10. The repair order is a legal document because

kirill [66]

Answer:

From the main bearings, the oil passes through feed-holes into drilled passages in the crankshaft and on to the big-end bearings of the connecting rod.

3 0

2 years ago

Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate o

topjm [15]

Answer:

Heat transfer = 2.617 Kw

Explanation:

Given:

T1 = 300 k

T2 = 440 k

h1 = 300.19 KJ/kg

h2 = 441.61 KJ/kg

Density = 1.225 kg/m²

Find:

Mass flow rate = 1.225 x [1.3/60]

Mass flow rate = 0.02654 kg/s

mh1 + mw = mh2 + Q

0.02654(300.19 + 240) = 0.02654(441.61) + Q

Q = 2.617 Kw

Heat transfer = 2.617 Kw

4 0

3 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago

Technician A says that the first step in diagnosing engine condition is to perform a thorough visual inspection. Technician B sa

mojhsa [17]

Answer:

Technician A

Explanation: Technician A is correct. Technician B is wrong, as an oil leak can trickle down onto other engine components, away from where the leak actually is.

5 0

3 years ago

Luids of viscosities 1  =0.1 N.s/ms and 2  =0.15 N.s/m2 are contained between two plates (each plate 1

Fantom [35]

Answer:

a) 1 / 3 N

b) 5/3 m/s

Explanation:

For constant speed V at the interface of the two fluids the net force is zero.

Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.

$t_{top} = t_{bottom}\\where\\t = u.\frac{dU}{dt} \\Hence\\(\frac{V - 1}{h_{2} })*u_{top} = (\frac{V}{h_{1} })*u_{bottom} \\\\0.3V = 0.5\\\\V = 5/3 m/s$

For force of top plate

$F_{top} = u_{top} * \frac{V-U}{h_{2} }*A=( 0.15)*(\frac{5/3 -1}{0.3 } )*(1)\\F_{top} = \frac{1}{3} N$

5 0

3 years ago