Answer:
47.10 m/s
Explanation:
Diameter of the horizontal pipe, d = 7 cm = 0.07 m
Area of the horizontal pipe, A =
or
A = = 0.00384 m²
Head between the water surface and the pipe level, z = 4 m
Head added by the pump, hp = 8 - 200Q²
where, Q is the discharge
losses, hL = 5 m
now,
applying the Bernoulli's theorem between the water surface and the pipe outlet,
we have
where,
P₁ = pressure at the water surface = 0 (as atmospheric pressure only)
V₁ = Velocity at the free surface = 0
ρ is the density of the water
P₂ = pressure at the outlet = 0 (as atmospheric pressure only)
V₂ = Velocity at the outlet
on substituting the values, we get
or
or
also,
Q = A × V₂
thus,
or
V₂ = 47.10 m/s
Is this vehicle currently in shape to safely get me where I need to be
Answer:
(a)
(b)Q= -35.69 KW
Explanation:
Given:
We know that foe air
Mass flow rate for air =2.3 kg/s
(a)
By mass balancing
(b)
Now from first law for open(nozzle) system
Δh=
Q=-15.52 KJ/s
⇒ KW
Q= -35.69 KW
If heat will loss from the system then we will take negative and if heat will incoming to the system we will take as positive.
<u>Solution and Explanation:</u>
Step 1 Temperature from T =27C to T = 0C constant volume process
Step 2 At 0C volume changes from V1 to V1/2 , isothermally
Step 3 Temperature changes from 0C to 27C , constant volume process
Step 4 Volume changes from V/2 to V1 isothermally at 27C
From the description of the problem it is clear that the problem is a cyclic process , which involves 2 isothermal process at 27C and 0C and two constant volume process at volume coressponding to the volume of ideal gas at the given conditions. since it is a cyclic proess (comes back to the initial state) are zero because they are sate functions and in a cyclic process the change in state functions are zero,
The first law can be written as with appropriate sign convention , since \
Qnet = Wnet
work is zero for the two constant volume process , contribution to work comes only from isothermal process ,and for ideal gas it is given as
It is given
, so this amount of work has to be done on the system , since Q= W , Q = -2221.91 Joules of heat is to be transfered from the system.