Answer:
I. 3.316 kW
II. 1.218 kW
III. 2.72
Explanation:
At state 1, the enthalpy and entropy are determined using the given data from A-13.
At P1 = 200kpa and T1 = 0,
h1 = 253.07 kJ/kg
s1 = 0.9699 kJ/kgK
At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus
h2(s) = 295.95 kJ/kg
The actual enthalpy is then gotten by
h2 = h1 + [h2(s) - h1]/n
h2 = 253.07 + [295.95 - 253.07]/0.88
h2 = 253.07 + 48.73
h2 = 301.8 kJ/kg
h3 = h4 = 120.43 kJ/kg
Heating load is determined from energy balance, thus,
Q'l = m'(h1 - h4)
Q'l = 0.025(253.07 - 120.43)
Q'l = 0.025 * 132.64
Q'l = 3.316 kW
Power is determined by using
W' = m'(h2 - h1)
W'= 0.025(301.8 - 253.07)
W'= 0.025 * 48.73
W'= 1.218 kW
The Coefficient Of Performance is Q'l / W'
COP = 3.316/1.218
COP = 2.72
Answer:
c. V2 equals V1
Explanation:
We can answer this question by using the continuity equation, which states that:
(1)
where
A1 is the cross-sectional area in the first section of the pipe
A2 is the cross-sectional area in the second section of the pipe
v1 is the velocity of the fluid in the first section of the pipe
v2 is the velocity of the fluid in the second section of the pipe
In this problem, we are told that the pipe has a uniform cross sectional area, so:
A1 = A2
As a consequence, according to eq.(1), this means that
v1 = v2
so, the velocity of the fluid in the pipe does not change.
Answer:b
Explanation:
We know power delivered by Pump is

where
=Density of fluid
=Flow rate
=acceleration due to gravity
=Change in Elevation
If
is increased by 4 time then


So power increases by four times.
C. Both; require a wheel alignment after replacement