Question

Q 1 A. Transfer temperature degree and pressures below:
1. 700 mm lily to KN/m^2 and Bar.
2. 200 mm water to N/m^2 and Bar.
3.1500 PSI to Bar.
4.200°F to Siles.
5. - 273° Siles to F.
6. 150° F to kelven.
* Need to know:
Water density = 1000 kg/m^3.
Lily density =13600 Kg/m^3.

Answer 1

Solutions:

1. Given density of lily = 13600 Kg/m³

Pressure exerted by 700 mm of lily is given as: P = density*acceleration due to gravity*height of column

P = 13600*9.8*700*10^-3 = 93296 N/m²

So, 700 mm lily equals to 93.296 KN/m².

1 KN/m² = 0.01 bar

So, 700 mm lily equals to 0.93296 bar.

2.  Pressure exerted by 200 mm water

P = 1000*9.8*0.2 = 1960 N/m²

In bar, P equals to 1960*10^-5 = 0.0196 bar.

3. 1 psi = 0.0689 bar

1500 psi = 1500*0.0689 = 103.85 bar

4. 200 F

(F-32)/9 = C/5

Putting F=200, we get C= 93.33

So, 200^oF = 93.33^oC.

5. Putting C = -273 in the above equation, we get F = -459.4

So, -273 ^oC = -459.4 ^oF

6. Putting F=150, we get C= 65.55

Temperature in kelvin = temperature in celsius + 273

So, 150 ^oF = 65.55+273 = 338.55 K.