Power consumption of phototubes A phototube1 (or "photoemissive cell") is a simple vacuum tube device that works by the photoelectric effect; it produces a current when light strikes the photocathode.
For it to work, a potential difference is generally applied between the cathode and the anode of 15V, I only need to know the current to calculate the power.
Is the current it consumes equal to the current it generates?
I search the data sheet but did not find the power consumed.

1as opposed to a photomultiplier tube
 A: An old fashion photo-tube (These days you might use a photo-transistor.), has a plate (the cathode) and an anode in a vacuum. A small fraction of the photons striking the plate cause electrons to be ejected into the vacuum.  The applied voltage produces an electric field which accelerates these free electrons toward the anode. This flow constitutes the current, which depends on the intensity of the light, and can be quite small.  This current times the voltage determines the power, which can also be quite small.
A:  source
The phototube is a two terminal device; it has a photocathode that emits electrons, and an anode that collects them. In a circuit, there will be an applied voltage difference and a measured current, and assuming DC voltage and a constant intensity of light, the total power consumed electrically will indeed be given by
$$P = IV.$$
See also answers to Biasing network for phototubes and photomultipliers and this excellent, thorough phototube manual at worldradiohistory.com
As the device has a capacitance, if either the bias or light intensity is modulated at high frequency you'll have to do a more thorough analysis of resistive vs reactive power.
As you mention, the photoelectric effect applies; electrons are emitted when the energy of the photons are greater than the work function (See also Methods of measurement based on photoemission)
While pure, clean metal surfaces may have fairly high work functions requiring UV photons to emit electrons, specially formulated photocathode materials can have work functions as low as 1 volt allowing direct detection of infrared photons out past 1200 nm in some special cases.
It's interesting how the power dissipation is divided between the cathode and anode. If for example your bias is 20 volts and the work function is 2 eV, then assuming 100% quantum efficiency (one photoelectron out for every photon in) roughly 10% of the electrical power is dissipated in the photocathode, (they don't call it work function for nothing) and roughly 90% is dissipated in the anode which the accelerated 8 eV electrons hit it and loose their energy inside it.


 
source bottom left: "RCA 935 Phototube - notice the large cathode for better light collection, bottom right: "Infrared-sensitive Phototube from the 1940s" (from here)
below: Source: How the Phototube vacuum tube works - with DIY example


Figure  7.5.2 : This old image from the 1954 Bulletin on Narcotics from United Nations Office on Drugs and Crime shows how a vacuum phototube is biased for the collection of photoelectron current.

