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3 years ago

How many D-cell batteries would it take to power a human for 1 day?

Engineering

1 answer:

lara31 [8.8K]3 years ago

6 0

Answer:

it would take approximately 232 to 258 D cell batteries to power a human for 1 day.

Explanation:

Estimate the recommended daily food intake (in Food Calories).

An average adult man requires between 2000 to 3000 calories per day.

Estimate the daily energy (in joule) a human needs.

As we know 1 food calorie is equal to 4.184 Joules of energy

2000*4.184 to 3000*4.184

8368 to 12552 Joules

But for engineering calculations 1 food calorie is equal to 1000 engineering calories, so

8368*1000 to 12552*1000

8368000 to 12552000 Joules

Estimate the voltage (in volt) of one D cell battery.

The voltage of a D cell battery is around 1.5 Volts

Estimate the charge (in amp-hours) of one D cell battery.

The amp-hours of a D cell battery varies with the manufacturing company, the typical amp-hours are in the range of 6 to 10 amp-hours.

Estimate the energy of one D cell battery (in watt-hour).

Energy in watt-hour is given by

voltage*amp-hour

1.5*6 to 1.5*10

9 to 15 watt-hour

Estimate the number of D-cell batteries it takes to power a human for 1 day.

First let us calculate the energy in a D cell battery,

1 watt-hour is equal to 3600 Joules

9*3600 to 15*3600

32400 to 54000 Joules

The number of D cell batteries required is found by dividing the energy need of a human by the energy stored in a D cell battery.

12552000/54000 to 8368000/32400

232 to 258 batteries

Therefore, it would take approximately 232 to 258 D cell batteries to power a human for 1 day.

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3 years ago

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

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$h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}$

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5 0

3 years ago

In this assignment, you will write a user interface for your calculator using JavaFX. Your graphical user interface (GUI) should

Zolol [24]

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

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import javafx.scene.Group;

import javafx.scene.Scene;

import javafx.scene.layout.VBox;

import javafx.scene.layout.HBox;

import javafx.scene.control.TextField;

import javafx.scene.control.Button;

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Scene scene = new Scene(root, 200, 294);

mainBox.getChildren().add(inpBox);

HBox rowOne = new HBox();

Button btn7 = new Button("7");

btn7.setMinWidth(50);

btn7.setMinHeight(50);

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btn8.setMinWidth(50);

btn8.setMinHeight(50);

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btn9.setMinWidth(50);

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Button btn5 = new Button("5");

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btn6.setMinWidth(50);

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Button btnMul = new Button("*");

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Button btn1 = new Button("1");

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btn2.setMinWidth(50);

btn2.setMinHeight(50);

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btnSub.setMinWidth(50);

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rowThree.getChildren().addAll(btn1,btn2,btn3,btnSub);

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btnEq.setMinHeight(50);

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4 0

3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

3 years ago