Two Pumfsdeterminepump arun3)both are runningconnectedin seriesin systetheoperationfointifサuy2) only fumpErunPumpAkfmi= 50 49-5 48-346.544 4239-5 36.532-52575too125 150 175225Pump Ahimt 4 39-539383733634 32 3658 2552575100125150175 200 225250SYStem !374043-549-5 50-5 545 59-5 6680130175z00a tam?/hrto 2550구5t00+25deluxe

Pump 1: ∆hm 50 49.5 48.5 46.5 44 42 39.5 36.5 32.5 Qm3/h 0 25 75 100 125 150 175 200 225…

Answered: Two Pumfs connected in series in syste…

Two Pumfs connected in series in syste determine pump 3run 3)both are running the operation Point if 1) outy 2) onty fump&run

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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Hi can yousolve by using bisection method E=((hf)/(ehf/kT-1)+(½)hf h=6.626*10^-27 k=1.38*10^-16 T=310E=3.97*10^-14a=10^12b=10^14
The horizontal force due to water acting on the bolt (Fh): Fh=ρ×A1×V1×V1--V2 Fh=1000×π4×0.0752×1.254×1.254+11.286( Fh)=69.472 This is the only part I did not understand how to get. Can you please explain more? I arrived with this equation, -P1A1 - P2A2 (cancelled since atmos. Pres.) +Fh = rho A1v1 (v2-v1) Why did P1A1 disappear in your solution? How did you come up with (v1 - - v2) ? What is the control volume? Why did you not do a summation of forces is equal to change in momentum but instead go immediately to getting the horizontal force acting on the bolt?
ᴛʜᴇ ꜱʜᴀꜰᴛ ᴡʜᴏꜱᴇ ᴛᴏʀQᴜᴇ ᴠᴀʀɪᴇꜱ ꜰʀᴏᴍ 2000 ᴛᴏ 6000 ɪɴ-ʟʙ ʜᴀꜱ 1 1/2 ɪɴᴄʜ ɪɴ ᴅɪᴀᴍᴇᴛᴇʀ ᴀɴᴅ 60,000 ᴘꜱɪ ʏɪᴇʟᴅ ꜱᴛʀᴇɴɢᴛʜ. ᴄᴏᴍᴘᴜᴛᴇʀ ꜰᴏʀ ᴛʜᴇ ꜱʜᴀꜰᴛ ᴍᴇᴀɴ ᴀᴠᴇʀᴀɢᴇ ꜱᴛʀᴇꜱꜱ. ᴀ. 6036 ᴘꜱɪ ʙ. 6810 ᴘꜱɪ ᴄ. 5162 ᴘꜱɪ ᴅ. 5550 ᴘꜱɪ
Answers (a) er=15.30-250.2/r^2 , etheta=15.30+250.2/r^2 (b) delta c = 8.12x10^-4 in
A vertical jet thru a nozzle supports a 150.90N weight. The velocity and diameter of the jet are 17.90m/s and 3.90 cm respectively. Find the distance of load from the nozzle in meters. Group of answer choices 13.79 31.79 17.93 71.39I need an answer for this asap :(
ꜱᴜᴘᴘᴏʀᴛ ᴛʜᴇ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ᴛʜᴇ ɴᴇᴄᴇꜱꜱᴀʀʏ ꜱᴄʜᴇᴍᴀᴛɪᴄ ᴅɪᴀɢʀᴀᴍꜱ , ᴘᴠ ᴀɴᴅ ᴛꜱ ᴅɪᴀɢʀᴀᴍꜱᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴘᴏᴡᴇʀ ʀᴇQᴜɪʀᴇᴅ ʙʏ ᴀ ꜱʏꜱᴛᴇᴍ ᴏꜰ ᴏɴᴇ ᴄᴏᴍᴘʀᴇꜱꜱᴏʀ ꜱᴇʀᴠɪɴɢ ᴛᴡᴏ ᴇᴠᴀᴘᴏʀᴀᴛᴏʀꜱ . ᴏɴᴇᴇᴠᴀᴘᴏʀᴀᴛᴏʀ ᴄᴀʀʀɪᴇꜱ ᴀ ʟᴏᴀᴅ ᴏꜰ 35 ᴋᴡ ᴀᴛ 10 ᴅᴇɢ ᴄ ᴀɴᴅ ᴛʜᴇ ᴏᴛʜᴇʀ ᴀ ʟᴏᴀᴅ ᴏꜰ 70 ᴋᴡ ᴀᴛ -5 ᴅᴇɢ ᴄ .ᴀʙᴀᴄᴋ ᴘʀᴇꜱꜱᴜʀᴇ ᴠᴀʟᴠᴇ ʀᴇᴅᴜᴄᴇꜱ ᴛʜᴇ ᴘʀᴇꜱꜱᴜʀᴇ ɪɴ ᴛʜᴇ 10 ᴅᴇɢ ᴄ ᴇᴠᴀᴘᴏʀᴀᴛᴏʀ ᴛᴏ ᴛʜᴀᴛ ᴏꜰ ᴛʜᴇ 5 ᴅᴇɢ ᴄᴇᴠᴀᴘᴏʀᴀᴛᴏʀ . ᴛʜᴇ ᴄᴏɴᴅᴇɴꜱɪɴɢ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ɪꜱ 37 ᴅᴇɢ ᴄ . ɪꜰ ᴛʜᴇ ʀᴇꜰʀɪɢᴇʀᴀɴᴛ ɪꜱ ᴀᴍᴍᴏɴɪᴀ ,ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴄᴏᴘ .ᴀ) 6.33 ʙ) 5.33 ᴄ) 4.33 ᴅ) 3.33
ɪɴ ᴀɴ ᴀᴍᴍᴏɴɪᴀ ʀᴇꜰʀɪɢᴇʀᴀᴛᴏʀ , ᴛʜᴇ ᴘʀᴇꜱꜱᴜʀᴇ ɪɴ ᴛʜᴇ ᴇᴠᴀᴘᴏʀᴀᴛᴏʀ ɪꜱ 267. 58 ᴋᴘᴀ ᴀɴᴅ ᴛʜᴇᴀᴍᴍᴏɴɪᴀ ᴀᴛ ɪꜱ 0.14 ᴅʀʏ ᴡʜɪʟᴇ ᴀᴛ ᴛʜᴇ ᴇxɪᴛ ɪꜱ 0.86 ᴅʀʏ . ᴅᴜʀɪɴɢ ᴄᴏᴍᴘʀᴇꜱꜱɪᴏɴ ᴛʜᴇ ᴡᴏʀᴋ ᴅᴏɴᴇ ᴘᴇʀ ᴋɢᴏꜰ ᴀᴍᴍᴏɴɪᴀ ɪꜱ 20 ,000 ᴋɢ .ᴍ . ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴛʜᴇ ᴠᴀᴘᴏʀ ᴇɴᴛᴇʀɪɴɢ ᴛʜᴇ ᴄᴏᴍᴘʀᴇꜱꜱᴏʀ ᴘᴇʀ ᴍɪɴᴜᴛᴇɪꜰ ᴛʜᴇ ʀᴀᴛᴇ ᴏꜰ ᴀᴍᴍᴏɴɪᴀ ᴄɪʀᴄᴜʟᴀᴛɪᴏɴ ɪꜱ 6.54 ᴋɢ. ᴛʜᴇ ᴄᴏᴍᴘʀᴇꜱꜱᴏʀ ʜᴀꜱ ᴀ ᴠᴏʟᴜᴍᴇᴛʀɪᴄ ᴇꜰꜰɪᴄɪᴇɴᴄʏ ᴏꜰ 90%ᴀɴᴅ ɪᴛ ʀᴜɴꜱ ᴀᴛ 150 ʀᴘᴍ .ᴛʜᴇ ʀᴀᴛɪᴏ ᴏꜰ ꜱᴛʀᴏᴋᴇ ᴛᴏ ʙᴏʀᴇ ɪꜱ 1 .ʟᴀᴛᴇɴᴛ ᴇɴᴛʜᴀʟᴘʏ ᴏꜰ ᴀᴍᴍᴏɴɪᴀ ᴀᴛ ʙᴀꜱᴇᴅᴏɴ ɪᴛꜱ ᴇᴠᴀᴘᴏʀᴀᴛɪɴɢ ᴘʀᴇꜱꜱᴜʀᴇ ɪꜱ 320 ᴋᴄᴀʟ ᴘᴇʀ ᴋɢ ᴀɴᴅ ɪᴛꜱ ʟᴀᴛᴇɴᴛ ꜱᴘᴇᴄɪꜰɪᴄ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴀᴍᴍᴏɴɪᴀ ʙᴀꜱᴇᴅᴏɴ ɪᴛꜱ ᴇᴠᴀᴘᴏʀᴀᴛɪɴɢ ᴘʀᴇꜱꜱᴜʀᴇ ɪꜱ 0.436 ᴍ3 ᴘᴇʀ ᴋɢ .ꜱᴘᴇᴄɪꜰɪᴄ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴀᴍᴍᴏɴɪᴀ .ꜱᴜᴘᴘᴏʀᴛ ʏᴏᴜʀ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ᴛʜᴇ ɴᴇᴄᴇꜱꜱᴀʀʏ ꜱᴄʜᴇᴍᴀᴛɪᴄ ᴅɪᴀɢʀᴀᴍꜱ , ᴘʜ ᴀɴᴅ ᴛꜱ ᴅɪᴀɢʀᴀᴍꜱ thank youOption:ᴀ) 2.23 ᴄ) 2.46ʙ) 3.23 ᴅ) 4.62
ᴀɴ ᴀᴍᴍᴏɴɪᴀ ꜱɪᴍᴘʟᴇ ꜱᴀᴛᴜʀᴀᴛɪᴏɴ ᴄʏᴄʟᴇ ᴏᴘᴇʀᴀᴛᴇꜱ ᴇᴠᴀᴘᴏʀᴀᴛᴏʀ ᴀɴᴅ ᴄᴏɴᴅᴇɴꜱᴇʀ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇꜱ ᴏꜰ -20°ᴄ ᴀɴᴅ 35°ᴄ,ʀᴇꜱᴘᴇᴄᴛɪᴠᴇʟʏ. ᴛʜᴇ ꜱʏꜱᴛᴇᴍ ɪꜱ ᴛᴏ ʙᴇ ᴜꜱᴇᴅ ɪɴ ᴘʀᴏᴅᴜᴄɪɴɢ 6000 ᴋɢ ᴏꜰ ɪᴄᴇ ᴀᴛ -12°ᴄ ꜰʀᴏᴍ ᴡᴀᴛᴇʀ ᴀᴛ 29°ᴄ ɪɴ 10 ʜᴏᴜʀꜱ. ᴀꜱꜱᴜᴍɪɴɢ ʟᴏꜱꜱᴇꜱ ᴛᴏ ʙᴇ 30 ᴘᴇʀ ᴄᴇɴᴛ ᴏꜰ ᴛʜᴇ ʜᴇᴀᴛ ᴛᴏ ʙᴇ ᴀʙꜱᴏʀʙᴇᴅ ꜰʀᴏᴍ ᴛʜᴇ ᴡᴀᴛᴇʀ, ᴅᴇᴛᴇʀᴍɪɴᴇ (ᴀ) ᴛʜᴇ ᴍᴀꜱꜱ ꜰʟᴏᴡ ʀᴀᴛᴇ, (ʙ) ᴛʜᴇ ʜᴇᴀᴛ ʀᴇᴊᴇᴄᴛᴇᴅ ᴀᴛ ᴛʜᴇ ᴄᴏɴᴅᴇɴꜱᴇʀ, ᴀɴᴅ (ᴄ) ᴛʜᴇ ᴘᴏᴡᴇʀ ʀᴇQᴜɪʀᴇᴅ ʙʏ ᴛʜᴇ ᴄᴏᴍᴘʀᴇꜱꜱᴏʀ.
For the following core plugs, gas floods were carried out using nitrogen. The oil produced from plugs 1 and 2 for the gas floods was 9.0 and 6.9 cm3, respectively. What is the residual oil saturation (Sorg) in these two plugs? Plug No. Initial Plug Saturations (%) ∅ Bulk Volume ρgrain ρo ρw ρg So Sw Sg (%) (cm3) (g/cm3) (g/cm3) (g/cm3) (g/cm3) 1 64.64 35.36 0 38.27 63.05 2.719 0.723 1.0216 0.001 2 71.93 28.07 0 34.63 51.05 2.724 0.723 1.0216 0.001
please show step-by-step calculations. answer choices are 0.34, 0.38, 0.42, 0.54, 0.85, 2.20, 2.58, 2.67, 3.11, 3.15, 1484, 1803, 2087, 2101, 2126
I am using eq 3.117 in Shigley's 10ed: m*x-double dot+b*x-dot+k*x=(1/2)C_x*(e_c)^2 Since x-double dot=X-dot=0, equations becomes: k*x=(1/2)C_x*(e_c)^2 then plug in eq 3.110, C_x=(n*E_0*w)/d Then solve for x. Does this look correct?
How to solve these two equations, to find IJ (kN) and ID (kN)? x: -IJ*cos37.5º -ID*cos56.9º+(-26.3kN)*cos37.5º)=0 y: IJ*sin37.5º -ID*sin56.9º-10kN-(-26.3kN)*sin37.5º=0