P1/P0)^((k-1)/k)
T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K
The ideal gas equation:
P0 * v0 / T0 = P1 * v1 / T1
v1 = P0 * v0 * T1 / (T0 * P1)
v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg
V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3
The work of this transformation is:
L1 = P1*V1 - P0*V0
L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg
Q1 = 0 because it is an isentropic process.
Then the second transformation. It is at constant pressure.
P2 = P1 = 105 kPa
The enthalpy is raised in 52 kJ
Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh
And the idal gas equation is:
P1 * v1 / T1 = P2 * v2 / T2
T2 = T1 * P2 * v2 / (P1 * v1)
Replacing:
Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2
Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)
v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)
The Cv of air is 0.7 kJ/kg
v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg
V2 = 2.25 * 0.2 = 0.45 m^3
T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K
The heat exchanged is Q = Δh = 52 kJ
The work is:
L2 = P2*V2 - P1*V1
L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ
The total work is
L = L1 + L2
L = -14.7 + 9.45 = -5.25 kJ
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Answer:
1) g=31.87ft/s^2
2)m=120
W=119.64lbf
Explanation:
first part
the weight of a body with mass is calculated by the following equation
W=mg
we convert 120lb to slug
m=120lbx1slug/32.147lb=3.733slug
solving for g
g=W/m
g=119/3.733=31.87ft/s^2
second part
the mass is the same
m=120lb=3.733slug
Weight
W=3.733slug*32.05ft/s^2=119.64lbf
What impact does modulus elasticity have on the structural behavior of a mechanical design?
Answer with Explanation:
The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:
1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.
2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.
3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity the resistance it offer's to shear forces and the torques is more.
While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.
Refrigerant R-12 is used in a Carnot refrigerator
operatingbetween saturated liquid and vapor during the heat
rejectionprocess. If the cycle has a high temperature of 50 deg C
and a lowtemperature of -20 deg C, find the heat transferred from
therefrigerated space, the work required, the coefficient
ofperformance and the quality at the beginning of the heat
additioncycle.
Answer:
Heat transferred from the refrigerated space = 95.93 kJ/kg
Work required = 18.45 kJ/kg
Coefficient of performance = 3.61
Quality at the beginning of the heat addition cycle = 0.37
Explanation:
From figure
[tex] Q_H [/tex] is heat rejection process
[tex] Q_L [/tex] is heat transferred from the refrigerated space
[tex] T_H [/tex] is high temperature = 50 °C + 273 = 323 K
[tex] T_L [/tex] is low temperature = -20 °C + 273 = 253 K
[tex] W_{net} [/tex] is net work of the cycle (the difference between compressor's work and turbine's work)
Coefficient of performance of a Carnot refrigerator [tex] (COP_{ref}) [/tex] is calculated as
[tex] COP_{ref} = \frac{T_L}{T_H - T_L} [/tex]
[tex] COP_{ref} = \frac{253 K}{323 K - 253 K} [/tex]
[tex] COP_{ref} = 3.61 [/tex]
From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table
[tex] Q_H = 122.5 kJ/kg [/tex]
From coefficient of performance definition
[tex] COP_{ref} = \frac{Q_L}{Q_H - Q_L} [/tex]
[tex] Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L[/tex]
[tex] Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)} [/tex]
[tex] Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)} [/tex]
[tex] Q_L = 95.93 kJ/kg [/tex]
Energy balance gives
[tex] W_{net} = Q_H - Q_L [/tex]
[tex] W_{net} = 122.5 kJ/kg - 95.93 kJ/kg [/tex]
[tex] W_{net} = 26.57 kJ/kg [/tex]
Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)
[tex] x = \frac{s_1 - s_f}{s_g - s_f} [/tex]
From figure
[tex] s_1 = s_4 = 1.165 kJ/(K kg) [/tex]
Replacing with table values
[tex] x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)} [/tex]
[tex] x = 0.37 [/tex]
Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get
[tex] h_1 = h_f + x \times (h_g - h_f) [/tex]
[tex] h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg [/tex]
[tex] h_1 = 241.58 kJ/kg [/tex]
By energy balance, [tex] W_{t} [/tex] turbine's work is
[tex] W_{t} = |h_1 - h_4| [/tex]
[tex] W_{t} = |241.58 kJ/kg - 249.7 kJ/kg| [/tex]
[tex] W_{t} = 8.12 kJ/kg [/tex]
Finally, [tex] W_{c} [/tex] compressor's work is
[tex] W_{c} = W_{net} + W_{t}[/tex]
[tex] W_{c} = 26.57 kJ/kg + 8.12 kJ/kg[/tex]
[tex] W_{c} = 34.69 kJ/kg [/tex]
What is 220 C in degrees Fahrenheit (F)?
Answer:
428°F
Explanation:
The equation to convert degrees Celsius to degrees fahrenheit is
°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32
°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)
What is the function of air preheater? How does air preheating save fuel?
Answer:
Air preheater:
Air preaheater is a heat exchanger which take heat from flue gases and increase the temperature of the air before entering into the combustion process.
As we know that ,flue gases contains high temperature and that temperature can be used by air preheater and it will reduce the amount of heat addition in the combustion process and that leads to increase the efficiency of plant.When heat addition reduces for the same out put of power it means that less amount of fuel is required for the same out put of power as without using air preheater. The main purpose of air preheater is to increase the thermal efficiency.
By increasing the temperature of air it will reduce the amount of fuel reduce for the combustion.
What are the parameters that affect life and drag forces on an aerofoil?
Answer:
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Explanation:
Drag force:
Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.
Drag force given as
[tex]F_D=\dfrac{1}{2}\rho\ A\ V^2[/tex]
So we can say that drag force depends on following properties
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotational speed of 620 rpm. Determine the torque of the input shaft shaft, in N-m.
Answer:
Torque at input shaft will be 176.8695 N-m
Explanation:
We have given input power [tex]P_{IN}=42.6KW=42.6\times 10^3W[/tex]
Angular speed = 2300 rpm
For converting rpm to rad/sec we have multiply with [tex]\frac{2\pi }{60}[/tex]
So [tex]2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec[/tex]
We have to find torque
We know that power is given by [tex]P=\tau \omega[/tex], here [tex]\tau[/tex] is torque and [tex]\omega[/tex] is angular speed
So [tex]42.6\times 10^3=\tau \times 240.855[/tex]
[tex]\tau =176.8695N-m[/tex]
So torque at input shaft will be 176.8695 N-m
A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.
Answer:
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
Explanation:
Given that
Diameter(d)=62 mm
Thickness(t)= 300 μm=0.3 mm
Internal pressure(P)=100 KPa
Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .
The hoop stress
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
Longitudinal stress
[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]
Now by putting the values
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
[tex]\sigma _h=\dfrac{100\times 62}{2\times 0.3}[/tex]
[tex]\sigma _h=10.33MPa[/tex]
[tex]\sigma _l=5.16MPa[/tex]
So the principle stress are
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is 120 MPa?
Answer:
119.35 mm
Explanation:
Given:
Inside diameter, d = 100 mm
Tensile load, P = 400 kN
Stress = 120 MPa
let the outside diameter be 'D'
Now,
Stress is given as:
stress = Load × Area
also,
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-d^2)[/tex]
or
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
thus,
400 × 10³ N = 120 × [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
or
D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]
or
D = 119.35 mm
Answer:
D =119.35 mm
Explanation:
given data:
inside diameter = 100 mm
load = 400 kN
stress = 120MPa
we know that load is given as
[tex]P = \sigma A[/tex]
where:
P=400kN = 400000N
[tex]\sigma = 120MPa[/tex]
[tex]A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)[/tex]
[tex]A=\frac{1}{4} \pi (D^2 - 10000)[/tex]
putting all value in the above equation to get the required diameter value
[tex]400 = 120*\frac{1}{4} \pi (D^2 - 10000)[/tex]
solving for
D =119.35 mm
For two 0.2 m long rotating concentric cylinders, the velocity distribution is given by u(r) = 0.4/r - 1000r m/s. If the diameters are 2 cm and 4 cm, respectively, calculate the fluid viscosity if the torque on the inner cylinder is measured to be 0.0026 N*m.
Answer:
5.9*10^-3 Pa*s
Explanation:
The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.
τ = μ * du/dr
u(r) = 0.4/r - 1000*r
The derivative is:
du/dr = -1/r^2 - 1000
On the radius of the inner cylinder this would be
u'(0.02) = -1/0.02^2 - 1000 = -3500
So:
τ = -3500 * μ
We don't care about the sign
τ = 3500 * μ
That is a tangential force per unit of area.
The area of the inner cylinder is:
A = h * π * D
And the torque is
T = F * r
T = τ * A * D/2
T = τ * h * π/2 * D^2
T = 3500 * μ * h * π/2 * D^2
Then:
μ = T / (3500 * h * π/2 * D^2)
μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. Determine the minimum required wire diameter assuming a factor of safety 5 and that the yield strength of the steel is 900MPa.
Answer:
11.6 mm
Explanation:
With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:
σadm = strength / fos
σadm = 900 / 5 = 180 MPa
The stress is the load divided by the section:
σ = P / A
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
[tex]d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm[/tex]
If 3.7 grams of a gas contains 3.7 × 10^22 molecules, what is the molar mass of this gas in units of g/mol?
Answer:
Molar mass of the gas will be 60.65 gram/mole
Explanation:
We have given mass of gas = 3.7 gram
Gas contains [tex]3.7\times 10^{22}[/tex]
We know that any gas contain [tex]6.022\times 10^{23}[/tex] molecules in 1 mole
So number of moles [tex]=\frac{3.7\times 10^{22}}{6.022\times 10^{23}}=0.061[/tex]
We know that number of moles [tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]
So [tex]0.061=\frac{3.7}{molar\ mass}[/tex]
Molar mass = 60.65 gram/mole
A room with dimensions of 3 x 10 x 20 m is estimated to have outdoor air brought in at an infiltration rate of 1/4 volume changes per hour. Determine the infiltration rate in m^3/s.
Answer:
The infiltration rate is of 0.042 m^3/s.
Explanation:
The total volume of the room is:
V = 3 * 10 * 20 = 600 m^3
If the air infiltration rate is of 1/4 volume per hour:
v = V/4 * 1 hour /3600 seconds
Replacing:
v = 600/4 * 1/3600 = 0.042 m^3/s
The infiltration rate would then be of 0.042 m^3/s.
A heat pump with a 2 kW motor is used to heat a building at 30 deg C. The building loses heat at a rate of 0.5 kW per degree difference to the colder ambient at T amb. The heat pump has a coefficient of performance that is 50 % of a carnot heat pump. What is the maximum ambient temperature for which the heat pump is sufficient?
Answer:
T=5.3° C
Explanation:
Given that
Power input to the pump = 2 KW
Building loses heat rate = 0.5 KW/C
So rate of heat transfer = 0.5(273+30-T)
rate of heat transfer = 0.5(303-T)
T=Ambient temperature
Building temperature = 30° C
We know that ,heat pump is used to heat the building.
COP of pump = 0.5 COP of Carnot heat pump
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{273+30}{303-T}[/tex]
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{303}{303-T}[/tex]
[tex]COP\ of\ pump=\dfrac{303-T}{Power}[/tex]
[tex]COP\ of\ pump=0.5\times \dfrac{303-T}{2}[/tex] -----1
And also
[tex]COP\ of\ pump=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex] ----2
So from now equation 1 and 2
[tex]\dfrac{303-T}{4}=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]
So T= 278.38 K=5.3° C
T=5.3° C
Ambient temperature =5.3° C.
a valueable preserved biological specimen is weighed by suspeding it from a spring scale. it weighs 0.45 N when it is suspendedin air and 0.081 N when it is suspended in a bottle of alchol what is its dencity?
Answer:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is [tex]1.557\times10^{-7}[/tex]m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
Here, v is velocity, [tex]\nu[/tex] is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
[tex]Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}[/tex]
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
Define the hydraulic diameter for a rectangular duct
Answer with Explanation:
Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.
It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.
The hydraulic diameter is 4 times the hydraulic radius of any section.
For a rectangular duct as shown in the attached figure
[tex]R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)} [/tex]
Where
[tex]D_{h}[/tex] is the hydraulic diameter of the duct with depth 'd' and width 'b'
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