2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
1 answer:
Answer:
coupling is in tension
Force = -244.81 N
Explanation:
Diameter of Hose ( D1 ) = 35 mm
Diameter of nozzle ( D2 ) = 25 mm
water gage pressure in hose = 510 kPa
stream leaving the nozzle is uniform
exit speed and pressure = 32 m/s and atmospheric
<u>Determine the force transmitted by the coupling between the nozzle and hose </u>
attached below is the remaining part of the detailed solution
Inlet velocity ( V1 ) = V2 ( D2/D1 )^2
= 32 ( 25 / 35 )^2
= 16.33 m/s
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Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
B.197 gpm and 12.4 L/s
Explanation:
Given that
Load Q = 404.5 KW
Water inlet temperature= 6.1 °C
Water outlet temperature= 13.9°C
We know that specific heat for water
Now from energy balance
by putting the values
(1 Kg/s = 15.85 gal/min)
We can say that
We know that
12.38=1000 x volume flow rate
So
volume flow rate = 12.38 L/s
So the option B is correct.