How long does a car battery last?

If the resistance in a series circuit is 12 K Ohms, and the amperage is 0. 0063 amps, what is the voltage?

kkurt [141]

Answer:

The resistance “seen” by the 12-Volt battery is 500 ohms . The total resistance is the sum of the resistors R1 and R2. Each resistor will drop a calculable amount of voltage. This voltage is given by Ohm's Law.

3 0

2 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

10 Khz is equivalent to all except<br> a 1 Mhz<br> b. 10 KHZ<br> c..01 Mhz<br> d. 10000 Hz

xenn [34]

Answer: A 1 MHz

Explanation:

10Khs equals to 10,000Hz/10Khz/0.01Mhz/0.00001Ghz

4 0

3 years ago

8.17 Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 10 MPa, 4808C, and the condenser pre

katrin2010 [14]

Answer:

attached below

Explanation:

8 0

3 years ago

In a paragraph of 125 words, distinguish the engineering design process from the reverse engineering process, citing at least th

Yuki888 [10]

The difference between the engineering design process and the reverse engineering process are there, but very subtle. The engineering design process is a process based on careful planning and months of designing to create a blueprint for a certain project. The reverse engineering process is the base of learning to create something by working backwards on a previously made inspiration for your project. Three differences are that that the engineering design process doesn't require any physical learning and is more based on mental and written learning (unlike the reverse engineering process). Another difference includes that in the reverse engineering process you don't need to take anything apart and learn how things work based on the on hands aspect of this variety of engineering. The final difference is that the engineering design project is made fresh from your own thoughts and not based off of a similar project.

7 0

3 years ago

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