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2 years ago

Which of the following describes what happened when engineers created a separate pump for each actuator?

Engineering

1 answer:

Archy [21]2 years ago

4 0

Answer:

Technician A says the use of a larger input piston gives the system a mechanical advantage or increase in the force available to do work. Technician B says the multiplication of force through a hydraulic system is directly proportional to the difference in the piston sizes throughout the system.

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Who play 1v1 lol unblocked games 76

nikklg [1K]

Answer:

I did not what is it about?

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2 years ago

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A three-story school has interior column bays that are spaced 25 ft apart in both directions. If the loading on the flat roof is

Lady_Fox [76]

Answer:

Explanation:

Floor Load:

Lo= 50psf

At= 25x25 = 625 square feet

$L= Lo(0.25 +15/\sqrt{KuAt)}$

$L=50(0.25+15/\sqrt{(4)(625)}$ = 13.1psf

%reduction= 13.1/50 = 26%

Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k

7 0

2 years ago

thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat

eduard

Answer:

$q_{in} = 528.6\,\frac{kJ}{kg}$

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

$q_{in} = h_{g} - h_{mix}$

The specific enthalpies are:

Liquid-Vapor Mixture:

$h_{mix} = 2217.2\,\frac{kJ}{kg}$

Saturated Vapor:

$h_{g} = 2745.8\,\frac{kJ}{kg}$

The thermal energy per unit mass required to heat the steam is:

$q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}$

$q_{in} = 528.6\,\frac{kJ}{kg}$

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2 years ago

Anytime scaffolds are assembled or __________, a competent person must oversee the operation.

egoroff_w [7]

D. made because made is a synonym for assembled

4 0

2 years ago

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A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th

trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

- The mass flow rate of steam through the boiler

- The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

E_6 + E_2 = E_3

flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

y*h_6 + (1-y)*h_3 = h_3

y*2810 + (1-y)*192.8 = 721

Compute y: y = 0.2018

- Heat produced by the boiler q_b:

q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b: q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

q_c = (1-y)*(h_8 - h_1)

q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c: q_c = 1834.26 KJ/ kg

- Net power output w_net:

w_net = q_b - q_c

w_net = 3303.58 - 1834.26

w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

flow(m) = P / w_net

compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

u = 1 - (q_c / q_b)

u = 1 - (1834.26/3303.58)

u = 44.48 %

5 0

2 years ago