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oksian1 [2.3K]
2 years ago
6

Problematic substances mechanical engeneering???

Engineering
1 answer:
Sonja [21]2 years ago
6 0

Answer:

The climate crisis. ...

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Enlisting the youth

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A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
Explain when it is appropriate to use Tier II?
garik1379 [7]

????? what do u mean tier 2

3 0
2 years ago
"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state.
Harlamova29_29 [7]

Answer:

See attached images for the diagrams and tables

3 0
3 years ago
Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc
aksik [14]

Answer:

a) \eta = 13.455\%, b) E_{day} = 812.716\,kJ, c) C_{month. total} = 19.505\, USD, d) t = 40.588\,years

Explanation:

a) The area of the solar panel is:

A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}

A = 1.858\,m^{2}

The energy potential is determined herein:

\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})

\dot E_{o} = 1858\,W

The efficiency of the solar panel is:

\eta = \frac{\dot E}{\dot E_{o}}\times 100\%

\eta = \frac{250\,W}{1858\,W}\times 100\%

\eta = 13.455\%

b) The energy generated by the solar panel is presented below:

E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )

E_{day} = 812.716\,kJ

c) The energy generated per month and per panel is:

E_{month} = 30\cdot E_{day}

E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ}  \right)

E_{month} = 6.773\,kWh

Monthly energy savings due to the use of 18 panels are:

C_{month, total} = 18\cdot E_{month}\cdot c

C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )

C_{month. total} = 19.505\, USD

d) The payback of the solar energy system is:

t = \frac{9500\,USD}{12\cdot (19.505\,USD)}

t = 40.588\,years

6 0
3 years ago
Given A = {1,2,42,57,99,538,677}, B = {1,5,6,7,{2,3} , and C ={1,{7}, {8,9}}, answer the following questions, use the proper not
professor190 [17]

Answer:

c < 7

Explanation:

The cardinal indicates the number or quantity of the elements of a set, be this finite or infinite quantity.

Given, A = {1, 2, 42, 57, 99, 538, 677}  

B = {1, 5, 6, 7, {2, 3} }

C = {1, {7}, {8, 9} }

Let's say the cardinality of the subset is: c

Inequality that represents the cardinality of any proper subset of A is: c < 7

Hope this helps!

8 0
3 years ago
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