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3 years ago

Physical properties of minerals

2 answers:

6 0

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

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sesenic [268]3 years ago

4 0

All minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

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The heat transfer rate due to free convection from a vertical surface, 1 m high and 0.6 m wide, to quiescent air that is 20 K co

andreyandreev [35.5K]

Answer:

The ratio of heat transfer rate is 0.88

Explanation:

Given;

Case1 :

height of vertical surface, L = 1 m

width of vertical surface, w = 0.6 m

Case 2:

height of vertical surface, L = 0.6 m

width of vertical surface, w = 1 m

At an assumed film temperature of air = 300 K

then, read off from heat transfer table, temperature inverse β, surface area flow rate v, and Pr, to determine Rayleigh number for the two cases.

β = 1/300 = 0.00333 K⁻¹

v = 15.89 x 10⁻⁶ m²/s

Pr = 0.69

Case 1, L = 1 m

$R_a = \frac{g\beta TL^3P_r}{v^2}$

$R_a = \frac{9.8*0.00333* 20*1^3*0.69}{(15.89x10^{-6})2} \\\\R_a = 1.784 *10^9$

Case 2, L = 0.6 m

$R_a = \frac{g\beta TL^3P_r}{v^2} \\\\R_a = \frac{9.8*0.00333* 20*0.6^3*0.69}{(15.89*10^{-6})^2}\\\\ R_a = 3.853 *10^8$

From the values of Rayleigh numbers above, case 1 is Turbulent flow while case 2 is laminar flow

Thus: C₁ = 0.1, n₁ = ¹/₃

C₂ = 0.59, n₂ = 1/4

Ratio of heat transfer rate is given as:

$\frac{q_1}{q_2} = \frac{h_1 \delta T}{h_2 \delta T} \\\\\frac{q_1}{q_2} = \frac{h_1}{h_2} \\\\But, \frac{hL}{k} = CR_a^n L, \ \ h=\frac{k}{L}(CR_a^n L)\\\\\frac{q_1}{q_2} = \frac{C_1R_a_1^n L_2}{C_2R_a_2^n L_1} = \frac{0.1(1.784*10^9)^{\frac{1}{3}} *0.6}{0.59(3.853*10^8)^{\frac{1}{4}} *1} \\\\\frac{q_1}{q_2} = \frac{72.76}{82.66} = 0.88$

Therefore, the ratio of heat transfer rate is 0.88

4 0

3 years ago

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

sergeinik [125]

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

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

6 0

3 years ago

Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pre

dezoksy [38]

The answer is 6because I just took the test!

5 0

2 years ago

9 b. A sign (including the post and base) weighs 40 pounds and is

tigry1 [53]

Answer:

The weight should be added to the base of the sign to keep it from tipping is 65.6 lb

Explanation:

Given data:

A sigh weighs 40 pounds

Suported by an 18 in x 18 in square

Force of the wind 13.2 lb

Questions: Will the sign tip over, if yes, how much evelnly distributed weight should be added to the base of the sign to keep it from tipping, W = ?

The sign and the post have a length of 6 ft. You need to calculate the distance from the edge to the middle point:

18/2 = 9 in = 0.75 ft

Force acting in the base (40 lb):

$F=\frac{40*0.75}{6} =5lb$

The weight should be added to the base:

$(40+W)*\frac{0.75}{6} =13.2\\W=65.6lb$

8 0

4 years ago

Which of the following has special properties that allow forces and pressure to be distributed evenly?

Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
Therefore, when pressure is applied to them, it permeates evenly on all parts.
Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.

7 0

3 years ago