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 ₂Pin I₁Hence,=36605286.93(Pin-Pout) = PinVin28 lbm29 ft³Substitute 0.075×-(Pin-Pout) = 0.075xComments (2)PoutPinftfor Pin and 2 for Vand 6.93 forS28 lbm29 ft³= 3.7×10+ psiTherefore, the pressure difference is 3.7×10+ftx2 (6.93-1)x;Sps1PoutPinlbf s²ft²·X32.2 lbm ft 144 in²I B(0)AOengine at 800 ft/s and leaves at 1500 ft/s. The mass of fuel added to the exhaustgas may be neglected. How many pounds of air per second flow through theengine?In a steady-state methane-air flame at approximately atmospheric pressure, thetemperature is raised from 70 to 3200°F. The incoming air-gas mixture and theproducts of combustion may both be considered ideal gases with a molecularweight 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 pressuredifference from one side of the flame to the other? This problem and itsconsequences are discussed elsewhere [9].7.7. A pipe is delivering a liquid in steady laminar flow, as described in Chap. 6. Whatis the ratio of the momentum in the outflowing stream to the momentum whichwould exist if the flow were of uniform velocity over the entire cross section of thepipe?7.8. The pipe bend in Fig. 7.20 is attached to the rest of the piping system by twoflexible hoses which transmit no forces. The fluid enters in the x direction andleaves in the-y direction. The flow rate is 500 lbm/s, and the inlet and outletvelocities are each 100 ft/s. The cross-sectional areas at points 1 and 2 are both0.5 ft². The pressure at point 1 is 50 psig and at point 2 is 40 psig. Calculate F, andFy, the x and y components, respectively, of the force in the pipe support.FlowyF₂FIGURE 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 outgoinggas mixture respectively and F is the force which is pressure difference between inlet andoutlet.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 theincoming gas flow.Force in the system is related as;A(Pin-Pout)Here, P and Pout are the pressures at inlet and outlet respectively.CommentStep 2 of 5The relation between density and velocity is given by the relation;VoutPoutPinV=V₁outPoutPinSubstitute 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 = 0Pin AVin Vin -Vinout +A(P₁-P) = 0PinAV₂ [V₂. =))+²2 -1)=(PV2 Pout -1 = (P-Pout)PinThe relation between density and temperature is given by the relation;Pout - TT₂=Pin T₁CommentConvert temperature in Fahrenheit to temperature in Rankine using the relation;T[°R]=T[°F]+460Step 3 of 5Substitute 68°F for the temperature in Fahrenheit;T[°R]=68+460T = 528°RThe value of T₁ is 528°RCommentStep 4 of 5(5)

Answered: am having trouble understanding this…

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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