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vodomira [7]
3 years ago
15

Identify an object in your house that contains a physical system and list three questions you could use to define the system

Engineering
1 answer:
jonny [76]3 years ago
4 0

Answer:

ALL CAREFULLY ANSWERED CORRECTLY

Explanation:

1) A loaf of Bread PHYSICAL SYSTEM

✓ How can the environment affect the edibility of the bread

✓ What are the constituents that makes up the bread

✓ What process is involved in these constituents mixing to form the loaf.

2) The law of thermodynamics makes us to understand that when heat/energy passes through a system, the systems internal energy changes with respect to the conservation of energy law. That is energy lost = energy gained. Typically, ice would melt in a cup of hot tea because of the thermal energy in the molecules of the hot tea. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The temperature of the ice raises due to the kinetic energy added to it and it melts to water.

3) The theory of systems view the world as a complex system of interconnected parts. If we consider the society; (financial systems, political systems, etc) we will agree that they individually have their own components and it's the summation of this components that makes the system, this implies that system thinking could be applicable in this kinda of systems as long as they are made up of components.

4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.

5) ~ Nature shows its splendid soul

~Never ceases to leave us in amazement

~And we are in love

You might be interested in
How to design a solar panel<br>​
artcher [175]

Answer:

#1) Find out how much power you need

#2 Calculate the amount of batteries you need.

#3 Calculate the number of solar panels needed for your location and time of year.

#4 Select a solar charge controller.

#5 Select an inverter.

#6 Balance of system

Explanation: To design solar panel, consider the following steps

1.) Find the power consumption demands

The first step in designing a solar PV system is to find out the total power and energy consumption of all loads that need to be supplied by the solar PV system as follows:

Calculate total Watt-hours per day for each appliance used.

 Add the Watt-hours needed for all appliances together to get the total Watt-hours per day which must be delivered to the appliances.

Calculate total Watt-hours per day needed from the PV modules.

Multiply the total appliances Watt-hours per day times 1.3 (the energy lost in the system) to get the total Watt-hours per day which must be provided by the panels.

2. Size the PV modules

Different size of PV modules will produce different amount of power. To find out the sizing of PV module, the total peak watt produced needs. The peak watt (Wp) produced depends on size of the PV module and climate of site location. We have to consider panel generation factor which is different in each site location. For Thailand, the panel generation factor is 3.43. To determine the sizing of PV modules, calculate as follows:

2.1 Calculate the total Watt-peak rating needed for PV modules

Divide the total Watt-hours per day needed from the PV modules (from item 1.2) by 3.43 to get the total Watt-peak rating needed for the PV panels needed to operate the appliances.

Calculate the number of PV panels for the system

Divide the answer obtained in item 2.1 by the rated output Watt-peak of the PV modules available to you. Increase any fractional part of result to the next highest full number and that will be the 

number of PV modules required.

Result of the calculation is the minimum number of PV panels. If more PV modules are installed, the system will perform better and battery life will be improved. If fewer PV modules are used, the system may not work at all during cloudy periods and battery life will be shortened.

3. Inverter sizing

An inverter is used in the system where AC power output is needed. The input rating of the inverter should never be lower than the total watt of appliances. The inverter must have the same nominal voltage as your battery.

For stand-alone systems, the inverter must be large enough to handle the total amount of Watts you will be using at one time. The inverter size should be 25-30% bigger than total Watts of appliances. In case of appliance type is motor or compressor then inverter size should be minimum 3 times the capacity of those appliances and must be added to the inverter capacity to handle surge current during starting.

For grid tie systems or grid connected systems, the input rating of the inverter should be same as PV array rating to allow for safe and efficient operation.

4. Battery sizing

The battery type recommended for using in solar PV system is deep cycle battery. Deep cycle battery is specifically designed for to be discharged to low energy level and rapid recharged or cycle charged and discharged day after day for years. The battery should be large enough to store sufficient energy to operate the appliances at night and cloudy days. To find out the size of battery, calculate as follows:

     4.1 Calculate total Watt-hours per day used by appliances.

     4.2 Divide the total Watt-hours per day used by 0.85 for battery loss.

     4.3 Divide the answer obtained in item 4.2 by 0.6 for depth of discharge.

     4.4 Divide the answer obtained in item 4.3 by the nominal battery voltage.

     4.5 Multiply the answer obtained in item 4.4 with days of autonomy (the number of days that you need the system to operate when there is no power produced by PV panels) to get the required Ampere-hour capacity of deep-cycle battery.

Battery Capacity (Ah) = Total Watt-hours per day used by appliancesx Days of autonomy

(0.85 x 0.6 x nominal battery voltage)

5. Solar charge controller sizing

The solar charge controller is typically rated against Amperage and Voltage capacities. Select the solar charge controller to match the voltage of PV array and batteries and then identify which type of solar charge controller is right for your application. Make sure that solar charge controller has enough capacity to handle the current from PV array.

For the series charge controller type, the sizing of controller depends on the total PV input current which is delivered to the controller and also depends on PV panel configuration (series or parallel configuration).

According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.3

Solar charge controller rating = Total short circuit current of PV array x 1.3

5 0
3 years ago
This problem has been solved!
lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

A)

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0
3 years ago
What do you guys like in engineering
Drupady [299]

Answer:

building lol and actually workin

Explanation:

3 0
3 years ago
Read 2 more answers
In what way is a spud wrench different from a pipe wrench?<br><br> See picture
maks197457 [2]
A spud wrench and pipe is for install pipes
6 0
3 years ago
Read 2 more answers
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
3 years ago
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