[Fwd: Re: [Tutor] Syntax error]
Roland Casselbrant
roland.casselbrant@bonetmail.com
Tue Feb 18 19:01:01 2003
This is a multi-part message in MIME format.
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HEre is the complete code:
hhe =3D 5000.0
vvu =3D 5000.0
stringA =3D 'Nettoenergibehovet: '
stringB =3D 'Nettoenergibehov minus vattenuppvarmning: '
stringC =3D 'Kompenserat for elpannans verkningsgrad: '
stringD =3D 'Effektbehovet vid dimensionerande temperatur: '
stringE =3D 'Till detta tillkommer arsmedeleffekt f=F6r varmvattenuppvarm=
ning: '
stringF =3D 'Dimensionerande effektbehov for flerbostadshuset i exemplet=20
blir da: '
stringZ =3D ' '
stringG =3D ' Effektbehov for eluppvarmt hus '
stringH =3D 'Program skrivet av Roland Casselbrant'
stringI =3D '*************************************'
stringJ =3D 'Efter gradkorrigering av varmebehovet har vi: '
stringK =3D 'Ett normalar blir elanvandningen f=F6r smahuset i exemplet=20
alltsa: '
print stringZ
print stringZ
print stringG
print stringH
print stringI
print stringZ
totel =3D input("Total elaanvandning per ar: [kWh] ")
pannverk =3D input("Elpannans armedelverkningsgrad: [%] ")
totyta =3D input("Total uppvarmd bostadsyta: [m2] ")
byggar =3D input("Byggar: ")
graddar =3D input("Graddagar aktuellt ar: ")
gradnorm =3D input("Graddagar normalar: ")
dut =3D input ("Dimensionerande utetemperatur: [grader Celsius] ")
# F=F6rst m=E5ste man r=E4kna bort den del av f=F6rbrukningen
# som inte =E4r klimatberoende
# d.v.s. hush=E5llsel och varmvattenuppv=E4rmning.
# Vi r=E4knar h=E4r med 5000 kWh hush=E5llsel och 5000 kWh
# f=F6r varmvattenuppv=E4rmning.
# Uppv=E4rmningsbehovet blir d=E5:
print stringA, totel
print stringZ
nettoel =3D totel-vvu-hhe
print stringB, nettoel
print stringZ
# Nu =E4r inte hela denna energim=E4ngd "nyttig" utan vi m=E5ste kompense=
ra
# f=F6r elpannans verkningsgrad. S=E5ledes =E4r nettoenergibehovet
# f=F6r uppv=E4rmning:
panna =3D pannverk/100.0
uppvarm =3D nettoel*panna
print stringC, uppvarm
print stringZ
# Effektbehovet vid dimensionerande temperatur l=F6ser vi ut
# fr=E5n ekvationen:
efbh =3D ((uppvarm/graddar*24) * (20-(dut))
print stringD, efbh
print stringZ
# Till detta tillkommer =E5rsmedeleffekt f=F6r varmvattenuppv=E4rmning.
mdff =3D (vvu/8760)
print stringE, mdff
print stringZ
# Dimensionerande effektbehov f=F6r flerbostadshuset i exemplet blir d=E5=
:
deff =3D efbh+mdff
print stringF, deff
print stringZ
# Graddagskorrigering av v=E4rmebehovet g=E5r till s=E5 att man tar
# kvoten av graddagarna
# f=F6r normal=E5r och graddagarna f=F6r aktuell period och multiplicerar
# med det
# klimatberoende nettoenergibehovet f=F6r uppv=E4rmning enligt f=F6ljande=
:
gradkorr =3D ((gradnorm/graddar)*uppvarm
print stringJ, gradkorr
print stringZ
# Ett normal=E5r blir elanv=E4ndningen f=F6r sm=E5huset i exemplet allts=E5=
:
totalel =3D ((gradkorr/panna)+(vvu+hhe))
print stringK, totalel
print stringZ
EOF/roland.casselbrant@bonetmail.com
-------- Original Message --------
Subject: Re: [Tutor] Syntax error
Date: Tue, 18 Feb 2003 16:06:45 -0700
From: Bob Gailer <ramrom@earthling.net>
To: roland.casselbrant@bonetmail.com, tutor@python.org
At 10:48 PM 2/18/2003 +0100, Roland Casselbrant wrote:
>The following lines is getting Syntax errors and I can=B4t understand w=
hy?
Please copy the session and any associated program to the body of an e-ma=
il
so we can see exactly the sequence of operations and the error messages.
Bob Gailer
mailto:ramrom@earthling.net
303 442 2625
--=20
Roland Casselbrant
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Seen it all, done it all, can't remember most of it
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
WWW: www.casselbrant.net
E-mail: roland.casselbrant@bonetmail.com
Mobile phone: +0046-(0)70-4785052
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