melinda is using a rectangular brass bar in a sculpture she is creating. the brass bar has a length that is 4 more than 3 times
2 answers:
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The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×
m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.
= 150×
/1
= 0.15
/ Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13
/ Kg.
= 350+(400-350)
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
=
= 0.15
/ Kg
In saturated water labels for = 40°C.
= 0.001008
/ Kg
= 19.515
/ Kg
The final state is two phase region <
<
.
In saturated water labels for = 40°C.
=
= 7.3851 kPa
= 7.3851 kPa
Answer:
2
Explanation:
So for solving this problem we need the local heat transfer coefficient at distance x,
We integrate between 0 to x for obtain the value of the coefficient, so
Substituing
The ratio of the average convection heat transfer coefficient over the entire length is 2
Answer:
volumetric flow rate =
Velocity in pipe section 1 =
velocity in pipe section 2 = 12.79 m/s
Explanation:
We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.
The density of water is = 997 kg/m³
density = mass/ volume
since we are given the mass, therefore, the volume will be mass/density
25/997 =
volumetric flow rate =
Average velocity calculations:
<em>Pipe section A:</em>
cross-sectional area =
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
<em>Pipe section B:</em>
cross-sectional area =
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)