Engineering

Chemical EngineeringQ&A Library6.30. A fireman is using a high-pressure stream of water from a hose to combat a ragingforest fire. At one end, the hose has an inside diameter of dconnected to a high-pressure water reservoir at P1 300 psi. At the other end is anozzle with diameter d2 3 cm that is exposed to atmospheric pressure P2= 1 atmTo a good approximation, flowing water in this process can be modeled as having a5 cm and is3constant temperature and constant mass density p = 1 g/cm = 1,000 kg/m2.incompressible liquid, the following equation well describesMoreover, for anisothermal enthalpy changes:Δh xυΔΡwhere v is the volume per mole or molecule (depending(a) How much faster is the velocity of the exitingon the basis of h)water at the nozzle than that offactor (e.g., u2=the entering stream at the reservoir? Express your answer as a1.5 u1).(b) If the hose is well insulated, find the exit velocity of the stream of water, in m/s.(c) Find the exit volumetric flow rate of water, in gal/s.(d) Instead, you suspect that the hose is not well insulated because you find thatthe actual, measured exit velocity is 90% of what you calculated in part (b).The inlet velocity, however, remains the same. This suggests that frictionallosses result in a dissipation of heat to the environment as the water maintainsconstant temperature. Find the heat lost in kJ per kg of waterQuestion

Asked Jun 3, 2019

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This problem is (6.30) from a book "Thermodynamics and Statistical Mechanics An Integrated Approach by M. Scott Shell"

Step 1

The information provided in the question is as follows:

Inside diameter of the hose is *d _{1 }*= 5 cm

Nozzle diameter is *d _{2 }*= 3 cm

Pressure at the one end of the hose (let it be called as point 1) that is connected to a reservoir is *P _{1}* = 300 psi

The unit conversion for pressure is given as

1 atm = 14.7 psi

1 atm = 101325 Pa

So, *P _{1}* = 206785 Pa

At point 2, nozzle is exposed to the atmospheric pressure, i.e., *P _{2}* = 1 atm = 101325 Pa

Mass density *ρ *and temperature remains constant.

The isothermal enthalpy change is represented by relation *Δh* = *vΔP, *where *v* is the volume per unit mass, which can be taken as the reciprocal of mass density.

Step 2

Since the given system is an open system the energy balance applied between point 1 and 2 is given by equation (1) for a steady flow process.Here, *Q* is the heat transfer, *W* is the shaft work, *ΔH* is enthalpy change taking place, *ΔK.E* is change in kinetic energy and *ΔP.E* is change in potential energy.

Assuming the system is well insulated therefore, *Q *= 0. Also, no shaft work is obtained therefore, *W= *0. Since the hose and the nozzle are at the same level, therefore, *z _{1 }= z_{2}* .Hence,

Step 3

(a) According to the equation of continuity, the relation between the velocity of water in the hose (*u1*) and the nozzle (*u2*) is given by equation (3). In this equation *A1* is the area of cross section of the hose and *A2* is the area of cross section of the nozzle. *A1...*

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