Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:
![\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7D%20%7D%7B%5Cgamma%20%7D%20%2B%5Cfrac%7BV_%7B1%7D%5E%7B2%7D%20%20%7D%7B2g%7D%20%2Bz_%7B1%7D%20%3D%5Cfrac%7BP_%7B2%7D%20%7D%7B%5Cgamma%20%7D%20%2B%5Cfrac%7BV_%7B2%7D%5E%7B2%7D%20%20%7D%7B2g%7D%20%2Bz_%7B2%7D%2Bh_%7Bi%7D%20-h_%7Bpump%7D%20%2C%20if%20V_%7B1%7D%20%3D0%2Cz_%7B2%7D%20%3D0%5C%5Ch_%7Bpump%7D%20%3D%5Cfrac%7BV_%7B2%7D%5E%7B2%7D%7D%7B2%7D%20%2Bh_%7Bi%7D-z_%7B1%7D)
For the pipe 1, the flow velocity is:
![V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }](https://tex.z-dn.net/?f=V_%7B1%7D%20%3D%5Cfrac%7BQ%7D%7B%5Cfrac%7B%5Cpi%20D%5E%7B2%7D%20%7D%7B4%7D%20%7D)
Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m
![V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s](https://tex.z-dn.net/?f=V_%7B1%7D%20%3D%5Cfrac%7B0.018%7D%7B%5Cfrac%7B%5Cpi%20%2A0.06%5E%7B2%7D%20%7D%7B4%7D%20%7D%20%3D6.366m%2Fs)
The Reynold´s number is:
![Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B%5Crho%20%2AV%2AD%7D%7Bu%7D%20%3D%5Cfrac%7B999.1%2A6.366%2A0.06%7D%7B1.138x10%5E%7B-3%7D%20%7D%20%3D335339.4)
![\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cepsilon%20%7D%7BD%7D%20%3D%5Cfrac%7B0.00026%7D%7B0.06%7D%20%3D0.0043)
Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:
![h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m](https://tex.z-dn.net/?f=h_%7B1%7D%20%3D%5Cfrac%7BV_%7B1%7D%5E%7B2%7D%20%20%7D%7B2g%7D%20%28k_%7BL%7D%2B%5Cfrac%7BfL%7D%7BD%7D%20%20%29%3D%5Cfrac%7B6.366%5E%7B2%7D%20%7D%7B2%2A9.8%7D%20%2A%280.5%2B%5Cfrac%7B0.0294%2A20%7D%7B0.06%7D%20%29%3D21.3m)
For the pipe 2, the flow velocity is:
![V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%5Cfrac%7B0.018%7D%7B%5Cfrac%7B%5Cpi%20%2A0.03%5E%7B2%7D%20%7D%7B4%7D%20%7D%20%3D25.46m%2Fs)
The Reynold´s number is:
![Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B%5Crho%20%2AV%2AD%7D%7Bu%7D%20%3D%5Cfrac%7B999.1%2A25.46%2A0.03%7D%7B1.138x10%5E%7B-3%7D%20%7D%20%3D670573.4)
![\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cepsilon%20%7D%7BD%7D%20%3D%5Cfrac%7B0.00026%7D%7B0.03%7D%20%3D0.0087)
The head of pipe 1 is:
![h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m](https://tex.z-dn.net/?f=h_%7B2%7D%20%3D%5Cfrac%7BV_%7B2%7D%5E%7B2%7D%20%20%7D%7B2g%7D%20%28k_%7BL%7D%2B%5Cfrac%7BfL%7D%7BD%7D%20%20%29%3D%5Cfrac%7B25.46%5E%7B2%7D%20%7D%7B2%2A9.8%7D%20%2A%280.5%2B%5Cfrac%7B0.033%2A36%7D%7B0.03%7D%20%29%3D1326.18m)
The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:
![h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m](https://tex.z-dn.net/?f=h_%7Bpump%7D%20%3D%5Cfrac%7B25.46%5E%7B2%7D%20%7D%7B2%2A9.8%7D%20%2B1347.48-36%3D1344.55m)
The required pumping power is:
![P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW](https://tex.z-dn.net/?f=P%3DQ%5Crho%20%2Ag%2Ah_%7Bpump%7D%20%20%3D0.018%2A999.1%2A9.8%2A1344.55%3D236965.16W%3D236.96kW)
Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
Answer:
C
Explanation:
I COULD be wrong, i'm not sure but im confident its c
The weight of the specimen in SSD condition is 373.3 cc
<u>Explanation</u>:
a) Apparent specific gravity = ![\frac{A}{A-C}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA-C%7D)
Where,
A = mass of oven dried test sample in air = 1034 g
B = saturated surface test sample in air = 1048.9 g
C = apparent mass of saturated test sample in water = 975.6 g
apparent specific gravity =
= ![\frac{1034}{1034-675 \cdot 6}](https://tex.z-dn.net/?f=%5Cfrac%7B1034%7D%7B1034-675%20%5Ccdot%206%7D)
Apparent specific gravity = 2.88
b) Bulk specific gravity ![G_{B}^{O D}=\frac{A}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7BA%7D%7BB-C%7D)
![G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7B1034%7D%7B1048.9-675%20%5Ccdot%206%7D)
= 2.76
c) Bulk specific gravity (SSD):
![G_{B}^{S S D}=\frac{B}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BS%20S%20D%7D%3D%5Cfrac%7BB%7D%7BB-C%7D)
![=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209%7D%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D)
= 2.80
d) Absorption% :
![=\frac{B-A}{A} \times 100 \%](https://tex.z-dn.net/?f=%3D%5Cfrac%7BB-A%7D%7BA%7D%20%5Ctimes%20100%20%5C%25)
![=\frac{1048 \cdot 9-1034}{1034} \times 100](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-1034%7D%7B1034%7D%20%5Ctimes%20100)
Absorption = 1.44 %
e) Bulk Volume :
![v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}](https://tex.z-dn.net/?f=v_%7Bb%7D%3D%5Cfrac%7B%5Ctext%20%7B%20weight%20of%20dispaced%20water%20%7D%7D%7BP%20%5Comega%20t%7D)
![=\frac{1048 \cdot 9-675 \cdot 6}{1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D%7B1%7D)
= ![373.3 cc](https://tex.z-dn.net/?f=373.3%20cc)
Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 ×
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85
)²× 9
× sin18 × cos18
Qd = 94.305 ×
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 ×
× π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 ×
m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 ×
- 85.2 ×
Qx = 9.10
m³/s