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2 years ago

What is the purpose of O-ring and valve seals in a cylinder head?

Engineering

1 answer:

Andrews [41]2 years ago

8 0

Answer:

its to show the shape is flat and only flat at the botom and top and you can set it up ther way and it wlll still look the same.

Explanation:

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Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

Furkat [3]

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

$W = \dot{m}c_p \Delta T$

Where

$C_p =$ Specific heat

$\dot{m}$ = Mass flow rate

$\Delta T$ = Change in Temperature

Our values are given as

$C_p = 1.005kJ/kgK \rightarrow$ Specific heat of air

$T_1 = 50\°C$

$\dot{m} = 2kg/s$

$W = 8kW$

From our equation we have that

$W = \dot{m}c_p \Delta T$

$W = \dot{m}c_p (T_2-T_1)$

Rearrange to find $T_2$

$T_2 = \frac{W}{\dot{m}c_p}+T_1$

Replacing

$T_2 = \frac{8}{2*1.005}+(50+273)$

$T_2 = 326.98K \approx 53.98\°C$

Therefore the exit temperature of air is 53.98°C

6 0

3 years ago

The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes

kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value