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zzz [600]
3 years ago
10

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?
Engineering
1 answer:
blagie [28]3 years ago
6 0

Answer:

a) L = 1.5

L_q = 0.9

W =  \dfrac{1 }{8 } \, hour

W_q =  \dfrac{3}{40 } \, hour

P = \dfrac{3}{5 }

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

L = \dfrac{\lambda }{\mu -\lambda }

μ = Service rate

L_q = The number of customers waiting in line

L_q = p\times L

W = The time spent waiting including being served

W = \dfrac{1 }{\mu -\lambda }

W_q = The time spent waiting in line

W_q = P \times W

P = The system utilization

P = \dfrac{\lambda }{\mu  }

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;

L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5

P = \dfrac{12 }{20 } = \dfrac{3}{5 }

L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9

W = \dfrac{1 }{20 -12 } =  \dfrac{1 }{8 } \ hour

W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour

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

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

P = \dfrac{12 }{40  } = \dfrac{3}{10}

W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour

λ = 12 trucks/hour

Total cost = mC_s + \lambda WC_w

m = 1

C_s = GH¢ = 1300

C_w = 400

Total cost with the old backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with the new backacter is given as follows;

1 \times 1300 + 12 \times \dfrac{1}{38}  \times 400 = \$ 1,426.32

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

\therefore W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

Total cost with the one backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with two backacters is given as follows;

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

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

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8 0
3 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
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Answer:

1) The three possible assumptions are

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b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

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The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

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T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0
3 years ago
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