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Tpy6a [65]
3 years ago
11

Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific vol

ume is one-tenth of the specific volume at state 1. Applying the ideal gas model and assuming variations in specific heat, determine (a) T2, in °C, and (b) the work, in kJ/kg.
Engineering
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg

Explanation:

Solution

T₁ = 35°C = 308

the first step to take is to Use the Table A-17: Ideal gas  properties for air:

Now,

At T₁ = 308 K

V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)

So,

V₁ =219.818 kJ/kg

Thus,

Vr₁ = 596 + [308-305/310-305] (572.3 - 596)

= 581.78

so,

Vr₂/Vr₁ = 1/10

Vr₂ =58.178

Applying Table A-17, at Vr₂ = 58.178

Then,

(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)

T₂ = 747.5 and K = 474.5 °C

V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)

so,

V₂ =549.996 kJ/kg

Hence,

(b) q - w = v₂ -v₁

= 0 -w  = 549.996- 219.818

w = 330.178 kJ/kg

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Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air
lara31 [8.8K]

Answer:

Magnesium sulphate is in two forms

Magnesium anhydrate and magnesium hepta hydrate

Density of MgSO4= 2.66 g/ml

Density of MgSO4. 7H2O= 1.68 g/ml

M. W of MgSO4= 120 kg/kmol

In order to form 1000 Kg 20wt% solution

A) the Kg of water required before addition of crystals for MgSO4 =(1000) (0.80)= 800 kg

Average density= (0.20) (2.66) +(0.80) (1) = 1.332 g/ml = 1332 Kg/m3

Volume of water required= (800/1332) = 600.60 litres

For hepta hydrate

Average density= (0.20) (1.68) +(0.80) (1) = 1.136 g/ml=

1136 kg/m 3

MgSO4. 7 H2O, mole fraction of water present =

7×18/(120+(7×18) )=0.512 = 51.2 mol%= 13.6 wt%

1000 kg of (20 wt%) solution required

Amount of pure MgSO4 needed= 1000(.20) = 200 Kg

But it contains 13.6% water hence total weight of

MgSO4. 7H2O= 200/(1-0.136) = 231.48 kg

Amount of water required= (1000-231.48) = 768.51 Kg

Volume of water required= 768.51/1136= 676.51 Litres

B) diameter of tank= 0.30 m

V= 3.142(d2)(H) /4

For pure MgSO4 , volume of water required=600.60 litres

H = 8.49 m

For hepta hydrate volume of water required=676.51 litres

H= 9.57 m

C) height of tank after addition of crystals

Total volume in case of pure MgSO4=

1000/(1332) = 0.7507= 750.7 litres

H= 10.62 m

Total volume of hepta hydrate =

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H= 12.45 m

D) after dissolution of crystals

The volume after dissolution in case of pure MgSO4

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H=8.49 m

in case of hepta hydrate

The volume after dissolution= ( volume of water + volume of water in MgSO4. 7H2O)

= (676.51+(231.48×0.136) = 707.99 Litres

H= 10.01 m

Part B question

A) Weight of house= 1×105 lb= 2.20 ×105

Force applied by house= (2.20×105) (9.81) = 2.15×106 N

Pressure of ballon= 1.05 atm= 1.063×105 Pa

Diameter= 9.5 inch= 0.2413 m

Area= (3.142) (0.2413) 2/4= 0.0457 m2

F= P×A= 4857.91 N

To float the house

Both forces should be equal

So force applied by 1 ballon= 4857.91 N

Total force required=2.15×106 N

Number of ballons required= (2.15×106) /4857.91=442.57=443 ballons

B) the ballon pressure must be greater than atmospheric pressure

If both pressures are equal then there will be no air flow

If outside pressure is higher, then air would flow from outside to inside of ballon causing deflation, hence pressure inside the ballon must be higher than outside

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