Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K
![\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}](https://tex.z-dn.net/?f=%5CDelta%20%5Cbar%7Bt%7D_%7Ba%7D%20%3D%5Cfrac%7Bt_%7BA_%7B1%7D%7D%2Bt_%7BA_%7B2%7D%7D%7D%7B2%7D%20-%20%5Cfrac%7Bt_%7BB_%7B1%7D%7D%2Bt_%7BB_%7B2%7D%7D%7D%7B2%7D)
Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;
![\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K](https://tex.z-dn.net/?f=%5CDelta%20%5Cbar%7Bt%7D_%7Ba%7D%20%3D%5Cfrac%7B1100%2B733.33%7D%7B2%7D%20-%20%5Cfrac%7B300%2B850%7D%7B2%7D%20%3D%20341%5Ctfrac%7B2%7D%7B3%7D%20%5C%2C%20K%20%3D%20341.67%20%5C%2C%20K)
Hence, from;
, we have
5912500 = 90 × A × 341.67
![A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B5912500%20%20%7D%7B90%20%5Ctimes%20341.67%7D%20%3D%20192.3%20%5C%2C%20m%5E2)
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
C. Multipoint fuel injection
Explanation:
A fuel injection system can be defined as a system found in the engine of most automobile cars, used for the supply of a precise amount of fuel or fuel-air mixture to the cylinders in an internal combustion engine through the use of an injector.
There are different types of fuel injection system and these includes;
I. Central-point injection.
II. Throttle (single point) body injection.
III. Gasoline direct injection.
IV. Multipoint (port) fuel injection.
Multipoint fuel injection is a type of fuel injection system that operates with fuel injectors located only in the intake manifold near each intake valve and sprays fuel toward the valve. As a result, it allows for the supply of a precise amount of fuel and as such creating a better air-fuel ratio for automobile cars.
Answer:
Bore = 7 cm
stroke = 6.36 cm
compression ratio = 10.007
Explanation:
Given data:
Cubic capacity of the engine, V = 245 cc
Clearance volume, v = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
V = ![\frac{\pi}{4}D^2L](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7DD%5E2L)
or
V = ![\frac{\pi}{4}\frac{D^3}{1.1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cfrac%7BD%5E3%7D%7B1.1%7D)
on substituting the values, we have
245 = ![\frac{\pi}{4}\frac{D^3}{1.1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cfrac%7BD%5E3%7D%7B1.1%7D)
or
D = 7.00 cm
Now,
we have
D/L = 1.1
thus,
L = D/1.1
L = 7/1.1
or
L= 6.36 cm
Now,
the compression ratio is given as:
![\textup{compression ratio}=\frac{V+v}{v}](https://tex.z-dn.net/?f=%5Ctextup%7Bcompression%20ratio%7D%3D%5Cfrac%7BV%2Bv%7D%7Bv%7D)
on substituting the values, we get
![\textup{compression ratio}=\frac{245+27.2}{27.2}](https://tex.z-dn.net/?f=%5Ctextup%7Bcompression%20ratio%7D%3D%5Cfrac%7B245%2B27.2%7D%7B27.2%7D)
or
Compression ratio = 10.007
Answer:
IDK
Explanation:
same thing is happening to me