True/False

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given :

$PV^{1.3}=C$

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

$\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}$

where $r_{c}$ is compression ratio = $\frac{v_{c}+v_{s}}{v_{c}}$

$r_{c}$ = $1+\frac{v_{s}}{v_{c}}$

where $v_{c}$ is compression volume

$v_{s}$ is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, $v_{30}=v_{c}+0.7v_{s}$

and at 70% compression, 30% of the swept volume remains

∴ $v_{70}=v_{c}+0.3v_{s}$

We know,

$\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}$

$\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}$

$\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\$

$1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}$

$v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}$

$0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}$

$0.2218v_{s} = 0.594v_{c}$

$v_{c}=0.3734 v_{s}$

∴ $r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}$

Therefore, compression ratio is $r_{c}$ = 3.678

Now efficiency, $\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}$

$\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}$

$\eta =0.32342$ , this is the ideal efficiency

Therefore actual efficiency, $\eta_{act} =0.5\times \eta _{ideal}$

$\eta_{act} =0.5\times 0.32342$

$\eta_{act} =0.1617$

Therefore total power required = 1 kW x 3600 J

= 3600 kJ

∴ we know efficiency, $\eta=\frac{W_{net}}{Q_{supply}}$

$Q_{supply}=\frac{W_{net}}{\eta _{act}}$

$Q_{supply}=\frac{3600}{0.1617}$

$Q_{supply}=22261.78 kJ$

Therefore fuel required = $\frac{22261.78}{45000}$

= 0.4947 kg/hr

5 0

4 years ago

Problem 8. Define a function gs1_error(n) that accepts as input a non-negative integer n and outputs the error of the Gauss-Seid

Fofino [41]

Answer:

The MATLAB code will be:

close all;

clc;

clear all;

a = [20, 1, -2; 3, 20, -1; 2,-3, 20];

b = [17; -18; 25];

tol = 1e-6;

[M,N] = size(a);

if M~=N

error('A is not a square Matrix');

end

x = [0,0,0];

xx(1,:) = x;

n = 100;

error = 0.00001;

for k = 2:n

for i = 1:N

s = 0;

for j = 1:N

if j ~= i

s = s + a(i,j) * x(j);

end

x(i) = (1/a(i,i)*(b(i) - s));

end

xx(k,:) = x;kk = k;

Error = abs(max(xx(kk,:)- xx(k-1,:)))

if Error<error

break;

end

fprintf ('The roots of equation are: ')

x

fprintf ('The number of iterations are: ')

k

fprintf ('The error is: ')

Error

Explanation:

7 0

3 years ago

If a plus sight of 12.03 ft is taken on BM A, elevation 312.547 ft, and a minus sight of 5.43 ft is read on point X, calculate t

DochEvi [55]

Answer:

Therefore, height of instrument is 324.577 ft

Therefore, elevation of point x is 330 m

Explanation:

Given that

Plus sight on BM = 12.03 ft

Minus sight is = 5.43 ft

Elevation = 312.547 ft

Height of instrument is H.I

H.I = elevation on bench mark + plus sight

= 312.547 + 12.03 = 324.577 ft

Therefore, height of instrument is 324.577 ft

Elevation at point x is = H.I - minus sight

= 324.577 - (- 5.43)

= 330.00 m

Therefore, elevation of point x is 330 m

3 0

4 years ago

The two basic categories of electrical switches are manual and ??

Ganezh [65]

Answer:

are Manual and Automatic

4 0

3 years ago

Read 2 more answers

Is a water proof material used around tubs and

Leviafan [203]

I think it’s hard board

4 0

3 years ago

Read 2 more answers