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Travka [436]
3 years ago
9

We have considered heat and work to be path-dependent. However, if all heat transfer with surroundings is performed using a reve

rsible heat transfer device (some type of reversible Carnot-type device), work can be performed by the heat transfer device during heat transfer to the surroundings. The net heat transferred to the surroundings and the net work done will be independent of the path. Demonstrate this by calculating the work and heat interactions for the system, the heat transfer device, and the sum for each of the following paths where the surroundings are at Tsurr = 273 K. The state change is from state 1, P1 = 0.1 MPa, T1 = 298 K and state 2, P2 = 0.5 MPa and T2 which will be found in part (a). CP = 7R/2. a. Consider a state change for an ideal gas in a piston/cylinder. Find T2 by an adiabatic reversible path. Find the heat and work such that no entropy is generated in the universe. This is path a. Sketch path a qualitatively on a P-V diagram. b. Now consider a path consisting of step b, an isothermal step at T1, and step c, an isobaric step at P2. Sketch and label the step on the same P-V diagram created in (a). To avoid generation of entropy in the universe, use heat engines/pumps to transfer heat during the steps. Calculate the WEC and WS as well as the heat transfer with the surroundings for each of the steps and overall. Compare to part (a) the total heat and work interactions with the surroundings. c. Now consider a path consisting of step d, an isobaric step at P1, and step e, an isothermal step at T2. Calculate the WEC and WS as well as the heat transfer with the surroundings for each of the steps and overall. Compare to part (a) using this pathway and provide the same documentation as in (b).
Engineering
1 answer:
NeX [460]3 years ago
3 0

Answer:

path-dependent. However, if all heat transfer with surroundings is performed using a reversible heat transfer device (some type of reversible Carnot-type device), work can be performed by the heat transfer device during heat transfer to the surroundings. The net heat transferred to the surroundings and the net work done will be independent of the path. Demonstrate this by calculating the work and heat interactions for the system, the heat transfer device, and the sum for each of the following paths where the surroundings are at Tsurr = 273 K. The state change is from state 1, P1 = 0.1 MPa, T1 = 298 K and state 2, P2 = 0.5 MPa and T2 which will be found in part (a). CP = 7R/2. a. Consider a state change for an ideal gas in a piston/cylinder. Find T2 by an adiabatic reversible path. Find the heat and work such that no entropy is generated in the universe. This is path a. Sketch path a qualitatively on a P-V diagram. b. Now consider a path

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zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

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Read 2 more answers
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
Which one of the following questions about population growth is the only TRUE statement?A) The size of a population can never ex
tatyana61 [14]

Answer:

Explanation:

5

6 0
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