Datums and benchmarks are used for _____

Pure butanol is to be fed into a semibatch reactor containing pure ethyl acetate to produce butyl acetate and ethanol. The react

marissa [1.9K]

Answer:

See the pictures attached

Explanation:

All the explanation is given in the pictures

3 0

3 years ago

What is the weight of a glider with a mass of 5.3 grams? (Hint: watch your units!)

Vinvika [58]

Answer: 0.053

Explanation:

So, we convert 5.3 grams into kilograms.

5.3g = 0.0053 kg (Since 1kg equals 1000g)

On Earth, gravity is 10 N/kg.

Weight = mass x gravity

Weight = 0.0053kg x 10 N/kg

Weight = 0.053 Newtons (On Earth)

6 0

3 years ago

A compressor receives air at 290 K, 95 kPa and shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate o

aev [14]

Answer:

P2 = 3.9 MPa

Explanation:

Given that

T₁ = 290 K

P₁ = 95 KPa

Power P = 5.5 KW

mass flow rate = 0.01 kg/s

solution

with the help of table A5

here air specific heat and adiabatic exponent is

Cp = 1.004 kJ/kg K

and k = 1.4

so

work rate will be

W = m × Cp × (T2 - T1) ..........................1

here T2 = W ÷ ( m × Cp) + T1

so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290

T2 = 838 k

so final pressure will be here

P2 = P1 × $(\frac{T2}{T1})^\frac{k}{k-1}$ ..............2

P2 = 95 × $(\frac{838}{290})^\frac{1.4}{1.4-1}$

P2 = 3.9 MPa

3 0

3 years ago

Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 320 K to p2 = 11 bar. The initial volum

tekilochka [14]

Answer:

$W_{12}=-53.9056KJ$

Part A:

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

Gas is ideal
System is closed system.
K.E and P.E is neglected
Process is polytropic

Since Process is polytropic so $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

Where n=1.25

Since Process is polytropic :

$\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}$

$V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3$

Now, $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

$W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})$

$W_{12}=-53.9056KJ$

We will now calculate mass (m) and Temperature T_2.

$m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\m=0.30338Kg$

$T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\T_{2} =555.14K$

Part A:

According to energy balance::

$Q=mc_{v}(T_{2}-T_{1})+W_{12}$

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

$Q=0.30338*0.657(555.14-320)+(-53.9056)$

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

From Table A-23:

$u_{1} at 320K = 7526 KJ/Kg$

$u_{2} at 555.14K = 15567.292$ (By interpolation)

$Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}$

$Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)$

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

7 0

4 years ago

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$