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Anna007 [38]
3 years ago
3

A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 27 ft wide and 93 ft long

. When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

Engineering
1 answer:
Monica [59]3 years ago
8 0

Answer:

a) for equilibrium

sum of F vertical is zero

F_{vertical} =0

so that

W_{b} =F_{B}=r_{h2o} *(submerged volume)\\=(62.4lb/ft^3)(5ft*27ft*93ft)\\=783432 lb

b)

F_{vertical} =0\\W_{b}+ W_{mg} =r_{h2o} *(submerged volume)\\W_{g} =(62.4lb/ft^3)*(7ft*27ft*93ft)-783432\\W_{g} =313372lb

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Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at
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Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

h_{f} = 173.358 \\h_{fg} = 2402.522

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

h_{2a} =  489.752\\h_{2b} =  313.2

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a}  - h_{4a}) - (h_{2a}  - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26  - 2241.448938 ) - (489.752  - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3a} -  h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4a} -  h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.06485

Part b) @ 4 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b}  - h_{4b}) - (h_{2b}  - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14  - 2405.54119 ) - (313.12  - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3b} -  h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4b} -  h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.038275

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A motor and a control system are connected to the hoist.

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When the steel cable is wound up again, the mine cage and skips are elevated.

The sheave wheel is a pulley wheel that stands above the mining shaft in Part 2. The hoist rope travels over the sheave wheel and down the mine shaft.

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Part 3: The head frame is the framework that holds the sheave wheel in place. When lifting the heavy mine cage, it must be robust enough to maintain the sheave wheel in place.

The head frame's left "legs" slope towards the hoist. This is due to the cable's strain dragging the entire frame in that direction. The sloping legs keep the head frame from tipping or collapsing.

Part 4: The cage and the jumps. Miners and equipment are transported up and down the mine in the cage. Skips are attached beside or beneath the cage.

Skips are used to transport ore and waste materials from mines.

<h3>What is a design brief?</h3>

A design brief, also called as a creative brief, is a program management document that identifies the scope, scale, and key aspects of your impending design project.

Learn more about design brief:

brainly.com/question/21422013

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liquid water enters a cooling tower operating at a steady state 40ºC with a mass flow rate of 105 kg/h. Cooled water at 25ºC exi
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