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4 years ago

I need solution for this question pleaseSelect the right answer

Engineering

2 answers:

Dafna11 [192]4 years ago

8 0

Answer:

1.E aluminum

2.E all

3. Either d or e leaning to e

Ostrovityanka [42]4 years ago

3 0

Explanation:

the 7th one answer is beacause mercury is bad at sharing electrons

the 8th one's answer is Rhodium

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The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h

marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

Inner Diameter d_i = 32 mm

Outer Diameter d_o = 39 mm

E_b = 93 GPa

α_b = 15.1*10^-6 / °C

- Ceramic Core:

Outer Diameter d_o = 32 mm

E_c = 310 GPa

α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

б_b = - 60 MPa

- The force equilibrium can be written as:

б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

A_b is the cross sectional area of the shell

A_c is the cross sectional area of the core

б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0

б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0

б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

б_c = 29.121 MPa , б_b = - 60 MPa

- The total strains in both brass shell and ceramic core is given by:

ξ_b = α_b*ΔT + б_b / E_b

ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

ξ_b = ξ_c

α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

ΔT = 62.11°C

8 0

3 years ago

A. An 80-percent efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at ar ate of 1.5 ft3 /

ss7ja [257]

Answer:

A) mechanical power used to overcome frictional effects in piping = 2.373 HP

B) efficiency = 66.82%

Pressure difference = 147 KPa

Explanation:

We are given;

efficiency of pump; η = 80% = 0.8

power input of pump;P_pump,in = 20 hp

rate being pumped; Q = 1.5 ft³/s = 0.04248 m³/s

free surface of pool; Δz = 80 ft = 24.384 m

mechanical pumping power used to deliver water is;

W_u = η × P_pump,in

W_u = 0.8 × 20

W_u = 16 Hp

Now, change in total mechanical energy of water will be equal to the change in potential energy.

Thus, it is expressed as;

ΔE_mech = ρ × Q × g × Δz

Where ρ is density of water = 1000 kg/m³

Thus;

ΔE_mech = 1000 × 0.04248 × 9.81 × 24.384 = 10,161.5150592 J/s

Converting to HP gives;

ΔE_mech = 13.627 HP

Now, mechanical power used to overcome frictional effects is given by;

W_frict = W_u - ΔE_mech

W_frict = 16 - 13.627

W_frict = 2.373 HP

B) We are now given;

W_elect,in = 15.4 KW

Volumetric flow rate; q = 70 l/s = 0.07 m³/s

Height; h = 15 m

Mass flow rate; m' = qρ

Where ρ is density of water = 1000 kg/m³.

m' = 0.07 × 1000

m' = 70 kg/s

Now,

ΔE_mech = workdone through height of 15m

Thus;

ΔE_mech = m'gh

ΔE_mech = 70 × 9.8 × 15

ΔE_mech = 10,290 W = 10.29 Kw

efficiency of the pump–motor unit = ΔE_mech/(W_elect,in) = 10.29/15.4 = 0.6682 or 66.82%

Pressure difference = ρgh = 1000 × 9.8 × 15 = 147,000 Pa = 147 KPa

3 0

3 years ago

... is an actual sequence of interactions (i.e., an instance) describing one specific situation; a ... is a general sequence of

Alexandra [31]

Answer:

Scenario
Use case
Scenarios
Scenarios
Use case

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.