A horizontal pipe has an abrupt expansion from D1 5 8 cm to D2 5 16 cm. The water velocity in the smaller section is 10 m/s and
the flow is turbulent. The pressure in the smaller section is P1 5 410 kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure P2, and estimate the error that would have occurred if Bernoulli’s equation had been used.
1 answer:
- Answer: Explanation: Application of the bernoulli's equation comes in from conservation of mass flow. The cross sectional area of the two pipes are calculated. from A = πD²/4 The velocity of water from conservation of mass flow is also calculated ; V2 = Ac1V1/Ac2 The Loss coefficient is then calculated from KL = (1 - Ac1²/Ac2²)² Then the head Loss (hL) is calculated The detailed calculated and appropriate steps is as shown in the attached files.
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Answer:
Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1
State2
So, the total mass of the aire entered is
At this point we need to obtain the properties through the tables, so
For Specific Internal energy,
For Specific enthalpy
For the second state the Specific internal Energy (6bar, 350K)
At the end we make a Energy balance, so
No work done there is here, so clearing the equation for Q
The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu