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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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Answer:

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4 0
3 years ago
Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h
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Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

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η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

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A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
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Answer:

\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

Given that tank is insulated so\dot{Q}=0 and no mass is leaving so

\dot{m_e}=0

\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt

m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

Mass conservation m_2-m_1=m_e-m_i

m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

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2.Due to presence of notch

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