am having trouble understanding this problem specifically where i highlighted are we using SG 0.075. what r the units?? attaced is solutiom from manual
am having trouble understanding this problem specifically where i highlighted are we using SG 0.075. what r the units?? attaced is solutiom from manual
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
i am having trouble understanding this problem specifically where i highlighted
are we using SG 0.075. what r the units??
attaced is solutiom from manual
![Substitute the value of T, and T, in equation (5);
Pout ₂
Pin I₁
Hence,
=
3660
528
6.93
(Pin-Pout) = PinVin
28 lbm
29 ft³
Substitute 0.075×-
(Pin-Pout) = 0.075x
Comments (2)
Pout
Pin
ft
for Pin and 2 for Vand 6.93 for
S
28 lbm
29 ft³
= 3.7×10+ psi
Therefore, the pressure difference is 3.7×10+
ft
x2 (6.93-1)x;
S
ps1
Pout
Pin
lbf s²
ft²
·X
32.2 lbm ft 144 in²
I B
(0)
A
O
engine at 800 ft/s and leaves at 1500 ft/s. The mass of fuel added to the exhaust
gas may be neglected. How many pounds of air per second flow through the
engine?
In a steady-state methane-air flame at approximately atmospheric pressure, the
temperature is raised from 70 to 3200°F. The incoming air-gas mixture and the
products of combustion may both be considered ideal gases with a molecular
weight of 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow.
If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure
difference from one side of the flame to the other? This problem and its
consequences are discussed elsewhere [9].
7.7. A pipe is delivering a liquid in steady laminar flow, as described in Chap. 6. What
is the ratio of the momentum in the outflowing stream to the momentum which
would exist if the flow were of uniform velocity over the entire cross section of the
pipe?
7.8. The pipe bend in Fig. 7.20 is attached to the rest of the piping system by two
flexible hoses which transmit no forces. The fluid enters in the x direction and
leaves in the-y direction. The flow rate is 500 lbm/s, and the inlet and outlet
velocities are each 100 ft/s. The cross-sectional areas at points 1 and 2 are both
0.5 ft². The pressure at point 1 is 50 psig and at point 2 is 40 psig. Calculate F, and
Fy, the x and y components, respectively, of the force in the pipe support.
Flow
y
F₂
FIGURE 7.20](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63f5865a-7713-44c4-be73-1c5678c4385b%2Fa93000c4-e097-414f-adb5-e36b89c67881%2Fqnmg4y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Substitute the value of T, and T, in equation (5);
Pout ₂
Pin I₁
Hence,
=
3660
528
6.93
(Pin-Pout) = PinVin
28 lbm
29 ft³
Substitute 0.075×-
(Pin-Pout) = 0.075x
Comments (2)
Pout
Pin
ft
for Pin and 2 for Vand 6.93 for
S
28 lbm
29 ft³
= 3.7×10+ psi
Therefore, the pressure difference is 3.7×10+
ft
x2 (6.93-1)x;
S
ps1
Pout
Pin
lbf s²
ft²
·X
32.2 lbm ft 144 in²
I B
(0)
A
O
engine at 800 ft/s and leaves at 1500 ft/s. The mass of fuel added to the exhaust
gas may be neglected. How many pounds of air per second flow through the
engine?
In a steady-state methane-air flame at approximately atmospheric pressure, the
temperature is raised from 70 to 3200°F. The incoming air-gas mixture and the
products of combustion may both be considered ideal gases with a molecular
weight of 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow.
If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure
difference from one side of the flame to the other? This problem and its
consequences are discussed elsewhere [9].
7.7. A pipe is delivering a liquid in steady laminar flow, as described in Chap. 6. What
is the ratio of the momentum in the outflowing stream to the momentum which
would exist if the flow were of uniform velocity over the entire cross section of the
pipe?
7.8. The pipe bend in Fig. 7.20 is attached to the rest of the piping system by two
flexible hoses which transmit no forces. The fluid enters in the x direction and
leaves in the-y direction. The flow rate is 500 lbm/s, and the inlet and outlet
velocities are each 100 ft/s. The cross-sectional areas at points 1 and 2 are both
0.5 ft². The pressure at point 1 is 50 psig and at point 2 is 40 psig. Calculate F, and
Fy, the x and y components, respectively, of the force in the pipe support.
Flow
y
F₂
FIGURE 7.20
![Consider momentum balance equation for the steady flow in x-direction;
m(Vin -Vout)+F=0
(1)
Here, m is the mass flow rate, V₁ and Vout are the velocities of the incoming and outgoing
gas mixture respectively and F is the force which is pressure difference between inlet and
outlet.
The mass flow rate is given by the relation;
Pin AV in
(2)
Here is the density of the gas at the inlet, A is the area and Vin is the velocity of the
incoming gas flow.
Force in the system is related as;
A(Pin-Pout)
Here, P and Pout are the pressures at inlet and outlet respectively.
Comment
Step 2 of 5
The relation between density and velocity is given by the relation;
Vout
Pout
Pin
V=V₁
out
Pout
Pin
Substitute equations (2), (3), (4) in equation (1);
m (Vin -Vout)+F=0
(0.₂))
(3)
Substitute equations (2), (3), (4) in equation (1);
m (Vin -Vout) + F = 0
Pin AVin Vin -Vinout +A(P₁-P) = 0
Pin
AV₂ [V₂. =))+²
2 -1)=(
PV2 Pout -1 = (P-Pout)
Pin
The relation between density and temperature is given by the relation;
Pout - T
T₂
=
Pin T₁
Comment
Convert temperature in Fahrenheit to temperature in Rankine using the relation;
T[°R]=T[°F]+460
Step 3 of 5
Substitute 68°F for the temperature in Fahrenheit;
T[°R]=68+460
T = 528°R
The value of T₁ is 528°R
Comment
Step 4 of 5
(5)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63f5865a-7713-44c4-be73-1c5678c4385b%2Fa93000c4-e097-414f-adb5-e36b89c67881%2Fnczw14_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider momentum balance equation for the steady flow in x-direction;
m(Vin -Vout)+F=0
(1)
Here, m is the mass flow rate, V₁ and Vout are the velocities of the incoming and outgoing
gas mixture respectively and F is the force which is pressure difference between inlet and
outlet.
The mass flow rate is given by the relation;
Pin AV in
(2)
Here is the density of the gas at the inlet, A is the area and Vin is the velocity of the
incoming gas flow.
Force in the system is related as;
A(Pin-Pout)
Here, P and Pout are the pressures at inlet and outlet respectively.
Comment
Step 2 of 5
The relation between density and velocity is given by the relation;
Vout
Pout
Pin
V=V₁
out
Pout
Pin
Substitute equations (2), (3), (4) in equation (1);
m (Vin -Vout)+F=0
(0.₂))
(3)
Substitute equations (2), (3), (4) in equation (1);
m (Vin -Vout) + F = 0
Pin AVin Vin -Vinout +A(P₁-P) = 0
Pin
AV₂ [V₂. =))+²
2 -1)=(
PV2 Pout -1 = (P-Pout)
Pin
The relation between density and temperature is given by the relation;
Pout - T
T₂
=
Pin T₁
Comment
Convert temperature in Fahrenheit to temperature in Rankine using the relation;
T[°R]=T[°F]+460
Step 3 of 5
Substitute 68°F for the temperature in Fahrenheit;
T[°R]=68+460
T = 528°R
The value of T₁ is 528°R
Comment
Step 4 of 5
(5)
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