Tahák Fyzika 20 - hydromechanika, termodynamika

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:Hydromechanika

:

:Nadoba valcoveho tvaru, prurez S1, voda do vysky h. U dna kruh. otvor S2. ?Cas, h=h/2.

:v=§(2.g.y)

:nadoba dV1=-S1 dy

:vytece dV2=S2.v dt

:dV1=dV2

:dt=-S1/(S2.§(2.g).§y)dy

:t=int h,h/2 (nad)=

:=S1/S2.§(h/g).(§2-1)

:

:Venturiho vodomer, dp=1.5E4 Pa. Hlavni potrubi d1=0.06 m, zuzene d2=0.03 m. ?Prutok za s, za h.

:z Bernoulliho r.

:1/2.ro.v^2+p=konst.

:p1-p2=1/2.ro(v2^2-v1^2)

:z r. kontinuity

:v2=S1.v1/S2

:v1^2=2.dp/(ro(d1^4/d2^4-1))=1.41m/s

:Qv=S1.v1

:V=Qv.t=3.99 l/s

:

:Do nadrze priteka voda proudem 21 s^-1. Na dne otvor S=5 cm^2. Vyska ustaleni?

:Qv=S.v=S.§(2gh)

:h=81.5 cm

:

:Fvz=V.ro kapaliny.g

:objem koule=1/6.pi.d^3

:

:?Sila vody na ctver. stenu akvaria a=20 cm. Vyska pusobiste?

:F=int p.dS=int 0,a

:(ro.g.(a-x).a)dx=

:=ro.g.a^3/2=39.2 N

:

:pusobiste

:int 0,x0 (x0-x)dF=

:=int x0,a (x-x0)dF

:dF=p.dS, dS=a.dx

:p=ro.g.x

:int 0,x0 (x.(x0-x))dx=

:=int x0,a (x.(x-x0))dx

:x0=2/3.a=6.67 cm od dna

:

:stavidlo, sirka a=180 cm, h=120 cm

:dS=a.dx

:F=int p.dS=

:=int 0,h (ro.g.x.a)dx=

:=ro.g.a.h^2/2=12.7 kN

:

:F k nadzvednuti stavidla?

:dS=3.2 dx

:Fv=35.28kN, F=13.33kN

:

:Vytokova rychlost

:Ep=Ek

:ro.g.h.V=1/2.ro.v^2.V

:v=§(2.g.h)

:

:Kapilarni deprese rtuti?

:ro=13.6, sigma=O=0.433N/m, r=0.5mm, V=120st

:h.ro.g=2.O.cos(V)/r

:h=-6.5mm

:

:lih

:O=h.ro.g.r/2=2.35E-2N/m

:

:Nadoba a,b,c,otvor S. Doba vyprazdneni?

:nadoba dV1=a.b dx

:vytok dV2=S.v dt

:v=§(2.g.x)

:dV1=dV2

:dt=a.b/(S.§(2gx))dx

:t=int 0,c (nad)

:t=a.b.§(2c)/(s.§g)

:Pist ve strikacce S1=1cm^2, vytok. otvor S2=1mm^2.

:Doba vytekani pri F=5N a posunu l=4cm

:1/2.ro.v1^2+F/S1=

:=1/2.ro.(S1.v1/S2)^2+10^5

:t=l/v1=0.4s

:

:Teplotni roztaznost

:

:dl=l0.alfa.dt

:sigma=E.eps

:eps=dl/(l0+dl)

:

:

:Idealni plyn

:

:Hustota vzduchu za normal podminek ro=1.293

:?ro pri p=0.2MPa, t1=20st

:pV=nRT

:ro=M.p/(R.T)

:ro1=ro0.p1.T0/(p0.T1)=2.41kgm^-3

:Stlaceny plyn pri 20st a 5MPa. p2=?, polovic plynu vypustime.

:p1.V=m1/M.R.T1

:p2.V=m2/M.R.T2

:m2=m1/2

:p=p1.T2/(2.T1)=2.4MPa

:

:Spojime-li nadobku s 8l kysliku s prazdnou. V2=?

:V=p1/p2.V1=24l obe

:V2=24-8=16l

:

:

:Vedeni tepla

:

:Dve soustredne koule R1,R2. Mezi nimi latka o vodivosti L.

:kulovou plochou r mezi koulemi projde za cas t

:Q=-L.S.t.dt/dr=

:=-L.4.pi.r^2.t.dt/dr

:Q.dr/r^2=-4.pi.L.t dt

:int R1,R2 resp. t1,t2

:Q=4.pi.L.t.R1.R2.(t1-t2)/(R2-R1)

:rozlozeni teploty mezi koulemi

:int R1,r resp t1,t

:Q=4.pi.L.t.R1.r.(t1-t)/(r-R1)

:porovname vysledne vztahy

:r.(t1-t)/(r-R1)=R2.(t1-t2)/(R2-R1)

:t=...

:

:Ocel tyc 50cm,2cm^2,300st - jeden konec. Druhy konec v ledu 20min.

:Q=L.t.S.dT/d

:m=Q/l=0.0255kg

:

:Cu tyc 20cm+Fe tyc 10cm

:200st ---- 20st

:j=?, Tstyk=?

:j=-L.grad T

:grad T=dt/tloustka

:j1=j2

:-L<Cu>.(t1-t)/l1=

:=-L<Fe>.(t-t2)/l2

:t=158.84

:j=8.2E4W/m^2

:

:Kolik prejde tepla pres cihl stenu?

:1h, 0.5m^2, tl. d=45cm, 20st, -8st, 0.54 W/(m.K)

:Q=L.t.S.dt/d=6E4J

:Termodynamika

:

:Carnotuv stroj

:ni1=30%. dT=?, ni2=40%

:chladic t2=5st

:ni=A/Q1=(Q1-Q2)/Q1=

:=(T1-T2)/T1

:dT=T2/(1-ni2)-T2/(1-ni1)=66st

:

:Zmena entropie? Smichame vodu1 50g,100st a vodu2 10g,10st.

:t=85st

:dS=dQ/T=m.c.dT/T

:S1=int T1,T (m1.c.dT/T)=m1.c.ln(T/T1)

:S2=int T2,T (m2.c.dT/T)=m2.c.ln(T/T2)

:deltaS=1.25J/K

:

:Zmena entropie 5g dusiku? Zahrejeme ho z -20 na 10st.

:dS=dQ/T

:izochoricky

:S12=int 1,2 (m.cv.dT/T)=

:=m.cv.ln(T2/T1)=0.415J/K

:izobaricky

:dS=(n.Cv.dT+p.dV)/T=

:=(n.Cp.dT)/T

:p.dV=n.R.dt

:Cp=Cv+R

:S=int 1,2 (m.cp.dT)/T

:S=m.cp.ln(T2/T1)=0.582J/K

:

:Kolik tepla se musi odebrat pri izotermickem stlaceni?

:m=50g O2, t=0st

:p1=0.2,p2=0.6MPa

:Ł=parc.der.

:ŁQ=ŁA=p.dV

:p=n.R.T/V

:ŁQ=n.R.T/V dV

:Q=int V1,V2 (n.R.T.1/V)dV

:T=konst.,plati

:Q=n.R.T.ln (V2/V1)

:V2/V1=p2/p1

:Q=3896J

:

:Ve valci vysky l1=35cm je vzduch t1=27st a p1=0.1MPa. Adiabaticke stlaceni, posun o l2=10cm. H=1.4, p=?, T=?

:p.V^H=konst.

:p2=p1.V1^H/V2^H

:p2=0.16MPa

:p.V/T=konst.

:T2=p2.V2.T1/(p1.V1)

:T2=343K

:

:Vzduch m=0.5kg, t1=27st jsme stlacili z p1=0.1MPa na 5.p1. A=? Cv=20.8J/mol.K, M=0.029kg/mol

:izotermicky prubeh

:A=n.R.T.ln(p2/p1)

:A=69211.4J

:adiabaticke stlaceni

:ŁA=-ŁU

:A=n.cv.T1(1-T2/T1)

:pV^H=konst

:(V2/V1)^H=p1/p2

:p.V/T=konst.

:V2/V1=p1.T2/(T1.p2)

:p1/p2=(p1/p2.T2/T1)^H

:T2/T1=(p1/p2)^(1-1/H)

:H=1+R/Cv=1.4

:A=n.Cv.T1.(1-(p1/p2)^(1-1/H))

:A=62811J

:

:Cyklicky proces-2 izochor,2 izoterm. A=?

:A=cykl.int p.dV=A12+A23+A34+A41

:izoch dV=0, proto

:A23=0, A41=0

:A=int V1,V2 (n.R.T/V)dV

:A12=n.R.T1.ln(V2/V1)

:A34=n.R.T2.ln(V1/V2)

:A=n.R.ln(V2/V1).(T1-T2)

:

:Cyklus--2 izobar, 2 izochor A=?

:A12=int V1,V2 (p2.dV)

:A12=p2.(V2-V1)

:A23=p1.(V1-V2)

:A=(p2-p1).(V2-V1)

:

:Carnotuv dej, dokazat dS=0

:dS=int V1,V2 (dQ/T1)+0+

:+int V2,V4 (dQ/T2)+0

:A=n.R.T.ln(V2/V1)

:V2/V1=V3/V4

:dS=n.R.ln(V2/V1)+

:+n.R.ln(V4/V3)=0

:

:Voda 1l, 17st + olovo 0.5kg, 300st. Zmena entropie?

:t=21.29st

:voda S=

:int T1,T (m.c.dT/T)=

:=m.c.ln(294.3/290)

:olovo S=

:int T2,T (m.c.dT/T)=

:=m.c.ln(294.3/573)

:S=18.6J/K

:

:Led 100g, 0st premenime na paru 100st. Zmena entropie?

:S=int Q1 (ŁQ/T)+.+int Q3 (ŁQ/T)

:S=Q1/T1+int T1,T2 (m.c.dT/T)+Q3/T2

:S=lt.m/T1+m.c.ln(T2/T1)+lv.m/T2

:S=857.7 J/K

:

:Zmena entropie? 5g dusiku

:t=-10st, izoterm. V1=5l, V2=2l

:Q=n.R.T.ln(V2/V1)

:S=Q/T=-1.36J/K

:

:

:Soust. latek s 1 a 2 fazemi

:

:Clausius-Clapeyron

:dp/dT=l/(T.(v2-v1))

:dT=nova-znama teplota

:T=nova teplota

:

:tlak pary pri 85st, lv=a-b.T

:pri t0=100st p0=1,013E5Pa

:uvazime v1<<v2

:dp/dT=(a-b.T)/(v2.T)

:v2=V/m=R.T/(M.p)

:dp/p=M.(a-b.T)/(R.T^2)

:p=p0.e^(-M/R.(a.(1/T-1/T0)+b.ln(T/T0)))

:p=58148Pa

:

:T=?,p=? trojneho bodu arzenu

:dana zavislost nasyc. par na teplote

:p1=p2

:k1.e^(-a1/T)=k2.e^(-a2/T)

:T=a2-a1/ln (k2/k1)

:T=1090.5K P=3.6MPa

:

:Nadoba 2m^3, 29st, 4kg kysliku. p=?

:van der Waals

:(p+a/V^2)(V-b)=n.R.T

:p=0.157MPa

:T=331K, dp=0.015MPa

:

:Olovo taje pri 0.1MPa, 327st. lt=? dp=0.1MPa, dt=0.008st,v2=1.034.v1

:v1=1/11340

:l=dp.T.0,034.v1/dT

:l=22.6E3J/kg

:

:Teplota tani ledu pri 1MPa?

:dp=p-pn=898700

:Tn=273K

:v1=1/ro ledu

:v2=1/ro vody

:dT=dp.Tn.dv/lt=-0.066st

:

:p=? var vody pri 99st

:dp=dT.l/(t.dv)

:pn-dp=0.0977MPa

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