Question list ✔Question 6 K During processing, a polymer is pumped through the plant. The pump must produce enough power to move the material a height of 10 meters [m] at a volumetric flowrate of 0.2 cubic meters per second [m³/s]. Assume the specific gravity of the polymer is 0.6262. If the pump is 75% efficient, what is the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower [hp]? Click the icon to view the table of common derived units in the Sl system. Click the icon to view the conversion table. Click the icon to view density of water.

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter17: Energy In Thermal Processes: The First Law Of Thermodynamics
Section: Chapter Questions
Problem 76P
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✔ Question 6
✔Question 7
Question 8
Question 9
Density of water
= 1 lb₂
= 1 g/cm³
= 1000 kg/m³
= 62.4 lb/ft³
= 1.94 slug/ft³
Print
Done
K
During processing, a polymer is pumped through the plant. The pump must produce enough power to move the material a height of 10 meters [m] at a volumetric flowrate of 0.2 cubic meters per second [m³/s]. Assume
the specific gravity of the polymer is 0.6262. If the pump is 75% efficient, what is the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower [hp]?
Click the icon to view the table of common derived units in the Sl system.
Click the icon to view the conversion table.
Click the icon to view density of water.
The power rating is
hp. (Round your answer to one dicimal place.)
- X ore Info
Length
1 m = 3.28 ft
1 km = 0.621 mi
1 in = 2.54 cm
1 mi = 5,280 ft
1 yd = 3 ft
Force
1 N = 0.225 lb,
Print
Power
1 W = 3.412 BTU/h
= 0.00134 hp
= 14.34 cal/min
= 0.7376 ft lb/s
Done
X
More Info
Dimension
Force (F)
Energy (E)
Power (P)
Pressure (P)
Voltage (V)
SI Unit
newton [N]
joule [J]
watt [W]
pascal [Pa]
volt [V]
1 N=1
Base SI Units
kg m
s²
1J=1Nm = 1
1W=1==1
S
m
N
1 Pa = 1 ==1
2
kg m²
1 V=1 =1
s²
kg m²
s³
3
S
Print
W
kg m²
A s³ A
3
kg
2
ms
Derived from
F=ma
Force = mass times acceleration
E=Fd
Energy = force times distance
E
P=
ī
Power = energy per time
F
P=Ā
Pressure = force per area
P
V=
Voltage = power per current
Done
-
X
Transcribed Image Text:Question list ✔ Question 6 ✔Question 7 Question 8 Question 9 Density of water = 1 lb₂ = 1 g/cm³ = 1000 kg/m³ = 62.4 lb/ft³ = 1.94 slug/ft³ Print Done K During processing, a polymer is pumped through the plant. The pump must produce enough power to move the material a height of 10 meters [m] at a volumetric flowrate of 0.2 cubic meters per second [m³/s]. Assume the specific gravity of the polymer is 0.6262. If the pump is 75% efficient, what is the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower [hp]? Click the icon to view the table of common derived units in the Sl system. Click the icon to view the conversion table. Click the icon to view density of water. The power rating is hp. (Round your answer to one dicimal place.) - X ore Info Length 1 m = 3.28 ft 1 km = 0.621 mi 1 in = 2.54 cm 1 mi = 5,280 ft 1 yd = 3 ft Force 1 N = 0.225 lb, Print Power 1 W = 3.412 BTU/h = 0.00134 hp = 14.34 cal/min = 0.7376 ft lb/s Done X More Info Dimension Force (F) Energy (E) Power (P) Pressure (P) Voltage (V) SI Unit newton [N] joule [J] watt [W] pascal [Pa] volt [V] 1 N=1 Base SI Units kg m s² 1J=1Nm = 1 1W=1==1 S m N 1 Pa = 1 ==1 2 kg m² 1 V=1 =1 s² kg m² s³ 3 S Print W kg m² A s³ A 3 kg 2 ms Derived from F=ma Force = mass times acceleration E=Fd Energy = force times distance E P= ī Power = energy per time F P=Ā Pressure = force per area P V= Voltage = power per current Done - X
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