Question 1 (20 points) Convert the following quantities in the required unitss ?
ID: 1022575 • Letter: Q
Question
Question 1 (20 points) Convert the following quantities in the required unitss ?/ uired unit% / x ' bar 15 bar 1 bar a) 1.5 MPa-15. bar , 100 °F-.... K c) d) 150°C= e) 5.0106 cal..kWh R 40 MW= f) g) Btu/s Ibr 3000 dyn= 1.2 g/cm 3-.... kg/ h) 800 kg/m3 = Ibm/ft3 cm3/s i) j) 30 L/h = estion 2 (10 points) A dead-weight gauge is used to measure pressures up to 3500 atm. The piston diameter is 0.95 cm. What is the mass (in kg) of the weights required? Question 3 (15 points) The reading on a mercury manometer at 20 °C (open to the atmosphere at one end) is 28.71 inches. The local gravity is 32.186 f/s. The local atmospheric pressure is 101.29 kPa. The density of mercury at 20 °C is 13.495 g/cm3. What is the absolute pressure being measured, both in kPa and in psia? Question 4 (10 points) The gravity on Mars can be measured with a spring, which suspends a mass of 0.40 kg. On earth, where the local acceleration of gravity is 9.81 m/s, the spring extends 1.55 cm. When the instrument package is landed on Mars, the spring is extended only 0.59 cm. What is the acceleration of gravity on Mars?Explanation / Answer
Q1.
a)
1.5MPa = bar
1 MPa = 10^6 Pa
1 bar = 100000 Pa = 10^5 Pa
then; the ratio
(10^6)/(10^5) = 10 MPa = 1 bar
1.5 MPa = x
10 MPa = 1 bar
x = 10*1.5 = 15 bar
1.5 MPa = 15 bar
b)
300 mbar = lbm/in2,
note that 32 lbm/in2 = 1 lbf/in2
and since lbf/in2 = pounds per square inch, aka PSI
300 mbar --> bar
P = 0.03 bar
1 bar = 14.5038 psi
0.03 bar = x psi
x = 14.5038*0.03 = 0.4351132 psi
1 psi = 32 lbm/in2
0.4351132 psi = x
x = 0.4351132 *32 = 13.9236 lbm/in2
then
300 mbar = 13.9236 lbm/in2
c)
100 °F = K
First, change F to C, then C to K
C = (F - 32) / 1.8
C = (100-32)/1.8 = 37.7777 C
then
C to K, just add 273.15
K = C + 273.15
K = 37.7777 + 273.15 = 310.927 K
d)
150 °C = R
First, change C to F
F = 1.8C + 32
F = 150*1.8 + 32 = 302 F
Then, from F to R, just add 459.67
R = F + 459.67
R = 302+459.67
R = 761.67
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