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3 years ago

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When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more

popular? How do you know?

1 answer:

irakobra [83]3 years ago

8 0

Answer:

how many people were asked though

Explanation:

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What is the approximate theoretical maximum efficiency of a heat engine receiving heat at 627°C and rejecting heat to 27°c? a)-2

AysviL [449]

Answer:

The correct option is (e) η = 0.66

Explanation:

given data:

temperature at entry is 627 degree C

temperature at exit is 27 degree C

the efficiency of a engine is given as η

$\eta = (1-\frac{T_{L}}{T_{H}})*100$

where $T_{L}$ is temperature in kelvin at exit point

$T_{H}$ is temperature in kelvin at entry point

$\eta = (1-\frac{27+273}{627+273})*100$

η = 66.66 %

5 0

3 years ago

What are three reasons why you should make your first prototype on a smaller scale?

erastovalidia [21]

Here’s a few -
Save time.
Save money.
Understand the problem better.
Get feedback quickly.
Procrastinate less and get started quickly.
Build a better product through many iterations.

Hope this helps !

8 0

3 years ago

Read 2 more answers

Which of the following is an example of a pulley?

alexira [117]

Answer:

<em><u>Door knob</u></em>

Explanation:

When the knob on one side of the door is turned, the shaft retracts the spring-loaded latch that holds the door closed. Without the knob in place, more force would be needed to turn the shaft and retract the latch.

3 0

3 years ago

Read 2 more answers

A soil sample has a moist unit weight of 117 pcf, a moisture content of 17 percent, and soil particles with a specific gravity o

nikklg [1K]

Answer:

(A) Dry unit weight. = 15.71 kN/m³ =100.91 pcf

(B) Porosity = 40.82 %

(C) Degree of saturation = 66.52 %

(D) Weight of water, in pounds per cubic foot, to be added to reach full saturation = 81.22 pcf

Explanation:

(A) γd = $\frac{\gamma }{1+w}$ = $\frac{18.379}{1+.17} = 15.71$ KN/m³

(B) $\gamma _d =\frac{G_{s}* \gamma _w }{1+e}$ therefore $1+e = \frac{G_{s}* \gamma _w }{\gamma _d }$ $\frac{2.7*9.81}{15.71} =$ 1.69

Therefore e = 0.69 and

Porososity n = $\frac{e}{1+e}$ = $\frac{0.69}{1+0.69}$ × 100% = 40.82 %

(C) $S_{e} = w*G_{s}$ therefore S = $\frac{w*G_{s}}{e}$ = $\frac{0.17*9.81}{0.69}$ ×100 = 66.52 %

(D) $\gamma _{sat} =\frac{(G_{s}+e) \gamma_{w} }{1+e}$ = $\frac{(2.7+0.69)9.81}{1+0.69}$ = 19.68 kN/m³

The required amount of water is found from γ =ρ×g

ρ mass of water = $\frac{\gamma}{g} = \frac{(19.68-18.38)\frac{kN}{m^{3} } }{9.81 kg\frac{m}{s^{2} } } *\frac{1000 N}{kN} *\frac{9.81 kg\frac{m}{s^{2} } }{N} = 1301 \frac{kg}{m^{3} }$

1 kg/m³ = 0.062 pcf therefore 1301 kg/m³ = 1301 ×0.062 pcf or 81.22 pcf

6 0

3 years ago

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

ivolga24 [154]

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg, s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291 - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ = 1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683 + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

$y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284$

The turbine control volume is given as follows;

$\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2} \right ) + \left (1 - y \right )\left (h_{2} - h_{3} \right )$

= (3507.41 - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

$\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6} \right ) + \left (1 - y \right )\left (h_{5} - h_{4} \right )$

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 - 151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

$\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7} \right )$

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

$\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}$

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, $\dot{m_{1}}$ , into the first turbine stage is given as follows;

$\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}$

$\dot{m_{1}}$ = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

$\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}$

$\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5} \right ]$

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

4 0

3 years ago