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;
}
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of 2770 J/kg·K, and a temperature of 250°C. The ambient air temperature is 25°C and the surroundings temperature is 25°C. If the convection coefficient is 6.4 W/m2·K and the emissivity of the plate is 0.42, determine the time rate of change of the plate temperature, , when the plate temperature is 250°C. Evaluate the magnitude of the heat losses by convection and by radiation.
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.
Explanation: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.
Learn more about Heat Transfer here:https://brainly.com/question/34419089
#SPJ3
Consider an aircraft traveling at high speed. At a point on its wing, the local shear stress is 312 N/m2, and the local conductive heat transfer to the wing is 450 kW/m2. Calculate the air velocity and temperature gradients normal to the surface assuming that the surface has a temperature of 330 K.
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:
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor that has an efficiency of 96.3 percent. Determine the reduction in the heat gain of the room due to higher efficiency under full-load conditions.
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]
Two capacitors with capacitances of 16 nF and 24 nF, respectively, are connected in parallel. This combination is then connected to a battery. If the charge on the 16 nF capacitor is 56 nC, what is the charge on the 24 nF capacitor
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.
Explanation: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:
V = Q/C = 56 nC / 16 nFV = 3.5 VSince 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:
Q = CV = 24 nF × 3.5 VQ= 84 nCThe charge on the 24 nF capacitor is 84 nC.
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: a) 3mm
b) 5400mm^3/min
c) 4000000chips/min
Explanation:
Wheel diameter(D) =150mm
Infeed(W)=0.06mm
Wheel speed(V)=1600m/min
Work speed(Vw)=0.3m/s
Cross feed(d)=5mm
Number of active grits per area of wheel surface =50grits/cm^2
Average length per chip(Lc)=?
Metal removal rate(Rmr)=?
Number of chips formed per unit time(nc)=?
a) Lc=(Dd)^0.5
Lc=(150*0.06)^0.5
Lc=3mm
b) Rmr=VwWd
Rmr=(0.3m/s)*(10^3mm/m)*(5mm)*(0.06mm)
Rmr=5400mm^3/min
c) nc=VWc
nc=(1600m/min)(10^3)(5mm)(50grits/cm^2)(10^-2)
nc=4,000,000chips/min.