Wheel diameter = 150 mm, and infeed = 0.06 mm in a surface grinding operation. Wheel speed = 1600 m/min, work speed = 0.30 m/s, and crossfeed = 5 mm. The number of active grits per area of wheel surface = 50 grits/cm2. Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work.

Answer:

4.536hp

Explanation:

The decrease in the heat gain of the room is determined from difference in electrical inputs:

[tex]Q = W_{shaft} (\frac{1}{n_{1} } - \frac{1}{n_{2} })\\Q = (75hp)*(\frac{1}{0.91 } - \frac{1}{0.963 })\\\\Q = 4.536 hp[/tex]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

  q=(struct node *)malloc(sizeof(struct node)*count);

  dfa->initialStateID = -1;

  dfa->presentStateID = -1;

  dfa->totalNumOfStates = 0;

  return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

  int tID;

  for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

       if (pPresentState->transitions[tID].condition(c))

      {

          dfa->presentStateID = pPresentState->transitions[tID].toStateID;

          return;

      }

  }

  dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{  

  newState->ID = pDFA->numOfStates;

  pDFA->states[pDFA->numOfStates] = newState;

  pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

  DFAState* state = dfa->states[fromStateID];

  state->transitions[state->numOfTransitions].toStateID = toStateID;

  state->numOfTransitions++;

}

void reset(DFA* dfa)

{

  dfa->presentStateID = dfa->initialStateID;

}

Answer:

The air velocity is 1442.3m/s

The temperature gradient is 0.00311K/m

Explanation:

Rate of heat transfer = local conductive heat transfer × area

Rate of heat transfer = force × distance/time (distance/time = velocity)

Therefore, rate of heat transfer = force × velocity

Force (F) × velocity (v) = local conductive heat transfer coefficient (k) × Area (A)

F/A × v = k

Shear stress = F/A = 312N/m^2

k = 450kW/m^2 = 450×1000W/m^2 = 450000W/m^2

312 × v = 450000

v = 450000/312 = 1442.3m/s

Air velocity (v) = 1442.3m/s

Temperature gradient = Temperature (T)/distance (s)

From equations of motion

v^2 = u^2 + 2gs

u = 0m/s, v = 1442.3m/s, g = 9.8m/s^2

1442.3^2 = 2×9.8×s

s = 2080229.29/19.6 = 106134.15m

Temperature gradient = 330K/106134.15m = 0.00311K/m

Answer:

The solution is shown in the images attached with the answer.

Explanation:

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

The time rate of change of the hotplate's temperature is 0.0062 K/s. The magnitude of the heat losses by convection and radiation is 64.23 W.

The question is asking for the time rate of change of the plate temperature when it is at 250°C and the magnitude of the heat losses by convection and by radiation.

First, transform the initial plate temperature from Celsius to Kelvin, so 250°C = 523.15 K.

The air temperature is also given in Celsius, which is 25°C = 298.15 K.

Next, we calculate the heat loss due to convection using the formula Q_conv = h * A * (T_plate - T_air), where h is the convection coefficient, A is the surface area of the plate, and T_plate and T_air are the temperatures of the plate and the air, respectively.

Substituting the given values, we get: Q_conv = 6.4 W/m^2.k * 0.25 m * 0.25 m * (523.15 K - 298.15 K) = 1.80 W.

The heat loss due to radiation can be calculated using the Stefan-Boltzmann law: Q_rad = ε * σ * A * (T_plate^4 - T_surrounding^4), where ε is the emissivity, σ is the Stefan-Boltzmann constant (5.67 * 10^-8 W/m^2.K^4), and T_surrounding is the surrounding temperature.

Again plugging in the given values, we get the heat loss due to radiation as Q_rad = 0.42 * 5.67 * 10^-8 W/m^2.K^4 * 0.25 m * 0.25 m * (523.15 K^4 - 298.15 K^4) = 62.43 W.

So, the total heat loss Q = Q_conv + Q_rad = 1.80 W + 62.43 W = 64.23 W.

To find the time rate of change of the temperature, we use the formula: dT/dt = Q / (m*C), where dT/dt is the time rate of change of the plate temperature, m is the mass, and C is the specific heat. Substituting the values, we get: dT/dt = 64.23 W / (3.75 kg * 2770 J/kg.K) = 0.0062 K/s.

https://brainly.com/question/34419089

#SPJ3

Answer:

charge on the 24 nF capacitor is 84 nC

Explanation:

given data

capacitance Q1 = 16 nF

capacitance Q2 = 24 nF

charge on the 16 nF capacitor C1 = 56 nC

solution

we get here capacitor in parallel have same voltage

V1 = V2   ...............1

here voltage V1 = [tex]\frac{Q1}{C1}[/tex]

[tex]\frac{Q1}{C1}[/tex] = [tex]\frac{Q2}{C2}[/tex]    .....................2

put here value we get

[tex]\frac{56}{16} = \frac{Q2}{24}[/tex]

Q = 84 nC

so charge on the 24 nF capacitor is 84 nC

The charge on the 24 nF capacitor, when paired in parallel with a 16 nF capacitor connected to the same battery, is determined to be 84 nC.

When two capacitors are connected in parallel and then to a battery, they both will have the same voltage across them. Given that the charge (Q) on a capacitor is equal to the product of its capacitance (C) and the voltage (V), and assuming the 16 nF capacitor has a charge of 56 nC, the charge on the 24 nF capacitor can be determined using the formula Q = CV.

First, find the voltage using the 16 nF capacitor:

Since the capacitors are in parallel, the voltage across the 24 nF capacitor is also 3.5 V. Therefore, the charge on the 24 nF capacitor is:

The charge on the 24 nF capacitor is 84 nC.