A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks are used for this purpose. If it was found out that on an average of 12 trucks per hour arrived and the average loading time was 3 minutes for each truck. A truck must queue until it is loaded. The backacter’s daily all-in rate is GH¢ 1000 and that of the truck is GH¢ 400.
a) Compute the operating characteristics: L, Lq, W, Wq, and P.

b) The company is considering replacing the backacter with a bigger one which will have an average service rate of 1.5 minutes to serve trucks waiting to have their schedules improved. As a manager, would you recommend the new backacter if the daily all-in rate is GH¢ 1300.

c) The site management is considering whether to deploy an extra backwater to assist the existing one. The daily all-in-rate and efficiency of the new backwater is assumed to be the same as that of the existing backwater. Should the additional backwater be deployed?

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: 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:

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:

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]

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:

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

                          Hope this helped!  Have a great day!

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

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:

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)