TikZ:How to draw the pump schematic drawing

Question

Here My first Tikz picture: help on positioning and suggestions we have a nice drawing similar to this image.

How to draw the pump element in the above image? How to draw the anticlock arrow?

Like turbine is it possible to make \tikzset for pump

\tikzset{turb/.style={draw,trapezium,shape border rotate=90,inner sep=1pt,minimum width=2.5cm,trapezium stretches=true,trapezium angle=80,on grid,below right= of evaporatore}}

---

\documentclass{standalone} 
\usepackage{tikz}
\usetikzlibrary{positioning,shapes.geometric,decorations.pathmorphing,decorations.pathreplacing,decorations.shapes,decorations.markings,patterns,calc,fit,arrows}

\begin{document}

\begin{tikzpicture}[>=latex',auto,inner sep=2mm,node distance=2cm and 3cm]

%set styles for the axis between turbine and pump and for the boxes

\tikzset{box1/.style={draw,minimum width=2.5cm,rectangle,thick}}
\tikzset{deco/.style={decoration={markings,
                       mark=at position #1 with {\arrow{>}}},
                       postaction={decorate}}}
\tikzset{turb/.style={draw,trapezium,shape border rotate=90,inner sep=1pt,minimum width=2.5cm,trapezium stretches=true,trapezium angle=80,on grid,below right= of evaporatore}}   

% draw nodes
\node[box1] (evaporatore)  {Boiler};
\node[turb] (turbina) {Turbine};
\node[box1,on grid,below left=of turbina] (condensatore){Condenser};
%\node[draw,circle,on grid,below left= of evaporatore] (pompa) {Pump};
\node[draw,circle,on grid,below left= of evaporatore] (pompa) {Pump}; 
\draw (pompa.70) |- ++(2.5mm,5mm) coordinate (mid)  -| (pompa.east);


\begin{scope}[>=triangle 45]
 \draw [deco=0.6]  (evaporatore)                -| (turbina.top right corner);
 \draw [deco=0.6]  (turbina.bottom left corner) |- (condensatore);
 \draw [deco=0.4]  (condensatore)               -| (pompa);
 \draw [deco=0.6]  (mid)                      |- (evaporatore);
 \end{scope}

%draw the "shaft"

\path(pompa) to node[]{shaft} (turbina);
\draw[pattern=north east lines] ($(pompa.east)+(0,-3pt)$) rectangle ($(turbina.west)+(0,3pt)$);
\draw[pattern=north east lines] ($(turbina.east)+(0,-3pt)$) rectangle ++(1,6pt);
\draw[->] (25:1) arc (10:320:0.2cm and 0.75cm) ;
%
\end{tikzpicture}
\end{document}

Pump element changed (thanks to percusse). Instead of node, coordinate is used... \draw (pompa.70) |- ++(2.5mm,5mm) coordinate (mid) -| (pompa.east);

Answer 1

The arrow is simply a long arc :

\begin{tikzpicture}
\draw[-latex] (15:2.5mm) arc (15:345:2.5mm and 7.5mm);
\end{tikzpicture}

gives

also the pump is a node with extension:

\begin{tikzpicture}
\node[draw,circle] (p) {Pump}; 
\draw (p.70) |- ++(1mm,5mm) -| (p.east);
\end{tikzpicture}

Answer 2

Something like that

\documentclass[11pt]{scrartcl}
\usepackage{tikz}
\usetikzlibrary{arrows}

\begin{document}
\begin{tikzpicture}
\draw (0,-0.4) rectangle (2,0.4);
\draw[->] (25:1) arc (30:320:0.2cm and 0.75cm) ;

\end{tikzpicture}
\end{document}

Update with your example

\documentclass{standalone} 
\usepackage{tikz}
\usetikzlibrary{positioning,shapes.geometric,decorations.pathmorphing,decorations.pathreplacing,decorations.shapes,decorations.markings,patterns,calc,fit,arrows}

\begin{document}

\begin{tikzpicture}[>=latex',auto,inner sep=2mm,node distance=2cm and 3cm]

%set styles for the axis between turbine and pump and for the boxes

\tikzset{box1/.style={draw,minimum width=2.5cm,rectangle,thick}}
\tikzset{deco/.style={decoration={markings,
                       mark=at position #1 with {\arrow{>}}},
                       postaction={decorate}}}
\tikzset{turb/.style={draw,trapezium,shape border rotate=90,inner sep=1pt,minimum width=2.5cm,trapezium stretches=true,trapezium angle=80,on grid,below right= of evaporatore}}   

% draw nodes
\node[box1] (evaporatore)  {Boiler};
\node[turb] (turbina) {Turbine};
\node[box1,on grid,below left=of turbina] (condensatore){Condenser};
%\node[draw,circle,on grid,below left= of evaporatore] (pompa) {Pump};
\node[draw,circle,on grid,below left= of evaporatore] (pompa) {Pump}; 
\draw (pompa.70) |- ++(2.5mm,5mm) coordinate (mid)  -| (pompa.east);


\begin{scope}[>=triangle 45]
 \draw [deco=0.6]  (evaporatore)                -| (turbina.top right corner);
 \draw [deco=0.6]  (turbina.bottom left corner) |- (condensatore);
 \draw [deco=0.4]  (condensatore)               -| (pompa);
 \draw [deco=0.6]  (mid)                      |- (evaporatore);
 \end{scope}

%draw the "shaft"

\path(pompa) to node[]{shaft} (turbina);
\draw[pattern=north east lines] ($(pompa.east)+(0,-3pt)$) rectangle ($(turbina.west)+(0,3pt)$);
\draw[pattern=north east lines] ($(turbina.east)+(0,-3pt)$) rectangle ++(1,6pt);
\draw[->] ([shift={(0.75,3pt)}]turbina.east) arc (10:350:0.2cm and {3pt/sin(10)}) ; 

\end{tikzpicture}
\end{document}  

Explanations for the last arrow

You need to determine the origin point. The point of reference here is turbina.east.

You can use calc library or the old method with shift as given below

($ (turbina.east)+(0.75,3pt)$) or ([shift={(0.75,3pt)}]turbina.east)

The origin point is on the top of the last rectangle. Then you need to choice the start angle (see the good answer from PolGab to understand how to get the end angle).

Answer 3

here is an example with PSTricks. The code (http://tug.org/PSTricks/main.cgi?file=Examples/Gallery/Gallery) can be run with xelatex to get directly a pdf output

Answer 4

The calculation of angles and radii of arc is feasible if the height of the node is fixed.

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\foreach \he in {1cm,1.3cm}
{
  \begin{tikzpicture}
    \node [minimum width=2cm,minimum height=\he,draw] (a){Test};
    \foreach \angle in {25,30,35} {
      \draw[-stealth]
      (a.north)
      arc[start angle=\angle,end angle=360-\angle,x radius=\he/4,y radius=\he/sin(\angle)/2]
      ;
    }
  \end{tikzpicture}
}
\end{document}