Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,
VEB=0.7V and VA=0.​

Answer:

a) [tex]L = 1.5[/tex]

[tex]L_q = 0.9[/tex]

[tex]W = \dfrac{1 }{8 } \, hour[/tex]

[tex]W_q = \dfrac{3}{40 } \, hour[/tex]

[tex]P = \dfrac{3}{5 }[/tex]

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

[tex]L = \dfrac{\lambda }{\mu -\lambda }[/tex]

μ = Service rate

[tex]L_q[/tex] = The number of customers waiting in line

[tex]L_q = p\times L[/tex]

W = The time spent waiting including being served

[tex]W = \dfrac{1 }{\mu -\lambda }[/tex]

[tex]W_q[/tex] = The time spent waiting in line

[tex]W_q = P \times W[/tex]

P = The system utilization

[tex]P = \dfrac{\lambda }{\mu }[/tex]

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

[tex]L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5[/tex]

[tex]P = \dfrac{12 }{20 } = \dfrac{3}{5 }[/tex]

[tex]L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9[/tex]

[tex]W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour[/tex]

[tex]W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour[/tex]

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

[tex]P = \dfrac{12 }{40 } = \dfrac{3}{10}[/tex]

[tex]W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]

[tex]W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour[/tex]

λ = 12 trucks/hour

Total cost = [tex]mC_s + \lambda WC_w[/tex]

m = 1

[tex]C_s[/tex] = GH¢ = 1300

[tex]C_w[/tex] = 400

Total cost with the old backacter is given as follows;

[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]

Total cost with the new backacter is given as follows;

[tex]1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32[/tex]

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

c)

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

[tex]\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]

Total cost with the one backacter is given as follows;

[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]

Total cost with two backacters is given as follows;

[tex]2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32[/tex]

The additional backacter will increase the total costs, therefore, it should not be deployed.

Answer:

  4.94 kW

Explanation:

The heat energy produced by the fuel in one hour is ...

  (2.29 L/h)(0.749 kg/L)(43.7 MJ/kg) = 74.954677 MJ/h

Then the power output is ...

  (74.954677 MJ/h)(1 h)/(3600 s) = 20.8207 kJ/s

Multiplying this heat energy by the efficiencies of the processes involved, the output power is ...

  (20.8207 kW)(0.25)(0.95) = 4.94 kW

Answer:Just multiply 90 by itself 46 times

Explanation:

do it

Answer:

i) VT = 52.16

VR = 147.85

ii) VT = 61

VR = 138.99

Explanation:

The step by step solution is been done, please check the attached file below to see it

Answer:

[tex]\| \vec A \| = 6.163\,m[/tex]

Explanation:

Sean A, B y C vectores coplanares tal que:

[tex]\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})[/tex], [tex]\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})[/tex] y [tex]\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})[/tex]

Donde [tex]\| \vec A \|[/tex], [tex]\| \vec B \|[/tex] y [tex]\| \vec C \|[/tex] son las normas o magnitudes respectivas de los vectores A, B y C, mientras que [tex]\theta_{A}[/tex], [tex]\theta_{B}[/tex] y [tex]\theta_{C}[/tex] son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

[tex]\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}[/tex]

[tex]\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}[/tex]

Asimismo, se sabe que [tex]\| \vec B \| = 5\,m[/tex], [tex]\theta_{B} = 60^{\circ}[/tex], [tex]\vec A \,\bullet \,\vec B = 30\,m^{2}[/tex], [tex]\vec B\, \bullet\, \vec C = 35\,m^{2}[/tex], [tex]\|\vec A \| = \| \vec C \|[/tex] y [tex]\theta_{C} = \theta_{A} + 25^{\circ}[/tex]. Entonces, las ecuaciones quedan simplificadas como siguen:

[tex]30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})[/tex]

[tex]35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})][/tex]

Es decir,

[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]

[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})][/tex]

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

[tex]\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}[/tex]

[tex]\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}[/tex]

Es decir,

[tex]\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}[/tex]

[tex]\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}[/tex]

Las nuevas expresiones son las siguientes:

[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]

[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})][/tex]

Ahora se simplifican las expresiones, se elimina la norma de [tex]\vec A[/tex] y se desarrolla y simplifica la ecuación resultante:

[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]

[tex]35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})[/tex]

[tex]\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}[/tex]

[tex]30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})[/tex]

[tex]122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}[/tex]

[tex]35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}[/tex]

[tex]\tan \theta_{A} = \frac{35.41}{65.6}[/tex]

[tex]\tan \theta_{A} = 0.540[/tex]

Ahora se determina el ángulo de [tex]\vec A[/tex]:

[tex]\theta_{A} = \tan^{-1} \left(0.540\right)[/tex]

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

[tex]\theta_{A, 1} \approx 28.369^{\circ}[/tex] y [tex]\theta_{A, 2} \approx 208.369^{\circ}[/tex]

Ahora, la magnitud de [tex]\vec A[/tex] es:

[tex]\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}[/tex]

[tex]\| \vec A \| = 6.163\,m[/tex]

Answer:

The uniform pressure for the necessary axial force  is  W = 945 N

The uniform wear for the necessary axial force is  W = 970.15 N

Explanation:

Solution

Given that:

r₁ = 0.1 m

r₂ = 0.05m

μ = 0.35

p = 12 N or kW

N = 1500 rpm

W = 1000 N

The angular velocity is denoted as  ω= 2πN/60

Here,

ω = 2π *1500/60 = 157.07 rad/s

Now, the power transferred becomes

P = Tω this is the equation (1)

Thus

12kW = T * 157.07 rad/s

T = 76.4 N.m

Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :

T = nμW R this is the equation (2)

The frictional surface of the mean radius is denoted by

R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]

=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]

R is =0.077 m

Now, we replace this values and put them into the equation (2)

It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m

n = 2.809 = 3

The number of pair surfaces is = 3

Secondly, we determine the uniform wear.

So, the mean radius is denoted as follows:

R = r₁ + r₂/ 2

=0.1 + 0.05/2

=0.075 m

Now, we replace the values and put it into the equation (2) formula

76. 4 N.m = n *0.35* 1000 N * 0.075 m

n= 2.91 = 3

Again, the number of pair surfaces = 3

However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.077 m

W = 945 N

Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.075 m

W = 970. 15 N

Answer:

Gear Y would turn Counter-Clockwise do to the opposite force created from gear X.

                          Hope this helped!  Have a great day!

Answer:

(a) 6.91 mm (b) 160 MPa

Explanation:

Solution

Given that:

E = 200 GPa

The rod length = 48 mm

P =P¹ = 6 kN

Recall that,

1 kN = 10^3 N

1 m =10^3 mm

I GPa = 10^9 N/m²

Thus

The rod deformation is stated as follows:

δ = PL/AE-------(1)

σ = P/A----------(2)

Now,

(a) We substitute the values in equation and obtain the following:

48 * 10 ^⁻3 m =  (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]

Thus, we simplify

A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²

A =0.0375 * 10 ^⁻3 m²

A =37.5 mm²

A = π/4 d²

Thus,

d² = 4A /π

After inserting the values we have,

d = √37.5 * 4/3.14 mm

= 6.9116 mm

or d = 6.91 mm

Therefore, the smallest that should be used is 6.91 mm

(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)

Thus,

σ = P/A

σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²

σ= 160 MPa

Note: I MPa = 10^6 N/m²

Hence the the corresponding normal stress is σ= 160 MPa

The correct question is;

A particular table in a relational database contains 100,000 rows, each of which requires

200 bytes of memory. Estimate the time in milliseconds to to insert a new row into the

table when each of the following indices on the related attribute is used. Assume a page

size of 4K bytes and a page access time of 20 ms.

a. No index (heap file)

b. A clustered, non-integrated B+ tree index, with no node splitting

required. Assume that each index entry occupies 100 bytes. Assume that the

index is 75% occupied and the actual data pages are 100% occupied. Assume

that all matching entries are in a single page.

Answer:

A) 20 ms

B) 120 ms

Explanation:

A) Append (at the end of file). Just one IO, i.e., 20 ms

B) Now, when we assume that each entry in the index occupies 100 bytes, then an index page can thus hold 40 entries. Due to the fact that the data file occupies 5000 pages, the leaf level of the tree must contain at least 5000/40 pages which is 125 pages.

So, the number of levels in the tree (assuming page 75% occupancy in the

index) is (log_30 (125)) + 1 = 3. Now, if we assume that the index is clustered and not integrated with the data file and all matching entries are in a single

page, then 4 I/O operations and 80ms are required to retrieve all matching

records. Two additional I/O operations are required to update the leaf page

of the index and the data page. Hence, the time to do the insertion is

120ms.

Answer:

screw thrust = ML[tex]T^{-2}[/tex] 

Explanation:

thrust of a screw propeller is given by the equation = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]T^{-1}[/tex]

Reynolds number is dimensionless so it cancels out

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L[tex]T^{-1}[/tex]

fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]

N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M[tex]L^{-3}[/tex][tex](LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1} L}{LT^{-1} }[/tex]

= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex]

screw thrust = ML[tex]T^{-2}[/tex] 

This is the dimension for a force which indicates that thrust is a type of force

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

Answer:

V(t) = XLI₀sin(π/2 - ωt)

Explanation:

According to Maxwell's equation which is expressed as;

V(t) = dФ/dt ........(1)

Magnetic flux Ф can also be expressed as;

Ф = LI(t)

Where

L = inductance of the inductor

I = current in Ampere

We can therefore Express Maxwell equation as:

V(t) = dLI(t)/dt ....... (2)

Since the inductance is constant then voltage remains

V(t) = LdI(t)/dt

In an AC circuit, the current is time varying and it is given in the form of

I(t) = I₀sin(ωt)

Substitutes the current I(t) into equation (2)

Then the voltage across inductor will be expressed as

V(t) = Ld(I₀sin(ωt))/dt

V(t) = LI₀ωcos(ωt)

Where cos(ωt) = sin(π/2 - ωt)

Then

V(t) = ωLI₀sin(π/2 - ωt) .....(3)

Because the voltage and current are out of phase with the phase difference of π/2 or 90°

The inductive reactance XL = ωL

Substitute ωL for XL in equation (3)

Therefore, the voltage across inductor is can be expressed as;

V(t) = XLI₀sin(π/2 - ωt)

Answer:

The value of relative density is 75 % while that of degree of saturation is 54.82%.

Explanation:

Given Data:

Bulk density of Sandy Soil=[tex]\gamma_b=18.8\ kN/m^3[/tex]

Void Ratio in Densest state=[tex]e_{max}=0.88[/tex]

Void Ratio in Loosest state=[tex]e_{min}=0.48[/tex]

Water content=[tex]w=12\%[/tex]

To Find:

Relative Density=[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%[/tex]

Degree of Saturation=[tex]S=\dfrac{w\times G_s}{e}[/tex]

Now all the other values are given except e. e is calculated as follows

e is termed as In situ void ratio and is given as

[tex]e=\dfrac{\gamma_w \times G_s-\gamma_d}{\gamma_d}[/tex]

Here

γ_w is the density of water whose value is 1

G_s is the constant whose value is 2.65

γ_d is the dry density of the sandy soil which is calculated as follows:

[tex]\gamma_d=\dfrac{\gamma_b}{1+\dfrac{w}{100}}[/tex]

Putting values

[tex]\gamma_d=\dfrac{18.8}{1+\dfrac{12}{100}}\\\gamma_d=16.78\ kN/m^3=1.678 g/cc \\[/tex]

Putting this value in the equation of e gives

[tex]e=\dfrac{1 \times 2.65-1.678}{1.678}\\e=0.579=0.58[/tex]

So the value of Relative density is given as

[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%\\D_R=\dfrac{0.88-0.58}{0.88-0.48} \times 100 \%\\D_R=75 \%[/tex]

So the value of relative density is 75 %

Now the value of degree of saturation is given as

[tex]S=\dfrac{w\times G_s}{e}\\S=\dfrac{12\times 2.65}{0.58}\\S=54.82 \%[/tex]

The value of degree of saturation is 54.82%.

Answer:

The relative density = 0.83 which is equivalent to 83%

The degree of saturation, S = 0.58 which gives 58% saturation

Explanation:

The parameters given are;

Water content W% = 12%

Bulk unit weight, γ = 18.8 kN/m³

Void ratio of  [tex]e_{min}[/tex]  = 0.48

Void ratio of  [tex]e_{max}[/tex] = 0.88

[tex]G_S[/tex] = Constant (As learnt from an answer to the question on the current page) = 2.65 for Sandy soil

[tex]\gamma =\dfrac{W}{V} = \dfrac{W_{w}+W_{s}}{V}[/tex]

Where, V = 1 m³

W = 18.8 KN

Bulk unit weight, γ = [tex]\gamma_d[/tex] × (1 + W)

∴ 18.8 =  [tex]\gamma_d[/tex] × (1 + 0.12)

[tex]\gamma_d[/tex] = 18.8/ (1.12) = 16.79 kN/m³

[tex]\gamma_d =\dfrac{W_s}{V} = \dfrac{W_{s}}{1} = 16.79 \, kN/m^3[/tex]

[tex]W_s[/tex] = 16.79 kN

∴ [tex]W_w[/tex] = 18.8 kN - 16.79 kN = 2.01 kN

[tex]m_w = 2.01/9.81 = 0.205 \, kg[/tex]

Volume of water = 0.205 m³

[tex]\gamma = \dfrac{GS \times \gamma _{w}\times \left (1+w \right )}{1 + e} = \dfrac{GS \times 9.81\times \left (1+0.12 \right )}{1 + e} =18.8[/tex]

e + 1 = 0.58×GS = 0.58×2.65 =

e = 1.54 - 1 = 0.55

The relative density is given by the relation;

[tex]Relative \ density, Dr=\dfrac{e_{max} - e}{e_{max} - e_{min}}[/tex]

[tex]Relative \ density, Dr=\dfrac{0.88 - e}{0.88 - 0.48} = \dfrac{0.88 - 0.55}{0.4} = 0.83[/tex]

The relative density = 0.83

The relative density in percentage = 0.83×100 = 83%

S·e = GS×w = 0.12·2.65

S×0.55 = 0.318

The degree of saturation, S = 0.58

The degree of saturation, S in percentage = 58%.

Answer:

The correct answer is (I) 290.81 W (II) 83.413 W (III) It is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy

Explanation:

Solution

Recall that:

Pressure (p) = 1.7 bar

Temperature (T₁) = 400°C which is = 673k

The velocity (v) = 50.0 m/s.

The pipe diameter (D)= 8.2 cm approximately  8.2 * 10 ^⁻2 m

The molecular air weight  (M)= 29 g/mol

Suppose Air is seen as an ideal gas

where pv = mrT

p =(m/v) r T = p = ρrT

So,

r = the characteristics of gas constant (R/m)

p = pressure

R =The universal gas constant

T = temperature

ρ = density which is (kg/m³)

R  is 8.314 J/mole -k

Then

1.7 * 10 ^5 = ρ * (8.314 /29) * 673

The density ρ = 881.09 g/ m³

(I) The mass flow rate = ρAV

thus,

m = 881.09 *π/4 ( 8.2 * 10^⁻2)² * 50

Therefore m = 232.65 g/s

We already know that

k= 1/2  mv²

k =1/2 *232.65/1000 * (50)²

so at 400°C k = 290.81 W

(II) Now in solving the  process of the constant pressure we recall that

P = ρrT

Air is cooled  to 250°C

p/r = ρT this is constant

So,

ρ₁T₂ = ρ₂T₂

881.09 * 673 = ρ₂ * 523

ρ₂ = 1133.79 g/m³

Thus,

m = ρ₂AV = 1133.79 * π /4 (8.2 * 10^ ⁻2) * 50

Hence m = 299.37 g/s

now,

k =1.2 mv² = 1.2 *(299.37)/1000 * (50)²

At 250°C, k = 374.22 W

Thus,

Δk = k ( 250°c) - k ( 400°c)

Δk = 374.22 - 290.81

Therefore,

Δk=83.413 W

(III) The steady state formula is given below

Q = W + ΔkE +ΔPE + ΔH

Now,

W = work (shaft)

Q =The rate of transfer of heat

ΔkE = The change in kinetic energy

ΔPE= The change in potential energy

ΔH =Change in enthalphy

For no shaft work, W =0

The horizontal pipe ΔPE = 0

Therefore,

The rate of heat transfer is explained as follows:

Q =ΔkE + ΔH

Because of the enthalphy,  Q is not equal to ΔkE

Finally, it is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy.

Answer:

ESTABLISHED

Explanation:

What is TCP?

A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.  

When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".

The most common socket states are:

LISTEN:

Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.

ESTABLISHED:

When a TCP connection is up and running meaning that both sides can send and receive data.

CLOSED:

The CLOSED state means that there is no TCP connection.

There are a total of 11 TCP socket states:

1. LISTEN

2. SYN-SENT

3. SYN-RECEIVED

4. ESTABLISHED

5. FIN-WAIT-1

6. FIN-WAIT-2

7. CLOSE-WAIT

8. CLOSING

9. LAST-ACK

10. TIME-WAIT

11. CLOSED

Answer:

a. 20.265 MN

b. 0.555 MNm

c. 403.44 KW

Explanation:

Given:-

- The width ( w ) = 1000 mm

- Original thickness ( to ) = 10 mm

- Final thickness ( t ) = 5 mm

- The radius of the rollers ( R ) = 600 mm

- The peripheral speed of the roller ( v ) = 0.12

- Deformation efficiency ( ε ) = 55%

Find:-

a) the roller force ( F )

b) the roller torque ( T )

c) the performance on the pair of rollers. ( P )

Solution:-

- The process of flat rolling entails a pair of compressive forces ( F ) exerted by the rollers on the steel sheet that permanently deforms.

- The permanent deformation of sheet metal is seen as reduced thickness.

- We will assume that the compressive force ( F ) acts normal to the point of contact between rollers and metal sheet.

- The roll force ( F ) is defined as:

                                 [tex]F =L*w*Y_a_v_g[/tex]

Where,

                     L: The projected length of strip under compression

                     Y_avg: The yielding stress of the material = 370 MPa

- The projected length of strip under compression is approximated by the following relation:

                               [tex]L = \sqrt{R*( t_o - t_f )} \\\\L = \sqrt{0.6*( 0.01 - 0.005 )} \\\\L = 0.05477 m[/tex]

- The Roll force ( F ) can be determined as follows:

                            [tex]F = (0.05477)*(1 )*(370*10^6 )\\\\F = 20.265 MN[/tex]

- The roll torque ( T ) is given by the following relation as follows:

                               [tex]T = \frac{L}{2} * F\\\\T = \frac{0.05477}{2} * 20.265\\\\T = 0.555 MNm[/tex]

- The rotational speed of the rollers ( N ) is determined by the following procedure:

                               [tex]f = \frac{v}{2\pi* R} = \frac{0.12}{2*\pi 0.6} = 0.03181818 \frac{rev}{s} \\\\N = f*60 = 1.9090 rpm[/tex]

- The power consumed by the pair of rollers ( P ) is given by:

                              [tex]P = \frac{2\pi * F * L * N}{e*60,000} KW \\\\P = \frac{2\pi * ( 20.265*10^6) * (0.05477) * (1.90909 ) }{60,000*0.55} KW\\\\P = 403.44 KW[/tex]

Answer:

The power produced by the turbine is 74655.936 kW.

Explanation:

A turbine is a device that operates at steady-state. Let suppose that turbine does not have heat interactions with surroundings, as well as changes in potential and kinetic energies are neglictible. Power output can be determined by First Law of Thermodynamics:

[tex]-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]

[tex]\dot W_{out} = \dot m\cdot (h_{in}-h_{out})[/tex]

Let suppose that water enters as saturated vapor and exits as saturated liquid. Specific enthalpies are, respectively:

[tex]h_{in} = 2748.1\,\frac{kJ}{kg}[/tex]

[tex]h_{out} = 225.94\,\frac{kJ}{kg}[/tex]

The power produce by the turbine is:

[tex]\dot W_{in} = \left(29.6\,\frac{kg}{s} \right)\cdot \left(2748.1\,\frac{kJ}{kg} - 225.94\,\frac{kJ}{kg} \right)[/tex]

[tex]\dot W_{in} = 74655.936\,kW[/tex]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

Answer:

(The diagram of the question is given in Attachment 1)

The largest load which can be applied is:

P=67.62 kN

Explanation:

All the forces are shown in the diagram in Attachment 2.

∑ F(y)= 0 (Upwards is positive)

[tex]F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P[/tex]

∑ F(x)= 0 (Towards Right is positive)

[tex]N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P[/tex]

∑ F(y)= 0 (Upwards is positive)

[tex]F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P[/tex]

The cross sectional area of a-a:

[tex]A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}[/tex]

σ = N(a-a)/A(a-a)

Substitute values:

[tex]160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN[/tex]

Т= V(a-a)/A(a-a)

Substitute values:

[tex]60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN[/tex]

To satisfy both the condition, we have to choose the lower value of P.

P=67.62 kN

Answer:

a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s

b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2

c) m' = 2.915 kg/s

Explanation:

Given:-

- The inlet pressure, Pi = 1.0 MPa

- The inlet temperature, Ti = 750 K

- Inlet velocity is negligible

- Steady, Isentropic Flow

- The specific heat ratio of air, k = 1.4

- Exit Mach number, Mae = 2

- The throat area, Ath = 20 cm^2

- Gas constant of air, R = 0.287 KJ / kg.K

Find:-

(a) the throat conditions (static pressure, temperature, density, and mach number)

b) the exit plane conditions

c) the mass flow rate

Solution:-

- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).

- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).

- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:

                             [tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]

- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:

                             [tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]

- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:

                            [tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]

                            [tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]

- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:

                           [tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]

- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:

                        [tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]  

- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:

                         [tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]

Answer:

  forever (no solution)

Explanation:

If the figure you're working with is the one shown below, no amount of mixing will reduce the solid content to 35%.

The vessel contains 40% solids.

The incoming feed is ...

  (100 kg/h)×(60% solids) = 60 kg solids/h

out of a total influx of material of ...

  (100 kg/h +70 kg/h) = 170 kg/h

That means the solids content of the inflow is ...

  (60 kg)/(170 kg) = 0.352941 ≈ 35.3%

The solids content cannot ever be less than 35.3%. The problem has no solution.

_____

We suggest you discuss this question with your teacher to see how they would solve it.

I'd say number 4, number 3 looks like an exhaust valve

Answer:

The entropy change of nitrogen during this process. is - 0.32628 kJ/K.

Explanation:

Solution

Given that:

A piston cylinder device contains =0.78 kg of nitrogen gas

Temperature = 37°C

The  nitrogen gas constant of R = 0.2968 kJ/kg.K

At room temperature cv = 0.743 kJ/kg.K

Now,

We assume that at specific condition the nitrogen can be treated as an ideal gas

Nitrogen has a constant volume specific heat at room temperature.

Thus,

From the polytropic relation, we have the following below:

T₂/T₁ =(V₁/V₂)^ n-1 which is,

T₂ = T₁ ((V₁/V₂)^ n-1

= (310 K) (2)^1.3-1 = 381.7 K

So,

The entropy change of nitrogen is computed as follows:

ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)

= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))

= 0.57954 * 0.2080 +  (-0.2057)

= 0.12058 + (-0.2057) = -0.32628

Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.

Answer:

[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]

Explanation:

Given that :

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)

i.e

k = 0.14 W/mK

∝ = 6.35 × 10⁻⁸ m²/s

L = 0.01 m

[tex]B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857[/tex]

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number [tex]F_o > 0.2[/tex]

⇒  [tex]\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)[/tex]

From Table 5.1 ; at [tex]B_1[/tex] = 14.2857

[tex]C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad[/tex]

[tex]\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}][/tex]

[tex]In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}[/tex]

[tex]-1.9399=-0.001350 *t_f[/tex]

[tex]t_f = \dfrac{-1.9399}{-0.001350}[/tex]

[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

Answer: The summation of the reaction forces is equal to the applied load.

Explanation:

When solving an indeterminate structure, it is important for one to satisfy the force-displacement requirements, compatibility, and the equilibrium of the structure.

With regards to the above question, for a statically indeterminate axially loaded member, the summation of the reaction forces will be equal to applied load. Here, the summation of the reactive forces will then be zero. This is a condition that is necessary to solve the unknown reaction forces.

Answer:

No. of Machines Required = 47

Explanation:

First, we calculate the number of parts that can be manufactured by one machine during the shift. For that purpose, we use following formula:

No. of Parts Manufactured by 1 Machine = Total Operating Time/Time taken to perform Operation

where,

Total Operating Time = 8 h - 1 h = (7 h)(60 min/h) = 420 min

Time taken to perform operation = 20 min

Therefore,

No. of Parts Manufactured by 1 Machine = 420 min/20 min

No. of Parts Manufactured by 1 Machine = 21

Now, we will calculate the no. of non-defective parts manufactured by 1 machine. Since, it is given that one-fifth of the parts manufactured by machine are defective. Therefore, the non-defective parts will be: 1 - 1/5 = 4/5 (four-fifth).

No. of Non- Defective Parts Manufactured by 1 Machine = N = (4/5)(21)

N = 16.8 = 16 (Since, the 17th part will not be able to complete in time)

So, the no. of machines required to produce 750 non-defective parts is given by:

No. of Machines Required = No. of non-defective parts required/N

No. of Machines Required = 750/16

No. of Machines Required = 46.9

No. of Machines Required = 47 (Since, 46 machines will not be able complete the job)