A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it
1 answer:
You might be interested in
Answer:
a) -0.0934 kJ/kg. K
b) The direction of flow is from right to left.
Explanation:
A free flow diagram of the horizontal insulated duct is as shown below.
NOW,
Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:
Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.
If the value is less than zero; then we conclude that the assumption is wrong.
Then, the flow is said to be in the opposite direction
Formula for the change in specific entropy can be calculated as:
where;
are specific entropies
R = universal gas constant
= pressure at location 1
= pressure at location 2
We obtain the specific properties of air at temperature at = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)
= 1.8279 kJ/kg.K
We also obtain the specific properties of air at temperature = 22°C + 273) K = 295 K
From the table A- 22
= 1.68515 kJ/kg . K
R =
0.95 bar
0.8 bar
Now replacing our values into equation (2) from above; we have;
Equating our result to equation (1)
Therefore , our assumption is wrong and the direction of flow is said to be from right to left.
We therefore conclude that the direction of flow is from right to left.
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
